# Vector fields on C*-algebras, semigroups of endomorphisms and gauge groups

Innocenti V. Maresin

###### Abstract

What is a vector field on a C*-algebra is defined. Its relation to semigroups
of *-endomorphisms was researched. Some results given about those vector
fields and semigroups. There are also various constructions of semigroups
including one parametrized by the cone of the future in (3+1)-dimensional
Minkowski space. Physical interpretations will be presented in a separate
paper.

This work is supported by Russian Foundation for Basic Research,
grant 10-01-00178.

###  0.1 Introduction

The formalism of C*-algebras is known to be used for quantization and quantum
observables for a long time. Several ideas of differential geometry, such as
bundles, differential forms and integrals, admit their translations to
algebraic language. This paper presents one concept which was apparently
missing: vector fields. In Section 1 we define real vector fields on a
C*-algebra, a continuous semigroup of *-endomorphisms of the same algebra, the
relation between them, and basic examples. In Section 2 we present some less
trivial constructs, such as the commutator of vector fields and a semigroup
parametrized by several real variables. Section 3 gives the construction of
vector fields via representations of the algebra. Section 4 gives some
implications to semigroups, if there exist ones generated by vector fields
from Section 3.

###  0.2 Basic facts about C*-algebras

C*-algebra ${\mathcal{A}}$ is a complex Banach algebra with an antilinear
involution, which reverses the order of multiplication: ${(a\hskip
1.25ptb)}^{*}={b}^{*}\hskip 1.25pt{a}^{*}$. The norm must satisfy:

$||a||^{2}=||{a}^{*}\hskip 1.25pta||$

${{\mathcal{A}}_{herm}}$ is a real linear subspace of self-conjugate
(Hermitian) elements.

A *-endomorphism does not have to keep the norm. Though, one can prove that a
*-endomorphism cannot increase the norm of an element:

$||E({a}^{*}\hskip 1.25pta)||=||E(a)||^{2},\ ||E({a}^{*}\hskip 1.25pta\hskip
1.25pt{a}^{*}\hskip 1.25pta)||=||E(a)||^{4},\ ||E(({a}^{*}\hskip
1.25pta)^{4})||=||E(a)||^{8},\ \cdots$

##  1 Vector fields and real-parametrized endomorphism semigroups

1.1. A real-parametrized semigroup (or group) of *-endomorphisms is a family
of *-endomorphisms parametrized by non-negative (or real) numbers, such that:
$E_{\zeta}\operatorname{\circ}E_{\eta}=E_{\zeta+\eta}$ and
$E_{0}=\mathrm{Id}$. These semigroups will be referred simply as “endomorphism
semigroups”. Possible example is
$E_{t}(q)=\exp(-itg)q\exp(itg),g\in{{\mathcal{A}}_{herm}}$, and it is actually
the only possible example in finite dimensions. If the algebra is infinite-
dimensional, then other semigroups may exist, and for a semigroup not defined
by a conjugation with an exponent we cannot compute the derivative
${\frac{d}{dt}}E_{t}(q)$ for any element $q$. For most constructions of a
semigroup (it’s not a theorem), though, we can find such *-subalgebra dense in
${\mathcal{A}}$ that *-endomorphisms of a semigroup preserve it and
aforementioned derivative is defined for any $q$ from the subalgebra. It is
obvious that such semigroup, restricted to this subalgebra, satisfies the
following ordinary differential problem:

${\scriptstyle\forall
q\in{{\mathcal{A}}_{V}}:}{\frac{d}{dt}}E_{t}(q)=V(E_{t}(q))\,,{\scriptstyle\
\forall t:}E_{t}:{{\mathcal{A}}_{V}}\to{{\mathcal{A}}_{V}}\mbox{\small{} a
bounded linear map,}\ E_{0}=\mathrm{Id},\hskip 16.00008pt\mbox{(!)}$

where ${{\mathcal{A}}_{V}}$ is a *-subalgebra dense in ${\mathcal{A}}$ and
$V:{{\mathcal{A}}_{V}}\to{\mathcal{A}}$ is a fixed (unbounded) linear
operator. Another way to express this is a differential equation
$\dot{E}=V\operatorname{\circ}E$ where the differentiation is understood in
the strong operator topology (not norm topology) on bounded linear operators
on ${{\mathcal{A}}_{V}}$. Are there necessary conditions on $V$? First, it
must commute with *, because any *-endomorphism has to preserve this
operation. Second, by differentiation of the condition (!) we obtain the
product rule:

$\forall u,v\in{{\mathcal{A}}_{V}}:V(u\hskip 1.25ptv)=u\hskip
1.25ptV(v)+V(u)\hskip 1.25ptv\hskip 30.00005pt\mbox{(\\.{ })}$

In other words, the operator $V$ must be a differentiation on
${{\mathcal{A}}_{V}}\,$.

1.2. Definition: Let ${{\mathcal{A}}_{V}}$ be a dense *-subalgebra, and
$V:{{\mathcal{A}}_{V}}\to{\mathcal{A}}$ is a linear operator satisfying the
equation (˙). Then $V$ is referred to as a vector field on ${\mathcal{A}}$. If
more precision is required, the one can say that $V$ is a vector field on
${\mathcal{A}}$ defined in the subalgebra ${{\mathcal{A}}_{V}}\,$. A vector
field which commutes with * will be referred to as a real vector field.

1.3. Theorem: Given a real vector field $V$ defined in
${{\mathcal{A}}_{V}}\,$, an $\varepsilon>0$ and a solution of the equation (!)
for $t\in[0,\varepsilon)$. Then:

  * •

The solution is a family of *-endomorphisms of ${{\mathcal{A}}_{V}}$ (and,
hence, of ${\mathcal{A}}$ by continuity).

  * •

The solution, in given conditions, is unique.

  * •

The solution extends to any $t\in[0,+\infty)$.

  * •

The solution is continuous by both $q$ (on all ${\mathcal{A}}$) and $t$. In
other words, ${E^{V}}_{t}$ is continuous in the (strong) operator topology.

Note: The restriction that solutions are looked for in an operator space, not
as individual trajectories, allows to bypass uniqueness problems of ordinary
differential equations.

Proof. From the property (˙) of $V$ follows a differential equality for a
solution $E$ of (!):

$\forall u,v\in{{\mathcal{A}}_{V}}:{\frac{d}{dt}}E_{t}(u\hskip
1.25ptv)={\frac{d}{dt}}(E_{t}(u)\hskip 1.25ptE_{t}(v))$

By integration of it by $t$ from $t=0$ we get that $E_{t}(u\hskip 1.25ptv)$
always equals to $E_{t}(u)\hskip 1.25ptE_{t}(v)$ in ${{\mathcal{A}}_{V}}$,
which in turn extends to all ${\mathcal{A}}$ by continuity. Conservation of *
is proved similarly from the condition that $V$ is real. Suppose two solutions
$E$ and $\tilde{E}$. Then, for any $\kappa\in{\mathbb{C}}$: $(1-\kappa)\hskip
1.25ptE+\kappa\hskip 1.25pt\tilde{E}$ is also a solution of (!), each of which
we already know to be a family of endomorphisms. For given $t$ the dependence
on $\kappa$ can either define a complex line or a fixed endomorphism (if there
is no dependence). But complex lines of *-endomorphisms are impossible: either
because they preserve *, or because each of them has the norm not greater than
1. Hence, for any $t$: $E_{t}=\tilde{E}_{t}$.

Extension to all $[0,+\infty)$ is made by composition, from the Archimedean
property of real numbers and uniqueness of the solution. Continuity by $t$ for
any $q\in{\mathcal{A}}$ can be proved from density of ${{\mathcal{A}}_{V}}$
and $|E_{t}(q)|\leq|q|$ inequality, for example, by direct verification of
$\varepsilon$–$\delta$ definition of continuity. []

Hence, for any real vector field $V$ there are only two possibilities: either
it defines a semigroup of endomorphisms of ${\mathcal{A}}$, or the equation
(!) has no solution anywhere for $t>0$. The former case will be referred to as
“$V$ generates a semigroup”, or “$V$ is semigroup-generating”, and this
semigroup will be denoted as ${E^{V}}$. Informally, ${E^{V}}_{t}$ is the
exponential of $t\hskip 1.25ptV\,$.

The only obvious property of the set of semigroup-generating real vector
fields defined in ${{\mathcal{A}}_{V}}$ is that this set is a cone.

1.4. Proposition: If a vector field $V$ generates a semigroup, then it
commutes with all its endomorphisms:

$\forall t\geq 0:\ V\operatorname{\circ}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ t}}}={{E^{\displaystyle V}}_{{}_{\displaystyle\
t}}}\operatorname{\circ}V\,.$

Proof: First of all, sides of this equality belong to the same space of
operators from ${{\mathcal{A}}_{V}}$ to ${\mathcal{A}}$, because ${E^{V}}$
preserves ${{\mathcal{A}}_{V}}\,$. From (!):

$V\operatorname{\circ}{{E^{\displaystyle V}}_{{}_{\displaystyle\
t}}}=\frac{d}{d\tau}{{E^{\displaystyle V}}_{{}_{\displaystyle\
\tau}}}|_{\tau=t}={{E^{\displaystyle V}}_{{}_{\displaystyle\
t}}}\operatorname{\circ}\left(\frac{d}{d\tau}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ \tau}}}|_{\tau=0}\right)={{E^{\displaystyle
V}}_{{}_{\displaystyle\ t}}}\operatorname{\circ}V$

where the middle equality follows from the fact that ${E^{V}}$ is a real-
parametrized semigroup of (linear) endomorphisms. The derivative is understood
as pointwise on elements of the *-subalgebra ${{\mathcal{A}}_{V}}$. Or, more
“scientifically”, in the strong operator topology from ${{\mathcal{A}}_{V}}$
to ${\mathcal{A}}\,$. []

1.5. Example: let $M$ be a real ${\mathcal{C}}^{1}$-smooth manifold. We’ll
denote as ${\mathcal{C}}_{0}(M)$ the algebra of continuous complex-valued
functions vanishing at infinity with pointwise operations multiplication and *
(as the complex conjugation). “Vanishing at infinity” means that
$|f(\cdot)|\geq\varepsilon$ must be compact for any $\varepsilon>0$ – this
property is important to get a C*-algebra on a manifold which is not compact.
Then any vector field $V_{M}$ on $M$ translates to the vector field on
${\mathcal{C}}_{0}(M)$ defined in the *-subalgebra ${\mathcal{C}}_{0}^{1}(M)$
of continuously differentiable functions (vanishing at infinity with the
derivative, if needed) by tautological pointwise scalar product

$V_{{\mathcal{C}}^{1}}(f)(\xi):=(df)(\xi)\hskip 1.25ptV_{M}(\xi)$

of the 1-form of derivative of $f$ and the given vector field on $M$. If the
differential equation $\dot{\xi}=V(\xi)$ has for $t>0$ the family of solutions
depending continuously on the initial condition $x(0)$, namely,
$F:M\times[0,+\infty)\to M$ such that

$\forall\xi_{0},t:\frac{\partial}{\partial t}F(\xi_{0},t)=V(F(\xi_{0},t));\
F(\xi_{0},0)=\xi_{0}\,,$

then the corresponding endomorphism semigroup of ${\mathcal{C}}_{0}(M)$ is
expressed as:

$({E^{V}}_{t}(f))(\xi)=f(F(\xi,t))$

This admits generalizations beyond ${\mathcal{C}}^{1}$-smoothness (say, to
functions with cusps in a point where $V$ is not defined) with appropriate
tuning of the *-subalgebra.

1.6. Theorem: for the commutative C*-algebra ${\mathcal{C}}_{0}({\mathbb{R}})$
of continuous functions vanishing at infinity and its *-subalgebra
${\mathcal{C}}_{0}^{1}({\mathbb{R}})$ of continuously differentiable
functions, the set of semigroup-generating real vector fields defined in
${\mathcal{C}}_{0}^{1}({\mathbb{R}})$ is not convex.

Proof:

$V_{1}=(\sqrt[3]{x}+1)\frac{d}{dx},\ V_{2}=(\sqrt[3]{x}-1)\frac{d}{dx}$

Both vector fields generate semigroups, and even groups.

For $x>-1$ it may be expressed with a ${\mathcal{C}}^{1}$-substitution:

$y:=\int\frac{1}{V_{1}}=\int\frac{dx}{\sqrt[3]{x}+1}=3(\ln(\sqrt[3]{x}+1)-\sqrt[3]{x}+\frac{1}{2}x^{\frac{2}{3}}),\
y\in{\mathbb{R}}$

which gives $V_{1}={d}/{dy}$ and $({E^{V_{1}}}_{t}(f))(y)=f(y+t)$. Note that
the ${\mathcal{C}}^{1}$-smoothness of $y$ relatively to $x$ in $x=y=0$ is not
obvious in the given expression, but can be verified by differentiation by
$x$. The similar expression (infinitely smooth this time) exists for the
action of ${E^{V_{1}}}$ on $x<-1$. $V_{2}$ gives same substitutions save for
replacement of $x$ with $-x$ and $x>-1$ with $x<1$.

Red: graphs $y=V(x)\hskip 1.125ptdx$ for $V_{1},V_{2}\,$.
Purple: integral curves for $V_{1}\,$.

Indeed, the field

$\frac{V_{1}+V_{2}}{2}=\sqrt[3]{x}\frac{d}{dx}$

is not a semigroup-generating, since the differential equation
$\dot{x}=\sqrt[3]{x}$ with the initial condition $x(0)=0$ has many solutions
at $t>0$, for example $x=\pm(\frac{2}{3}t)^{\frac{3}{2}}$, where the solution
with the “+” sign is continuous with solutions for $x(0)>0$ and one with the
“$-$” sign is continuous with solutions for $x(0)<0$. This obliges
$({E^{(V_{1}+V_{2})/2}}_{t}(f))(0)$ to take values of $f$ simultaneously from
positive and negative points, hence such endomorphism is not possible. []

Certainly, the same is true for the algebra ${\mathcal{C}}_{0}$ of any real
interval (of whatever type: closed or open).

$y=V(x)\hskip 1.125ptdx$ for the red curve. $y=t$ for purple integral curves.

Replacement of ${\mathcal{C}}^{1}$ with ${\mathcal{C}}^{k}$ or
${\mathcal{C}}^{\infty}$ does not change much.

The set of semigroup-generating vector fields is not convex even in the
simplest case of infinite-dimensional C*-algebra and the most natural choice
of a dense *-subalgebra: smooth functions. Henceforth, one should not expect
that this set would be convex in more complicated algebras. This theorem means
that if we are, for some reason, interested in semigroup, then it cannot be
constructed easily from generic real vector fields, even if they generate
semigroups.

##  2 Endomorphism semigroups and gauge groups

2.1. Note that any element $g$ of the algebra defines the vector field,
defined in all ${\mathcal{A}}$, by adjoint representation as a Lie algebra:
$\operatorname{ad}_{g}(\cdot):={[{g},{\cdot}]}$. Somewhat counter-intuitively,
skew-Hermitian ($g+{g}^{*}=0$, i.e. imaginary) elements define real vector
fields,111 Apparently, the source of well-knows disparateness in
real/imaginary factors between mathematicians (who follow algebraic intuition)
and physicists (who might follow differential-geometric intuition).  and, vice
versa, Hermitian elements define “imaginary” (i.e. exchanging Hermitian and
skew-Hermitian elements) vector fields, because the commutator of two
Hermitian elements is skew-Hermitian.

Any real vector field on ${\mathcal{A}}$ defined by $\operatorname{ad}_{g}$
(i.e. $g$ must be skew-Hermitian: $g+{g}^{*}=0$) generates a group of
*-endomorphisms. If the algebra has “1”, then this group can be obtained via
exponentiation. A C*-algebra is a Banach algebra, and we can compute the
exponent of an element, which of skew-Hermitian element will be unitary:
${(\exp(g))}^{*}\hskip 1.25pt\exp(g)=1=\exp(g)\hskip 1.25pt{(\exp(g))}^{*}\,$.
Unitary elements define (internal) automorphisms of the algebra by
conjugation;222 If there is no “1” in ${\mathcal{A}}$, then such automorphisms
also exist, although this is not a conjugation in such case.  see below.

2.2. From here onwards we’ll assume that ${\mathcal{A}}$ is a unital algebra.
Let ${\mathcal{G}}$ be a closed subset of unitary elements, i.e. such $u$ that
${u}^{*}\hskip 1.25ptu=1=u\hskip 1.25pt{u}^{*}$ , which forms a group under
multiplication. Then $\operatorname{Ad}_{u}(\cdot):=u\hskip 1.25pt\cdot\hskip
1.25pt{u}^{*}$ is its action on ${\mathcal{A}}$ by C*-automorphisms. Denote as
${\mathfrak{g}}$ the tangent space of ${\mathcal{G}}$ in 1. It is easy to
check that it is a closed linear subspace,333 Over real numbers
${\mathfrak{g}}$ is a closed subspace of skew-Hermitian elements
($g+{g}^{*}=0$). Of course, the complex tangent space is the complexification
of the real tangent space.  that ${{\mathfrak{g}}^{\mathbb{C}}}$ is closed
under the commutator operation, but it is not, generally, a *-subalgebra.

Though, exponents of elements of ${{\mathfrak{g}}^{\mathbb{R}}}$ belong to
${\mathcal{G}}$, because it is topologically closed.

2.3. Definition: Let $E$ be an endomorphism semigroup, and ${\mathcal{G}}$ be
a closed connected group of unitary elements. If $E_{t}$ preserves the tangent
space ${\mathfrak{g}}\ $444 It is not important whether $E$ has to preserve
real or complex tangent space ${\mathfrak{g}}$, because any skew-Hermitian
element always maps to a skew-Hermitian one by a *-endomorphism.  for any
$t>0$, then we’ll refer to ${\mathcal{G}}$ as a gauge group for $E$.

Note that a *-endomorphism (if it is not an isomorphism) usually does not map
${\mathcal{G}}$ to ${\mathcal{G}}$, even if it preserves its tangent space
${\mathfrak{g}}$.

Definition: Let $V$ be a real vector field generating an endomorphism
semigroup ${E^{V}}$, ${\mathcal{G}}$ be its gauge group, and let the
intersection ${\mathfrak{g}}\cap{{\mathcal{A}}_{V}}$ be dense in
${\mathfrak{g}}$. Then we’ll refer to ${\mathcal{G}}$ as a gauge group for the
vector field $V$.

In this case, for any $g\in{\mathfrak{g}}\cap{{\mathcal{A}}_{V}}$ the
derivative ${\frac{d}{dt}}{E^{V}}_{t}(g)|_{t=0}$ (which equals to $V(g)$ by
1.3.) belongs to ${\mathfrak{g}}$, which means that $\forall
g\in{\mathfrak{g}}\cap{{\mathcal{A}}_{V}}:V(g)\in{\mathfrak{g}}$. The author
currently does not know whether this condition on $V$ is sufficient to have a
gauge group ${\mathcal{G}}$.

2.4. Theorem: Let $V$ be a semigroup-generating real vector field with a gauge
group ${\mathcal{G}}$. Then:

  * •

The family $\operatorname{Ad}_{u}\operatorname{\circ}{E^{V}}_{t}$ of
*-endomorphisms, where $u\in{\mathcal{G}},\ t\geq 0\,$, is a semigroup.

  * •

For every $g\in{\mathfrak{g}}$ the sum of vector fields
$V+\operatorname{ad}_{g}$ generates an endomorphism semigroup which lies
within aforementioned two-parameters semigroup.

Proof: Due to associativity of the composition, to prove the first statement
it’s sufficient to demonstrate that

$\forall t\geq 0,\ \forall\tilde{u}\in{\mathcal{G}}:\exists
u\in{\mathcal{G}}:{\operatorname{Ad}_{\displaystyle\,u}}\operatorname{\circ}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ t}}}={{E^{\displaystyle V}}_{{}_{\displaystyle\
t}}}\operatorname{\circ}{\operatorname{Ad}_{\displaystyle\,\tilde{u}}}$

Since ${\mathcal{G}}$ is connected, we can check it only in a neighborhood of
1, say, for $||{\tilde{u}}-1||<1\,$ where natural logarithm is defined by its
Taylor series:

$\ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots$

After a trivial observation that logarithm of $\tilde{u}$ has to belong to
${\mathfrak{g}}$ it becomes obvious that $u=\exp({E^{V}}_{t}(\ln{\tilde{u}}))$
is the solution – it directly follows from the fact that ${E^{V}}_{t}$ is a
*-endomorphism. We also got that the dependence of $u$ on both ${\tilde{u}}$
and $t$ is continuous.

Now we can realize that the semigroup generated by $V+\operatorname{ad}_{g}$
is expressed as

${{E^{\displaystyle\ V+\operatorname{ad}_{g}}}_{{}_{\displaystyle\
t}}}={\operatorname{Ad}_{\displaystyle\,\exp(\int_{0}^{t}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ t-\tau}}}(g)\hskip 1.25ptd\tau)_{+}}}\
\operatorname{\circ}{{E^{\displaystyle V}}_{{}_{\displaystyle\ t}}}\hskip
30.00005pt\mbox{(\char 126\relax)}$

where $\exp(\ldots)_{+}$ is the ordered exponential.[3] For $0\leq s\leq t\,$,
it satisfies the identity:

$\exp(\int_{0}^{t}{{E^{\displaystyle V}}_{{}_{\displaystyle\
t-\tau}}}(g)\hskip 1.25ptd\tau)_{+}=\exp(\int_{0}^{s}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ s-\tau}}}(g)\hskip 1.25ptd\tau)_{+}\hskip
1.25pt\exp(\int_{0}^{t-s}{{E^{\displaystyle V}}_{{}_{\displaystyle\
t-\tau}}}(g)\hskip 1.25ptd\tau)_{+}$

On the other hand, since ${E^{V}}_{s}$ is an endomorphism of the algebra, we
have:

${\operatorname{Ad}_{\displaystyle\,\exp(\int_{0}^{t-s}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ t-\tau}}}(g)\hskip
1.25ptd\tau)_{+}}}={\operatorname{Ad}_{\displaystyle\,\exp({{E^{\displaystyle
V}}_{{}_{\displaystyle\ s}}}\,\int_{0}^{t-s}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ t-s-\tau}}}(g)\hskip 1.25ptd\tau)_{+}}}=$
$={{E^{\displaystyle V}}_{{}_{\displaystyle\
s}}}\operatorname{\circ}{\operatorname{Ad}_{\displaystyle\,\exp(\int_{0}^{t-s}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ t-s-\tau}}}(g)\hskip 1.25ptd\tau)_{+}}}$

With it, we immediately see that compositions of *-endomorphisms from (~)
behave as they must do it in a semigroup. So, ordering in the “abnormal”
direction, higher semigroup-parameter factors to right and lower to left, is
necessary to put factors with semigroup parameter close to $t$ adjacent to the
${E^{V}}_{t}$ endomorphism.

Initial condition from (!) is evident. The only thing we need to complete the
proof of (~) is to compute its derivative. Since we already know that the
formula (~) defines a semigroup, we can do it only at $t=0\,$:

${\frac{d}{dt}}\left({\operatorname{Ad}_{\displaystyle\,\exp(\int_{0}^{t}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ t-\tau}}}(g)\hskip 1.25ptd\tau)_{+}}}\
\operatorname{\circ}{{E^{\displaystyle V}}_{{}_{\displaystyle\
t}}}\right)_{t=0}=$
$={\operatorname{ad}_{\displaystyle\,{\frac{d}{dt}}\exp(\int_{0}^{t}{{E^{\displaystyle
V}}_{{}_{\displaystyle\ t-\tau}}}(g)\hskip
1.25ptd\tau)_{+}|_{t=0}}}+V={\operatorname{ad}_{\displaystyle\,g}}+V\ .$

[]

2.5. Commutator of vector fields cannot be defined straightly since we defined
the codomain of $V$ to be a broader space than its domain. Though, if there
are two fields $Y:{\mathcal{A}}_{Y}\to{\mathcal{A}}$ and
$Z:{\mathcal{A}}_{Z}\to{\mathcal{A}}$ and there exists such dense *-subalgebra
${\mathcal{A}}_{YZ}$ that is a subset of both ${\mathcal{A}}_{Y}$ and
${\mathcal{A}}_{Z}$, $Y$ maps it into ${\mathcal{A}}_{Z}$ and $Z$ maps it into
${\mathcal{A}}_{Y}$, then the commutator may be defined in this narrower
*-subalgebra:

${[{Y},{Z}]}:=Y\operatorname{\circ}Z-Z\operatorname{\circ}Y$

The (˙) condition (see 1.1.) for ${[{Y},{Z}]}$ is satisfied:

${[{Y},{Z}]}(u\hskip 1.25ptv)=Y(u\hskip 1.25ptZ(v)+Z(u)\hskip
1.25ptv)-Z(u\hskip 1.25ptY(v)+Y(u)\hskip 1.25ptv)=$ $={\scriptstyle u\hskip
0.875ptY(Z(v))+Y(u)\hskip 0.875ptZ(v)+Z(u)\hskip 0.875ptY(v)+Y(Z(u))\hskip
0.875ptv-u\hskip 0.875ptZ(Y(v))-Z(u)\hskip 0.875ptY(v)-Y(u)\hskip
0.875ptZ(v)-Z(Y(u))\hskip 0.875ptv\ }=$ $=u\hskip
1.25pt{[{Y},{Z}]}(v)+{[{Y},{Z}]}(u)\hskip 1.25ptv\,$

Since real vector fields commute with *, it is obvious that the commutator of
two real vector fields is real; see notes in 2.1.

2.6. Lemma: Let $V_{1}$ be a semigroup-generating real vector field defined in
a dense *-subalgebra ${{\mathcal{A}}_{V}}$. Let $V_{2}$ be a vector field such
that the commutator ${[{V_{1}},{V_{2}}]}$ is defined in some (narrower) dense
*-subalgebra ${\mathcal{A}}_{V_{1}V_{2}}$ and is a bounded operator. Then:

${[{{{E^{\displaystyle V_{1}}}_{{}_{\displaystyle\
t}}}},{V_{2}}]}=\int\limits_{0}^{t}{{E^{\displaystyle
V_{1}}}_{{}_{\displaystyle\
\tau}}}\operatorname{\circ}{[{V_{1}},{V_{2}}]}\operatorname{\circ}{{E^{\displaystyle
V_{1}}}_{{}_{\displaystyle\ t-\tau}}}\hskip 1.25ptd\tau$

Note that if ${[{V_{1}},{V_{2}}]}={\operatorname{ad}_{\displaystyle\,c}}$ for
certain $c\in{\mathcal{A}}$, then

$\int\limits_{0}^{t}{{E^{\displaystyle V_{1}}}_{{}_{\displaystyle\
\tau}}}\operatorname{\circ}{[{V_{1}},{V_{2}}]}\operatorname{\circ}{{E^{\displaystyle
V_{1}}}_{{}_{\displaystyle\ t-\tau}}}\hskip
1.25ptd\tau={\operatorname{ad}_{\displaystyle\,\int\limits_{0}^{t}{{E^{\displaystyle
V_{1}}}_{{}_{\displaystyle\ \tau}}}(c)\hskip
1.25ptd\tau}}\operatorname{\circ}{{E^{\displaystyle
V_{1}}}_{{}_{\displaystyle\ t}}}$

because ${E^{V_{1}}}$ consists of endomorphisms of the algebra.

Proof: First we have to ensure that the operator family
${[{{E^{V_{1}}}_{t}},{V_{2}}]}$ is continuous by $t$ in the strong operator
topology from ${\mathcal{A}}_{V_{1}V_{2}}$ to ${\mathcal{A}}\,$. It is obvious
for the ${E^{V_{1}}}_{t}\operatorname{\circ}{V_{2}}$ term, but for
${V_{2}}\operatorname{\circ}{E^{V_{1}}}_{t}$ we have to check that
${E^{V_{1}}}_{t}$ preserves ${\mathcal{A}}_{V_{1}V_{2}}$, which follows from
the boundness of ${[{V_{1}},{V_{2}}]}$.555 More strictly, given
${\mathcal{A}}_{V_{1}V_{2}}\,$ might have to be extended to some broader
*-algebra, such that both ${V_{1}}\operatorname{\circ}{V_{2}}$ and
${V_{2}}\operatorname{\circ}{V_{1}}$ are defined and the (new) algebra is
preserved by ${E^{V_{1}}}$.  Then, the formula can be checked by interpolation
between ${V_{2}}\operatorname{\circ}{E^{V_{1}}}_{t}$ and
${E^{V_{1}}}_{t}\operatorname{\circ}{V_{2}}$ by
${E^{V_{1}}}_{\tau}\operatorname{\circ}{V_{2}}\operatorname{\circ}{E^{V_{1}}}_{t-\tau}$
(say, for $\tau=k\hskip 1.25ptt/n,\ k=1\ldots n-1$, like in Riemann integral),
using continuity of ${E^{V_{1}}}_{t}$ by $t$. []

2.7. Theorem: In all conditions of 2.6., let $V_{2}$ generate its semigroup,
let both semigroups have a gauge group ${\mathcal{G}}$, and $\exists
c\in{{\mathfrak{g}}^{\mathbb{R}}}:{[{V_{1}},{V_{2}}]}={\operatorname{ad}_{\displaystyle\,c}}\,$.666
Note that “$\operatorname{ad}$” of any element of ${\mathcal{A}}$ is bounded.
Then ${E^{V_{1}}}_{t}\operatorname{\circ}{E^{V_{2}}}_{s}$ and
${E^{V_{2}}}_{s}\operatorname{\circ}{E^{V_{1}}}_{t}$ differ only by left
composition with “$\operatorname{Ad}$” of an element of ${\mathcal{G}}$.

From this theorem, remembering 2.4., immediately follows that the family
$\operatorname{Ad}_{u}\operatorname{\circ}{E^{V_{2}}}_{s}\operatorname{\circ}{E^{V_{1}}}_{t}$
of *-endomorphisms, where $u\in{\mathcal{G}},\ t\geq 0,\ s\geq 0\,$, is a
semigroup. In a more abstract sense, this means a principal bundle[4] over
$[0,+\infty)\times[0,+\infty)$ with a semigroup structure corresponding to
addition of 2-vectors, with the representation by *-endomorphisms and with the
left action777 The standard definition of a principal bundle requires a right
action, but for this construct, a left action is better suited.  of
${\mathcal{G}}$. It is also possible to prove the statement which generalizes
the second statement of the theorem of 2.4., but we’ll made such
generalisation in Section 4 for a more general case.

Proof: Aforementioned “$\operatorname{Ad}$” term can be obtained by solving a
differential equation (by $s$) in ${\mathcal{G}}$, using Lemma 2.6. and the
fact that $V_{2}$ preserves the gauge algebra. []

##  3 Constructions of vector fields

This section will present possible constructions of vector fields via
representations of the algebra. Although in simple cases these fields generate
semigroups, the section does not contain theorems about it.

3.1. Assume that ${\mathcal{A}}$ has a representation on $L^{2}({\mathbb{R}})$
(with variable $x\in{\mathbb{R}}$). The unbounded skew-Hermitian operator
${\mathrm{d}}=\frac{d}{dx}$ of differentiation by $x\,$888 In quantum-
mechanical parlance, ${\mathrm{d}}=i\hskip 1.25pt\hat{p}/\hbar$.  is defined
on ${\mathcal{C}}^{1}$-smooth functions, which are dense in $L^{2}$. Suppose
that in some dense *-subalgebra the operator $V(q)={{\mathrm{d}}}^{*}\hskip
1.25ptq+q\hskip 1.25pt{\mathrm{d}}={[{q},{{\mathrm{d}}}]}\,$ is defined. It
means that aforementioned operator on $L^{2}({\mathbb{R}})$ is bounded and
belongs to the representation. For example, such *-subalgebra would be
${\mathcal{C}}^{1}_{0}({\mathbb{R}})$ for
${\mathcal{A}}={\mathcal{C}}_{0}({\mathbb{R}})\,$. From the product rule for
$V$ we see that it is a vector field. If ${\mathcal{A}}$ is translation-
invariant, then $V$ generates a group of *-endomorphisms, called “shifts”.

3.2. The “translation” vector field $V(q)={{\mathrm{d}}}^{*}q+q{\mathrm{d}}$
may be expressed in a matrix form:

${\mathrm{J}}:=\left(\begin{matrix}\mathbf{1}\\\
{\mathrm{d}}\end{matrix}\right);\ \ \ \
{}{{\mathrm{J}}}^{*}=\left(\begin{matrix}\mathbf{1}&{{\mathrm{d}}}^{*}\end{matrix}\right);\
\ \ \ {}V(q)={{\mathrm{J}}}^{*}\hskip 1.25pt\left(\begin{matrix}0&q\\\
q&0\end{matrix}\right)\hskip 1.25pt{\mathrm{J}}$

Or, more complicated:

$S:=\left(\begin{matrix}0&\mathbf{1}\\\ \mathbf{1}&0\end{matrix}\right);\ \ \
\ {}V(q)={{\mathrm{J}}}^{*}\hskip 1.25ptS\hskip
1.25pt\operatorname{{\mathrm{diag}}}(q,q)\hskip 1.25pt{\mathrm{J}}$

From skew-Hermitianness of “${\mathrm{d}}$” we have ${{\mathrm{J}}}^{*}\hskip
1.25ptS\hskip 1.25pt{\mathrm{J}}=0$.

Now let us think that instead of $\operatorname{{\mathrm{diag}}}(q,q)$ we have
another representation $\pi$ of ${\mathcal{A}}$, in another Hilbert space
${\mathcal{D}}$, which commutes with certain operator $S$. Suppose that
${\mathrm{J}}:{\mathcal{H}}\to{\mathcal{D}}$ (transition operator) is such
unbounded operator with dense domain that the operator product

$V(q)={{\mathrm{J}}}^{*}\hskip 1.25ptS\hskip 1.25pt\pi(q)\hskip
1.25pt{\mathrm{J}}$

for any $q\in{{\mathcal{A}}_{V}}$ belongs to our (default) representation of
${\mathcal{A}}$ in ${\mathcal{H}}$.

This bears some resemblance to Alain Connes’[2] definition of differentiation,
by commutator with the operator $F$, although we, contrary, try to avoid
commutators.

It is easy to check that conditions

${{\mathrm{J}}}^{*}\hskip 1.25ptS\hskip 1.25pt{\mathrm{J}}=0;\hskip
40.00006pt\forall u,v\in{{\mathcal{A}}_{V}}:{[{u},{{{\mathrm{J}}}^{*}}]}\hskip
1.25ptS\hskip 1.25pt{[{{\mathrm{J}}},{v}]}=0$

are sufficient to make $V$ a vector field. Note that
${[{u},{{{\mathrm{J}}}^{*}}]}$ means
$u\operatorname{\circ}{{\mathrm{J}}}^{*}-{{\mathrm{J}}}^{*}\operatorname{\circ}\pi(u)$,
an operator from ${\mathcal{D}}$ to ${\mathcal{H}}$, and similarly
${[{{\mathrm{J}}},{v}]}:{\mathcal{H}}\to{\mathcal{D}}\,$. Also, if $S$ is
self-conjugated, then $V$ is real.

3.3. What appears after replacement of one-dimensional “${\mathrm{d}}$” with
something like Dirac operator? Suppose that ${\mathrm{J}}$ depends linearly on
the parameter from ${\mathbb{C}}^{2}$. Say, there are two operators
${\mathrm{J}}_{0},{\mathrm{J}}_{1}$ and we define
${\mathrm{J}}_{\phi}:=\phi^{L}\hskip 1.25pt{\mathrm{J}}_{L}\,.$ Then,

${V_{\phi\otimes{\phi}^{*}}}(q)={{\mathrm{J}}_{\phi}}^{*}\hskip 1.25ptS\hskip
1.25pt\pi(q)\hskip 1.25pt{\mathrm{J}}_{\phi}\,;\ \phi\in{\mathbb{C}}^{2}$

The use of tensor product symbol is motivated by an observation that the
vector field depends on parameters bilinearly:

$V_{L{L^{\prime}}}(q)={{{{\mathrm{J}}}^{*}}_{L^{\prime}}}\hskip 1.25ptS\hskip
1.25pt\pi(q)\hskip 1.25pt{{\mathrm{J}}_{L}}\,,\mbox{ or,}$
$V_{X}(q)=X^{L{L^{\prime}}}\hskip
1.25pt{{{{\mathrm{J}}}^{*}}_{L^{\prime}}}\hskip 1.25ptS\hskip
1.25pt\pi(q)\hskip 1.25pt{{\mathrm{J}}_{L}}\mbox{ where }X\mbox{ is a }2\times
2\mbox{ matrix.}$

We can get a 4-dimensional family of vector fields at the price of only
2-dimensional, because ${\mathrm{J}}$ and ${{\mathrm{J}}}^{*}$ are essentially
the same thing. Though, (for self-conjugated $S$) only fields parametrized by
Hermitian matrices will be real. Note that $\phi\otimes{\phi}^{*}$ defines a
$2\times 2$ Hermitian matrix with rank not greater than 1 and non-negative
trace. We postulated the dependence of ${\mathrm{J}}$ on 2 complex numbers,
and got a vector field depending on a $2\times 2$ matrix, which gave the
structure of (1+3)-dimensional Minkowski space known in Special Relativity.
This construction of the family of vector fields is speculative, but it leads
to interesting situation if we suppose that, in some theory, we got that
namely $V_{\phi\otimes{\phi}^{*}}={{\mathrm{J}}_{\phi}}^{*}\hskip
1.25ptS\hskip 1.25pt\pi(q)\hskip 1.25pt{\mathrm{J}}_{\phi}$ are semigroup-
generating, but not other fields in this 4-dimensional complex-linear family.

##  4 Endomorphism semigroup parametrized by a cone

4.1. The construction from the previous section gives the ${\mathbb{C}}^{2}$
of vector fields of the type $\phi\otimes{\phi}^{*}$, which is actually the
cone over a Riemann sphere, due to independence of the phase of $\phi$.
Suppose that all of those fields generate a semigroup.999 This is the case,
for example, if we defined translations in Minkowski space and our
representation is defined on functions on the cone of the past, or another
domain containing the whole cone of the past of each of its points. A shift
semigroup acts on a domain by filling “overflowed” space with zeros (not
explained in this version of the paper).  Let us also suppose that those
fields and semigroups has ${\mathcal{G}}$ as a gauge group and commutators of
those fields lie in the gauge algebra.

Obviously, this construction corresponds to the geometry of the cone of the
past in the Minkowski space, where 4-coordinates and $2\times 2$ matrices
correspond as:

$X=\left(\begin{matrix}x^{0}+x^{3}&x^{1}-i\hskip 1.25ptx^{2}\\\ x^{1}+i\hskip
1.25ptx^{2}&x^{0}-x^{3}\end{matrix}\right)=x^{0}\hskip
1.25pt{\mathrm{I}}+\sum\limits_{k=1,2,3}x^{k}\hskip 1.25pt\sigma_{k}\,,\mbox{
where }\sigma_{k}\mbox{ are Pauli matrices.}$

Here, the semigroup of translations cannot generated by two 1-parameter
semigroups, so the result from 2.7. is not applicable. BTW, it cannot be
generated by any finite number of 1-parameter semigroups.

How can we generalize 2.7. for this geometry?

4.2. The spinor representation possibly gave some insight about origins of
semigroup-generating vector fields parametrized by the light cone (i.e. by
null directions) in the Minkowski space, but it is useless when we have to
prove something about semigroups. From here onwards we’ll forget anything
about spinors and think we have $n$ fields $V_{k}\,$ and some cone $\Phi$ in
${\mathbb{R}}^{n}$ such that all fields $x^{k}\hskip 1.25ptV_{k}\,,\ x\in\Phi$
generate a semigroup. For the example from 4.1. it would be

$\Phi=\left\\{\,x\in{\mathbb{R}}^{4}\,|\,x^{0}\geq 0\,,\
(x^{0})^{2}-(x^{1})^{2}-(x^{2})^{2}-(x^{3})^{2}=0\,\right\\}$

Let all these semigroups have the same gauge group ${\mathcal{G}}$.101010 This
version of the paper lacks an explanation how gauge groups may be related to
the spinor construction from Section 3. To be fixed.  The condition about
commutators takes the form

${[{V_{k}},{V_{l}}]}={\operatorname{ad}_{\displaystyle\,c_{kl}}},\
c_{kl}\in{\mathfrak{g}}$

due to bilinearity of the commutator. Then:

  * •

There exists a principal bundle $F\to{\hat{\Phi}}$ (where $\hat{\Phi}$ is a
convex hull) with the structure group ${\mathcal{G}}$ and with a semigroup
structure on it which corresponds to vector addition in $\hat{\Phi}$.

  * •

There exists its representation ${\mathcal{E}}$ by *-endomorphisms of
${\mathcal{A}}$ which include all semigroups ${E^{\,x^{k}\hskip
0.875ptV_{k}}},\ x\in\Phi$ such that ${E^{\,x^{k}\hskip 0.875ptV_{k}}}_{t}$ is
parametrized by the point of $F$ corresponding to the point $t\hskip
1.25ptx^{k}$ of $\Phi$.

  * •

The fiber over the 0 point is the group ${\mathcal{G}}$ itself with its group
structure matching the semigroup structure of the bundle. So, endomorphisms
from the same fiber differ only by left composition with an
$\operatorname{Ad}_{u}$.

  * •

All (other) vectors $x\in\hat{\Phi}$ have their semigroups ${E^{\,x^{k}\hskip
0.875ptV_{k}}}$ included to $F$ by the same way.

Proof: By applying the theorem of 2.7. to two-dimensional subspaces of
${\mathbb{R}}^{n}$ generated by pairs of vectors from $\Phi$ we can construct
the bundle over such subspaces. Where subspaces intersect, consistency is
provided by existence and uniqueness of the corresponding real-parametrized
endomorphism semigroup. Iterating this construction $n-1$ times (or directly
applying 2.7.-like reasonings to $n$-tuples of vectors from $\Phi$) we extend
it to all $\hat{\Phi}$.

The vector fields $V_{k}$ (or, in other words, the field linearly parametrized
by an $n$-dimentional real space) actually define the connection on our
bundle, where $c_{kl}$ is its 2-form of curvature. The last (fourth) statement
is a manifestation of the fact that such connection can be integrated along
the path. []

## References

  * [1] Dixmier, Jaques. Les C*-algèbres et leurs représentations. Paris, 1969
  * [2] Connes, Alain. Noncommutative Geometry. Academic Press, San Diego, CA, 1994, 661 p., ISBN 0-12-185860-X.
  * [3] http://en.wikipedia.org/wiki/Ordered_exponential
  * [4] http://en.wikipedia.org/wiki/Principal_bundle