Kuzenko:2015rfx, Antoniadis:2017jsk, Antoniadis:2019gbd, cecotti:1987].

We start with the following $\mathcal{N}=2$ superfield,

$\mathcal{W}(y,\theta,\tilde{\theta})=X(y,\theta)+\sqrt{2}i\tilde{\theta}W(y,\theta)-\tilde{\theta}^{2}G(y,\theta)~{},\qquad
y^{\mu}=x^{\mu}+i\theta\sigma^{\mu}\bar{\theta}+i\tilde{\theta}\sigma^{\mu}\bar{\tilde{\theta}}~{},$
(1)

which is chiral with respect to both supersymmetries:

$\bar{D}_{\dot{\alpha}}\mathcal{W}=\bar{\tilde{D}}_{\dot{\alpha}}\mathcal{W}=0~{}.$
(2)

Since we are ultimately interested in partial ${\cal N}=2\to{\cal N}=1$
supersymmetry breaking, we will mostly use ${\cal N}=1$ superfields associated
to the $\theta$ Grassmann variables to describe manifest supersymmetry, while
we use the $\tilde{\theta}$ variable for the hidden non-linearly realized
supersymmetry. The ${\cal N}=1$ superfields $X$, $W_{\alpha}$, and $G$ of eq.
(1) are chiral under the manifest ${\cal N}=1$ supersymmetry. Under the
additional hidden ${\cal N}=1$ supersymmetry, they transform as follows:

$\displaystyle\tilde{\delta}X$ $\displaystyle=$
$\displaystyle\sqrt{2}i\epsilon W~{},$ (3a) $\displaystyle\tilde{\delta}W$
$\displaystyle=$
$\displaystyle\sqrt{2}\sigma^{\mu}\bar{\epsilon}\partial_{\mu}X+\sqrt{2}i\epsilon
G~{},$ (3b) $\displaystyle\tilde{\delta}G$ $\displaystyle=$
$\displaystyle-\sqrt{2}\partial_{\mu}W\sigma^{\mu}\bar{\epsilon}~{}.$ (3c)

The superfield (1) has $16+16$ independent off-shell components and is
reducible. It contains the degrees of freedom of an $\mathcal{N}=2$ vector and
tensor multiplet. To reduce the degrees of freedom and describe an irreducible
${\cal N}=2$ off-shell vector multiplet, we impose the following conditions on
the ${\cal N}=1$ components of $\cal W$:

  * (i)

First that $W_{\alpha}$ is the field-strength superfield of an $\mathcal{N}=1$
vector multiplet satisfying,

$D^{\alpha}W_{\alpha}-\bar{D}_{\dot{\alpha}}\bar{W}^{\dot{\alpha}}=0~{},$ (4)

  * (ii)

and that

$G=\frac{1}{4}\bar{D}^{2}\bar{X}~{}.$ (5)

The latter condition can easily be seen to be consistent since it is
straightforward to verify that $\frac{1}{4}\bar{D}^{2}\bar{X}$ transforms in
the same way as $G$ given in (3c). Therefore we can impose (5) without
violating $\mathcal{N}=2$ supersymmetry.

Since $\mathcal{W}$ is chiral with respect to both sets of supersymmetries, we
can consider the following Lagrangian,

$\displaystyle\mathcal{L}^{\mathcal{N}=2}_{\mathcal{W}^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\int d^{2}\theta
d^{2}\tilde{\theta}\,\mathcal{W}^{2}+c.c.=\frac{1}{4}\int
d^{2}\theta\Big{(}W^{2}-\frac{1}{2}X\bar{D}^{2}\bar{X}\Big{)}+c.c.~{}.$ (6)

On the other hand, the $\mathcal{N}=2$ Maxwell theory written in terms of the
$\mathcal{N}=1$ chiral superfields $X$ and $W_{\alpha}$ is given by

$\displaystyle\mathcal{L}^{\mathcal{N}=2}_{\text{ Maxwell}}$ $\displaystyle=$
$\displaystyle\int d^{2}\theta d^{2}\bar{\theta}\,\bar{X}X+\frac{1}{4}\int
d^{2}\theta\,W^{2}+\frac{1}{4}\int d^{2}\theta\,\bar{W}^{2}~{},$ (7)
$\displaystyle=$ $\displaystyle\frac{1}{4}\int
d^{2}\theta\Big{(}W^{2}-\frac{1}{2}X\bar{D}^{2}\bar{X}\Big{)}+c.c.+{\rm
total~{}derivative}~{}.$

We see that these two Lagrangians are the same, confirming that the extra
constraint imposed on $\mathcal{W}$ is correct. The off-shell $\mathcal{N}=2$
vector multiplet can therefore be described in term of the following
$\mathcal{N}=2$ superfield

$\mathcal{W}(y,\theta,\tilde{\theta})=X(y,\theta)+\sqrt{2}i\tilde{\theta}W(y,\theta)-\frac{1}{4}\tilde{\theta}^{2}\bar{D}^{2}\bar{X}(y,\theta)~{},$
(8)

where $X$ and $W_{\alpha}$ are $\mathcal{N}=1$ chiral and vector multiplets,
respectively. Their component expansion reads:

$\displaystyle W_{\alpha}$ $\displaystyle=$
$\displaystyle-i\lambda_{\alpha}+\theta_{\alpha}{\sf
D}-i(\sigma^{\mu\nu}\theta)_{\alpha}F_{\mu\nu}+\theta^{2}(\sigma^{\mu}\partial_{\mu}{\bar{\lambda}})_{\alpha}~{},$
(9a) $\displaystyle X$ $\displaystyle=$ $\displaystyle
x+\sqrt{2}\theta\chi-\theta^{2}\,{\sf F}~{}.$ (9b)

Following [Rocek:1997hi] (see also [Kuzenko:2015rfx, Antoniadis:2017jsk,
Antoniadis:2019gbd]), we break $\mathcal{N}=2$ supersymmetry by considering a
Lorentz and ${\cal N}=1$ invariant condensate with a non-trivial dependence on
the hidden Grassmann variables
$\langle\mathcal{W}\rangle=\mathcal{W}_{\text{def}}\propto\tilde{\theta}^{2}\neq
0$, such that

$\displaystyle\mathcal{W}$ $\displaystyle\rightarrow$
$\displaystyle{\mathcal{W}_{\text{new}}}=\langle\mathcal{W}\rangle+\mathcal{W}=\mathcal{W}+\mathcal{W}_{\text{def}}~{},$
(10a) $\displaystyle{\mathcal{W}_{\text{new}}}$ $\displaystyle=$
$\displaystyle
X+\sqrt{2}i\tilde{\theta}W-\frac{1}{4}\tilde{\theta}^{2}\Big{(}\bar{D}^{2}\bar{X}+\frac{2}{\kappa}\Big{)}~{}.$
(10b)

The hidden supersymmetry transformations of the ${\cal N}=1$ components of the
deformed ${\cal N}=2$ vector multiplet turn out to be

$\displaystyle\tilde{\delta}X$ $\displaystyle=$
$\displaystyle\sqrt{2}i\epsilon W~{},$ (11a) $\displaystyle\tilde{\delta}W$
$\displaystyle=$
$\displaystyle\frac{i}{\sqrt{2}\kappa}\epsilon+\frac{i}{2\sqrt{2}}\epsilon\mkern
1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern
1.5mu^{2}\bar{X}+\sqrt{2}\sigma^{\mu}\bar{\epsilon}\partial_{\mu}X~{}.$ (11b)

Assuming the model under consideration preserves the manifest ${\cal N}=1$
supersymmetry, which implies $\langle\mkern
1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu^{2}X\rangle=0$, the
explicit non-linear $\kappa$-dependent term in the transformation of the
fermionic $W_{\alpha}$ signals the spontaneous partial breaking ${\cal
N}=2\to{\cal N}=1$ of the hidden supersymmetry.

To describe the Maxwell-Goldstone multiplet for the partial breaking ${\cal
N}=2\to{\cal N}=1$, we impose the following nilpotent constraint on the
deformed ${\cal N}=2$ superfield ${\mathcal{W}_{\text{new}}}$ [Rocek:1997hi]:

$({\mathcal{W}_{\text{new}}})^{2}=0~{}.$ (12)

Once reduced to ${\cal N}=1$ superfields, following the expansion (10b), this
constraint implies the Bagger-Galperin constraint [Bagger:1996wp]

$\frac{1}{\kappa}X=W^{2}-\frac{1}{2}X\bar{D}^{2}\bar{X}~{},$ (13)

which can be solved to eliminate $X$ in terms of $W^{2}=W^{\alpha}W_{\alpha}$
and its complex conjugate
$\bar{W}^{2}=\bar{W}_{\dot{\alpha}}\bar{W}^{\dot{\alpha}}$ as

$X=\kappa
W^{2}-\kappa^{3}\bar{D}^{2}\Big{[}\frac{W^{2}\bar{W}^{2}}{1+\mathcal{A}+\sqrt{1+2\mathcal{A}-\mathcal{B}^{2}}}\Big{]}~{}.$
(14)

Here we have introduced

$\mathcal{A}=\frac{\kappa^{2}}{2}(D^{2}W^{2}+\bar{D}^{2}\bar{W}^{2})=\overline{\mathcal{A}}~{},\qquad\mathcal{B}=\frac{\kappa^{2}}{2}(D^{2}W^{2}-\bar{D}^{2}\bar{W}^{2})=-\overline{\mathcal{B}}~{}.$
(15)

For later use, we denote the lowest components of the composite superfields
$\cal A$ and $\cal B$ as

$A=\mathcal{A}|_{\theta=0}~{},\qquad B=\mathcal{B}|_{\theta=0}~{}.$ (16)

We will not repeat the derivation of (14) which can be found in the original
paper [Bagger:1996wp], and was reviewed and slightly modified in section 1 for
our analysis in two dimensions.

The ${\cal N}=1$ supersymmetric BI action can be constructed using the
following ${\cal N}=1$ (anti-)chiral Lagrangian linear in $X$:

$\mathcal{L}_{\kappa}=\frac{1}{4\kappa}\Big{(}\int d^{2}\theta X+\int
d^{2}\bar{\theta}\bar{X}\Big{)}~{}.$ (17)

The second hidden supersymmetry eq. (11a) written in terms of the
unconstrained real vector multiplet $V$, where $W_{\alpha}=-1/4\mkern
1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu^{2}D_{\alpha}V$, takes
the form

$\tilde{\delta}X=-\frac{1}{4}\sqrt{2}i\epsilon^{\alpha}\bar{D}^{2}D_{\alpha}V~{}.$
(18)

Using the fact that $D^{2}\mkern
1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern
1.5mu^{2}D_{\alpha}\propto\partial_{\alpha{\dot{\alpha}}}D^{2}\mkern
1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu^{\dot{\alpha}}$, one can
immediately see that the supersymmetry variation of $\mathcal{L}_{\kappa}$ in
(17) is a total derivative. Therefore this supersymmetric BI action is
invariant under the second hidden non-linear supersymmetry.

Using the solution (14), the supersymmetric BI Lagrangian takes the explicit
form

$\displaystyle\mathcal{L}_{\kappa}$ $\displaystyle=\frac{1}{4\kappa}\int
d^{2}\theta\Big{(}\kappa
W^{2}-\kappa^{3}\bar{D}^{2}\Big{[}\frac{W^{2}\bar{W}^{2}}{1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}}\Big{]}\Big{)}+c.c.\,$
$\displaystyle=\frac{1}{4}\int d^{2}\theta\,W^{2}+\frac{1}{4}\int
d^{2}\bar{\theta}\,\bar{W}^{2}+2\kappa^{2}\int d^{2}\theta
d^{2}\bar{\theta}\,\frac{W^{2}\bar{W}^{2}}{1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}}~{},$
(19)

which makes it clear that the supersymmetric BI is a non-linear deformation of
the free ${\cal N}=1$ Maxwell theory. This supersymmetric extension of BI was
first constructed by Bagger and Galperin in [Bagger:1996wp]. In this work when
we refer to the supersymmetric BI theory, we will always mean the Bagger-
Galperin action.

We can easily calculate the flow under the $\kappa^{2}$ coupling constant,

$\frac{\partial\mathcal{L}_{\kappa}}{\partial\kappa^{2}}=2\int d^{2}\theta
d^{2}\bar{\theta}\,\frac{W^{2}\bar{W}^{2}}{1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}}\frac{1}{\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}}~{}.$
(20)

Our goal is now to show that the right hand side of this flow equation on-
shell is the specific supercurrent bilinear (14) that we introduced earlier.
This will establish a supercurrent-squared flow for the supersymmetric BI
action.

Before turning to the core of this analysis let us recall that at the leading
order in $\kappa^{2}$, the fact that $D=4$ ${\cal N}=1$ BI satisfies a
supercurrent-squared flow was already noticed in [cecotti:1987]. This result
was also highlighted recently in the introduction of [Chang:2018dge]. In fact,
note that in the free limit $\alpha=\kappa=0$, the Lagrangian (19) becomes the
${\cal N}=1$ supersymmetric Maxwell theory. Its supercurrent multiplet is

${\mathcal{J}}_{\alpha\dot{\alpha}}=-4W_{\alpha}\bar{W}_{\dot{\alpha}}~{},\qquad\mathcal{X}=0~{},$
(21)

where $\mathcal{X}=0$ because super-Maxwell theory is scale invariant. The
supersymmetric $T^{2}$ deformation operator (14) is then simply given by

$\mathcal{O}_{T^{2}}=\frac{1}{16}{\mathcal{J}}_{\alpha\dot{\alpha}}{\mathcal{J}}^{\alpha\dot{\alpha}}-\frac{5}{8}\mathcal{X}\bar{\mathcal{X}}=W^{2}\bar{W}^{2}~{},$
(22)

and to leading order (20) turns into [cecotti:1987]

$\frac{\partial\mathcal{L}_{\kappa}}{\partial\kappa^{2}}=\int d^{2}\theta
d^{2}\bar{\theta}\,W^{2}\bar{W}^{2}+\mathcal{O}(\kappa^{2})=\int d^{2}\theta
d^{2}\bar{\theta}\,\mathcal{O}_{T^{2}}+\mathcal{O}(\kappa^{2})~{}.$ (23)

This shows that the supercurrent-squared flow equation is satisfied at this
order. The rest of this section is devoted to demonstrating the full non-
linear extension of this result. First, we are going to look at the bosonic
truncation of (19) and (20).

####  2 Bosonic truncation

In the pure bosonic case the gauginos are set to zero in (9a),
$\lambda=\bar{\lambda}=0$, and $W^{2},\bar{W}^{2}$ only have
$\theta^{2},\bar{\theta}^{2}$ components, so $\mathcal{A},\mathcal{B}$ can
only contribute to the lowest components:

$A=\mathcal{A}|_{\theta=0}=2\kappa^{2}\Big{(}F^{2}-2{\sf
D}^{2}\Big{)}~{},\qquad B=\mathcal{B}|_{\theta=0}=2\kappa^{2}iF\tilde{F}~{}.$
(24)

Therefore the supersymmetric BI Lagrangian reduces to

$\mathcal{L}=\frac{1}{8\kappa^{2}}\Big{[}1-\sqrt{1+4\kappa^{2}\Big{(}F^{2}-2{\sf
D}^{2}\Big{)}-4\kappa^{4}\Big{(}F\tilde{F}\Big{)}^{2}}\Big{]}~{}.$ (25)

The auxiliary field can be set to ${\sf D}=0$ using its ${\rm EOM}$, and the
Lagrangian is then equivalent to the bosonic BI Lagrangian (1) with the
identification $\alpha^{2}=8\kappa^{2}$. This immediately implies that on-
shell the bosonic truncation of the supersymmetric BI satisfies a $T^{2}$ flow
equation driven by the $O_{T^{2}}$ operator (6), as we discussed in (10a). A
similar conclusion holds for the complete supersymmetric model of (19) and
(20).

####  3 Supersymmetric Born-Infeld as a supercurrent-squared flow

The supercurrent for the supersymmetric BI action (19) was computed in
[Kuzenko:2002vk] for $\kappa^{2}=\frac{1}{2}$. To simplify notation, we will
also consider the special case $\kappa^{2}=\frac{1}{2}$ in our intermediate
computations. The $\kappa$-dependence can be restored easily and will appear
in the final formulae.

We can straightforwardly use the results of [Kuzenko:2002vk] for our
supercurrent-squared flow analysis. The FZ multiplet was computed for a class
of models described by the following Lagrangian,

$\displaystyle{\mathcal{L}}$ $\displaystyle=$ $\displaystyle\frac{1}{4}\int
d^{2}\theta\,W^{2}+\frac{1}{4}\int
d^{2}\bar{\theta}\,\bar{W}^{2}+\frac{1}{4}\int d^{2}\theta
d^{2}\bar{\theta}\,W^{2}\bar{W}^{2}\Lambda(u,\bar{u})~{},$ (26)

where

$u=\frac{1}{8}D^{2}W^{2}~{},\qquad\bar{u}=\frac{1}{8}\bar{D}^{2}\bar{W}^{2}~{}.$
(27)

The action (19) turns out to be given by the following choice of
$\Lambda(u,\bar{u})$

$\Lambda(u,\bar{u})=\frac{4}{1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}}~{},$
(28)

where

$\mathcal{A}=2(u+\bar{u})~{},\qquad\mathcal{B}=2(u-\bar{u})~{}.$ (29)

Following [Kuzenko:2002vk], we also introduce the composite superfields

$\Gamma(u,\bar{u})=\frac{\partial(u\Lambda)}{\partial
u}~{},\qquad\bar{\Gamma}(u,\bar{u})=\frac{\partial(\bar{u}\Lambda)}{\partial\bar{u}}~{},$
(30)

which, in the case of interest to us where (28) holds, satisfy

$\displaystyle\Gamma+\bar{\Gamma}-\Lambda$ $\displaystyle=$
$\displaystyle\frac{4}{\Big{(}1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{)}\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}}~{},$
(31a) $\displaystyle\bar{u}\Gamma+u\bar{\Gamma}$ $\displaystyle=$
$\displaystyle 1-\frac{1}{\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}}~{}.$ (31b)

The supercurrents will also be functionals of the following composite

$\displaystyle iM_{\alpha}$ $\displaystyle=$ $\displaystyle
W_{\alpha}\Bigg{[}1-\frac{1}{4}\bar{D}^{2}\Bigg{(}\bar{W}^{2}\Big{(}\Lambda+\frac{1}{8}D^{2}(W^{2}\frac{\partial\Lambda}{\partial
u})\Big{)}\Bigg{)}\Bigg{]}~{},$ (32a) $\displaystyle=$ $\displaystyle
W_{\alpha}\Big{(}1-2\bar{u}\Gamma\Big{)}+W\bar{W}(\cdots)+W^{2}(\cdots)~{},$
(32b)

where $W\bar{W}(\cdots)$ denotes terms which are proportional to
$W_{\alpha}\bar{W}_{\dot{\alpha}}$, while $W^{2}(\cdots)$ denotes terms
proportional to $W^{2}$. We will use similar notation with ellipses denoting
quantities with bare fermionic terms that will not contribute to the
calculation because of nilpotency conditions.

With the ingredients introduced above, the FZ multiplet for the supersymmetric
BI action is given by [Kuzenko:2002vk]

$\displaystyle\mathcal{X}$ $\displaystyle=$
$\displaystyle\frac{1}{6}W^{2}\bar{D}^{2}\Big{(}\bar{W}^{2}(\Gamma+\bar{\Gamma}-\Lambda)\Big{)}~{},$
(33a) $\displaystyle\mathcal{J}_{\alpha\dot{\alpha}}$ $\displaystyle=$
$\displaystyle-2iM_{\alpha}\bar{W}_{\dot{\alpha}}+2iW_{\alpha}\bar{M}_{\dot{\alpha}}+\frac{1}{12}[D_{\alpha},\bar{D}_{\dot{\alpha}}]\Big{(}W^{2}\bar{W}^{2}\Big{)}\cdot\Big{(}\Gamma+\bar{\Gamma}-\Lambda\Big{)}$
(33b) $\displaystyle+W^{2}\bar{W}(\cdots)+\bar{W}^{2}W(\cdots)~{}.$

For our purposes, the superfields $X$ and
${\mathcal{J}}_{\alpha{\dot{\alpha}}}$ can be further simplified as follows:

$\displaystyle\mathcal{X}$ $\displaystyle=$
$\displaystyle\frac{1}{6}W^{2}\bar{D}^{2}\bar{W}^{2}\cdot\Big{(}\Gamma+\bar{\Gamma}-\Lambda\Big{)}+W^{2}\bar{W}(\cdots)~{},$
(34a) $\displaystyle=$
$\displaystyle\frac{2W^{2}\bar{D}^{2}\bar{W}^{2}}{3\Big{(}1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{)}}+W^{2}\bar{W}(\cdots)~{},$
(34b)

and

$\displaystyle\mathcal{J}_{\alpha\dot{\alpha}}$ $\displaystyle=$
$\displaystyle-4W_{\alpha}\bar{W}_{\dot{\alpha}}(1-\bar{u}\Gamma-u\bar{\Gamma})+\frac{1}{12}[D_{\alpha},\bar{D}_{\dot{\alpha}}]\Big{(}W^{2}\bar{W}^{2}\Big{)}\cdot\Big{(}\Gamma+\bar{\Gamma}-\Lambda\Big{)}$
(35b) $\displaystyle+W^{2}\bar{W}(\cdots)+\bar{W}^{2}W(\cdots)~{},$
$\displaystyle=$
$\displaystyle-\frac{4W_{\alpha}\bar{W}_{\dot{\alpha}}}{\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}}+\frac{2D_{\alpha}W^{2}\cdot\bar{D}_{\dot{\alpha}}\bar{W}^{2}}{3\Big{(}1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{)}\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}}$
$\displaystyle+W^{2}\bar{W}(\cdots)+\bar{W}^{2}W(\cdots)~{},$

where we used (31a).

The computation of $\mathcal{X}\bar{\mathcal{X}}$ is trivial and receives
contributions only from the square of the first term in (34b). The computation
of $\mathcal{J}^{2}$ is less trivial. It is obvious that the last two
complicated terms in the second line of (35b) make no contribution since all
the terms are proportional to $W\bar{W}$, and we have the nilpotency property
$W_{\alpha}W_{\beta}W_{\gamma}=0$. The square of the first term is easy to
compute, and it is proportional to $W^{2}\bar{W}^{2}$. Next we consider the
product between the first and second term in (35b) which leads to the
relation:

$W_{\alpha}\bar{W}_{\dot{\alpha}}\cdot
D^{\alpha}W^{2}\cdot\bar{D}^{\dot{\alpha}}\bar{W}^{2}=W^{2}(DW)\cdot\bar{W}^{2}(\bar{D}\bar{W})=0~{}.$
(36)

Remarkably, this cross term vanishes, as shown in Appendix 7.F. Therefore, we
have the on-shell relation that

$W^{2}\mkern 1.5mu\overline{\mkern-1.5muW\mkern-1.5mu}\mkern
1.5mu^{2}DW=0~{}.$ (37)

A simple physical interpretation of this condition is that the manifest
supersymmetry is preserved on-shell, implying that the auxiliary field ${\sf
D}\propto D^{\alpha}W_{\alpha}|_{\theta=0}$ has no vev, and is at least linear
in gaugino fields $\lambda_{\alpha}\propto W_{\alpha}|_{\theta=0}$. The
vanishing of this cross term can be compared with the pure bosonic case where
the cross terms in $T^{2}$ vanish because of the tracelessness property of the
free Maxwell stress tensor; see section 3. Finally, we compute the square of
the second term in (35b) which includes the following structure:

$D^{\alpha}W^{2}\cdot\bar{D}^{\dot{\alpha}}\bar{W}^{2}\cdot
D_{\alpha}W^{2}\cdot\bar{D}_{\dot{\alpha}}\bar{W}^{2}=W^{2}\bar{W}^{2}D^{2}W^{2}\bar{D}^{2}\bar{W}^{2}~{}.$
(38)

Here we have used
$(D_{\alpha}W_{\beta})(D^{\alpha}W^{\beta})=-\frac{1}{2}D^{2}W^{2}+W^{\beta}D^{2}W_{\beta}$
to simplify the result.

In summary, on-shell the contributions to the supercurrent-squared operator
${\mathcal{O}}_{T^{2}}$ defined in eq. (14) are given by

$\displaystyle\mathcal{J}^{2}$
$\displaystyle=-\frac{1}{8}\Bigg{\\{}\frac{16W^{2}\bar{W}^{2}}{\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}^{2}}+\frac{4W^{2}\bar{W}^{2}D^{2}W^{2}\bar{D}^{2}\bar{W}^{2}}{9\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}^{2}\Big{(}1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{)}^{2}}\Bigg{\\}}\,,$
(39) $\displaystyle\mathcal{X}\bar{\mathcal{X}}$
$\displaystyle=\frac{4}{9}\frac{W^{2}\bar{W}^{2}D^{2}W^{2}\bar{D}^{2}\bar{W}^{2}}{\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}^{2}\Big{(}1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{)}^{2}}\,.$
(40)

Adding these results gives the supersymmetric $T^{2}$ primary operator
${\mathcal{O}}_{T^{2}}$:

$\displaystyle\mathcal{O}_{T^{2}}=-\frac{1}{2}\Big{(}\mathcal{J}^{2}+\frac{5}{4}\mathcal{X}\bar{\mathcal{X}}\Big{)}$
$\displaystyle=\frac{W^{2}\bar{W}^{2}}{\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}^{2}}\Bigg{(}1-\frac{D^{2}W^{2}\bar{D}^{2}\bar{W}^{2}}{4\Big{(}1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{)}^{2}}\Bigg{)}\,,$
$\displaystyle=\frac{W^{2}\bar{W}^{2}}{\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}^{2}}\Big{(}1-\frac{\mathcal{A}^{2}-\mathcal{B}^{2}}{(1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}})^{2}}\Big{)}\,$
$\displaystyle=\frac{2W^{2}\bar{W}^{2}}{\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{(}{1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{)}}}\,.$
(41)

It is worth noting that the simplifications occurring in constructing
${\mathcal{O}}_{T^{2}}$ from the supercurrents are very similar to the bosonic
case of (10a).

Comparing with (20), we see that eq. (41) proves that the supersymmetric BI
action (19) is an on-shell solution of the flow equation

$\displaystyle\frac{\partial\mathcal{L}_{\kappa}}{\partial\kappa^{2}}$
$\displaystyle=$ $\displaystyle\int d^{2}\theta
d^{2}\bar{\theta}^{2}\,\frac{2W^{2}\bar{W}^{2}}{\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{(}{1+\mathcal{A}+\sqrt{1+2\mathcal{A}+\mathcal{B}^{2}}\Big{)}}}~{},$
(42a) $\displaystyle=$ $\displaystyle\int d^{2}\theta
d^{2}\bar{\theta}^{2}\,\mathcal{O}_{T^{2}}+\text{total derivatives}+{\rm
EOM}~{}.$

It therefore describes a supercurrent-squared deformation of the ${\cal N}=1$
free Maxwell Lagrangian. This result establishes a relationship between non-
linearly realized supersymmetry and supercurrent-squared flow equations in
$D=4$. We note that this analysis can be extended to the supersymmetric ModMax
theory, whose supercurrent-squared deformation is the corresponding
supersymmetric ModMax-BI theory [Ferko:2022iru].

Before closing this section, we should make a few comments regarding the on-
shell condition (37) used in establishing the supercurrent-squared flow
equation for the $D=4$ ${\cal N}=1$ BI action. First it is important to stress
that the flow equation is not satisfied by the supersymmetric BI action off-
shell. Second, we note that the specific combination of $\mathcal{J}^{2}$ and
$\mathcal{X}\bar{\mathcal{X}}$ studied is the unique choice for which (19)
satisfies a supercurrent-squared flow equation, even if only on-shell.

Such a non-trivial condition satisfied by the on-shell supersymmetric BI
action is intriguing and hints at the existence of appropriate (super)field
redefinitions which might lead to a different supersymmetric extension of BI
that satisfies the flow equation off-shell. For example, it is know that the
dependence of the off-shell extension on the auxiliary field ${\sf D}$ can be
modified by appropriate (super)field redefinitions, as well as redefinitions
of the full superspace Lagrangian. We refer to [Cecotti:1986gb,
GonzalezRey:1998kh, Kuzenko:2011tj, Bagger:1997pi] for a list of relevant
papers on this subject. Under field redefinitions, the hidden supersymmetry
will be modified but will remain a non-linearly realized symmetry of the
theory. The existence of an off-shell solution of the supercurrent-squared
flow is an interesting question for future research.

###  5 $D=4$ Goldstino Action From Supercurrent-Squared Deformation

In section 4 we showed that the Bagger-Galperin action for the $D=4$, ${\cal
N}=1$ supersymmetric BI theory satisfies a supercurrent-squared flow. It is
known that the truncation of this model to fermions describes a Goldstino
action for $D=4$, ${\cal N}=1$ supersymmetry breaking; see, for example,
[Hatanaka:2003cr, Kuzenko:2005wh, Kuzenko:2011tj]. The ${\cal N}=1$ non-
linearly realized supersymmetry arises as the non-linearly realized part of
the ${\cal N}=2\to{\cal N}=1$ breaking of the supersymmetric BI. We have shown
in sections 3 and 2 that the bosonic truncation of the supersymmetric BI
satisfies a $T^{2}$ flow equation. The same should be true for the fermionic
truncation. More generally, one might argue that $D=4$ ${\cal N}=1$ Goldstino
models could satisfy a sort of flow equation that organizes their expansion in
the supersymmetry breaking scale parameter.

Note that in the $D=2$ case, the intuition is similar. If we consider the
actions analyzed in section 1 that describe Goldstone models for partial $D=2$
${\cal N}=(4,4)\to{\cal N}=(2,2)$ supersymmetry breaking, one can immediately
argue that their fermionic truncation describes Goldstino actions possessing
non-linearly realized $D=2$ ${\cal N}=(2,2)$ supersymmetry. These, by
construction, are expected to satisfy a $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$-flow equation. In fact,
such an argument is in agreement with the very nice recent analysis of
[Cribiori:2019xzp] where a $D=2$ Goldstino model possessing ${\cal N}=(2,2)$
non-linearly realized supersymmetry was shown to satisfy the supercurrent-
squared flow equation for the FZ multiplet, which was given in (10) of Chapter
3.666We refer to [Farakos:2016zam] for a discussion of various models
possessing non-linearly realized $(2,2)$ supersymmetry. The model analyzed in
[Cribiori:2019xzp] is the analogue of the $D=4$ model of [Casalbuoni:1988xh,
Komargodski:2009rz] and related on-shell to the Goldstino model of
[Rocek:1978nb].777Note that the Goldstino models of [Casalbuoni:1988xh,
Komargodski:2009rz, Rocek:1978nb] were shown in [Kuzenko:2005wh,
Kuzenko:2011tj] to be identical to the fermionic truncation of the
supersymmetric BI action up to a field redefinition of the Goldstino. This
section is devoted to showing that these $D=4$ ${\cal N}=1$ Goldstino models
satisfy a supercurrent-squared flow driven by the operator
${\mathcal{O}}_{T^{2}}$ of the supersymmetric BI, in agreement with the
arguments given above.

####  1 $D=4$ Goldstino actions

The Volkov-Akulov (VA) action is the low energy description of supersymmetry
breaking. There are several representations of the Goldstino action that are
equivalent to the Volkov-Akulov form; see [Kuzenko:2011tj, Cribiori:2016hdz]
for comprehensive discussions. Here we will focus on two models, but we start
by reviewing a few general features of Goldstino actions.

The original VA action was obtained by requiring its invariance under the the
non-linear supersymmetry transformation [Volkov:1973ix]

$\delta_{\xi}\lambda^{\alpha}=\frac{1}{\kappa}\xi^{\alpha}-i\kappa(\lambda\sigma^{m}\bar{\xi}-\xi\sigma^{m}\bar{\lambda})\partial_{m}\lambda^{\alpha}~{}.$
(1)

Explicitly, the original Lagrangian was proven to be

$\mathcal{L}_{\rm VA}=-\frac{1}{2\kappa^{2}}\det
A=-\frac{1}{2\kappa^{2}}-\frac{i}{2}(\lambda\sigma^{m}\partial_{m}\bar{\lambda}-\partial_{m}\lambda\sigma^{m}\bar{\lambda})+\text{interactions}~{},$
(2)

where

$A_{m}{}^{a}=\delta_{m}{}^{a}-i\kappa^{2}\partial_{m}\lambda\sigma^{a}\bar{\lambda}+i\kappa^{2}\lambda\sigma^{a}\partial_{m}\bar{\lambda}~{}.$
(3)

The alternative representation of the Goldstino action that interests us was
originally introduced by Casalbuoni et al. in [Casalbuoni:1988xh], and later
rediscovered and made fashionable by Komargodski and Seiberg
[Komargodski:2009rz]. This model, which following recent literature we will
call the KS model, was constructed by imposing nilpotent superfield
constraints as a generalization of Roček’s seminal ideas for the Goldstino
model described in [Rocek:1978nb]. After integrating out an auxiliary field in
the KS model, described in more detail in the next section, the explicit form
of the Lagrangian is given by the following very simple combination of terms:

$\mathcal{L}_{\text{KS}}=-f^{2}-\frac{i}{2}(\psi\sigma^{m}\partial_{m}\bar{\psi}-\partial_{m}\psi\sigma^{m}\bar{\psi})-\frac{1}{4f^{2}}\partial^{\mu}\bar{\psi}^{2}\partial_{\mu}\psi^{2}-\frac{1}{16f^{6}}\psi^{2}\bar{\psi}^{2}\partial^{2}\psi^{2}\partial^{2}\bar{\psi}^{2}~{}.$
(4)

The action is invariant under a quite involved non-linearly realized
supersymmetry transformation whose explicit form can be found in
[Kuzenko:2010ef, Kuzenko:2005wh]. The Goldstino actions described by (2) and
(4) prove to be equivalent off-shell up to a field redefinition
[Kuzenko:2010ef, Kuzenko:2005wh].

####  2 $D=4$ KS Goldstino model as a supercurrent-squared flow

The goal in the rest of this section is to straightforwardly generalize the
analysis of [Cribiori:2019xzp] to $D=4$ and to show how the KS action
satisfies a flow equation arising from a $T^{2}$ deformation of the free
fermion action.

##### KS model

Let us start by reviewing the Goldstino model of [Casalbuoni:1988xh,
Komargodski:2009rz]. Consider the following Lagrangian

$\mathcal{L}_{\rm KS}=\int d^{4}\theta\,\bar{\Phi}\Phi+\int
d^{2}\theta\Big{(}f\Phi+\frac{1}{2}\Lambda\Phi^{2}\Big{)}+\int
d^{2}\bar{\theta}\Big{(}f\bar{\Phi}+\frac{1}{2}\bar{\Lambda}\bar{\Phi}^{2}\Big{)}~{},$
(5)

where $\Phi,\,\bar{\Phi}$ are $D=4$ ${\cal N}=1$ chiral and anti-chiral
superfields, satisfying the constraints
$\bar{D}_{\dot{\alpha}}\Phi=D_{\alpha}\bar{\Phi}=0$. The constant parameter
$f$, which describes the supersymmetry breaking scale, is real. The
superfields $\Lambda,\,\bar{\Lambda}$ are chiral and anti-chiral Lagrange
multipliers whose ${\rm EOM}$ yield the nilpotent constraints

$\Phi^{2}=\bar{\Phi}^{2}=0~{}.$ (6)

The equation of motion for $\Phi$ is

$\frac{1}{4}\bar{D}^{2}\bar{\Phi}=\Lambda\Phi+f~{},\qquad\frac{1}{4}D^{2}\Phi=\bar{\Lambda}\bar{\Phi}+f~{}.$
(7)

As a consequence, we also have

$\Phi\bar{D}^{2}\bar{\Phi}=4f\Phi~{},\qquad\bar{\Phi}D^{2}\Phi=4f\bar{\Phi}~{},$
(8)

where the nilpotent properties of (6) are used. Note that the constraints (6)
and (8) are the ones originally used by Roček to define his Goldstino model
[Rocek:1978nb]. These observations make manifest the on-shell equivalence of
the KS model with Roček’s Goldstino model in a simple superspace setting. The
off-shell equivalence of all these Golstino models up to field redefinitions,
including the VA action, was proven in [Kuzenko:2011tj].

The Lagrange multiplier in (5) imposes the nilpotent constraint $\Phi^{2}=0$
on the chiral superfield $\Phi$. This condition can be solved in terms of the
spinor field $\psi$ and the auxiliary field $F$ of the chiral multiplet,
[Casalbuoni:1988xh, Komargodski:2009rz]:

$\Phi=\frac{\psi^{2}}{2F}+\sqrt{2}\theta\psi+\theta^{2}F~{},$ (9)

which is sensible assuming that $F\neq 0$. Substituting back into (5) gives a
Lagrangian expressed in terms of $\psi$ and the auxiliary field $F$,

${\mathcal{L}}_{\rm
KS}=-\frac{i}{2}\psi\sigma^{\mu}\partial_{\mu}\bar{\psi}+\frac{1}{2}\bar{F}F+\frac{1}{8}\frac{\bar{\psi}^{2}}{\bar{F}}\partial^{2}\Big{(}\frac{\psi^{2}}{F}\Big{)}+fF+c.c.~{}.$
(10)

The auxiliary field can then be eliminated using its equation of motion, which
can be solved in closed form

$F=-f\Bigg{(}1+\frac{\bar{\psi}^{2}}{4f^{4}}\partial^{2}\psi^{2}-\frac{3}{16f^{8}}\psi^{2}\bar{\psi}^{2}\partial^{2}\psi^{2}\partial^{2}\bar{\psi}^{2}\Bigg{)}~{},$
(11)

together with the complex conjugate expression for $\bar{F}$. Plugging (11)
into (10) gives the Goldstino action (4) [Casalbuoni:1988xh,
Komargodski:2009rz].

#####  $D=4$ Goldstino action as a supercurrent-squared flow

One advantage of using the KS model compared to other Goldstino actions is the
relatively simple form of the action, thanks to the Lagrange multiplier, which
makes the computation of its supercurrent easier. The FZ multiplet resulting
from the action (5) is

$\displaystyle{\mathcal{J}}_{\alpha\dot{\alpha}}$
$\displaystyle=2D_{\alpha}\Phi\cdot\bar{D}_{\dot{\alpha}}\bar{\Phi}-\frac{2}{3}[D_{\alpha},\bar{D}_{\dot{\alpha}}](\Phi\bar{\Phi})=\frac{2}{3}D_{\alpha}\Phi\cdot\bar{D}_{\dot{\alpha}}\bar{\Phi}-\frac{2i}{3}\Big{(}\Phi\partial_{\alpha\dot{\alpha}}\bar{\Phi}-\bar{\Phi}\partial_{\alpha\dot{\alpha}}\Phi\Big{)}\,,$
(12) $\displaystyle\mathcal{X}$
$\displaystyle=4\Big{(}f\Phi+\frac{1}{2}\Lambda\Phi^{2}\Big{)}-\frac{1}{3}\bar{D}^{2}(\Phi\bar{\Phi})=\frac{8}{3}f\Phi+2\Lambda\Phi^{2}\,.$
(13)

The composite operators
${\mathcal{J}}^{\alpha\dot{\alpha}}{\mathcal{J}}_{\alpha\dot{\alpha}}$ and
$\mathcal{X}\bar{\mathcal{X}}$ are then

${\mathcal{J}}^{\alpha\dot{\alpha}}{\mathcal{J}}_{\alpha\dot{\alpha}}=\frac{64}{9}f^{2}\Phi\bar{\Phi}+\text{total
derivatives}+{\rm EOM}~{},$ (14)

and

$\mathcal{X}\bar{\mathcal{X}}=\frac{64}{9}f^{2}\Phi\bar{\Phi}+{\rm EOM}~{},$
(15)

where we used (6) and (7). The supercurrent-squared operator (14) then takes
the form

$\mathcal{O}_{T^{2}}=\frac{1}{16}{\mathcal{J}}^{\alpha\dot{\alpha}}{\mathcal{J}}_{\alpha\dot{\alpha}}-\frac{5}{8}\mathcal{X}\bar{\mathcal{X}}=-4f^{2}\Phi\bar{\Phi}+{\rm
EOM}+\text{total derivatives}~{}.$ (16)

The descendant operator ${\sf O}_{T^{2}}$ of eq. (15) becomes

${\sf O}_{T^{2}}=\int d^{2}\theta
d^{2}\bar{\theta}\;\mathcal{O}_{T^{2}}=-4f^{2}\int d^{2}\theta
d^{2}\bar{\theta}\;\Phi\bar{\Phi}=2f^{3}\int d^{2}\theta\,\Phi+2f^{3}\int
d^{2}\bar{\theta}\,\bar{\Phi}~{},$ (17)

where we used (8) in the last equality.

From (5), it is easy to see that the following relation holds:

$\frac{\partial\mathcal{L}_{\rm KS}}{\partial f}=\int d^{2}\theta\,\Phi+\int
d^{2}\bar{\theta}\,\bar{\Phi}~{}.$ (18)

By identifying the coupling constants,

$\gamma=-\frac{1}{4f^{2}}~{},$ (19)

it follows immediately that the KS action,

$S_{\gamma}=\int d^{4}x\,\mathcal{L}_{\rm KS}~{},$ (20)

satisfies the flow equation

$\frac{\partial S_{\gamma}}{\partial\gamma}=\int d^{4}xd^{2}\theta\,\Phi+\int
d^{4}xd^{2}\bar{\theta}\,\bar{\Phi}=\int d^{4}xd^{2}\theta
d^{2}\bar{\theta}\;\mathcal{O}_{T^{2}}=\int d^{4}x\,{\sf O}_{T^{2}}~{}.$ (21)

This proves that (5) satisfies a supercurrent-squared flow (or $T^{2}$ flow)
equation. Because on-shell the actions (4) and (5) are equivalent, and the
equation

$\int d^{2}\theta d^{2}\bar{\theta}\;\mathcal{O}_{T^{2}}={\sf O}_{T^{2}}$

holds, eq. (21) proves that the $D=4$ ${\cal N}=1$ Goldstino action arises
from a supercurrent-squared deformation.888The careful reader may find that
the flow can also be satisfied by other supercurrent-squared operators,
${\mathcal{J}}^{2}-r\bar{\mathcal{X}}\mathcal{X}$, with arbitrary $r$ because
of the linearity between ${\mathcal{J}}^{2}$ and
$\bar{\mathcal{X}}\mathcal{X}$. It is worth pointing out the same thing
happens in $D=2$ [Cribiori:2019xzp]. We stress that this is not the case for
the supercurrent-squared flow satisfied by the $D=4$ supersymmetric Born-
Infeld action.

###  6 Higher Dimensions and Connections to Amplitudes

In this thesis, we have primarily focused on two-dimensional quantum field
theories, where ${T\overline{T}}$ and related supercurrent-squared operators
can be unambiguously defined at the quantum level. For $D>2$, there appears to
be no complete argument showing that any of the proposed operators
$O_{T^{2}}^{[r]}$ of eq. (1), including the holographic operator of
[Taylor:2018xcy, Hartman:2018tkw], possess any particularly nice quantum
properties.

However, despite the absence of a well-defined quantum operator in higher
dimensions, the connection between non-linearly realized symmetries and our
$D=4$, $\mathcal{N}=1$ example suggests that operators of this form still have
some special properties (at least at the classical level). It would be very
interesting to understand whether any supersymmetric completion of our
descendant operator ${\sf O}_{T^{2}}$ of (15) could provide a well-defined
deformation at the quantum level. Such an extension would necessarily involve
contributions from other current-squared type operators which do not vanish
on-shell. This possibility seems most promising in theories with extended, and
perhaps even maximal, supersymmetry.

One interesting direction for future investigation concerns the relationship
between $T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$
deformations and amplitudes. In two dimensions, $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$ simply modifies the
$S$-matrix of the undeformed theory by a CDD factor [Dubovsky:2013ira], but
one might wonder about the $S$-matrices of higher-dimensional theories
deformed by generalizations of $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$. One hint is that
theories with non-linearly realized symmetries exhibit enhanced soft behavior
– indeed, in the case of non-linearly realized supersymmetry, there is a proof
that such symmetries generically lead to constraints on the soft behavior of
the $S$-matrix [Kallosh:2016qvo]. This fact which has been applied to the
Volkov-Akulov action [Kallosh:2016lwj], which also satisfies a $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$-like flow.

There are also examples involving purely bosonic theories. For instance, in
four dimensions, the Dirac action is the unique Lorentz-invariant Lagrangian
for a single scalar which is consistent with factorization, has one derivative
per field, and exhibits soft degree $\sigma=2$ for its scattering amplitudes
[Cheung:2014dqa]. Similarly, it has been shown that the Born-Infeld action for
a vector can be fixed by demanding enhanced soft behavior in a particular
multi-soft limit [Cheung:2018oki], which can be understood in the context of
T-duality and dimensional reduction [Elvang:2019twd]. Given the hints of a
deeper relationship between supercurrent-squared deformations, non-linearly
realized symmetries, and actions of Dirac or Born-Infeld type, it is natural
to ask whether such deformations enhance the soft behavior of scattering
amplitudes in a more general context.

##  Chapter 5 ${T\overline{T}}$ and Non-Abelian Gauge Theory

In this chapter, we will define a theory for a non-abelian gauge field in two
dimensions using the ${T\overline{T}}$ operator, following the discussion in
“A Non-Abelian Analogue of DBI from ${T\overline{T}}$” [Brennan:2019azg].

Although we focus on the bosonic formulation of ${T\overline{T}}$ rather than
the supercurrent-squared deformations developed in the preceding chapters, the
gauge theory which we define here is compatible with maximal supersymmetry.
This deformed theory shares some properties with the Dirac-Born-Infeld theory
which describes gauge fields living on a brane, although it does not reduce to
DBI even in the abelian case. We begin by reviewing some facts and motivation
about brane physics.

###  1 Background on Brane Physics

Imagine a $p$-brane embedded in an ambient $(D+1)$-dimensional Minkowski
space-time. Any such brane spontaneously breaks the Poincaré symmetry of the
ambient space-time:

$ISO(D,1)\,\rightarrow\,ISO(p,1).$ (1)

In particular, the breaking of translational symmetry guarantees the existence
of $D-p$ universal scalar fields on the brane world-volume, collectively
denoted $\phi$, which are Nambu-Goldstone (NG) bosons for the broken
translations. The physics of these modes is governed by the Dirac action,

$S_{\rm Dirac}=-T_{p}\int
d^{p+1}\sigma\,\sqrt{-\det\left(\eta_{\mu\nu}+\partial_{\mu}\phi\partial_{\nu}\phi\right)},$
(2)

with brane world-volume coordinates $\sigma$ and a single dimensionful
parameter $T_{p}$. The form of this action is fixed by the broken Lorentz
symmetries, which are non-linearly realized. There might also be a function of
any additional non-universal scalar fields multiplying this form, which we
will not consider here. This action can also equivalently be viewed as the
Nambu-Goto action for the brane in static gauge.

The Born-Infeld action, on the other hand, defines a non-linear interacting
extension of Maxwell theory with action:

$S_{\rm BI}=-T_{p}\int d^{p+1}\sigma\,\sqrt{-\det\left(\eta_{\mu\nu}+\alpha
F_{\mu\nu}\right)}.$ (3)

One of the striking physical differences between Born-Infeld theory and
Maxwell theory is the existence of a critical electric field determined by the
dimensionful parameter $\alpha$. In string theory, Born-Infeld theory
describes the leading interactions for the gauge-field supported on a D-brane
[Abouelsaood:1986gd]. In that context both $T_{p}$ and $\alpha$ are fixed in
terms of the fundamental string tension $\alpha^{\prime}$ with
$\alpha=2\pi\alpha^{\prime}.$

The combined Dirac-Born-Infeld (DBI) action is a complete description of the
physics of a single D-brane at leading order in string perturbation theory,
and under the assumption that acceleration terms like $\partial F$ or
$\partial^{2}\phi$ are negligible:

$S_{\rm DBI}=-T_{p}\int
d^{p+1}\sigma\,\sqrt{-\det\left(\eta_{\mu\nu}+\partial_{\mu}\phi\partial_{\nu}\phi+\alpha
F_{\mu\nu}\right)}.$ (4)

For multiple coincident branes, the abelian gauge symmetry is replaced by a
non-abelian symmetry, and the fields $(\phi,F)$ typically take values in the
adjoint representation of the gauge group. We will not assume any particular
representation for the scalar fields in this discussion. It is natural to pose
the following long considered question: what might replace
$(\ref{DBI_nonabelian})$ in the non-abelian theory? For the scalar fields
appearing in the induced metric of the Dirac action $(\ref{Dirac})$, one could
easily imagine making the replacement

$\partial_{\mu}\phi\partial_{\nu}\phi\,\rightarrow\,\Tr\left(D_{\mu}\phi
D_{\nu}\phi\right),$ (5)

where $\phi$ is now matrix-valued and $D$ is an appropriate covariant
derivative. For the Born-Infeld action of $(\ref{BI_nonabelian})$, however, an
interesting gauge-invariant replacement of this sort is not possible. In
$(\ref{replace})$ $\Tr$ denotes the trace over gauge indices. When needed, we
will use $\tr$ to denote the trace over Lorentz indices so that,

$\tr(F^{2})=F_{\mu\nu}F^{\nu\mu}.$ (6)

Indeed what one means by the Born-Infeld approximation, namely neglecting
acceleration terms like $DF$, is ambiguous. Unlike the abelian case,

$\commutator{D_{\mu}}{D_{\nu}}=-iF_{\mu\nu},$ (7)

so there is no clear cut way of truncating the full brane effective action by
throwing out acceleration terms.

With considerable hard work there is, however, some data known about brane
couplings beyond the two derivative non-abelian kinetic terms
$\Tr(F_{\mu\nu}F^{\mu\nu})$ at leading order in string perturbation theory.
This information is very nicely summarized in the thesis [Koerber:2004ze] to
which we refer for a more complete discussion. For comparative purposes with
our analysis, we note that the known $F^{4}$ terms are correctly captured by a
symmetrized trace prescription [Tseytlin:1997csa, Bergshoeff:2001dc]. Up to
overall scaling, they are given by

${\rm STr}\left(\tr F^{4}-\frac{1}{4}(\tr F^{2})^{2}\right),$ (8)

with

${\rm STr}\left(T_{1}T_{2}\ldots T_{n}\right)=\frac{1}{n!}\sum_{\rm\sigma\in
S_{n}}\Tr(T_{\sigma(1)}T_{\sigma(2)}\ldots T_{\sigma(n)}).$ (9)

This prescription is known to fail for higher derivative terms. What is
important for us is that $(\ref{f4terms})$ defines a single trace operator.

To define a non-abelian analogue of the abelian DBI theory
$(\ref{DBI_nonabelian})$, our approach will be to ${T\overline{T}}$ deform a
non-abelian gauge theory with scalar matter in two dimensions. In contrast to
the preceding parts of this thesis, in this chapter we will restrict to
bosonic theories for simplicity. A priori this approach has no connection to
either brane physics or string theory. We will do this in steps by first
recalling known results about deforming free scalars and Maxwell fields
[Cavaglia:2016oda, Bonelli:2018kik, Conti:2018jho], and then extending to
charged matter and non-abelian gauge theories. Other than also involving an
infinite collection of irrelevant operators, the reason this approach should
be viewed as giving a non-abelian analogue of DBI is that the
${T\overline{T}}$ deformation of a free scalar with parameter $\lambda$
already gives the Dirac action, as we reviewed in Chapter 1:

$S_{\lambda}=\int
d^{2}\sigma\frac{1}{2\lambda}\left(\sqrt{1+4\lambda|\partial\phi|^{2}}-1\right)~{}.$
(10)

This direct connection with brane physics is reason enough to suspect that
${T\overline{T}}$ applied to gauge-theory will give further insight into brane
physics.

The ${T\overline{T}}$ deformation of Maxwell theory, however, is already
different from the Born-Infeld theory of $(\ref{BI_nonabelian})$. This is the
reason we call the $T\bar{T}$-deformed theory an analogue rather than a
generalization of DBI; it does not reduce to DBI even in the case of abelian
gauge theory. The couplings are not given by the square-root structure of a
relativistic particle but rather by a hypergeometric function [Conti:2018jho].
This might not seem very exciting in two dimensions where pure gauge fields
have no propagating degrees of freedom, but that is no longer the case when we
add scalar fields, even in the abelian setting. For interesting recent
discussions of ${T\overline{T}}$-deformed gauge theories, see
[Santilli:2018xux, Ireland:2019vvj].

For the non-abelian theory defined using ${T\overline{T}}$, the $O(F^{4})$
terms are already very different from what is known about non-abelian BI
theory. Rather than involving a single trace operator like $(\ref{f4terms})$,
they involve double trace operators. The ${T\overline{T}}$-deformed theory is
quite remarkable because it has the following properties:

  * •

The theory is compatible with supersymmetry [Baggio:2018rpv, Chang:2018dge,
Jiang:2019hux, Chang:2019kiu, Coleman:2019dvf, Ferko:2019oyv, Jiang:2019trm].
Indeed, we have discussed at length in Chapters 2 and 3 that ${T\overline{T}}$
always preserves degeneracies in energy levels like those implied by
supersymmetry, and that one can often make the supersymmetry manifest via a
superspace flow construction. In particular, if one ${T\overline{T}}$ deforms
a maximally supersymmetric starting theory then this supersymmetry is
preserved!

  * •

The theory is believed to exist at the quantum level, unlike DBI which is an
effective theory with our present level of understanding.

  * •

The theory has a critical electric field like BI.

This is already quite surprising in the abelian case with uncharged matter.
Folklore suggests that some of these properties, like compatibility with
maximal supersymmetry, should only have been true for DBI. Indeed there are no
obvious reasons that the structures seen here should not emerge from string
theory, either in a closed or open string setting. In fact, the $D=10$ space-
time effective action for the type I/heterotic strings does contain a double
trace $F^{4}$ term, which is required for anomaly cancelation in $D=10$, or
more generally required by supersymmetry [Bergshoeff:1988nn]. In the heterotic
string, the term arises at tree-level and takes the schematic form:

$S_{\rm het}\sim\int d^{10}x\sqrt{g}e^{-2\phi}\left(\Tr
F^{2}\right)^{2}+\ldots.$ (11)

In the dual type I frame, relevant for a brane picture, the same coupling
arises from diagrams with Euler characteristic $-1$ [Tseytlin:1995fy,
Tseytlin:1995bi]. This leads us to suspect that the ${T\overline{T}}$ flow
equation is connected with corrections to two-dimensional beta functions from
higher orders in string perturbation theory.

As a final comment, we note in passing that a different way of deforming a
$2d$ gauge theory is via holography. For instance, $2d$ JT gravity can be
written in gauge theory variables as a BF theory, and one can then
${T\overline{T}}$-deform the $(0+1)$-dimensional dual to this BF theory. The
bulk interpretation of such a deformation is explored in [us:gravity].

###  2 Deforming Pure Gauge Theory

Before we go on to solve the ${T\overline{T}}$ equation to find a non-abelian
analogue of the DBI action, we will first illustrate the above techniques by
solving for the ${T\overline{T}}$-deformed Yang-Mills theory. That is, we
begin with an undeformed Lagrangian of the form

$\displaystyle k\mathcal{L}_{0}$
$\displaystyle=\,F_{\mu\nu}^{a}F^{\mu\nu}_{a}=\,\Tr\left(F_{\mu\nu}F^{\mu\nu}\right)$
(1)

where we will often suppress the trace for convenience and simply write
$F^{2}$ for $\Tr(F^{2})$, so that

$\displaystyle\mathcal{L}_{0}=\frac{1}{k}F^{2}.$ (2)

We retain an overall dimensionless constant $k$ in the Lagrangian; in most of
the calculations that follow, which we will suppress factors of $k$ by setting
$k=1$. The value of $k$ does not affect the equations of motion associated
with the action (2), but the sign of $k$ will be important for determining
critical value of the field strength $F^{2}$. To see the maximum allowed
electric field for the deformed theory in Minkowski signature, we will find
that we must take $k<0$ so that the undeformed action is positive (however,
all of our other results are valid in either Minkowski or Euclidean
signature). We will restore factors of $k$, replacing
$F^{2}\to\frac{1}{k}F^{2}$, when the sign is relevant.

Our goal is to find the deformed Lagrangian
$\mathcal{L}(\lambda)=f(\lambda,F^{2})$ which solves the flow equation

$\partial_{\lambda}{\mathcal{L}}_{\lambda}={\rm det}\,T_{\mu\nu}~{},$ (3)

with initial condition $\mathcal{L}(0)=\mathcal{L}_{0}$.

It is first convenient to rewrite this flow equation using the fact that
$2\times 2$ matrices $M$ satisfy

$\displaystyle\det(M)=\frac{1}{2}\left(\left(\tr
M\right)^{2}-\tr\left(M^{2}\right)\right)\,.$ (4)

Thus we can write the $T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern
1.5mu$ flow equation in a manifestly Lorentz-invariant way as

$\displaystyle\partial_{\lambda}\mathcal{L}_{\lambda}=\frac{1}{2}\left(\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}-T^{\mu\nu}T_{\mu\nu}\right).$
(5)

Notice that the stress-energy tensor is a single-trace operator in the
undeformed theory. In the $\lambda$ expansion, the leading order deformation
of the Lagrangian is therefore automatically a double trace operator.
Including further corrections in $\lambda$ will only generate higher order
multi-trace operators. In particular – as we show in Appendix 7.G and discuss
in more detail in Chapter 6 – a single-trace deformation like the leading
$F^{4}$ terms of non-abelian DBI in $(\ref{f4terms})$ will never be generated
from a ${T\overline{T}}$ flow beginning from an undeformed Lagrangian which is
only a function of $F^{2}$.

The details of the calculation of the stress tensor components
$T_{\mu\nu}^{(\lambda)}$ for the Lagrangian $\mathcal{L}(\lambda,F^{2})$ are
presented in Appendix 7.G, where we find the $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$ operator for an
arbitrary Lagrangian depending on a field strength $F_{\mu\nu}$ and a complex
scalar $\phi$. We can set the scalar $\phi$ to zero in the result of that
Appendix to find the flow equation for a pure gauge field Lagrangian. The
result, using the shorthand notation $x=F^{2}$, is

$\displaystyle\frac{df}{d\lambda}=f(x)^{2}-4f(x)x\frac{\partial f}{\partial
x}+4x^{2}\left(\frac{\partial f}{\partial x}\right)^{2}.$ (6)

Next we will present several methods for solving (6):

  1. 1.

by directly solving the flow equation (5) in a series expansion;

  2. 2.

by writing the solution implicitly in terms of a complete integral;

  3. 3.

and by dualizing the field strength $F^{2}$ to a scalar, deforming, and
dualizing back.

The same three methods will also be applied to the case of a gauge field
coupled to scalars in Section 3.

####  1 Series solution of flow equation

The differential equation (6) derived in the preceding subsection can be
brought into a simpler form by refining our ansatz to

$\displaystyle f(\lambda,F^{2})=F^{2}g(\lambda F^{2})$ (7)

for some new function $g$. For convenience, we define the dimensionless
variable $\chi=\lambda F^{2}=\lambda x$. Then the function $g$ satisfies the
differential equation

$\displaystyle\frac{\partial g}{\partial\chi}=\left(g(\chi)+2\chi
g^{\prime}(\chi)\right)^{2}~{}.$ (8)

One can solve this differential equation by making a series ansatz of the form
$g(\chi)=\sum_{n=0}^{\infty}c_{n}\chi^{n}$, determining the first several
coefficients $c_{n}$. To order $\chi^{6}$, the function $g(\chi)$ is given by

$\displaystyle
g(\chi)=1+\chi+3\chi^{2}+13\chi^{3}+68\chi^{4}+399\chi^{5}+2530\chi^{6}+\mathcal{O}(\chi^{7}).$
(9)

To determine the generating function, one can refer to an encyclopedia of
integer sequences [hypergeometricSequence] to find that $g$ can be written as
a generalized hypergeometric function,

$\displaystyle
g(\chi)={}_{4}F_{3}\left(\frac{1}{2},\frac{3}{4},1,\frac{5}{4};\frac{4}{3},\frac{5}{3},2;\frac{256}{27}\cdot\chi\right)~{}.$
(10)

Thus the full solution for the deformed Lagrangian can be written as

$\displaystyle\mathcal{L}(\lambda)$
$\displaystyle=F^{2}\cdot{}_{4}F_{3}\left(\frac{1}{2},\frac{3}{4},1,\frac{5}{4};\frac{4}{3},\frac{5}{3},2;\frac{256}{27}\cdot\lambda
F^{2}\right)$
$\displaystyle=\frac{3}{4\lambda}\left({}_{3}F_{2}\left(-\frac{1}{2},-\frac{1}{4},\frac{1}{4};\frac{1}{3},\frac{2}{3};\frac{256}{27}\cdot\lambda
F^{2}\right)-1\right).$ (11)

The functions on the first and second lines of (11) are equivalent because of
a hypergeometric functional identity. We will use the expression in the first
line, written in terms of ${}_{4}F_{3}$ rather than ${}_{3}F_{2}$, but we
include the second expression to make contact with the work of
[Conti:2018jho], where this expression was first derived. We also note that
the function (11) has appeared in an analogue of $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$ defined for $(0+1)$
dimensional theories [Gross:2019ach].

####  2 Implicit solution

Later it will be useful to solve the ${T\overline{T}}$ flow equation using a
different method. We begin with the differential equation (6), but this time
we make the ansatz

$\displaystyle f(\lambda,F^{2})=\frac{1}{\lambda}h(\lambda F^{2}).$ (12)

As before, we define the dimensionless variable $\chi=\lambda F^{2}$. In terms
of $h$, the differential equation becomes

$\displaystyle 4\chi^{2}\left(h^{\prime}(\chi)\right)^{2}-4\chi
h(\chi)h^{\prime}(\chi)-\chi h^{\prime}(\chi)+h(\chi)^{2}+h(\chi)=0.$ (13)

Equation (13) is quadratic in $h^{\prime}(\chi)$, so we can solve to find

$\displaystyle\frac{dh}{d\chi}=\frac{1+4h(\chi)-\sqrt{1-8h(\chi)}}{8\chi},$
(14)

where we have chosen the root which makes $h^{\prime}(\chi)$ finite as
$\chi\to 0$, assuming $\lim_{\chi\to 0}h(\chi)=0$.

We may separate variables in (14) to write

$\displaystyle\int\frac{8\,dh}{1+4h(\chi)-\sqrt{1-8h(\chi)}}=\int\frac{d\chi}{\chi}.$
(15)

The integrals can be evaluated in terms of logarithms; exponentiating both
sides then yields

$\displaystyle\chi=C\left(1-\sqrt{1-8h}\right)\left(3+\sqrt{1-8h}\right)^{3}.$
(16)

Equation (16) implicitly defines the solution $h(\chi)$ to the $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$ flow equation via the
roots of an algebraic equation.

We note that, choosing $C=\frac{1}{256}$, equation (16) is consistent with the
solution derived in the previous section. Recall that the two ansatzes we made
here and in subsection (1) are related by

$\displaystyle f(\lambda,F^{2})=F^{2}f(\chi)=\frac{1}{\lambda}h(\chi),$ (17)

so $h(\chi)=\chi f(\chi)$. Indeed, one can check that the function

$\displaystyle
h(\chi)=\chi\cdot{}_{4}F_{3}\left(\frac{1}{2},\frac{3}{4},1,\frac{5}{4};\frac{4}{3},\frac{5}{3},2;\frac{256}{27}\cdot\chi\right),$
(18)

satisfies the functional identity

$\displaystyle\chi=\frac{1}{256}\left(1-\sqrt{1-8h(\chi)}\right)\left(3+\sqrt{1-8h(\chi)}\right)^{3}.$
(19)

We therefore see that the hypergeometric (11) obtained earlier is, in fact, an
algebraic function that can be defined as a root of (19).111Other examples of
hypergeometric functions which can be expressed algebraically include those on
Schwarz’s list [Schwarz1873], which is summarized on Wikipedia.

####  3 Solution via dualization

The above result can also be derived in a different way. The details of this
procedure do not depend on the sign of our constant $k$ nor the signature, so
we will set $k=1$ and take Minkowski signature for concreteness. The
undeformed Lagrangian (2) is then

$\displaystyle\mathcal{L}_{0}=\frac{1}{k}F_{\mu\nu}F^{\mu\nu}=-2F_{01}^{2},$
(20)

which can be equivalently expressed by dualizing the field strength to a
scalar, as in

$\displaystyle\mathcal{L}_{0}$
$\displaystyle=\frac{1}{2}\phi^{2}+\phi\,\epsilon^{\mu\nu}F_{\mu\nu}$
$\displaystyle=\frac{1}{2}\phi^{2}+2\phi F_{01}.$ (21)

The equation of motion for $\phi$ arising from (21) is

$\displaystyle\frac{\delta\mathcal{L}_{0}}{\delta\phi}=\phi+2F_{01}=0,$ (22)

so $\phi=-2F_{01}$, and then replacing $\phi$ with its equation of motion
yields

$\displaystyle\mathcal{L}_{0}=\frac{1}{2}\left(-2F_{01}\right)^{2}+2\left(-2F_{01}\right)F_{01}=-2F_{01}^{2},$
(23)

which matches (20). On the other hand, (21) is easy to $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$ deform. After coupling
to gravity, one has

$\displaystyle
S_{0}[g]=\left(\frac{1}{2}\int\sqrt{-g}\,\phi^{2}\,d^{2}x\right)+\left(\int\phi\,\epsilon^{\mu\nu}F_{\mu\nu}\,d^{2}x\right),$
(24)

where the second term is purely topological and thus independent of the
metric. The undeformed Lagrangian, then, is a pure potential term
$V(\phi)=\phi^{2}$ for the boson $\phi$. The solution to the $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$ flow equation at finite
$\lambda$ for a general potential is well-known [Cavaglia:2016oda,
Bonelli:2018kik]; in this case, one finds

$\displaystyle\mathcal{L}(\lambda)$
$\displaystyle=\frac{\frac{1}{2}\phi^{2}}{1-\frac{\lambda}{2}\phi^{2}}+\phi\,\epsilon^{\mu\nu}F_{\mu\nu}$
$\displaystyle=\frac{\phi^{2}}{2-\lambda\phi^{2}}+2\phi F_{01}.$ (25)

We now integrate out $\phi$. The equation of motion for $\phi$ arising from
(25) is

$\displaystyle 0$
$\displaystyle=\frac{\delta\mathcal{L}(\lambda)}{\delta\phi}$
$\displaystyle=2\phi+F_{01}\left(2-\lambda\phi^{2}\right)^{2},$ (26)

or

$\displaystyle F_{01}=-\frac{2\phi}{(2-\lambda\phi^{2})^{2}}.$ (27)

To proceed, we series expand (27) in $\phi$ to find

$\displaystyle
F_{01}=-\frac{\phi}{2}-\frac{\lambda\phi^{3}}{2}-\frac{3\lambda^{2}\phi^{5}}{8}-\frac{\lambda^{3}\phi^{7}}{4}-\frac{5\lambda^{4}\phi^{9}}{32}+\mathcal{O}\left(\phi^{11}\right),$
(28)

and then apply the Lagrange inversion theorem to find a series expansion for
$\phi$ in terms of $F_{01}$, yielding

$\displaystyle\phi=-2F_{01}+8\lambda
F_{01}^{2}-72\lambda^{2}F_{01}^{5}+832\lambda^{3}F_{01}^{7}-10880\lambda^{4}F_{01}^{9}+\mathcal{O}\left(F_{01}^{11}\right).$
(29)

Substituting the expansion (29) into the action (25), and expressing the
result in terms of $F^{2}=F_{\mu\nu}F^{\mu\nu}=-2F_{01}^{2}$, gives

$\displaystyle\mathcal{L}(\lambda)=F^{2}\left(1+\lambda F^{2}+3\lambda
F^{4}+13\lambda^{3}F^{6}+68\lambda^{4}F^{8}+\cdots\right).$ (30)

The Taylor coefficients appearing in (30) are precisely those of the
hypergeometric (10). The procedure of iteratively solving (27) for $\phi$ and
substituting into (25), therefore, reproduces

$\displaystyle\mathcal{L}(\lambda)=F^{2}\cdot{}_{4}F_{3}\left(\frac{1}{2},\frac{3}{4},1,\frac{5}{4};\frac{4}{3},\frac{5}{3},2;\frac{256}{27}\cdot\lambda
F^{2}\right)$ (31)

which matches the solution which we derived by different methods above.

This procedure – dualizing the field strength $F^{2}$ to a scalar $\phi$,
${T\overline{T}}$ deforming the scalar action, and then dualizing back – is
closely related to an observation made in [Conti:2018jho]. There the authors
noted that, although the deformed Lagrangian (11) is quite complicated, the
corresponding Hamiltonian satisfies the simple relation

$\displaystyle\mathcal{H}_{\lambda}=\frac{\mathcal{H}_{0}}{1-\lambda\mathcal{H}_{0}},$
(32)

where the Hamiltonian is a function of the conjugate momentum

$\displaystyle\Pi^{1}=\frac{\partial\mathcal{L}_{\lambda}}{\partial\dot{A}_{1}}.$
(33)

The Legendre transform which converts the Lagrangian $\mathcal{L}_{\lambda}$
to the Hamiltonian $\mathcal{H}_{\lambda}$ is mathematically equivalent to the
process of dualizing the field strength $F_{01}$ to a scalar $\phi$.

####  4 Comparison to Born-Infeld

For the moment, we specialize to the abelian case where the Born-Infeld action
can be unambiguously defined. The Lagrangian (11) differs from the Born-Infeld
action in two dimensions. To order $\lambda^{4}$, our solution has the series
expansion

$\displaystyle\mathcal{L}(\lambda)=F^{2}+\lambda
F^{4}+3\lambda^{2}F^{6}+13\lambda^{3}F^{8}+68\lambda^{4}F^{10}+\mathcal{O}(\lambda^{5}).$
(34)

On the other hand, the Born-Infeld action (after normalizing the coefficient
of $F^{2}$ to match (34) at order $F^{2}$) has the expansion

$\displaystyle\frac{1}{2\lambda}\sqrt{1+4\lambda
F^{2}}=\frac{1}{2\lambda}+F^{2}-\lambda
F^{4}+2\lambda^{2}F^{6}-5\lambda^{3}F^{8}+14\lambda^{4}F^{10}+\mathcal{O}(\lambda^{5}).$
(35)

Although the Taylor coefficients for the Born-Infeld action and the
“hypergeometric action” differ, both exhibit a critical value for the electric
field. In the case of Born-Infeld, this is obvious; replacing
$F^{2}=-2F_{01}^{2}$, we see that the action

$\displaystyle\frac{1}{2\lambda}\sqrt{1-8\lambda F_{01}^{2}}$ (36)

is only real for

$\displaystyle F_{01}<\frac{1}{\sqrt{8\lambda}}.$ (37)

To see the critical electric field for the action $\mathcal{L}(\lambda)$
defined in (10), it is most convenient to use the implicit form (19):

$\displaystyle\lambda
F^{2}=\frac{1}{256}\left(1-\sqrt{1-8\lambda\mathcal{L}(\lambda)}\right)\left(3+\sqrt{1-8\lambda\mathcal{L}(\lambda)}\right)^{3}.$
(38)

The right side is maximized when $\mathcal{L}(\lambda)=\frac{1}{8\lambda}$,
where it takes the value $\frac{27}{256}$, which means that

$\displaystyle F^{2}<\frac{27}{256\lambda}.$ (39)

Recall that our Lagrangian (2) contained an overall constant to track signs;
to restore factors of $k$, we replace $F^{2}\to\frac{1}{k}F^{2}$. In Minkowski
signature, we should take $k<0$ so that
$\mathcal{L}_{0}=-\frac{1}{k}F_{\mu\nu}F^{\mu\nu}=-\frac{2}{k}F_{01}^{2}$ is
positive. Letting $k=-1$, we find

$\displaystyle F_{01}<\sqrt{\frac{27}{512\lambda}},$ (40)

which is a different critical value for the electric field than (37).

However, pure Yang-Mills theory in two dimensions has no propagating degrees
of freedom, so the difference between the expansions (34) and (35) does not
have much physical effect (at least in infinite volume). To detect the
difference between these theories, we should couple the gauge field to matter,
as we do in section 3.

###  3 Non-Abelian Analogue of DBI

In this section, we will consider an action for a Yang-Mills gauge field
$F_{\mu\nu}^{a}$ coupled to a scalar $\phi$ in some representation of the
gauge group. The undeformed Lagrangian is taken to be

$\displaystyle\mathcal{L}_{0}$
$\displaystyle=F_{\mu\nu}^{a}F^{\mu\nu}_{a}+|D\phi|^{2}$ $\displaystyle\equiv
F^{2}+|D\phi|^{2},$ (1)

where we again use the shorthand $F^{2}=\Tr\left(F_{\mu\nu}F^{\mu\nu}\right)$.
We have set the overall constant $k$, which appears in (2), equal to $1$ for
simplicity. If one were to retain this constant, however, one would need an
extra minus sign in (1) in Minkowski signature.

In what follows, we will also define $x=F^{2}$ and $y=|D\phi|^{2}$ for
convenience; here

$\displaystyle|D\phi|^{2}$
$\displaystyle=\left(D_{\mu}\phi\right)\left(D^{\mu}\phi\right)^{\ast},$
$\displaystyle D_{\mu}$ $\displaystyle=\partial_{\mu}-iA_{\mu},$ (2)

and gauge group indices will be suppressed.

At finite $\lambda$, we take a general ansatz of the form

$\displaystyle\mathcal{L}_{\lambda}=f(\lambda,x=F^{2},y=|D\phi|^{2}).$ (3)

The stress tensor components $T_{\mu\nu}^{(\lambda)}$ for the Lagrangian (3),
and the differential equation arising from (5), are worked out in Appendix
7.G. The resulting partial differential equation, equation (7), is copied here
for convenience:

$\displaystyle\frac{df}{d\lambda}=f^{2}-4fx\frac{\partial f}{\partial
x}-2fy\frac{\partial f}{\partial y}+4x^{2}\left(\frac{\partial f}{\partial
x}\right)^{2}+4xy\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}.$
(4)

Our goal in the following subsections will be to solve (7) by several methods,
just as we did in the case of pure gauge theory.

####  1 Series solution of flow equation

We know that (7) reduces to the Dirac action, (9), when the gauge field is set
to zero, and that it reduces to the hypergeometric action of Section 2, (11),
when the scalar is set to zero. Therefore, in the coupled case it is natural
to make an ansatz of the form

$\displaystyle f(\lambda,x,y)$
$\displaystyle=\frac{1}{2\lambda}\left(\sqrt{1+4\lambda
y}-1\right)+{}_{3}F_{4}\left(\frac{1}{2},\frac{3}{4},1,\frac{5}{4};\frac{4}{3},\frac{5}{3},2;\frac{256}{27}\cdot\lambda
x\right)\cdot x$
$\displaystyle\qquad+\sum_{n=3}^{\infty}\sum_{k=1}^{n}c_{n,k}\lambda^{n-1}x^{k}y^{n-k}.$
(5)

The sum on the final line of (5) allows for all possible couplings between
$F^{2}$ and $|D\phi|^{2}$, with the appropriate power of $\lambda$ required by
dimensional analysis. One can then determine the coefficients $c_{n,k}$ by
plugging the ansatz (5) into (7) and solving order-by-order in $\lambda$. The
result, up to coupled terms of order $\lambda^{8}$, is

$\displaystyle f(\lambda,x,y)$
$\displaystyle=\frac{1}{4\lambda}\left(\sqrt{1+16\lambda
y}-1\right)+{}_{3}F_{4}\left(\frac{1}{2},\frac{3}{4},1,\frac{5}{4};\frac{4}{3},\frac{5}{3},2;\frac{256}{27}\cdot\lambda
x\right)\cdot x$ $\displaystyle\quad-\lambda^{2}xy^{2}+\lambda^{3}\left(4\
xy^{3}-4\,x^{2}y^{2}\right)+\lambda^{4}\left(18\,x^{2}y^{3}-22\,x^{3}y^{2}-14\,xy^{4}\right)$
$\displaystyle\quad+\lambda^{5}\left(-140\,x^{4}y^{2}+104\,x^{3}y^{3}-65\,x^{2}y^{4}+48\,xy^{5}\right)$
$\displaystyle\quad+\lambda^{6}\left(-165\,xy^{6}+220\,x^{2}y^{5}-364\,x^{3}y^{4}+680\,x^{4}y^{3}-969\,x^{5}y^{2}\right)$
$\displaystyle\quad+\lambda^{7}\left(572\,xy^{7}-726\,x^{2}y^{6}+1120\,x^{3}y^{5}-2244\,x^{4}y^{4}+4788\,x^{5}y^{3}-7084\,x^{6}y^{2}\right)$
$\displaystyle\quad+\lambda^{8}\Big{(}-2002\,xy^{8}+2392\,x^{2}y^{7}-3160\,x^{3}y^{6}+5814\,x^{4}y^{5}-14630\,x^{5}y^{4}+35420\,x^{6}y^{3}$
$\displaystyle\qquad\qquad-53820\,x^{7}y^{2}\Big{)}.$ (6)

We were unable to find an closed-form expression for the function which
generates the couplings (6). However, it is interesting to study the
corrections in various approximations.

For instance, consider the coupled terms between $F^{2}$ and $|D\phi|^{2}$ to
leading order in the variable $y=|D\phi|^{2}$. Retaining only the couplings in
(6) proportional to $y^{2}$, one finds

$\displaystyle f(\lambda,x,y)$
$\displaystyle=\frac{1}{4\lambda}\left(\sqrt{1+16\lambda
y}-1\right)+{}_{3}F_{4}\left(\frac{1}{2},\frac{3}{4},1,\frac{5}{4};\frac{4}{3},\frac{5}{3},2;\frac{256}{27}\cdot\lambda
x\right)\cdot x-\lambda^{2}xy^{2}-4\lambda^{3}\,x^{2}y^{2}$
$\displaystyle\quad-2\lambda^{4}\,x^{3}y^{2}-140\lambda^{5}\,x^{4}y^{2}-969\lambda^{6}\,x^{5}y^{2}-7084\lambda^{7}\,x^{6}y^{2}-53820\lambda^{8}\,x^{7}y^{2}$
$\displaystyle\quad+\mathcal{O}\left(\lambda^{9},\lambda^{3}xy^{3}\right).$
(7)

These series coefficients resum into another hypergeometric function
[hypergeometricCoupledSequence],

$\displaystyle f(\lambda,x,y)$
$\displaystyle=\frac{1}{4\lambda}\left(\sqrt{1+16\lambda
y}-1\right)+{}_{3}F_{4}\left(\frac{1}{2},\frac{3}{4},1,\frac{5}{4};\frac{4}{3},\frac{5}{3},2;\frac{256}{27}\cdot\lambda
x\right)\cdot x$
$\displaystyle\quad-\lambda^{2}xy^{2}\,{}_{3}F_{2}\left(\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{2}{3},\frac{4}{3};\frac{256}{27}\cdot\lambda
x\right)+\mathcal{O}\left(\lambda^{3}xy^{3}\right).$ (8)

Defining the hypergeometric appearing in the correction term as

$\displaystyle
g(\chi)={}_{3}F_{2}\left(\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{2}{3},\frac{4}{3};\frac{256}{27}\cdot\chi\right),$
(9)

one can show that $g$ satisfies the functional relation

$\displaystyle\chi=\frac{g(\chi)-1}{g(\chi)^{4}}.$ (10)

The maximum of the function $\frac{g-1}{g^{4}}$ occurs when $g=\frac{4}{3}$,
at which this function takes the maximal value of $\frac{27}{256}$. Therefore,
the maximum value of $\chi$ for which the function (10) is defined is
$\chi=\frac{27}{256}$, giving a critical field strength

$\displaystyle F^{2}<\frac{27}{256\lambda}.$ (11)

We note that this is the same value of the critical electric field as that in
the uncoupled term involving ${}_{3}F_{4}$ in (8), which we saw in (40) for
the case of pure gauge theory. It is reasonable to expect that the value of
the critical electric field is modified if one includes corrections to all
orders in $|D\phi|^{2}$, as is the case for the Dirac-Born-Infeld action.

####  2 Implicit solution

We can instead solve (7) in terms of a complete integral. First, we refine our
ansatz to

$\displaystyle f(\lambda)$
$\displaystyle=\frac{1}{\lambda}g(\chi,\eta)\quad,\quad\chi=\lambda
x~{},~{}\eta=\lambda y~{}.$ (12)

After doing this, the differential equation becomes

$\displaystyle 0=4\chi^{2}\left(\frac{\partial
g}{\partial\chi}\right)^{2}+4\eta\chi\frac{\partial
g}{\partial\chi}\frac{\partial g}{\partial\eta}-4\chi g\frac{\partial
g}{\partial\chi}-\chi\frac{\partial g}{\partial\chi}-2\eta g\frac{\partial
g}{\partial\eta}-\eta\frac{\partial g}{\partial\eta}+g^{2}+g.$ (13)

Making a change of variables to

$\displaystyle p$ $\displaystyle=\log(\chi)\quad,\quad q=\log(\eta)~{},$ (14)

and writing $g(\chi,\eta)=w(p,q)$, the differential equation for $h$ becomes

$\displaystyle 0=4\left(\frac{\partial w}{\partial
p}\right)^{2}+4\frac{\partial w}{\partial p}\frac{\partial w}{\partial
q}-\left(4w+1\right)\frac{\partial w}{\partial
p}-\left(2w+1\right)\frac{\partial w}{\partial q}+w^{2}+w~{}.$ (15)

This can be solved by consulting a handbook of partial differential equations
(see, for instance, equation 15 in section 2.2.6 of [polyanin]). For any
partial differential equation of the form

$\displaystyle 0=f_{1}(w)\left(\frac{\partial w}{\partial
x}\right)^{2}+f_{2}(w)\frac{\partial w}{\partial x}\frac{\partial w}{\partial
y}+f_{3}(w)\left(\frac{\partial w}{\partial
y}\right)^{2}+g_{1}(w)\frac{\partial w}{\partial x}+g_{2}(w)\frac{\partial
w}{\partial y}+h(w),$ (16)

the solution $w(x,y)$ is given implicitly by the following complete integral:

$\displaystyle C_{3}$
$\displaystyle=C_{1}x+C_{2}y+\int\frac{2F(w)\,dw}{G(w)\pm\sqrt{G(w)^{2}-4F(w)h(w)}},$
$\displaystyle F(w)$
$\displaystyle=C_{1}^{2}f_{1}(w)+C_{1}C_{2}f_{2}(w)+C_{2}^{2}f_{3}(w),$
$\displaystyle G(w)$ $\displaystyle=C_{1}g_{1}(w)+C_{2}g_{2}(w).$ (17)

Our equation (15) is precisely of the form (16), after identifying the
independent variables $p\sim x$, $q\sim y$, and with the following functions:

$\displaystyle\begin{split}&f_{1}=f_{2}=4\quad,\quad f_{3}=0\quad,\quad
g_{1}=-4w-1~{},\\\ &g_{2}=-2w-1\quad,\quad h(w)=w^{2}+w.\end{split}$ (18)

Therefore, the functions $F$ and $G$ are

$\displaystyle F(w)$ $\displaystyle=4C_{1}^{2}+4C_{1}C_{2},$ $\displaystyle
G(w)$ $\displaystyle=C_{1}\left(-4w-1\right)+C_{2}\left(-2w-1\right)$
$\displaystyle=\left(-4C_{1}-2C_{2}\right)w-\left(C_{1}+C_{2}\right).$ (19)

Our solution, then, is

$\displaystyle C_{3}=C_{1}p+C_{2}q$ (20)
$\displaystyle+\int\frac{8\left(C_{1}^{2}+C_{1}C_{2}\right)\,dw}{-\left(4C_{1}+2C_{2}\right)w-\left(C_{1}+C_{2}\right)-\sqrt{\left(\left(4C_{1}+2C_{2}\right)w+\left(C_{1}+C_{2}\right)\right)^{2}-16\left(C_{1}^{2}+C_{1}C_{2}\right)\left(w^{2}+w\right)}}~{}.$

where we have taken the negative root in the denominator, appropriate if
$C_{1}+C_{2}<0$.

Choosing values of the constants $C_{1},C_{2},C_{3}$ in (2) gives an implicit
relation for the function $w(p,q)$ which solves the ${T\overline{T}}$ flow
equation. For instance, if we set $C_{2}=0$ and $C_{1}=-1$, equation (2)
becomes

$\displaystyle C_{3}+\log(\chi)=\int\frac{8\,dw}{4w+1-\sqrt{1-8w}},$ (21)

which reproduces the result (15) which we found in the case of pure gauge
theory. In this sense, our implicit solution is a generalization of the
technique of section 2 to the case where $|D\phi|^{2}\neq 0$.

The integral appearing in (2) can be computed explicitly in terms of
logarithms (or, equivalently, inverse hyperbolic tangents). The integral is of
the form

$\displaystyle I(w)=\int\frac{\alpha\,dw}{-\beta w-\gamma-\sqrt{\left(\beta
w+\gamma\right)^{2}-2\alpha\left(w^{2}+w\right)}},$ (22)

where

$\displaystyle\alpha$ $\displaystyle=8(C_{1}^{2}+C_{1}C_{2}),$
$\displaystyle\beta$ $\displaystyle=4C_{1}+2C_{2},$ $\displaystyle\gamma$
$\displaystyle=C_{1}+C_{2}.$ (23)

The result can be written as

$\displaystyle I(w)$
$\displaystyle=\frac{1}{2}\Bigg{(}(\gamma-\beta)\tanh^{-1}\left(\frac{\alpha(w+1)+(\beta-\gamma)(\gamma+\beta
w)}{(\beta-\gamma)\sqrt{\gamma^{2}+w^{2}\left(2\alpha+\beta^{2}\right)+2w(\alpha+\beta\gamma)}}\right)$
$\displaystyle\qquad-\gamma\tanh^{-1}\left(\frac{\gamma^{2}+w(\alpha+\beta\gamma)}{\gamma\sqrt{\gamma^{2}+w^{2}\left(2\alpha+\beta^{2}\right)+2w(\alpha+\beta\gamma)}}\right)$
$\displaystyle\qquad+\sqrt{2\alpha+\beta^{2}}\tanh^{-1}\left(\frac{\alpha+2\alpha
w+\beta(\gamma+\beta
w)}{\sqrt{2\alpha+\beta^{2}}\sqrt{\gamma^{2}+w^{2}\left(2\alpha+\beta^{2}\right)+2w(\alpha+\beta\gamma)}}\right)$
$\displaystyle\qquad+\left(\beta-\gamma\right)\log(w+1)+\gamma\log(w)\Bigg{)}.$
(24)

Exponentiating both sides then gives

$\displaystyle\exp\left(C_{3}-C_{1}p-C_{2}q\right)=\exp\left(I(w)\right).$
(25)

After simplifying the exponentials of the inverse hyperbolic tangents in (25),
the right side only involves rational functions and radicals. This relation,
therefore, gives an algebraic equation in $w$ whose roots are the solution to
the $T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$ flow.

By construction, a function $w(p,q)$ which satisfies (25) solves the
differential equation (15). However, this technique is more unwieldy than the
direct series solution for generating Taylor coefficients. The main utility of
this strategy is the conceptual result that the solution $w(p,q)$ is, in
principle, defined by the root of an equation which involves only radicals and
quotients, as we saw for the pure gauge theory case in (19).

####  3 Solution via dualization

One can also apply the dualization technique of section (2) to the coupled
action. Begin with the undeformed action

$\displaystyle\mathcal{L}_{0}=|D\phi|^{2}+F^{2}~{},$ (26)

where we put $k=1$ since the sign will not affect this calculation. Exactly as
before, this action is equivalent to

$\displaystyle\mathcal{L}_{0}=|D\phi|^{2}+\frac{1}{2}\chi^{2}+\chi\epsilon^{\mu\nu}F_{\mu\nu}~{},$
(27)

after integrating out the field $\chi$, although this form of the Lagrangian
hides some complexity because the covariant derivative $D$ is now non-local in
$\chi$. Ignoring this for the moment, we again note that the action coupled to
a background metric is of the form

$\displaystyle
S_{0}=\left(\int\sqrt{-g}\,d^{2}x\,\left(|D\phi|^{2}+\frac{1}{2}\chi^{2}\right)\right)+\left(\int
d^{2}x\,\chi\epsilon^{\mu\nu}F_{\mu\nu}\right)~{}.$ (28)

As far as the $T\mkern 1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$
deformation is concerned, (28) is simply the action of a complex scalar $\phi$
with a constant potential $V=\frac{1}{2}\chi^{2}$. The solution to the flow
equation at finite $\lambda$ is [Cavaglia:2016oda, Bonelli:2018kik]

$\displaystyle\mathcal{L}(\lambda)=\frac{1}{2\lambda}\sqrt{\frac{\left(1-\lambda\chi^{2}\right)^{2}}{\left(1-\frac{1}{2}\lambda\chi^{2}\right)^{2}}+2\lambda\,\frac{2|D\phi|^{2}+\chi^{2}}{1-\frac{1}{2}\lambda\chi^{2}}}-\frac{1}{2\lambda}\,\frac{1-\lambda\chi^{2}}{1-\frac{1}{2}\lambda\chi^{2}}+\chi\epsilon^{\mu\nu}F_{\mu\nu}~{}.$
(29)

As in section (2), one might hope to iteratively integrate out the auxiliary
field $\chi$ in (29) in order to express the result in terms of $F^{2}$. The
equation of motion for $\chi$ resulting from (29), after solving for $F_{01}$
(and assuming that $\lambda\chi^{2}<2$), is

$\displaystyle
F_{01}=\frac{\chi\left(-|D\phi|^{2}\lambda\left(2-\lambda\chi^{2}\right)-\sqrt{2|D\phi|^{2}\lambda\left(2-\lambda\chi^{2}\right)+1}-1\right)}{\left(2-\lambda\chi^{2}\right)^{2}\sqrt{2|D\phi|^{2}\lambda\left(2-\lambda\chi^{2}\right)+1}}~{}.$
(30)

Solving (30) by series inversion to give $\chi$ as a function of $F_{01}$,
then substituting back into (29), then determines the full $T\mkern
1.5mu\overline{\mkern-1.5muT\mkern-1.5mu}\mkern 1.5mu$ deformed action. The
result, up to order $F^{8}$ and using the shorthand $x=F^{2},y=|D\phi|^{2}$,
is

$\displaystyle\mathcal{L}(\lambda)=\frac{\sqrt{1+4\lambda
y}-1}{2\lambda}+\frac{2x\left(\sqrt{1+4\lambda y}+2\lambda
y\left(\sqrt{1+4\lambda y}+2\right)+1\right)}{\left(2\lambda
y+\sqrt{1+4\lambda y}+1\right)^{2}}$ $\displaystyle\quad+\frac{16\lambda
x^{2}}{\left(2\lambda y+\sqrt{1+4\lambda
y}+1\right)^{5}}\cdot\Bigg{[}2\lambda^{3}y^{3}\left(3\sqrt{1+4\lambda
y}+14\right)+\lambda^{2}y^{2}\left(17\sqrt{1+4\lambda y}+31\right)$
$\displaystyle\hskip 150.0pt+2\lambda y\left(4\sqrt{1+4\lambda
y}+5\right)+\sqrt{1+4\lambda y}+1\Bigg{]}$
$\displaystyle\quad+\frac{128\lambda^{2}x^{3}}{\left(2\lambda
y+\sqrt{1+4\lambda
y}+1\right)^{8}}\cdot\Bigg{[}4\lambda^{5}y^{5}\left(13\sqrt{1+4\lambda
y}+96\right)+2\lambda^{4}y^{4}\left(183\sqrt{1+4\lambda y}+496\right)$
$\displaystyle\hskip 150.0pt+8\lambda^{3}y^{3}\left(57\sqrt{1+4\lambda
y}+101\right)+2\lambda^{2}y^{2}\left(106\sqrt{1+4\lambda y}+145\right)$
$\displaystyle\hskip 150.0pt+6\lambda y\left(7\sqrt{1+4\lambda
y}+8\right)+3\left(\sqrt{1+4\lambda y}+1\right)\Bigg{]}$
$\displaystyle\quad+\frac{1024\lambda^{3}x^{4}}{\left(2\lambda
y+\sqrt{1+4\lambda
y}+1\right)^{11}}\cdot\Bigg{[}2\lambda^{7}y^{7}\left(323\sqrt{1+4\lambda
y}+3266\right)+\lambda^{6}y^{6}\left(8471\sqrt{1+4\lambda y}+30585\right)$
$\displaystyle\hskip 150.0pt+2\lambda^{5}y^{5}\left(9879\sqrt{1+4\lambda
y}+22795\right)+4\lambda^{4}y^{4}\left(4589\sqrt{1+4\lambda y}+8019\right)$
$\displaystyle\hskip 150.0pt+2\lambda^{3}y^{3}\left(4244\sqrt{1+4\lambda
y}+6093\right)+\lambda^{2}y^{2}\left(2083\sqrt{1+4\lambda y}+2577\right)$
$\displaystyle\hskip 150.0pt+26\lambda y\left(10\sqrt{1+4\lambda
y}+11\right)+13\left(\sqrt{1+4\lambda y}+1\right)\Bigg{]}~{}.$ (31)

We have checked by explicit computation that the series expansion (31) solves
the flow equation (7) to order $x^{4}$.

##  Chapter 6 ${T\overline{T}}$ Deformations of Gauge Theories in Arbitrary
Dimension

This chapter aims to extend the study of ${T\overline{T}}$ deformations of
pure gauge theories – which was discussed, in part, in Chapters 4 and 5 – to
general spacetime dimension. Because an unambiguous definition of the
${T\overline{T}}$ operator at the quantum level is available only in $d=2$,
the treatment of this chapter will be totally classical. We will view stress
tensor deformations as merely a mechanism for generating interesting one-
parameter families of classical Lagrangians, rather than an irrelevant
deformation by an operator which admits a definition by point-splitting.

Unlike the majority of this thesis, the current chapter contains new results
which do not appear in any previous publications. I thank Savdeep Sethi and
Gabriele Tartaglino-Mazzucchelli for useful discussions which prompted this
work.111I am also grateful to Mukund Rangamani, Julio Virrueta, Shruti
Paranjape, Brendan Halstead, and Luna Dole for fruitful conversations about
extensions of these results to the $p$-form and non-abelian settings.

###  1 Motivation and $3d$ Example

In the preceding chapters, we have seen that interesting gauge theories can be
built from ${T\overline{T}}$ flows in $d=2$ dimensions and in $d=4$
dimensions. In $d=2$, deforming the theory of a gauge field coupled to charged
matter gives a quantum-mechanically well-defined theory which is compatible
with maximal supersymmetry. This theory does not reduce to the Dirac-Born-
Infeld (DBI) action, even in the case of an abelian gauge group, and the pure
gauge part of the action is written in terms of a hypergeometric function. In
$d=4$, deforming the theory of an abelian gauge field using a
${T\overline{T}}$-like operator gives the Born-Infeld action, although in this
case we do not expect that the operator is well-defined by point-splitting
without imposing extended supersymmetry.

It is natural to ask whether interesting actions arise from deformations by
stress tensor bilinears in other numbers of spacetime dimensions besides $d=2$
and $d=4$. Although the analysis for $d=3$ is a special case of the one that
we will develop later in this chapter, we will include it explicitly here by
way of motivating some of the ideas which appear in the general case. In
particular, we will ask whether a Born-Infeld-type Lagrangian involving a
square root can be obtained from deforming the $d=3$ Maxwell Lagrangian by any
combination of stress tensor bilinears.

First consider a general Lagrangian depending on a field strength $F_{\mu\nu}$
and parameter $\lambda$. As we will prove later, in $d=3$ the only scalar
which can be constructed from $F_{\mu\nu}$ is $F^{2}=F^{\mu\nu}F_{\mu\nu}$.
Thus an arbitrary Lagrangian can be written as

$\displaystyle\mathcal{L}=f(\lambda,F^{2}).$ (1)

For convenience, let $x=F^{2}$. We can now compute the stress-energy tensor by
coupling to a background metric and varying with respect to the metric, where
our conventions will be given in (8). One finds

$\displaystyle T_{\mu\nu}=\eta_{\mu\nu}\,f-4\frac{\partial f}{\partial
x}\tensor{F}{{}_{\mu}^{\sigma}}F_{\nu\sigma}.$ (2)

We then compute the two combinations $T^{\mu\nu}T_{\mu\nu}$ and
$\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}$. The first is

$\displaystyle T^{\mu\nu}T_{\mu\nu}$
$\displaystyle=\left(\eta^{\mu\nu}\,f-4\frac{\partial f}{\partial
x}\tensor{F}{{}^{\mu}{}^{\sigma}}\tensor{F}{{}^{\nu}_{\sigma}}\right)\left(\eta_{\mu\nu}\,f-4\frac{\partial
f}{\partial x}\tensor{F}{{}_{\mu}^{\rho}}F_{\nu\rho}\right)$
$\displaystyle=3f^{2}-8f\frac{\partial f}{\partial
x}F^{2}+16\left(\frac{\partial f}{\partial
x}\right)^{2}F^{\mu\sigma}\tensor{F}{{}_{\sigma}^{\nu}}\tensor{F}{{}_{\mu}^{\rho}}F_{\rho\nu}.$
(3)

The last term,
$F^{\mu\sigma}\tensor{F}{{}_{\sigma}^{\nu}}\tensor{F}{{}_{\mu}^{\rho}}F_{\rho\nu}$,
naively appears to be a new scalar quantity constructed from $F_{\mu\nu}$
which defines an independent quantity in the action. However, in three
spacetime dimensions, one has the relation

$\displaystyle
F^{\mu\sigma}\tensor{F}{{}_{\sigma}^{\nu}}\tensor{F}{{}_{\mu}^{\rho}}F_{\rho\nu}=\frac{1}{2}\left(F_{\mu\nu}F^{\mu\nu}\right)^{2}.$
(4)

This is analogous to the fact that, in $d=2$ spacetime dimensions, one has
$\tensor{F}{{}_{\mu}^{\sigma}}F_{\sigma\nu}=-\frac{1}{2}\eta_{\mu\nu}\left(F_{\alpha\beta}F^{\alpha\beta}\right)$.
This was used in Appendix 7.G to obtain the general flow equation for two-
dimensional theories which was analyzed in Chapter 5; in both cases, we are
exploiting the fact that in low spacetime dimension ($d=2,3$), the number of
independent scalars that can be constructed from $F_{\mu\nu}$ is small.

Replacing $F_{\mu\nu}F^{\mu\nu}=x$, we then find

$\displaystyle T^{\mu\nu}T_{\mu\nu}=3f^{2}-8fx\frac{\partial f}{\partial
x}+8\left(\frac{\partial f}{\partial x}\right)^{2}x^{2}.$ (5)

The second combination is

$\displaystyle\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}$
$\displaystyle=\left(3f-4F^{2}\frac{\partial f}{\partial
x}\right)^{2}=9f^{2}-24fx\frac{\partial f}{\partial x}+16\left(\frac{\partial
f}{\partial x}\right)^{2}x^{2}\,.$ (6)

We would like to consider an arbitrary flow equation for the Lagrangian which
is built out of these stress tensor bilinears,

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=aT^{\mu\nu}T_{\mu\nu}-b\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}\,.$
(7)

Of course, we can absorb an overall factor of $a$ into the normalization of
$\lambda$ and then define the relative coefficient $-\frac{b}{a}$ as a new
parameter $r$. So it suffices to consider flow equations of the form

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=T^{\mu\nu}T_{\mu\nu}-r\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}\,.$
(8)

Using our expressions for the two stress tensor products derived above, the
general flow equation is

$\displaystyle\frac{df}{d\lambda}$
$\displaystyle=T^{\mu\nu}T_{\mu\nu}-r\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}$
$\displaystyle=3\left(1-3r\right)f^{2}-8\left(1-3r\right)fx\frac{\partial
f}{\partial x}+8\left(1-2r\right)\left(\frac{\partial f}{\partial
x}\right)^{2}x^{2},$ (9)

This flow equation may have interesting solutions for various values of the
relative coefficient $r$. For the moment, we will content ourselves by asking
whether it has any solutions of Born-Infeld type. In three spacetime
dimensions, one has

$\displaystyle\sqrt{-\det(\eta_{\mu\nu}+\alpha
F_{\mu\nu})}=\sqrt{1+\frac{1}{2}\alpha^{2}F^{2}}\,.$ (10)

Viewing $\lambda$ as playing the role of $\alpha^{2}$, we ask whether we can
ever get solutions of the form

$\displaystyle f(\lambda,x)=\frac{a}{\lambda}\left(\sqrt{1+b\lambda
x}-1\right)\,,$ (11)

which by construction is proportional to the Maxwell action as $\lambda\to
0$.222We briefly comment on dimensional analysis. Ordinarily, the free action
in $d$ spacetime dimensions is $S=-\frac{1}{4g^{2}}\int
d^{d}x\,F^{\mu\nu}F_{\mu\nu}$ and $F_{\mu\nu}$ has mass dimension $2$.
However, it is convenient to re-scale $F\to\frac{1}{g}F$ and eliminate the
prefactor of $\frac{1}{g^{2}}$ in the Lagrangian. In these conventions, the
combination $x=F^{2}$ has mass dimension $d$ so the field strength
$F_{\mu\nu}$ has mass dimension $\frac{d}{2}$. Similarly, the stress tensor
$T_{\mu\nu}$ has mass dimension $d$ in any number of dimensions, so stress
tensor bilinears like ${T\overline{T}}$ have mass dimension $2d$. Therefore,
by dimensional analysis of the flow equation
$\frac{\partial\mathcal{L}}{\partial\lambda}={T\overline{T}}$, we see that
$\lambda$ must have length dimension $d$. Thus in the case $d=3$ of interest
here, $\lambda$ has length dimension $3$. This means that the combination
$\lambda x$ under the square root in (11) is dimensionless, as required, and
the overall factor of $\frac{1}{\lambda}$ has the same dimension as $F^{2}$,
which is the appropriate dimension for a Lagrangian density.

By using the ansatz (11) in the general flow equation

$\displaystyle\frac{df}{d\lambda}=c_{1}f^{2}+c_{2}fx\frac{\partial f}{\partial
x}+c_{3}\left(\frac{\partial f}{\partial x}\right)^{2}x^{2}\,,$ (12)

one can see that (12) will have solutions of this square-root-type only if
$c_{3}=0$ and $c_{2}=-2c_{1}$, in which case the solution is

$\displaystyle f(\lambda,x)=\frac{1}{2\lambda
c_{1}}\left(\sqrt{1+4c_{2}\lambda x}-1\right)\,,$ (13)

where $c_{1},c_{2}$ are fixed by the initial condition, which in this case is
$f(0,x)=-\frac{x}{4}$. However, even if we choose $r=\frac{1}{2}$ so that the
unwanted final term in (1) drops out, the differential equation for our three-
dimensional flow is

$\displaystyle\frac{df}{d\lambda}=-\frac{3}{2}f^{2}+4fx\frac{\partial
f}{\partial x}.$ (14)

This does not have a square-root-type solution, since it is not of the form
(12) with $c_{2}=-2c_{1}$; instead we have $c_{2}=4$ and $-2c_{1}=3$. Thus the
Born-Infeld action cannot be obtained via any ${T\overline{T}}$-like
deformation of the Maxwell action in three spacetime dimensions.

The above analysis raises several questions. Can the Born-Infeld action be
realized as a ${T\overline{T}}$ flow in any other dimension $d>4$? Can we
obtain other interesting actions by applying ${T\overline{T}}$-like
deformations to seed theories other than the free Maxwell action? For
instance, perhaps we would like to begin with the Born-Infeld action as the
seed theory, rather than as a desired output of deformation, and see what
action is produced by the flow. What additional complications arise when
deforming non-abelian gauge theories? For instance, can one define non-abelian
analogues of BI in higher dimensions using ${T\overline{T}}$-like flows as
well?

The aim of this chapter is to present a framework for answering questions like
this by considering a ${T\overline{T}}$-like deformation of a quite general
gauge theory in arbitrary spacetime dimension. Throughout this chapter, we
will restrict to pure gauge theory without any charged matter. In the abelian
case, we will be able to explicitly derive a flow equation for a family of
${T\overline{T}}$-type deformations of such general theories. Interestingly,
the only case for which deformation of a free Maxwell theory yields a
Lagrangian of Born-Infeld type occurs for $d=4$. We will also discuss the more
subtle case of deformations for non-abelian gauge theories, where it is
immediate that one can never obtain a deformed Lagrangian of Born-Infeld type
because ${T\overline{T}}$ flows cannot generate terms with the appropriate
trace structure.

###  2 Deformations of General Abelian Gauge Theories

In this section, we will restrict to the simpler case of a $U(1)$ gauge field
$A_{\mu}$, which avoids complications from traces over group indices.

####  1 Signature and Sign Conventions

Unlike in the $3d$ example of the introductory section 1, for the remainder of
this chapter we will work in Euclidean signature in $d$ spacetime dimensions.
In this signature the field strength $F$ can be thought of as a $d\times d$
matrix regardless of the positioning of the indices, which are raised or
lowered with $\delta_{\mu\nu}$.

Our conclusions will be valid for either Euclidean or Minkowski signature, so
the choice of restricting to Euclidean signature is merely for simplicity; we
did not feel that the additional generality of treating both cases merited the
introduction of a constant $k=\pm 1$ and writing
$\mathcal{L}=\frac{k}{4}F_{\mu\nu}F^{\mu\nu}$ as in Chapter 5. We can afford
to be somewhat cavalier about the overall sign of the Lagrangian because our
goal is to understand deformations by stress tensor bilinears, and since
reversing the sign of the Lagrangian also flips the sign of $T_{\mu\nu}$, all
bilinears in $T_{\mu\nu}$ are insensitive to the sign of $\mathcal{L}$.
However, to avoid sign confusion and reiterate our conventions, we will state
a few formulas here.

We will take the Euclidean action for free Maxwell theory to be

$\displaystyle S_{\text{Euc}}=\frac{1}{4}\int
d^{d}x\,F_{\mu\nu}F^{\mu\nu}\equiv\frac{1}{4}\int d^{d}x\,F^{2}\,.$ (1)

The factors of the coupling $g$ have been absorbed into the field strength,
following the choice in footnote 2, so that $F_{\mu\nu}$ now has mass
dimension $\frac{d}{2}$. We emphasize that this is in contrast to the Lorentz
signature action,

$\displaystyle S_{\text{Lor}}=-\frac{1}{4}\int
d^{d}x\,F_{\mu\nu}F^{\mu\nu}\,,$ (2)

which was only used in the introductory discussion of Section 1 but will be
mostly discarded in favor of (1) from here on.

In this chapter, expressions like $F^{2n}_{\mu\nu}$ (which we will often write
with indices suppressed as $F^{2n}$) will always refer to the $d\times d$
matrix obtained by successively performing $2n$ matrix multiplications of
$F_{\mu\nu}$ by itself, leaving the first and last indices free:

$\displaystyle
F^{2n}_{\mu\nu}=\underbrace{F_{\mu\alpha_{1}}\tensor{F}{{}^{\alpha_{1}}_{\alpha_{2}}}\cdots\tensor{F}{{}^{\alpha_{2n-2}}_{\alpha_{2n-1}}}\tensor{F}{{}^{\alpha_{2n-1}}_{\nu}}}_{2n\text{
copies}}\,.$ (3)

Likewise, $\tr(M_{\mu\nu})=\delta^{\mu\nu}M_{\mu\nu}$ refers to the trace of
any matrix $M_{\mu\nu}$ with the Kronecker delta.333As in Chapter 5, we use
the notation $\tr$ for the trace over tensor indices and reserve $\Tr$ for the
trace over gauge indices in the non-abelian setting.

At the risk of being pedantic, we point out that this convention introduces a
potentially confusing sign in expressions like $\tr(F^{2})$. In our
conventions, the matrix $F^{2}_{\mu\nu}$ is defined according to (3) as

$\displaystyle F^{2}_{\mu\nu}=F_{\mu\rho}\tensor{F}{{}^{\rho}_{\nu}}\,,$ (4)

and therefore its trace is

$\displaystyle\tr(F^{2}_{\mu\nu})=\delta^{\mu\nu}F_{\mu\rho}\tensor{F}{{}^{\rho}_{\nu}}=F^{\nu\rho}F_{\rho\nu}\,.$
(5)

In particular, this means that

$\displaystyle\tr(F^{2})=F^{\nu\rho}F_{\rho\nu}=-F^{\rho\nu}F_{\rho\nu}=-F^{2}\,,$
(6)

using the notation $F^{2}=F^{\mu\nu}F_{\mu\nu}$ as above and in previous
chapters. For this reason, in our notation the Euclidean action
$S_{\text{Euc}}$ can be written as

$\displaystyle S_{\text{Euc}}=\frac{1}{4}\int d^{d}x\,F^{2}\,=-\frac{1}{4}\int
d^{d}x\,\tr(F^{2})\,.$ (7)

Our convention for the Hilbert stress energy tensor is

$\displaystyle T_{\mu\nu}=\frac{-2}{\sqrt{g}}\frac{\delta S}{\delta
g^{\mu\nu}}=g_{\mu\nu}\mathcal{L}-2\frac{\partial\mathcal{L}}{\partial
g^{\mu\nu}}\,.$ (8)

For example, the stress tensor associated with the Euclidean Lagrangian (1) is

$\displaystyle
T_{\mu\nu}^{\text{Euc}}=\frac{1}{4}g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}-F_{\mu\alpha}\tensor{F}{{}_{\nu}^{\alpha}}\,,$
(9)

while that for the Lorentzian Lagrangian (2) is

$\displaystyle
T_{\mu\nu}^{\text{Lor}}=-\frac{1}{4}g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}+F_{\mu\alpha}\tensor{F}{{}_{\nu}^{\alpha}}=-T_{\mu\nu}^{\text{Euc}}\,.$
(10)

So for instance (9) agrees with the Maxwell stress tensor (5) in Chapter 4.
But as we have emphasized, the stress tensor bilinears which enter into our
${T\overline{T}}$-like deformations are identical in the two cases:

$\displaystyle\left(T^{\text{Euc}}\right)_{\mu\nu}\left(T^{\text{Euc}}\right)^{\mu\nu}$
$\displaystyle=\left(T^{\text{Lor}}\right)_{\mu\nu}\left(T^{\text{Lor}}\right)^{\mu\nu}\,,$
$\displaystyle\tensor{\left(T^{\text{Euc}}\right)}{{}^{\mu}_{\mu}}\tensor{\left(T^{\text{Euc}}\right)}{{}^{\nu}_{\nu}}$
$\displaystyle=\tensor{\left(T^{\text{Lor}}\right)}{{}^{\mu}_{\mu}}\tensor{\left(T^{\text{Lor}}\right)}{{}^{\nu}_{\nu}}\,.$
(11)

One might object that, although the _deformations_ constructed from bilinears
in $T_{\mu\nu}$ are invariant under reversing the sign of $\mathcal{L}$, the
_solution_ to flow equations with initial condition $\mathcal{L}_{0}$ might
differ since $\mathcal{L}_{0}$ disagrees by a sign in the two signatures.
However, this ambiguity can be absorbed into a re-scaling of the flow
parameter $\lambda$. Since all of the analysis of this chapter will hold for
either sign of $\lambda$, we are free to modify a flow equation of the form

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=T^{\mu\nu}T_{\mu\nu}-r\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}$
(12)

by sending $\lambda\to\lambda^{\prime}=-\lambda$, so that

$\displaystyle\frac{\partial\left(-\mathcal{L}\right)}{\partial\lambda^{\prime}}=T^{\mu\nu}T_{\mu\nu}-r\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}\,.$
(13)

This operation allows us to map a solution to a flow equation with given
initial condition $\mathcal{L}_{0}$ to a solution with initial condition
$-\mathcal{L}_{0}$.

Of course, we have seen in the preceding chapters that the two signs of the
${T\overline{T}}$ flow parameter $\lambda$ lead to very different behavior.
Consider the example of deforming a $2D$ CFT. For a “good sign” choice, all of
the finite-volume energies in the deformed theory remain real, at least for
sufficiently small $\lambda$. The other “bad sign” choice generically leads to
a deformed theory where all but finitely many of the energies are complex. One
might therefore say that we should be more careful about making arguments
which require reversing the sign of $\lambda$, since this can lead to
pathologies in the quantum theory.

However, as we mentioned above, in this chapter we are concerned exclusively
with flow equations for classical Lagrangians. Indeed, since we work in
arbitrary dimension, our deforming operators are not even known to exist at
the quantum level. Thus it is perfectly acceptable to treat $\lambda$ as a
formal variable labeling a one-parameter family of Lagrangians and treat
either sign of this variable as equally valid.

This concludes the (somewhat tedious) discussion of our sign manifesto and
conventions for the present chapter. We now return to the more interesting
task of obtaining a flow equation for an arbitrary abelian gauge theory in $d$
dimensions.

####  2 Enumeration of Trace Structures

We must first parameterize the independent scalar quantities which can be
built from $F_{\mu\nu}$. This is obviously a dimension-dependent question. For
instance, in $d=4$, every scalar constructed from $F_{\mu\nu}$ can be
expressed in terms of two combinations

$\displaystyle x_{1}=F_{\mu\nu}F^{\nu\mu}=\tr(F^{2})\,,\qquad
x_{2}=F^{\mu\sigma}\tensor{F}{{}_{\sigma}^{\nu}}\tensor{F}{{}_{\nu}^{\rho}}F_{\rho\mu}=\tr(F^{4})\,.$
(14)

Equivalently, if we express the $4d$ Euclidean field strength tensor as

$\displaystyle F_{\mu\nu}=\begin{bmatrix}0&E_{x}&E_{y}&E_{z}\\\
-E_{x}&0&-B_{z}&B_{y}\\\ -E_{y}&B_{z}&0&-B_{x}\\\
-E_{z}&-B_{y}&B_{x}&0\end{bmatrix}_{\mu\nu}\,,$ (15)

then the two invariants satisfy

$\displaystyle x_{1}$
$\displaystyle=-2\left(\vec{E}^{2}+\vec{B}^{2}\right)\,,$ $\displaystyle
x_{2}$
$\displaystyle=\frac{1}{2}x_{1}^{2}-4\left(\vec{E}\cdot\vec{B}\right)^{2}\,.$
(16)

The fact that all scalars can be written in terms of $x_{1}$ and $x_{2}$ might
be more familiar if phrased in Lorenztian signature, where the two Lorentz
invariants that can be constructed from the field strength tensor in $d=4$ are
$\vec{E}^{2}-\vec{B}^{2}$ and $\vec{E}\cdot\vec{B}$, which are related to
$F\wedge\ast F$ and $F\wedge F$ respectively.

To begin the enumeration of independent scalars, we first recall a very basic
fact which follows from the Cayley-Hamilton theorem: every matrix satisfies
its own characteristic equation. That is, any $d\times d$ matrix $M$ obeys a
condition of the form

$\displaystyle
p(M)=M^{d}+c_{d-1}M^{d-1}+\cdots+c_{1}M+(-1)^{d}\det(M)I_{d}=0\,,$ (17)

where $I_{d}$ is the $d\times d$ identity matrix. The coefficients $c_{i}$ are
functions only of the traces $\mathrm{tr}\big{(}M^{l}\big{)}$ for
$l=1,\cdots,d$. Explicitly, they are given by

$\displaystyle
c_{i}=\sum_{\\{k_{l}\\}}\,\prod_{l=1}^{d}\,\frac{(-1)^{k_{l}+1}}{l^{k_{l}}k_{l}!}\left(\mathrm{tr}\,\big{(}M^{l}\big{)}\,\right)^{k_{l}}\,,$
(18)

where the sum runs over all sets of non-negative integers $k_{l}$ which
satisfy

$\displaystyle\sum_{l=1}^{d}lk_{l}=d-i\,.$ (19)

The condition (17) implies a limit on the number of independent trace
structures that a matrix can have, as we now review.444Another way to think
about this fact is that each $\tr(M^{k})$ is the sum
$\lambda_{1}^{k}+\cdots+\lambda_{d}^{k}$ where the $\lambda_{i}$ are
eigenvalues of $M$. Knowing the first $d$ such sums gives $d$ equations, which
is sufficient to determine all $d$ of the eigenvalues. From this data, any
trace $\tr(M^{n})$ can be computed. For antisymmetric matrices the eigenvalues
come in pairs $\pm\lambda$, except in odd dimensions where there is one
unpaired zero eigenvalue.

Suppose that we are given all of the traces

$\displaystyle\tr(M)\,,\,\tr\big(M^{2}\big{missing})\,,\,\cdots\,,\,\tr\big(M^{d}\big{missing})\,.$
(20)

We claim that all higher traces $\tr\big(M^{n}\big{missing})$ for $n>d$ can
then be determined. To see this, first note that $\det(M)$ is also known,
since we can take the trace of equation (17) and then solve for $\det(M)$ in
terms of the known traces:

$\displaystyle\det(M)=\frac{(-1)^{d+1}}{d}\left(\tr(M^{d})+c_{d-1}\tr(M^{d-1})+\cdots+c_{1}\tr(M)\right)\,.$
(21)

On the other hand, we can first multiply (17) by $M^{n-d}$ and then trace to
find that

$\displaystyle\tr(M^{n})=-\left(c_{d-1}\tr(M^{n-1})+\cdots+c_{1}\tr(M^{d-n+1})+(-1)^{d}\det(M)\tr(M^{d-n})\right)\,.$
(22)

Since all the coefficients $c_{i}$ and $\det(M)$ are known, we have therefore
solved for $\tr(M^{n})$ in terms of the traces $\tr\big(M^{k}\big{missing})$
for $k<n$. By induction, this shows that all traces $\tr(M^{n})$ for $n>d$ can
be determined from the known traces $\tr(M^{i})$ for $i=1,\cdots,d$.

The upshot of this theorem is that, in $d$ spacetime dimensions, there are
only $\lfloor\frac{d}{2}\rfloor$ independent trace structures
$\mathrm{tr}\big{(}F^{k}\big{)}$ for an antisymmetric $2$-tensor $F_{\mu\nu}$.
The reason for the floor function is because of a slight difference between
the cases of even and odd dimensions. For odd $d$, the determinant of any
antisymmetric matrix vanishes, as does the trace of any odd power of the
matrix. So in this case, if $d=2k+1$, the only independent trace structures
are

$\displaystyle\tr(F^{2})\,,\tr(F^{4})\,,\cdots\,,\tr F^{2k}\,.$ (23)

Therefore there are $k=\frac{d-1}{2}$ independent quantities. In even
dimension $d=2k$, where the determinant of $F$ may be non-zero, the
independent trace structures are

$\displaystyle\tr(F^{2})\,,\tr(F^{4})\,,\cdots\,,\tr F^{2k}\,.$ (24)

so there are $k=\frac{d}{2}$ independent quantities. Therefore, for the
general case, we need $\lfloor\frac{d}{2}\rfloor$ trace structures in order to
express an arbitrary scalar constructed from $F_{\mu\nu}$.

####  3 Deforming a General Lagrangian

In the previous subsection, we have seen that a general scalar quantity built
from an antisymmetric $2$-form $F_{\mu\nu}$ in $d$ dimensions can be expressed
as a function of the independent traces $\tr(F^{2})$, $\tr(F^{4})$, $\cdots$,
$F^{2k}$ where $k=\lfloor\frac{d}{2}\rfloor$. In particular, the Lagrangian is
a scalar and therefore takes this form, so the most general classical theory
of an abelian gauge field can be written

$\displaystyle\mathcal{L}(F)=\mathcal{L}\left(\tr(F^{2}),\tr(F^{4}),\cdots,\tr(F^{2k})\right)\,,$
(25)

For convenience, we will define a set of variables $x_{i}$ defined by

$\displaystyle x_{i}=\tr(F^{2i})\,,$ (26)

for $i=1,\cdots,k$. We now wish to compute the stress tensor associated with
the Lagrangian (25) so that we can construct the ${T\overline{T}}$
deformation. Using our convention (8) for the Hilbert stress tensor and
$g_{\mu\nu}=\delta_{\mu\nu}$ in Euclidean signature, one has

$\displaystyle
T_{\mu\nu}=\delta_{\mu\nu}\mathcal{L}-2\sum_{i=1}^{k}\frac{\partial\mathcal{L}}{\partial
x_{i}}\cdot\frac{\delta x_{i}}{\delta g^{\mu\nu}}\Bigg{|}_{g=\delta}\,.$ (27)

Restoring metric factors, each $x_{i}$ has the expansion

$\displaystyle x_{i}=g^{\alpha_{0}\beta_{0}}g^{\alpha_{1}\beta_{1}}\cdots
g^{\alpha_{2i-1}\beta_{2i-1}}F_{\beta_{0}\alpha_{1}}F_{\beta_{1}\alpha_{2}}\cdots
F_{\beta_{2i-2}\alpha_{2i-1}}F_{\beta_{2i-1}\alpha_{0}}\,.$ (28)

Computing the metric derivative thus yields $2i$ separate terms, one for each
factor of $g^{\alpha\beta}$:

$\displaystyle\frac{\delta x_{i}}{\delta g^{\mu\nu}}$
$\displaystyle=g^{\alpha_{1}\beta_{1}}\cdots
g^{\alpha_{2i-1}\beta_{2i-1}}F_{\nu\alpha_{1}}F_{\beta_{1}\alpha_{2}}\cdots
F_{\beta_{2i-2}\alpha_{2i-1}}F_{\beta_{2i-1}\mu}$
$\displaystyle\quad+g^{\alpha_{0}\beta_{0}}g^{\alpha_{2}\beta_{2}}\cdots
g^{\alpha_{2i-1}\beta_{2i-1}}F_{\beta_{0}\mu}F_{\nu\alpha_{2}}\cdots
F_{\beta_{2i-2}\alpha_{2i-1}}F_{\beta_{2i-1}\alpha_{0}}$
$\displaystyle\quad+\cdots+g^{\alpha_{0}\beta_{0}}g^{\alpha_{1}\beta_{1}}\cdots
g^{\alpha_{2i-2}\beta_{2i-2}}F_{\beta_{0}\alpha_{1}}F_{\beta_{1}\alpha_{2}}\cdots
F_{\beta_{2i-2}\mu}F_{\nu\alpha_{0}}\,.$ (29)

Each of the terms appearing in (3) can be brought into an identical form via a
re-labeling of indices. This must have been true on general grounds, since
each term in (3) is obtained by removing one metric factor
$g^{\alpha_{k}\beta_{k}}$ and replacing the corresponding indices as
$\alpha_{k}\to\mu$, $\beta_{k}\to\nu$. But since the entire expression $x_{i}$
is invariant under cyclic permutations of $k$, each such term is identical and
proportional to $F_{\mu\nu}^{2i}$.

However, one can also prove this claim explicitly. In each term, we simply
push the factor of $F$ with the $\nu$ index all the way to the left, push the
$F$ with the $\mu$ index to the right, and then cyclically permute the
indices. For example, in the second term we have

$\displaystyle g^{\alpha_{0}\beta_{0}}g^{\alpha_{2}\beta_{2}}\cdots
g^{\alpha_{2i-1}\beta_{2i-1}}F_{\beta_{0}\mu}F_{\nu\alpha_{2}}F_{\beta_{2}\alpha_{3}}\cdots
F_{\beta_{2i-2}\alpha_{2i-1}}F_{\beta_{2i-1}\alpha_{0}}$
$\displaystyle\quad=g^{\alpha_{0}\beta_{0}}g^{\alpha_{2}\beta_{2}}\cdots
g^{\alpha_{2i-1}\beta_{2i-1}}F_{\nu\alpha_{2}}F_{\beta_{2}\alpha_{3}}\cdots
F_{\beta_{2i-2}\alpha_{2i-1}}F_{\beta_{2i-1}\alpha_{0}}F_{\beta_{0}\mu}$
$\displaystyle\quad=g^{\alpha_{0}\beta_{0}}g^{\alpha_{1}\beta_{1}}\cdots
g^{\alpha_{2i-2}\beta_{2i-2}}F_{\nu\alpha_{1}}F_{\beta_{1}\alpha_{2}}\cdots
F_{\beta_{2i-3}\alpha_{2i-2}}F_{\beta_{2i-2}\alpha_{0}}F_{\beta_{0}\mu}$
$\displaystyle\quad=g^{\alpha_{1}\beta_{1}}\cdots
g^{\alpha_{2i-2}\beta_{2i-2}}g^{\alpha_{2i-1}\beta_{2i-1}}F_{\nu\alpha_{1}}F_{\beta_{1}\alpha_{2}}\cdots
F_{\beta_{2i-3}\alpha_{2i-2}}F_{\beta_{2i-2}\alpha_{2i-1}}F_{\beta_{2i-1}\mu}\,.$
(30)

In going from the first step to the second, we have simply commuted
$F_{\beta_{0}\mu}$ and $F_{\nu\alpha_{2}}$. To go to the third step, we
relabeled each $\alpha^{k}$ and $\beta^{k}$ to $\alpha^{k-1}$ and
$\beta^{k-1}$ for $k=2,\cdots,2i-1$. In the final step, we relabeled
$\alpha_{0},\beta_{0}$ to $\alpha_{i-1},\beta_{i-1}$. The final expression is
manifestly equivalent to the term on the first line of (3).

We note that this re-labeling procedure requires us to commute factors of
$F_{\beta\alpha}$ past one another. In the non-abelian context, to be
investigated in Section 5, the field strengths have non-trivial commutators so
that this calculation is more subtle.

Using a similar cyclic manipulation in each of the $2i$ terms to bring them
all into the same form, we obtain

$\displaystyle\frac{\delta x_{i}}{\delta g^{\mu\nu}}$
$\displaystyle=2ig^{\alpha_{1}\beta_{1}}\cdots
g^{\alpha_{2i-1}\beta_{2i-1}}F_{\nu\alpha_{1}}F_{\beta_{1}\alpha_{2}}\cdots
F_{\beta_{2i-2}\alpha_{2i-1}}F_{\beta_{2i-1}\mu}\equiv 2iF^{2i}_{\mu\nu}\,,$
(31)

where again we use the notation

$\displaystyle F^{2i}_{\mu\nu}=g^{\alpha_{1}\beta_{1}}\cdots
g^{\alpha_{2i-1}\beta_{2i-1}}F_{\mu\alpha_{1}}F_{\beta_{1}\alpha_{2}}\cdots
F_{\beta_{2i-2}\alpha_{2i-1}}F_{\beta_{2i-1}\nu}\,.$ (32)

Therefore, the stress tensor for a general Lagrangian is

$\displaystyle
T_{\mu\nu}=\delta_{\mu\nu}\mathcal{L}-4\sum_{i=1}^{k}i\frac{\partial\mathcal{L}}{\partial
x_{i}}F^{2i}_{\mu\nu}\,.$ (33)

Next we compute the appropriate bilinears. We have

$\displaystyle T^{\mu\nu}T_{\mu\nu}$
$\displaystyle=\left(\delta^{\mu\nu}\mathcal{L}-4\sum_{i=1}^{k}i\frac{\partial\mathcal{L}}{\partial
x_{i}}F^{2i,\mu\nu}\right)\left(\delta_{\mu\nu}\mathcal{L}-4\sum_{j=1}^{k}j\frac{\partial\mathcal{L}}{\partial
x_{j}}F^{2j}_{\mu\nu}\right)$
$\displaystyle=\mathcal{L}^{2}d-8\mathcal{L}\sum_{i=1}^{k}i\frac{\partial\mathcal{L}}{\partial
x_{i}}\tr(F^{2i})+16\sum_{i,j=1}^{k}ij\frac{\partial\mathcal{L}}{\partial
x_{i}}\frac{\partial\mathcal{L}}{\partial
x_{j}}F^{2i,\mu\nu}F_{\mu\nu}^{2j}\,.$ (34)

Meanwhile, the trace is

$\displaystyle\tensor{T}{{}^{\mu}_{\mu}}=\mathcal{L}d-4\sum_{i=1}^{k}i\frac{\partial\mathcal{L}}{\partial
x_{i}}\tr(F^{2i})\,,$ (35)

and its square is

$\displaystyle\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}=\mathcal{L}^{2}d^{2}-8\mathcal{L}d\sum_{i=1}^{k}i\frac{\partial\mathcal{L}}{\partial
x_{i}}\tr(F^{2i})+16\sum_{i,j=1}^{k}ij\frac{\partial\mathcal{L}}{\partial
x_{i}}\frac{\partial\mathcal{L}}{\partial x_{j}}\tr(F^{2i})\tr(F^{2j})\,.$
(36)

As we argued in the introduction, in order to treat flow equations by a
general linear combination of stress tensor bilinears, it is enough to
consider flow equations of the form

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=T^{\mu\nu}T_{\mu\nu}-r\left(\tensor{T}{{}^{\mu}_{\mu}}\right)^{2}\,.$
(37)

Deforming the Lagrangian by this linear combination gives

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}$
$\displaystyle=(d-rd^{2})\mathcal{L}^{2}-8\mathcal{L}(1-rd)\sum_{i=1}^{k}i\frac{\partial\mathcal{L}}{\partial
x_{i}}x_{i}+16\sum_{i,j=1}^{k}ij\frac{\partial\mathcal{L}}{\partial
x_{i}}\frac{\partial\mathcal{L}}{\partial
x_{j}}\left(x_{i+j}-rx_{i}x_{j}\right)\,.$ (38)

Although (38) is exact, it is not expressed as a closed flow equation using
only the set of independent trace structures $x_{1},\cdots,x_{k}$. This is
because term involving $x_{i+j}=\tr(F^{2(i+j)})$ introduces dependence on
traces $\tr(F^{2n})$ for integers $n>k$. These traces are not independent of
the lower traces, as we have argued, and must therefore be re-expressed in
terms of the other independent trace structures using the Cayley-Hamilton
theorem.

It is unwieldy, though possible, to perform this procedure and write a closed
flow equation using only the independent trace structures $x_{i}$ by using
equations (21), (22), and the definition (18) of the $c_{i}$. However, we will
prefer to leave this implicit and only perform the necessary substitutions as
needed in special cases later in this chapter.

Despite requiring the implicit definition of higher traces in terms of lower
ones, (38) reveals some general structure about these ${T\overline{T}}$-like
flows in arbitrary dimension, such as the dependence on $r$ and $d$ and the
introduction of dependence on higher traces via the final term. For instance,
one can see immediately that the flow equation simplifies for certain choices
of $r$ such as $r=\frac{1}{d}$, for which some of the terms on the right
vanish. One can also see an interesting structure in the large $d$ limit. If
we retaining only the leading term on the right side as $d\to\infty$, the flow
equation reduces to

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=-rd^{2}\mathcal{L}^{2}\,,$
(39)

which has the simple solution

$\displaystyle\mathcal{L}(\lambda)=\frac{\mathcal{L}_{0}}{1+d^{2}\mathcal{L}_{0}r\lambda}\,.$
(40)

For negative $\mathcal{L}_{0}$, such as $\mathcal{L}_{0}=-\frac{1}{4}F^{2}$ in
Lorentzian signature, or for positive $\mathcal{L}_{0}$ but negative $\lambda$
or $r$, we see that (40) has a singularity at finite $\lambda$. This is
reminiscent of the structure we saw in equation (35) of Chapter 3, namely

$\displaystyle V(\lambda,\phi)=\frac{V(0,\phi)}{1-\lambda V(0,\phi)}\,.$ (41)

In both cases, it appears that the ${T\overline{T}}$ deformation has led to a
pole in the Lagrangian for certain field configurations. As we stressed in
Chapter 3, this is a purely classical result, and one cannot necessarily draw
inferences about the structure of the quantum mechanical theory from this
observation. However, this structure suggests the possibility of a
modification of the vacuum structure in the theory – one might naively expect
such an infinite potential barrier to cut off different sectors of the Hilbert
space from one another. It would be interesting to study this possibility in
greater detail, perhaps in a simplified quantum mechanical (i.e.
$(0+1)$-dimensional) example.

Finally, we comment on one additional manipulation of (38) that may be useful
in obtaining series solutions for certain flows. The Lagrangian may always be
expanded as a power series in the variables $x_{k}=\tr(F^{2k})$. We recall
from footnote 2 that, in our conventions, the field strength $F_{\mu\nu}$ has
mass dimension $\frac{d}{2}$ and thus $x_{k}$ has mass dimension $kd$.
Likewise, $\lambda$ has length dimension $d$ so combinations like
$\lambda^{k}x_{k}$ are dimensionless. We will now consider the combinations of
terms that can appear in the power series expansion of the Lagrangian, subject
to the usual requirement that each term have mass dimension $d$.

The coefficient of $x_{1}=F^{2}$, the free kinetic term, already has mass
dimension $d$ and thus appears with no powers of $\lambda$. Additional terms
must be suppressed by powers of $\lambda$; for instance, $\lambda x_{2}$ and
$\lambda x_{1}^{2}$ have the appropriate dimension to appear in the
Lagrangian. More generally, a term with $2m$ factors of $F_{\mu\nu}$ must
appear multiplying $\lambda^{m-1}$. The total mass dimension of a term
$x_{1}^{n_{1}}x_{2}^{n_{2}}\cdots x_{k}^{n_{k}}$ is
$\left(n_{1}+2n_{2}+\cdots+kn_{k}\right)d$, and must therefore appear with a
factor of $\lambda^{n_{1}+2n_{2}+\cdots+kn_{k}-1}$. Thus the most general
expansion of the Lagrangian is

$\displaystyle\mathcal{L}=\sum_{N}c_{N}\lambda^{n_{1}+2n_{2}+\cdots+kn_{k}-1}x_{1}^{n_{1}}x_{2}^{n_{2}}\cdots
x_{k}^{n_{k}}\,.$ (42)

Here $N$ represents a multi-index of $k$ non-negative integers
$n_{1},\cdots,n_{k}$ (not all zero). If we are beginning from an undeformed
theory which is the ordinary Maxwell Lagrangian, the initial condition
$\mathcal{L}(\lambda=0)=-\frac{1}{4}x_{1}$ fixes the coefficient
$N_{0}=\\{1,0,0,\cdots,0\\}$ to be $c_{N_{0}}=-\frac{1}{4}$.

Such an expansion simplifies the problem of finding a series solution to our
master equation (38), since now the task is reduced to determining each of the
coefficients $c_{N}$. For instance, the left side of the flow equation is now

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=\sum_{N>N_{0}}c_{N}\left(n_{1}+2n_{2}+\cdots+kn_{k}-1\right)\lambda^{n_{1}+2n_{2}+\cdots+kn_{k}-2}x_{1}^{n_{1}}x_{2}^{n_{2}}\cdots
x_{k}^{n_{k}}\,,$ (43)

where now the notation $\sum_{N>N_{0}}$ means to sum over all multi-indices
$N=\\{n_{1},\cdots,n_{k}\\}$ except for $\\{1,0,\cdots,0\\}$. Likewise, each
of the $x_{i}$ derivatives appearing in the flow equation can be simply
evaluated as

$\displaystyle\frac{\partial\mathcal{L}}{\partial
x_{j}}=\sum_{N\,,\,n_{j}>0}c_{N}\,n_{j}\,\lambda^{n_{1}+2n_{2}+\cdots+kn_{k}-1}x_{1}^{n_{1}}x_{2}^{n_{2}}\cdots
x_{j}^{n_{j}-1}\cdots x_{k}^{n_{k}}\,,$ (44)

where similarly $\sum_{N,n_{j}>0}$ means to sum over all multi-indices
$N=\\{n_{1},\cdots,n_{k}\\}$ where $n_{j}$ is at least $1$.

After eliminating all of the higher trace structures $x_{n}$ for $n>k$ in the
master flow equation (38) using Cayley-Hamilton, one can then substitute (43)
on the left and (44) for each $x_{i}$ derivative on the right. The result,
though perhaps complicated, is in principle a perturbative expansion that
allows us to solve for the coefficients $c_{N}$ order-by-order in $\lambda$.

####  4 Low-Dimensional Examples

To illustrate the general formalism developed in the preceding subsection,
here we will develop the flow equations in $d=2$ and $d=3$. These cases are
especially simple since there is only one independent trace structure $x_{1}$,
so we need only express $x_{2}$ in terms of $x_{1}$ using the Cayley-Hamilton
theorem. Since we have studied these examples before, this will also provide a
usesful check of whether the formalism recovers previous results.

##### Two Dimensions

In two dimensions, the Cayley-Hamilton theorem reads

$\displaystyle M^{2}-(\tr(M))M+\det(M)I_{2}=0\,.$ (45)

For an antisymmetric matrix $M=F$, then, we have $\tr(F)=0$ and
$F^{2}=-\det(F)I_{2}$. Taking traces, one has

$\displaystyle\tr(F^{2})=-2\det(F)\,.$ (46)

If we first multiply by $F^{2}$ and then take traces, we instead find

$\displaystyle\tr(F^{4})=-\det(F)\tr(F^{2})=\frac{1}{2}\left(\tr(F)\right)^{2}\,.$
(47)

In terms of our variables $x_{i}=\tr(F^{2i})$, this means that

$\displaystyle x_{2}=\frac{1}{2}x_{1}^{2}\,.$ (48)

Using this, our general flow equation (38) with $d=2$ becomes

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=2\left(1-2r\right)\mathcal{L}^{2}-8\mathcal{L}(1-2r)x_{1}\frac{\partial\mathcal{L}}{\partial
x_{1}}+8\left(1-2r\right)x_{1}^{2}\left(\frac{\partial\mathcal{L}}{\partial
x_{1}}\right)^{2}\,.$ (49)

This should match the result that we derived in Chapter 5. Indeed, if $r=1$
the flow equation reduces to

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=-2\mathcal{L}^{2}+8\mathcal{L}x_{1}\frac{\partial\mathcal{L}}{\partial
x_{1}}-8x_{1}^{2}\left(\frac{\partial\mathcal{L}}{\partial
x_{1}}\right)^{2}\,.$ (50)

We may now multiply both sides by $-\frac{1}{2}$ and absorb this factor into a
re-scaling of $\lambda$ on the left side, which gives

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=\mathcal{L}^{2}-4\mathcal{L}x_{1}\frac{\partial\mathcal{L}}{\partial
x_{1}}+4x_{1}^{2}\left(\frac{\partial\mathcal{L}}{\partial
x_{1}}\right)^{2}\,.$ (51)

This matches (10), as expected. We note that we have re-scaled $\lambda$ by a
negative constant, which reverses the sign; this corresponds to the fact that
our convention for the “good sign” of $\lambda$ is opposite that in Chapter 5.
We will see a similar re-scaling by a constant $a=-8$ when comparing to Born-
Infeld in the next section.

##### Three Dimensions

In the case of $d=3$, the Cayley-Hamilton result (17) gives the identity

$\displaystyle
M^{3}-\left(\tr(M)\right)M^{2}+\frac{1}{2}\left(\left(\tr(M)\right)^{2}-\tr(M^{2})\right)M-\det(M)I_{3}=0\,.$
(52)

Let $M=F$ be antisymmetric. In odd dimensions, the determinant of any
antisymmetric matrix vanishes, and we manifestly also have $\tr(M)=0$, so the
condition becomes

$\displaystyle F^{3}-\frac{1}{2}\left(\tr(F^{2})\right)F=0\,.$ (53)

We may multiply through by $F$ and take the trace to find

$\displaystyle\tr(F^{4})=\frac{1}{2}\left(\tr(F^{2})\right)^{2}\,,$ (54)

which gives the same condition $x_{2}=\frac{1}{2}x_{1}^{2}$ which we found in
$d=2$. The flow equation, however, will differ because it explicitly depends
on $d$. The master equation (38) becomes

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}=3(1-3r)\mathcal{L}^{2}-8\mathcal{L}(1-3r)x_{1}\frac{\partial\mathcal{L}}{\partial
x_{1}}+8(1-2r)x_{1}^{2}\left(\frac{\partial\mathcal{L}}{\partial
x_{1}}\right)^{2}\,.$ (55)

This matches the result (1) which we derived earlier.

###  3 Comparison to Born-Infeld

In this section, we will examine the flow equation for gauge theories in
general dimension to see whether any such flow can yield the Born-Infeld
action from the Maxwell Lagrangian as the seed theory, besides the known case
in $d=4$.

####  1 Perturbative Comparison in $\lambda$

It will be sufficient to check whether Born-Infeld-type actions can satisfy a
${T\overline{T}}$-like flow equation to second order in $\lambda$. To make
this comparison, we first expand the Born-Infeld Lagrangian to the first few
orders. First note that, since $\det(M)=\det(M^{T})$ for any matrix $M$, we
have

$\displaystyle\det\left(\delta_{\mu\nu}+kF_{\mu\nu}\right)=\det\left(\delta_{\nu\mu}+kF_{\nu\mu}\right)=\det\left(\delta_{\mu\nu}-kF_{\mu\nu}\right)\,.$
(1)

Therefore, we can rewrite

$\displaystyle\sqrt{\det\left(\delta_{\mu\nu}+kF_{\mu\nu}\right)}$
$\displaystyle=\Big{[}\det\left(\delta_{\mu\nu}+kF_{\mu\nu}\right)\det\left(\delta_{\alpha\beta}-kF_{\alpha\beta}\right)\Big{]}^{\frac{1}{4}}$
$\displaystyle=\left[\det\left(\delta_{\mu\nu}-k^{2}F^{2}_{\mu\nu}\right)\right]^{\frac{1}{4}}\,.$
(2)

On the other hand, one has $\det(A)=\exp\left(\tr(\log(A))\right)$ for any
matrix $A$, so

$\displaystyle\left(\det\left(\delta_{\mu\nu}-k^{2}F^{2}_{\mu\nu}\right)\right)^{\frac{1}{4}}=\exp\left(\frac{1}{4}\tr\left(\log\left(\delta_{\mu\nu}-k^{2}F^{2}_{\mu\nu}\right)\right)\right)\,.$
(3)

In order to compare the Born-Infeld action to ${T\overline{T}}$-flows up to
order $\lambda^{2}$, we must Taylor-expand (3) to order $k^{6}$. This gives

$\displaystyle\sqrt{\det\left(\delta_{\mu\nu}+kF_{\mu\nu}\right)}$
$\displaystyle=1-\frac{k^{2}}{4}\tr(F^{2})+\frac{k^{4}}{4}\left(\frac{1}{8}\left(\tr(F^{2})\right)^{2}-\frac{1}{2}\tr(F^{4})\right)$
$\displaystyle\qquad-\frac{k^{6}}{6}\left(\frac{1}{64}\left(\tr(F^{2})\right)^{3}-\frac{3}{16}\tr(F^{2})\tr(F^{4})+\frac{1}{2}\tr(F^{6})\right)+\mathcal{O}\left(k^{8}\right)\,.$
(4)

Following the conventions in footnote 2, where the gauge coupling $g$ has been
absorbed into the field strength $F$, the combination
$\sqrt{\lambda}F_{\mu\nu}$ is dimensionless. Thus the appropriate form of a
Born-Infeld type Lagrangian which could potentially arise from a
${T\overline{T}}$ flow is

$\displaystyle\mathcal{L}_{\lambda}=\frac{1}{a\lambda}\left(-1+\sqrt{\det\left(\delta_{\mu\nu}+\sqrt{a\lambda}F_{\mu\nu}\right)}\right),$
(5)

where $a$ is arbitrary and the coefficient $\sqrt{a\lambda}$ in front of
$F_{\mu\nu}$ is chosen so that the Lagrangian reduces to
$\mathcal{L}_{0}=-\frac{1}{4}\tr(F^{2})=\frac{1}{4}F^{2}$ as $\lambda\to 0$.
Using (1) with $k=\sqrt{a\lambda}$, then, the expansion of
$\mathcal{L}_{\lambda}$ to order $\lambda^{2}$ is

$\displaystyle\mathcal{L}_{\lambda}$
$\displaystyle=-\frac{1}{4}\tr(F^{2})+\frac{a\lambda}{4}\left(\frac{1}{8}\left(\tr(F^{2})\right)^{2}-\frac{1}{2}\tr(F^{4})\right)$
$\displaystyle\qquad-\frac{a^{2}\lambda^{2}}{6}\left(\frac{1}{64}\left(\tr(F^{2})\right)^{3}-\frac{3}{16}\tr(F^{2})\tr(F^{4})+\frac{1}{2}\tr(F^{6})\right)+\mathcal{O}\left(\lambda^{3}\right)\,.$
(6)

Let’s now compare this to the Lagrangian one would get using the master flow
equation (38). Beginning with the initial condition
$\mathcal{L}_{0}=-\frac{1}{4}x_{1}$, we find

$\displaystyle\frac{\partial\mathcal{L}}{\partial\lambda}\Big{|}_{\lambda=0}$
$\displaystyle=\frac{1}{16}(d-rd^{2})x_{1}^{2}-\frac{1}{2}x_{1}^{2}(1-rd)+\left(x_{2}-rx_{1}^{2}\right)\,$
$\displaystyle=x_{2}+x_{1}^{2}\left(-\frac{1}{2}+\frac{d}{16}-\left(1-\frac{d}{2}+\frac{d^{2}}{16}\right)r\right)\,.$
(7)

Note that, when $d=4$, the quantity multiplying $r$ in the parentheses of (1)
vanishes, and the coefficient of $x_{1}^{2}$ simply becomes $-\frac{1}{4}$.
Assuming that $d\neq 4$, in order for the coefficients to match the order
$\lambda$ expansion of Born-Infeld we need

$\displaystyle a=-8\,,\qquad r=\frac{1}{d-4}\,.$ (8)

This scaling by $a=-8$ agrees with the factor of $\frac{1}{8}$ appearing in
(12). The overall sign difference arises from the fact that our convention for
the “good sign” of $\lambda$ in this chapter is the opposite of that which we
have previously used in two dimensions, for instance in Chapter 5. One can see
this by comparing equation (37) to equation (5). For the present purposes, it
is sufficient to assume $\lambda<0$ so that $a\lambda$ is positive.

Thus, to order $\lambda$, the deformed Lagrangian is

$\displaystyle\mathcal{L}=-\frac{1}{4}x_{1}+\lambda
x_{2}-\frac{\lambda}{4}x_{1}^{2}+\mathcal{O}(\lambda^{2})\,.$ (9)

Repeating the process by plugging this deformed Lagrangian into the master
equation and extracting the $\mathcal{O}(\lambda^{2})$ piece, we find

$\displaystyle\mathcal{L}\Big{|}_{\lambda^{2}}=\frac{\lambda^{2}}{2}\left(\frac{-12+d-32r+12rd-
rd^{2}}{8}\,x_{1}^{3}+\frac{20-d+32r-12rd+rd^{2}}{2}\,x_{1}x_{2}-16x_{3}\right)\,.$
(10)

Substituting the required value of $r=\frac{1}{d-4}$, this becomes

$\displaystyle\mathcal{L}\Big{|}_{\lambda^{2}}=\lambda^{2}\left(-\frac{1}{4}x_{1}^{3}+3x_{1}x_{2}-8x_{3}\right)\,.$
(11)

The coefficients of $x_{1}^{3},x_{1}x_{2},x_{3}$ in this term are
$-\frac{1}{4},3,$ and $-8$, respectively, which do not match the coefficients
in (1) when $a=-8$, which are $-\frac{1}{6}$, $2$, and $-\frac{16}{3}$. Thus
the Born-Infeld action can never be obtained by a flow involving stress tensor
bilinears in any number of dimensions where $\tr(F^{6})$ is independent of
lower traces, i.e. $d\geq 6$.

In $d<6$ dimensions, $x_{3}$ can be expressed in terms of $x_{1}$ and $x_{2}$
so checking the coefficients at order $\lambda^{2}$ requires eliminating
$x_{3}$ using the Cayley-Hamilton theorem. We have already checked most of the
cases $d<6$; in particular, $d=2$ gives the hypergeometric action discussed in
Chapter 5, we saw that $d=3$ does not give Born-Infeld in Section 1, and we
know from Chapter 4 that ${T\overline{T}}$ in four dimensions _does_ generate
a Born-Infeld type action. The only remaining case to check (besides $d=1$,
which is trivial) is $d=5$ spacetime dimensions, where $\tr(F^{2})$ and
$\tr(F^{4})$ are the independent trace structures. We perform this explicit
check in the next subsection.

####  2 Deformation in $d=5$

We will now handle one edge case which is not included in the preceding
analysis, namely $d=5$. In this case, the expression (10) for the
$\mathcal{O}(\lambda^{2})$ term in the ${T\overline{T}}$ flow is formally
correct. However, since the trace $x_{3}=\tr(F^{6})$ is not independent of the
lower $x_{i}$ in five dimensions, we must eliminate it in favor of lower
traces using the Cayley-Hamilton theorem. In five dimensions, (17) reads

$\displaystyle
M^{5}-(\tr(M))M^{4}+\frac{1}{2}\left(\left(\tr(M)\right)^{2}-\tr(M^{2})\right)M^{3}$
$\displaystyle\quad-\frac{1}{6}\left(\left(\tr(M)\right)^{3}-3\tr(M^{2})\tr(M)+2\tr(M^{3})\right)M^{2}$
$\displaystyle\quad+\frac{1}{24}\left(\tr(M)^{4}+8\tr(M^{3})\tr(M)+3\left(\tr(M^{2})\right)^{2}-6\tr(M^{2})\left(\tr(M)\right)^{2}-6\tr(M^{4})\right)M$
$\displaystyle\quad-\det(M)I_{5}=0\,.$ (12)

Letting $M=F$ be antisymmetric, which implies that
$\tr(F)=\tr(F^{3})=\det(F)=0$, this becomes

$\displaystyle
F^{5}-\frac{1}{2}\tr(F^{2})F^{3}+\frac{1}{24}\left(3\left(\tr\big(F^{2}\big{missing})\right)^{2}-6\tr\big(F^{4}\big{missing})\right)F=0\,.$
(13)

Multiplying both sides of (13) by $F$ and tracing, one finds

$\displaystyle\tr(F^{6})=\frac{1}{2}\tr(F^{2})\tr(F^{4})-\frac{1}{24}\left(3\left(\tr\big(F^{2}\big{missing})\right)^{2}-6\tr\big(F^{4}\big{missing})\right)\tr(F^{2})\,,$
(14)

or in terms of the variables $x_{i}$,

$\displaystyle x_{3}=\frac{3}{4}x_{1}x_{2}-\frac{1}{8}x_{1}^{3}\,.$ (15)

This means that the order $\lambda^{2}$ term arising from the
${T\overline{T}}$ flow is

$\displaystyle\mathcal{L}\Big{|}_{\lambda^{2}}=\frac{3\lambda^{2}}{4}\left(x_{1}^{3}-4x_{1}x_{2}\right)\,.$
(16)

We can compare this to the $\mathcal{O}(\lambda^{2})$ term in the expansion
(1) of Born-Infeld, after substituting the constraint (15), which yields

$\displaystyle\mathcal{L}\Big{|}_{\lambda^{2}}=\frac{\lambda^{2}}{2}\left(x_{1}^{3}-4x_{1}x_{2}\right)\,.$
(17)

Since (16) and (17) differ by an overall multiplicative factor, we conclude
that the Born-Infeld action is not generated by a ${T\overline{T}}$ flow in
five dimensions. This completes the proof that the only dimension in which the
Born-Infeld action arises from a ${T\overline{T}}$-like deformation of the
free Maxwell Lagrangian, for any choice of $r$, is in $d=4$.

####  3 Deformation in $d=4$

Although we have already seen in Chapter 4 that ${T\overline{T}}$ deformation
of the $4$-dimensional Maxwell action yields the Born-Infeld action, it is a
useful check of our general formalism to see how this emerges from the master
flow equation (38). In particular, since we have shown that this flow equation
does _not_ yield Born-Infeld in any $d>4$, it is instructive to see how the
numerical factors appearing in the Cayley-Hamilton theorem conspire to give a
deformed action of square-root type in $d=4$.

As we mentioned around equation (14), in four spacetime dimensions the two
independent scalars that can be constructed from $F_{\mu\nu}$ are

$\displaystyle x_{1}=F_{\mu\nu}F^{\nu\mu}=\tr(F^{2})\,,\qquad
x_{2}=F^{\mu\sigma}\tensor{F}{{}_{\sigma}^{\nu}}\tensor{F}{{}_{\nu}^{\rho}}F_{\rho\mu}=\tr(F^{4})\,.$
(18)

However, our master flow equation (38) will introduce dependence on two
additional trace structures, $x_{3}=\tr(F^{6})$ and $x_{4}=\tr(F^{8})$. First
we must express these in terms of lower trace structures. For a general
$4\times 4$ matrix $M$, the Cayley-Hamilton theorem (17) reads

$\displaystyle 0$
$\displaystyle=M^{4}-\left(\tr(M)\right)M^{3}+\frac{1}{2}\left(\left(\tr(M)\right)^{2}-\tr(M^{2})\right)M^{2}$
$\displaystyle\quad-\frac{1}{6}\left(\left(\tr(M)\right)^{3}-3\tr(M^{2})\tr(M)+2\tr(M^{3})\right)M+\det(M)I_{4}\,.$
(19)

If we specialize to the case where $M=F$, which is antisymmetric and thus has
vanishing trace (as does $F^{3}$), we have

$\displaystyle 0=F^{4}-\frac{1}{2}\tr(F^{2})F^{2}+\det(F)I_{4}\,.$ (20)

By taking traces, we can solve for the determinant as

$\displaystyle\det(F)=\frac{1}{8}\left(\tr\big(F^{2}\big{missing})\right)^{2}-\frac{1}{4}\tr(F^{4})\,.$
(21)

Similarly, if we first multiply (20) by $F^{2}$ and then take traces, we find

$\displaystyle\tr(F^{6})=\frac{1}{2}\tr(F^{2})\tr(F^{4})-\det(F)\tr(F^{2})\,,$
(22)

and if we first multiply (20) by $F^{4}$ and then take traces, one obtains

$\displaystyle\tr(F^{8})=\frac{1}{2}\tr(F^{2})\tr(F^{6})-\det(F)\tr(F^{4})\,.$
(23)

Expressed in terms of $x_{i}=\tr(F^{2i})$, these equations can be solved to
find

$\displaystyle x_{3}=-\frac{1}{8}x_{1}\left(x_{1}^{2}-6x_{2}\right)\,,\qquad
x_{4}=-\frac{1}{16}\left(x_{1}^{4}-4x_{1}^{2}x_{2}-4x_{2}^{2}\right)\,.$ (24)

The numerical factors appearing in (24) work out such that the general
argument that a ${T\overline{T}}$ flow disagrees with Born-Infeld at order
$\lambda^{2}$, presented in Section 1 for $d\geq 6$, fails in four dimensions.
This was already suggested by the term at leading order in $\lambda$, where
the coefficients automatically agreed in $d=4$ and required $r=\frac{1}{d-4}$
otherwise. However, we can also check that the $\mathcal{O}(\lambda^{2})$
piece works. Substituting in $d=4$ in equation (10), one finds

$\displaystyle\mathcal{L}_{{T\overline{T}}}\Big{|}_{\lambda^{2}}=-\frac{\lambda^{2}}{2}\left(x_{1}^{3}-8x_{1}x_{2}+16x_{3}\right)\,,$
(25)

and using our expression (24) for $x_{3}$, this is

$\displaystyle\mathcal{L}_{{T\overline{T}}}\Big{|}_{\lambda^{2}}=\frac{\lambda^{2}}{2}\left(x_{1}^{3}-4x_{1}x_{2}\right)\,,$
(26)

On the other hand, when $a=-8$, the $\mathcal{O}(\lambda^{2})$ term in (1) is

$\displaystyle\mathcal{L}_{BI}\Big{|}_{\lambda^{2}}$
$\displaystyle=-\frac{64\lambda^{2}}{6}\left(\frac{1}{64}\left(\tr(F^{2})\right)^{3}-\frac{3}{16}\tr(F^{2})\tr(F^{4})+\frac{1}{2}\Tr(F^{6})\right)$
$\displaystyle=\frac{\lambda^{2}}{2}\left(x_{1}^{3}-4x_{1}x_{2}\right)\,.$
(27)

So in this case, there is an exact match at order $\lambda^{2}$. In fact,
using the expressions (24), we can see how the all-orders flow equation