# Tree-Level Superstring Amplitudes: The Neveu-Schwarz Sector

Sergio Luigi Cacciatori  Università dell’Insubria, dipartimento di scienza e
alta tecnologia, via Valleggio 11, 22100, Como, Italy and INFN, via Celoria
16, 20100, Milano, Italy  ,  Samuel Grushevsky  Department of Mathematics and
Simons Center for Geometry and Physics, Stony Brook University, Stony Brook,
NY 11794-3651  and  Alexander A. Voronov  School of Mathematics, University of
Minnesota, Minneapolis, MN 55455 and Kavli IPMU (WPI), UTIAS, The University
of Tokyo, Kashiwa, Chiba 277-8583, Japan

(Date: February 22, 2024)

###### Abstract.

We present a complete computation of superstring scattering amplitudes at tree
level, for the case of Neveu-Schwarz insertions. Mathematically, this is to
say that we determine explicitly the superstring measure on the moduli space
$\mathcal{M}_{0,n,0}$ of super Riemann surfaces of genus zero with $n\geq 3$
Neveu-Schwarz punctures. While, of course, an expression for the measure was
previously known, we do this from first principles, using the canonically
defined super Mumford isomorphism [1]. We thus determine the scattering
amplitudes, explicitly in the global coordinates on $\mathcal{M}_{0,n,0}$,
without the need for picture changing operators or ghosts, and are also able
to determine canonically the value of the coupling constant. Our computation
should be viewed as a step towards performing similar analysis on
$\mathcal{M}_{0,0,n}$, to derive explicit tree-level scattering amplitudes
with Ramond insertions.

Research of the second author is supported in part by NSF grant DMS-21-01631.
Research of the third author is supported in part by World Premier
International Research Center Initiative (WPI Initiative), MEXT, Japan, and
Collaboration grant #585720 from the Simons Foundation.

###### Contents

  1. 1 Introduction
  2. 2 Basic conventions and constructions with SRSs and their moduli
  3. 3 The super Mumford form
    1. 3.1 Bases of cohomology
    2. 3.2 A formula for the super Mumford form
  4. 4 The super Kodaira-Spencer map over $\mathcal{M}_{0,n,0}$
    1. 4.1 The construction
    2. 4.2 Proof of the main theorem
    3. 4.3 The Čech cohomology computation
    4. 4.4 The dual map
    5. 4.5 The Berezinian of the dual map
  5. 5 Vertex-operator insertions
  6. 6 Tree-level scattering amplitudes in the physics literature
  7. 7 Conclusion
  8. A The super Mumford form over $\mathcal{M}_{0,n,0}$
  9. B Proof of 5: the Berezinian of the super Kodaira-Spencer map
  10. C The $z_{1}=\infty$ case

##  1\. Introduction

Bosonic string scattering amplitudes can be mathematically defined as
integrals over the space of all maps from Riemann surfaces with marked points
to a given spacetime manifold. To obtain an explicit expression for such an
integrand, one proceeds from first principles of string theory, but of course
eventually has to choose local coordinates on the moduli space. This
construction in particular relies on the _Mumford isomorphism_
$\det(\mathbb{E}_{1})^{\otimes 13}\cong\det(\mathbb{E}_{2})$ over the moduli
space $M_{g}$ of genus $g>1$ (bosonic) Riemann surfaces. Here
$\mathbb{E}_{k}\to M_{g}$ is the universal vector bundle of holomorphic
$k$-differentials, whose fiber over a Riemann surface $X\in M_{g}$ is
$H^{0}(X,\omega^{\otimes k})$, where $\omega=\mathcal{T}^{*}_{X}$ is the
holomorphic cotangent bundle of $X$. Algebro-geometrically, since
$\mathbb{E}_{2}$ is the cotangent bundle $\mathcal{T}^{*}_{M_{g}}$, the
Mumford isomorphism gives the expression
$\omega_{M_{g}}=c_{1}(\mathcal{T}^{*}_{M_{g}})=\det(\mathbb{E}_{2})\cong\lambda_{1}^{\otimes
13}$ for the canonical bundle $\omega_{M_{g}}$ of the moduli space, where
$\lambda_{1}:=c_{1}(\mathbb{E}_{1})=\det(\mathbb{E}_{1})\in\operatorname{Pic}(M_{g})$.
Since each space-time dimension contributes $\lambda_{1}^{\otimes 1/2}$ to the
integrand, in particular the Mumford isomorphism implies that bosonic string
theory could only make sense in 26 dimensions. We note that the bosonic
Mumford isomorphism is a statement that two vector bundles are isomorphic, but
it does not give a canonical isomorphism between them.

Bosonic string amplitudes were extensively studied in the 1980s, and various
expressions for them were obtained in arbitrary genus, see [2, 3] and also the
more recent literature [4, 5] summarizing these results and providing further
references. At tree level (i.e. no loops, i.e. genus zero Riemann surfaces)
explicit expressions for bosonic string amplitudes are well-known, see [6].

The mathematical foundations of the superstring scattering amplitudes were
reviewed and revisited by Witten [7], with further mathematical underpinnings
developed in [8, 9, 10]. Superstring scattering amplitudes can be
mathematically defined as integrals over the space of all maps from super
Riemann surfaces to a given space-time supermanifold. Recall that a _super
Riemann surface $($SRS$)$_ is a complex supermanifold $X$ of complex dimension
$(1|1)$, i.e. such that local charts at any point of $X$ are given by one even
and one odd coordinate, together with a _maximally non-integrable, odd
distribution_ $\mathcal{D}\subset\mathcal{T}_{X}$, i.e. a distribution
$\mathcal{D}$ of rank $(0|1)$ such that the Lie bracket of super vector fields
induces an isomorphism

$[-,-]\colon\thinspace\mathcal{D}\otimes\mathcal{D}\xrightarrow{\sim}\mathcal{T}_{X}/\mathcal{D}.$

This yields a canonical short exact sequence

(1) $0\to\mathcal{D}\to\mathcal{T}_{X}\to\mathcal{D}^{\otimes 2}\to 0.$

The _super Mumford isomorphism_ over the moduli space $\mathcal{M}_{g}$ of SRS
of genus $g>1$ is the isomorphism

(2) $\operatorname{Ber}(\mathbb{E}_{1/2})^{\otimes
5}=\operatorname{Ber}(\mathbb{E}_{3/2})\,,$

where $\mathbb{E}_{1/2}$ is now the virtual coherent sheaf over
$\mathcal{M}_{g}$, with fiber $H^{0}(X,\omega)-H^{1}(X,\omega)$ over
$X\in\mathcal{M}_{g}$, where
$\omega:=\operatorname{Ber}\mathcal{T}^{*}_{X}=\mathcal{D}^{*}$ now denotes
the canonical bundle of the SRS $X$, while $\mathbb{E}_{3/2}$ denotes the
bundle with fibers $H^{0}(X,\omega^{\otimes 3})$ (for which the $H^{1}$
vanishes automatically), and $\operatorname{Ber}$ denotes the Berezinian, also
known as superdeterminant.111Recall that on the bosonic Riemann surface
$X_{\operatorname{red}}$ underlying $X$, $\omega$ restricts to a square root
of the canonical bundle of $X_{\operatorname{red}}$, so it makes sense to talk
of $1/2$\- and $3/2$-differentials, by taking this choice of the line bundle
$\omega_{X_{\operatorname{red}}}^{\otimes 1/2}$. Since the cotangent bundle to
$\mathcal{M}_{g}$ is $\mathbb{E}_{3/2}$, this gives

(3) $\operatorname{Ber}(\mathbb{E}_{1/2})^{\otimes
5}=\omega_{\mathcal{M}_{g}}\,,$

which again is the reason why superstring theory can only be consistent in
$10=5\cdot 2$ dimensions (see below and [11, 8, 9, 7] for more details).
Crucially, unlike in the bosonic case, the super Mumford isomorphism (2) is
not just a statement that two line bundles over $\mathcal{M}_{g}$ are
abstractly isomorphic, but gives a canonical isomorphism between them, see
[1].

While the moduli space $M_{g}$ of genus $g>1$ bosonic Riemann surfaces is a
complex Deligne-Mumford stack of dimension $3g-3$, the moduli space
$\mathcal{M}_{g}$ of genus $g>1$ SRS is a complex superstack of dimension
$(3g-3\,|\,2g-2)$. For $g=0$ (resp. 1), to get a Deligne-Mumford stack as the
moduli space of bosonic Riemann surfaces, one needs to consider Riemann
surfaces together with at least $n=3$ (resp., $n=1$) marked points, whence
$\dim M_{0,n}=n-3;\ \dim M_{1,n}=n$. Unlike the bosonic case, where there is
only one notion of a marked point on a Riemann surface (which is a codimension
$1$ complex subspace, and equivalent to considering a puncture), there are two
differen notions of markings or punctures for super Riemann surfaces. A Neveu-
Schwarz (NS) marking on a super Riemann surface $X$ is just a point of $X$,
i.e. a subspace of complex dimension $(0|0)$ and codimension $(1|1)$, which
plays no role for the distribution $\mathcal{D}$. On the other hand, a Ramond
(R) marking is a subspace of codimension $(1|0)$, over which the maximal non-
integrability of the distribution $\mathcal{D}$ fails in a controlled way [7,
12, 13, 14].

We denote by $\mathcal{M}_{g,n,k}$ the moduli of genus $g$ super Riemann
surfaces with $n$ Neveu-Schwarz markings and $k$ Ramond markings, and then for
$g>1$ it is a superstack of dimension $(3g-3+n+k\,|\,2g-2+n+\tfrac{k}{2})$. To
write an explicit expression for superstring scattering amplitudes, one needs
to choose local coordinates on $\mathcal{M}_{g,n,k}$. We recall that locally
any (smooth or complex) supermanifold is split, i.e. is isomorphic to the
exterior algebra of a vector bundle on its underlying (bosonic) manifold. Thus
in a sufficiently small coordinate patch, $\mathcal{M}_{g,n,k}$ is modelled on
a vector bundle (quotiented by a finite group, as we are working with
orbifolds) over the moduli space
$SM_{g,n+k}=(\mathcal{M}_{g,n,k})_{\operatorname{red}}$ of Riemann surfaces
with a spin structure. However, as was shown by Donagi-Witten [15] (see also
Donagi-Ott [14] for the case with Ramond markings), for high enough genus,
globally $\mathcal{M}_{g,n,k}$ is not split and does not even admit a global
holomorphic projection to its reduced bosonic moduli space $SM_{g,n+k}$. In
physics literature [16], to write formulas on $\mathcal{M}_{g,n,k}$, one often
works in local coordinates, and then uses picture changing operators (PCOs) to
change to a different coordinate chart. To deal with the choices involved, one
introduces ghosts, and then integrates over them as well. In particular, this
approach led to spectacular computations of superstring scattering amplitudes
in genus 2 by D’Hoker and Phong [17].

Our focus in this paper is the moduli space of genus $0$ super Riemann
surfaces with only NS markings: $\dim\mathcal{M}_{0,n,0}=(n-3\,|\,n-2)$, for
$n\geq 3$, since the dimension of the automorphism group
$G:=\operatorname{OSp}(1|2)^{0}$ of the super projective line
${\mathbb{P}}^{1|1}$ with the odd distribution
$\mathcal{D}\subset\mathcal{T}_{{\mathbb{P}}^{1|1}}$ is equal to $(3|2)$. Then
$\mathcal{M}_{0,n,0}$ is the moduli space of $n$-tuples of labeled pairwise
distinct points
$(z_{1}|\zeta_{1}),\dots,(z_{n}|\zeta_{n})\in{\mathbb{P}}^{1|1}$, called _NS
punctures_ , up to the diagonal action of $G$. The group $G$ acts transitively
on triples of distinct bosonic points, and then further on pairs of odd
coordinates. Moreover, fixing the values of
$z_{1},z_{2},z_{3},\zeta_{1},\zeta_{2}$ to be
$z_{1}^{0},z_{2}^{0},z_{3}^{0},\zeta_{1}^{0},\zeta_{2}^{0}$ for some distinct
values $z_{1}^{0},z_{2}^{0},z_{3}^{0}\in{\mathbb{P}}^{1}$ gives a point in
each $G$-orbit in the configuration space of $n$-tuples of distinct points of
${\mathbb{P}}^{1|1}$. Each $G$-orbit contains a unique element with these
values, up to the action of the involution, which for the choice of
$\zeta_{1}^{0}=\zeta_{2}^{0}=0$ is the element of $G$ that fixes the bosonic
coordinate and changes the sign of the odd coordinate on ${\mathbb{P}}^{1|1}$,
see [18, § 2.12] and [19, § 5.1.2]. Thus, similarly to the bosonic case, there
are global coordinates $(z_{4},\dots,z_{n}\,|\,\zeta_{3},\dots,\zeta_{n})$ on
$\mathcal{M}_{0,n,0}$, which exhibit this supermoduli space as a dense open
subset of the superorbifold
$\mathbb{C}^{0|1}\times(\mathbb{C}^{1|1})^{n-3}/\mathbb{Z}_{2}\cong\mathbb{C}^{n-3\,|\,n-2}/\mathbb{Z}_{2}$
(provided one of $z_{1}^{0},z_{2}^{0},z_{3}^{0}$ is set to $\infty$), where
$\mathbb{Z}_{2}$ acts by the involution that changes the signs of all odd
coordinates simultaneously. Therefore it should be possible to write a global
expression for the tree-level superstring amplitudes, for the case of only NS
punctures, explicitly in these global coordinates, from first principles,
without introducing ghosts or picture changing operators.

When we talk about integrals, such as (9), over $\mathcal{M}_{0,n,0}$, we
treat them as improper integrals over a noncompact domain and ignore the
isotropy group $\mathbb{Z}_{2}$, since its only possible contribution to the
integral could be a factor of $2$. The choice of whether to divide or multiply
by 2 is left to the future reader, as the additivity properties of the Berezin
integral over superstacks are not yet well understood.

The first principles we mentioned above are canonical isomorphisms, such as
the super Mumford isomorphism (2), which simplifies in the case $g=0$. Over
$\mathcal{M}_{0,n,0}$, the super vector bundle $\mathbb{E}_{1/2}$ trivializes
canonically, and thus and thus so does $\operatorname{Ber}(\mathbb{E}_{1/2})$,
because $H^{0}({\mathbb{P}}^{1|1},\omega)=0$ and
$H^{1}({\mathbb{P}}^{1|1},\omega)=H^{0}({\mathbb{P}}^{1|1},\mathcal{O}_{{\mathbb{P}}^{1|1}})^{*}$
is canonically isomorphic to $\mathbb{C}$. However, in the presence of NS
markings, the canonical bundle $\omega_{\mathcal{M}_{g,n,0}}$ is identified
with $\operatorname{Ber}(H^{0}(\omega^{\otimes 3}(\operatorname{NS})))$, where
$\operatorname{NS}$ is the total NS divisor, rather than with
$\operatorname{Ber}(\mathbb{E}_{3/2})$, as in the case of no markings. With
this change, the super Mumford isomorphism (3) gets modified to a canonical
isomorphism

$\operatorname{Ber}(\mathbb{E}_{1/2})^{\otimes
5}\otimes\operatorname{Ber}(H^{0}(X,\omega^{\otimes
3}(\operatorname{NS})|_{\operatorname{NS}}))=\omega_{\mathcal{M}_{g,n,0}}\,,$

see [7, 12]. In the $g=0$ case, this simplifies to

(4) $\operatorname{Ber}(H^{0}(X,\omega^{\otimes
3}(\operatorname{NS})|_{\operatorname{NS}}))=\omega_{\mathcal{M}_{0,n,0}}\,.$

It is known that the line bundle $\omega^{\otimes 3}(\operatorname{NS})$ above
may be replaced by any other odd line bundle via the canonical isomorphism
(15), [1, 12]. As a consequence, we get the canonical isomorphism

(5)
$\operatorname{Ber}(H^{0}(X,\omega|_{\operatorname{NS}}))=\omega_{\mathcal{M}_{0,n,0}}\,,$

which will be used in the main theorem below.

In this paper we perform the computation of the tree-level NS scattering
amplitudes from first principles, not surprisingly recovering the known
formulas [20, 21, 22, 23] from the physics literature. The main ingredient is
the canonical determination of the holomorphic measure on
$\mathcal{M}_{0,n,0}$, which we deduce by using the canonical super Mumford
isomorphism; this also allows us to canonically fix the coupling constant. We
show that the choice of left-moving, conformal vertex-operator insertions
produces a section

(6)
$\left\langle\bigotimes_{j=1}^{n}V_{j}(z_{j}|\zeta_{j})\Big{[}\,\zeta[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{|}[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{]}\right\rangle$

of the line bundle $\operatorname{Ber}(H^{0}(X,\omega|_{\operatorname{NS}}))$
and compute its image under the isomorphism (5). Here we have uniformized
notation and assumed $z_{j}=z_{j}^{0}$ for $j\leq 3$ and
$\zeta_{j}=\zeta_{j}^{0}$ for $j\leq 2$. The vertex operators $V_{j}$
represent the incoming and outgoing particle states. These are vector-valued
functions, and the notation $\langle\ \rangle$ stands for the _vacuum
expectation value $(\\!$VEV$\,)$_, i.e. $\langle
O\rangle=\langle\Omega|O\Omega\rangle$, where $\Omega$ is the vacuum state,
$O$ any given operator having $\Omega$ in its domain, and $\langle|\rangle$
denotes the inner product on the Hilbert space of physical states.

######  Theorem 1.

The choice of conformal vertex-operator insertions (6) gives for any
$X\in\mathcal{M}_{0,n,0}$ a section of
$\operatorname{Ber}H^{0}(X,\linebreak[0]\omega|_{\operatorname{NS}})$. Then
the canonical isomorphism (5) gives a holomorphic section of the line bundle
$\omega_{\mathcal{M}_{0,n,0}}$ over the dense open subset of
$\mathcal{M}_{0,n,0}$ defined by forcing each NS puncture to stay within the
open chart with bosonic coordinate $z\neq\infty$. This section can be
expressed in global coordinates
$(z_{4},\dots,z_{n}\,|\,\zeta_{3},\dots\zeta_{n})$ on $\mathcal{M}_{0,n,0}$ as

(7) $\left\langle\prod_{j=1}^{n}V_{j}(z_{j}|\zeta_{j})\right\rangle d\nu\,,$

where

$d\nu=-\tfrac{1}{2^{n-2}}\left(z_{3}-z_{1}-\tfrac{1}{2}\zeta_{3}\zeta_{1}\right)\left(z_{3}-z_{2}-\tfrac{1}{2}\zeta_{3}\zeta_{2}\right)[\,dz_{4}\dots
dz_{n}\,|\,d\zeta_{3}\dots d\zeta_{n}]$

and $z_{j}=z^{0}_{j}$, $j=1,2,3$, and $\zeta_{a}=\zeta^{0}_{a}$, $a=1,2$, are
fixed finite values $($for any concrete purposes we can set
$\zeta^{0}_{1}=\zeta^{0}_{2}=0)$.

If instead we set $z_{1}=\infty$, or more concretely,
$(w_{1}|\eta_{1})=\left(\tfrac{1}{z_{1}}\Big{|}\tfrac{\zeta_{1}}{z_{1}}\right)=(0\,|\,\eta^{0}_{1})$,
and replace the vertex-operator insertion $\left\langle
V_{1}(z_{1}|\zeta_{1})\big{[}\,\zeta[dz|d\zeta]|_{\operatorname{NS}_{1}}\,\mid\,[dz|d\zeta]|_{\operatorname{NS}_{1}}\big{]}\right\rangle$
with

$\left\langle
V_{1}(0|\eta_{1})\big{[}\,\eta[dw|d\eta]|_{\operatorname{NS}_{1}}\mid[dw|d\eta]|_{\operatorname{NS}_{1}}\big{]}\right\rangle\,,$

then the corresponding canonical section of $\omega_{\mathcal{M}_{0,n,0}}$ on
the whole supermoduli space
$\mathcal{M}_{0,n,0}\subset\mathbb{C}^{n-3\,|\,n-2}/\mathbb{Z}_{2}$ becomes

(8) $\left\langle
V_{1}(0|\eta_{1})\prod_{j=2}^{n}V_{j}(z_{j}|\zeta_{j})\right\rangle d\nu\,,$

where

$d\nu=\tfrac{1}{2^{n-2}}\left(1+\tfrac{1}{2}\zeta_{3}\eta_{1}\right)\left(z_{3}-z_{2}-\tfrac{1}{2}\zeta_{3}\zeta_{2}\right)[\,dz_{4}\dots
dz_{n}\,|\,d\zeta_{3}\dots d\zeta_{n}]$

and $\eta_{1}=\eta_{1}^{0}$, $z_{j}=z^{0}_{j}$, $j=2,3$, and
$\zeta_{2}=\zeta^{0}_{2}$.

######  Remark 1.

This theorem has an immediate application to physical observables. Define the
tree-level amplitude for scattering $n\geq 3$ type II superstrings in the NS
sector as

(9)
${\mathcal{A}}_{n}=\int_{\mathcal{M}_{0,n,0}}\left\langle\prod_{j=1}^{n}V_{j}(z_{j}|\zeta_{j};\bar{z}_{j}|\bar{\zeta}_{j})\right\rangle
dm\,,$

where the bar indicates complex conjugation, the vertex operators combine
left- and right-movers, i.e., are smooth, rather than holomorphic, sections of
the corresponding bundles, and the measure is the product of the holomorphic
and antiholomorphic factors:

$dm=d\nu\cdot\overline{d\nu}\,,$

with $d\nu$ defined in 1. Thus we have gotten the standard expression for the
scattering amplitude, usually obtained in physics in an ad hoc way by
introducing ghosts and picture changing operators. We review this process in
Section 6 to reconcile it with our formulas. However, our formulas have the
advantage of also determining the closed-string coupling constant
$g_{c}=(1/2)^{2}=1/4$ canonically, since the super Mumford isomorphism is
defined canonically, not just up to scaling.

In (9) we have omitted the contribution of the dilaton field $\phi$, which
usually gives $g_{c}=\langle e^{\phi}\rangle$. If we were to include the
dilaton field, the coupling constant would have to be adjusted to
$g_{c}=\frac{1}{4}\langle e^{\phi}\rangle$.

While the holomorphic forms (7) and (8) in 1 are written as an expression in
terms of the chosen global coordinates on
$\mathcal{M}_{0,n,0}\hookrightarrow\mathbb{C}^{n-3|n-2}/\mathbb{Z}_{2}$, of
course, the actual forms (7) or (8) are well-defined on the moduli space,
independent of which coordinate system is chosen. The advantage of our
approach is that we are able to determine the holomorphic forms (7) and (8)
and measure $d\nu$ canonically from first principles, without any need for
PCOs or ghosts — including the determination of the coupling constant.

Beyond being a direct mathematical computation not involving ghosts or PCOs,
we hope that the current paper may be of reference value in assembling
together the many mathematical pieces of the construction. We also view our
computations as a stepping stone towards performing similar explicit
computations for tree-level superstring scattering amplitudes with Ramond
markings, where the procedure is quite different from the NS case [24]. For
the Ramond case, Ott and the third author [13] have recently obtained an
explicit geometric description of $\mathcal{M}_{0,0,k}$, which should enable a
similarly explicit determination of superstring scattering amplitudes.

Acknowledgments: The second author is grateful to Luis Álvarez-Gaumé for
sharing his knowledge of the subject and for literature references. The third
author thanks Mikhail Kapranov and Katherine Maxwell for useful discussions
and patience during the work on parallel projects. The first and third authors
would like to thank the Simons Center for Geometry and Physics for hospitality
in Spring 2023 during the “Supergeometry and supermoduli” program, where all
the authors started their attempt to reach a complete mathematical
understanding of genus zero superstring scattering amplitudes.

##  2\. Basic conventions and constructions with SRSs and their moduli

In this paper, we work with _super Riemann surfaces $($SRSs$)$ of genus zero_,
which all are isomorphic to the standard SRS of genus zero based on the
projective superspace
${\mathbb{P}}^{1|1}:=\operatorname{Proj}\mathbb{C}[u,v|\xi]$ over the ground
field $\mathbb{C}$ of complex numbers, see [18, 19]. We will write $X$ for
such an SRS, when it is essential to think of the _SRS, or superconformal,
structure_ , i.e. an odd, maximally non-integrable distribution
$\mathcal{D}\subset\mathcal{T}_{X}$, see below. The supermanifold
${\mathbb{P}}^{1|1}$ may be glued out of two affine open charts
$\mathbb{C}^{1|1}$ with coordinates $(z|\zeta)=(v/u\,|\,\xi/u)$ and
$(w|\eta)=(u/v\,|\,\xi/v)$ by the gluing functions

$w=1/z,\qquad\eta=\zeta/z.$

The _odd distribution_ $\mathcal{D}$ defining the SRS structure is generated
by the super vector field

(10) $D_{\zeta}:=\frac{\partial}{\partial\zeta}+\zeta\frac{\partial}{\partial
z}$

in the chart $(z|\zeta)$ and the super vector field

$\frac{\partial}{\partial\eta}-\eta\frac{\partial}{\partial w}$

in the chart $(w|\eta)$. Note that
$\tfrac{1}{2}[D_{\zeta},D_{\zeta}]=D_{\zeta}^{2}=\partial/\partial z$, and
this ensures the _maximal non-integrability_ of $\mathcal{D}$.

A remarkable feature of super Riemann surfaces is _dualism between points and
divisors_ , making the geometry of SRSs akin to that of classical Riemann
surfaces, where the fact that divisors are linear combinations of points plays
a prominent role. In particular, Neveu-Schwarz (NS) punctures have their
divisor avatars, and we will often switch between NS punctures and NS
divisors. What creates this dualism is the structure distribution
$\mathcal{D}$. Given a point $p$ on a SRS $X$, we can define a complex
subsupermanifold of dimension $0|1$ supported on $p$, i.e., a _prime divisor_
, with tangent space given by $\mathcal{D}|_{p}$. In a family of once-
punctured SRSs, represented by a universal SRS $\mathcal{C}$ over the moduli
space $\mathcal{M}$ with a section $p:\mathcal{M}\hookrightarrow\mathcal{C}$,
regarded as a subspace of the universal SRS $\mathcal{C}$, the corresponding
NS divisor is given as the total space, determined by its
$\mathcal{O}_{p}$-algebra $S(\mathcal{D}^{*}|_{p})$ of functions, of the odd
line bundle $\mathcal{D}|_{p}$ over ${p}$. The inclusion of this total space
is given by the projection of the conormal bundle
$\mathcal{T}_{\mathcal{C}/\mathcal{M}}^{*}\bigm{|}_{p}$ of $p$ in
$\mathcal{C}$ to $\mathcal{D}^{*}|_{p}$, which is the dual of the inclusion
$\mathcal{D}\hookrightarrow\mathcal{T}_{\mathcal{C}/\mathcal{M}}$. Conversely,
given a prime divisor, we can find a unique point in it so that the tangent
space to the divisor at this point is given by the distribution $\mathcal{D}$.
If $f=0$ is a local equation of a prime divisor and $D$ is a generator of
$\mathcal{D}$ near the divisor’s support, then the equations $f=0$ and $Df=0$
determine the corresponding point. In coordinates $(z|\zeta)$ in which the
structure distribution is generated by $D_{\zeta}$, see (10), the divisor
corresponding to a point $(z_{0}|\zeta_{0})$ is given by the equation
$z-z_{0}-\zeta\zeta_{0}=0$.

We will work with families $X\to S$ of SRSs of genus zero with NS punctures
over a base $S$. We will assume the base $S$ to be affine, just like the
supermoduli space. We will still denote the total space of the family by $X$.
We will also denote by $\Pi$ the parity change operator on a super vector
space or sheaf of such.

The _Berezinian $\operatorname{Ber}F$ of a free module_ $F$ of finite rank
over a supercommutative algebra is a rank-$(1|0)$ (or $(0|1)$, if $n$ is odd)
free module, defined as the set
$[e_{1},\dots,\mathfrak{e}_{m}\,|\,\varepsilon_{1},\dots,\varepsilon_{n}]$ of
equivalence classes of homogeneous bases
$\\{e_{1},\dots,e_{m}\,|\,\varepsilon_{1},\dots,\varepsilon_{n}\\}$ of $F$
modulo the relation

$[Me_{1},\dots,Me_{m}\,|\,M\varepsilon_{1},\dots,M\varepsilon_{n}]\sim(\operatorname{Ber}M)\cdot[e_{1},\dots,\mathfrak{e}_{m}\,|\,\varepsilon_{1},\dots,\varepsilon_{n}]$

for each invertible linear automorphism $M$ of the module $F$. We also add the
zero element to $\operatorname{Ber}F$ to make it a module. Here
$\operatorname{Ber}M$ is the _Berezinian, or superdeterminant_ , of $M$. If,
in a homogeneous basis, $M$ is represented by a block matrix

$M=\begin{pmatrix}A&B\\\ C&D\end{pmatrix}$

with respect to the decomposition $M=M_{0}\oplus M_{1}$ into the even and odd
parts, then

$\operatorname{Ber}M:=\det(A-BD^{-1}C)\det(D)^{-1}=\det(A)\det(D-CA^{-1}B)^{-1}.$

The Berezinian of a map of free modules is multiplicative:
$\operatorname{Ber}(M_{1}M_{2})=\operatorname{Ber}(M_{1})\operatorname{Ber}(M_{2})$.
Given a short exact sequence of free modules

$0\to F_{1}\to F_{2}\to F_{3}\to 0,$

we also get a canonical isomorphism of the Berezinians of the modules:

$\operatorname{Ber}F_{2}\xrightarrow{\sim}\operatorname{Ber}F_{1}\otimes\operatorname{Ber}F_{3}.$

A standard argument extends the construction of $\operatorname{Ber}F$ to
locally free modules and sheaves.

_Line bundles_ in the super setting will be divided into _even_ and _odd_
ones, depending on whether their rank is $1|0$ or $0|1$.

We will adopt the notation $\mathcal{L}^{k}$ for the $k$’th tensor power
$\mathcal{L}^{\otimes k}$ of a line bundle $\mathcal{L}$ for $k\in\mathbb{Z}$.
We will routinely use abbreviated notation for cohomology as well. For
example, $H^{i}(\omega^{k})$ will denote the cohomology space
$H^{i}(\mathcal{C}_{0,n,0},\omega_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}^{\otimes
k})$, where $\mathcal{C}_{0,n,0}\xrightarrow{p}\mathcal{M}_{0,n,0}$ is the
universal family of genus-$0$ SRSs with $n$ NS punctures. In full generality,
we should be working with the space
$H^{0}(\mathcal{M}_{0,n,0},R^{i}p_{*}\omega_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}^{\otimes
k})$, but it is indeed naturally isomorphic to
$H^{i}(\mathcal{C}_{0,n,0},\omega_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}^{\otimes
k})$, given that $\mathcal{M}_{0,n,0}$ is affine, being the
$\mathbb{Z}_{2}$-quotient of the complement of hyperplanes in
$\mathbb{C}^{0|1}\times(\mathbb{C}^{1|1})^{n-4}$. Likewise, the Berezinian of
cohomology $B(\mathcal{F})=\operatorname{Ber}H^{*}(\mathcal{F})$ will be
identified with
$\operatorname{Ber}H^{0}(\mathcal{M}_{0,n,0},\operatorname{Ber}(R^{\bullet}p_{*}\mathcal{F}))$.
See the details on the determinant and Berezinian of cohomology in [25, 26]
and [1, 12], respectively.

We will denote by $(z_{1}|\zeta_{1}),\dots,(z_{n}|\zeta_{n})$ the coordinates
of the NS punctures. To simplify notation, we will write $\omega$ for the
_relative dualizing sheaf_ $\omega_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}$,
and, when relevant, also for its restriction, also known as the canonical
bundle, to an individual SRS.

##  3\. The super Mumford form

We recall that the super Mumford isomorphism is a canonical isomorphism
$B(\omega^{3})=B(\omega)^{5}$ over the supermoduli space $\mathcal{M}_{g,0,0}$
for $g\geq 2$, [27, 1]. Here we will use the same construction to establish a
canonical isomorphism over $\mathcal{M}_{0,n,0}$:
$B(\omega^{3})=B(\omega)^{5}$. To start out, and also for further use, we
construct natural bases for the spaces of global sections of the powers of the
relative dualizing sheaf $\omega$.

###  3.1. Bases of cohomology

We denote by $t:=z[dz|d\zeta]^{-1}\in H^{0}(\mathcal{C}_{0,n,0},\omega^{-1})$
the generating section of $\omega^{-1}$, and consider, for any positive
integer $k$, the following short exact sequence on $\mathcal{C}_{0,n,0}$:

$0\to\omega^{k}\xrightarrow{\cdot
t}\omega^{k-1}\to\omega^{k-1}|_{\operatorname{div}t}\to 0.$

We make the maps in the associated long exact sequence of homology explicit
for $k=1,2,3$, by first writing out the bases of the relevant spaces, and then
identifying the maps on homology explicitly in these bases, so that $B_{k}(t)$
is the canonical isomorphism of the associated Berezinians
$B(\omega^{k-1})\simeq\Pi B(\omega^{k})\otimes
B(\omega^{k-1}|_{\operatorname{div}t})$. These, suitably modified in the
presence of NS markings, will be used for the computation of the super Mumford
form.

Case $k=1$. Here we get the exact sequence (of cohomology of sheaves over
$\mathcal{C}_{0,n,0}$)

$0=H^{0}(\omega)\to H^{0}(\mathcal{O})\to
H^{0}(\mathcal{O}|_{\operatorname{div}t})\to H^{1}(\omega)\to
H^{1}(\mathcal{O})=0$

The basis of $H^{0}(\mathcal{O})$ is $\\{1\,|\;\\}$ — this notation is to list
the even basis vectors followed by the odd basis vectors (none in this case)
after the dash. Since $H^{i}(\mathcal{O})=0$ for $i>0$, it follows that the
Berezinian
$B(\mathcal{O})=\operatorname{Ber}H^{0}(\mathcal{C}_{0,n,0};\mathcal{O})=\det
H^{0}(\mathcal{C}_{0,n,0};\mathcal{O})$ has as basis the section
$\operatorname{Ber}\\{1\,|\;\\}=\det\\{1\,|\;\\}$, which we will denote by
$[1\,|\;]$. Using relative super Serre duality, we see that
$H^{1}(\omega)=H^{0}(\mathcal{O})^{*}$ on each genus zero SRS $X$, which
globalizes since we have global coordinates on the affine superscheme
$\mathcal{M}_{0,n,0}$. Explicitly, the Serre dual to the even basis element
$1$ of $H^{0}(\mathcal{O})$ is given by
$\left\\{\left.\tfrac{\zeta}{z}[dz|d\zeta]\;\right|\ \right\\}$, and we
observe that
$B(\omega)=(\operatorname{Ber}(H^{1}(\omega)))^{-1}=\operatorname{Ber}(H^{0}(\mathcal{O}))=B(\mathcal{O})$
also has basis $[1\,|\;]$.

The basis of $H^{0}(\mathcal{O}|_{\operatorname{div}t})$ is
$\\{1|_{\operatorname{div}t}\,|\;\zeta|_{\operatorname{div}t}\\}$, and thus
$[1|_{\operatorname{div}t}\,|\;\zeta|_{\operatorname{div}t}]\in
B(\mathcal{O}|_{\operatorname{div}t})$ is the generating section of that
sheaf. We can thus compute

$B_{1}(t):=[1\,|\;]\otimes\Pi[1\,|\;]\otimes[1|_{\operatorname{div}t}\,|\;\zeta|_{\operatorname{div}t}]^{-1}\in
B(\mathcal{O})\otimes\Pi B(\omega)\otimes
B(\mathcal{O}|_{\operatorname{div}t})^{-1}\,.$

Case $k=2$. Here the exact sequence of cohomology is

$0=H^{0}(\omega^{2})\to H^{0}(\omega)\to
H^{0}(\omega|_{\operatorname{div}t})\to H^{1}(\omega^{2})\to H^{1}(\omega)\to
H^{1}(\omega|_{\operatorname{div}t})\to 0$

(where the last zero is simply for dimensional reasons). An explicit basis of
$H^{1}(\omega^{2})=H^{0}(\omega^{-1})^{*}$ can be written simply by thinking
of these as suitable volume forms with poles:

$\left\\{\frac{[dz|d\zeta]^{2}}{z}\;\left|\;\frac{\zeta[dz|d\zeta]^{2}}{z},\frac{\zeta[dz|d\zeta]^{2}}{z^{2}}\right\\}=\left\\{\frac{\zeta}{[dz|d\zeta]}\;\right|\;\frac{1}{[dz|d\zeta]},\frac{z}{[dz|d\zeta]}\right\\}^{*}\,.$

Thus the generating section of the Berezinian $B(\omega^{2})=B(\omega^{-1})$
is given by

$\left[\frac{\zeta}{[dz|d\zeta]}\;\left|\;\frac{1}{[dz|d\zeta]},\frac{z}{[dz|d\zeta]}\right.\right]$

The basis of $H^{0}(\omega|_{\operatorname{div}t})$ is given by
$\\{\zeta[dz|d\zeta]|_{\operatorname{div}t}\;|\;[dz|d\zeta]|_{\operatorname{div}t}\\}$,
and we finally compute

$B_{2}(t):=\Pi[1\,|\;]^{-1}\otimes\left[\frac{\zeta}{[dz|d\zeta]}\;\left|\;\frac{1}{[dz|d\zeta]},\frac{z}{[dz|d\zeta]}\right.\right]^{-1}\otimes\left[\zeta[dz|d\zeta]|_{\operatorname{div}t}\;\left|\;[dz|d\zeta]|_{\operatorname{div}t}\right.\right]\\\
\in\Pi B(\omega)^{-1}\otimes B(\omega^{2})^{-1}\otimes
B(\omega|_{\operatorname{div}t}).$

Case $k=3$. In this case a basis of $H^{1}(\omega^{3})=H^{0}(\omega^{-2})^{*}$
can be given as

$\left\\{\zeta,\zeta z,\zeta
z^{2}|z,z^{2}\right\\}\cdot\frac{[dz|d\zeta]^{3}}{z^{3}}\,,$

(where from now on we write such bases by separating out the common factor,
for readability) and we have the generating section

$\left[\frac{\zeta[dz|d\zeta]^{3}}{z},\frac{\zeta[dz|d\zeta]^{3}}{z^{2}},\frac{\zeta[dz|d\zeta]^{3}}{z^{3}}\;\left|\;\frac{[dz|d\zeta]^{3}}{z},\frac{[dz|d\zeta]^{3}}{z^{2}}\right.\right]\in
B(\omega^{3})=B(\omega^{-2}).$

Similarly by thinking of volume forms with poles, we see that
$\\{[dz|d\zeta]^{2}|_{\operatorname{div}t}\;|\;\zeta[dz|d\zeta]^{2}|_{\operatorname{div}t}\\}$
gives a basis of $H^{0}(\omega^{2}|_{\operatorname{div}t})$, and thus we
finally compute

$B_{3}(t):=\left[\frac{\zeta}{[dz|d\zeta]}\;\left|\;\frac{1}{[dz|d\zeta]},\frac{z}{[dz|d\zeta]}\right.\right]\\\
\otimes\Pi\left[\frac{\zeta[dz|d\zeta]^{3}}{z},\frac{\zeta[dz|d\zeta]^{3}}{z^{2}},\frac{\zeta[dz|d\zeta]^{3}}{z^{3}}\;\left|\;\frac{[dz|d\zeta]^{3}}{z},\frac{[dz|d\zeta]^{3}}{z^{2}}\right.\right]\\\
\otimes\left[[dz|d\zeta]^{2}|_{\operatorname{div}t}\;\left|\;\zeta[dz|d\zeta]^{2}|_{\operatorname{div}t}\right.\right]^{-1}\in
B(\omega^{2})\otimes\Pi B(\omega^{3})\otimes
B(\omega^{2}|_{\operatorname{div}t})^{-1}.$

###  3.2. A formula for the super Mumford form

We are now ready to determine explicitly the super Mumford isomorphism,
recalling that unlike the bosonic case, this is not just an abstract
isomorphism of line bundles, but in fact the canonical isomorphism given by
the element $\mu:=B_{3}(t)\otimes B_{2}(t)\otimes B_{1}(t)^{-2}\in
B(\omega^{3})\otimes B(\mathcal{O})^{-5}$, where to identify this product we
recall that $B(\mathcal{O})$ is canonically isomorphic to $B(\omega)$. To
obtain an expression for $\mu$, we manipulate the formulas for
$B_{1}(t),B_{2}(t),B_{3}(t)$ from the previous section. We obtain

$B_{2}(t)\otimes
B_{1}(t)^{-1}=\frac{\left[\zeta[dz|d\zeta]|_{\operatorname{div}t}\;\left|\;[dz|d\zeta]|_{\operatorname{div}t}\right.\right]\otimes[1|_{\operatorname{div}t}\;|\;\zeta|_{\operatorname{div}t}]}{[1\,|\;]^{3}\otimes\left[\frac{\zeta}{[dz|d\zeta]}\;\left|\;\frac{1}{[dz|d\zeta]},\frac{z}{[dz|d\zeta]}\right.\right]}$

Recall that restricting any even line bundles $\mathcal{L}$ and $\mathcal{M}$
on a SRS $X$ to the divisor $\operatorname{div}t$, we have canonical
isomorphisms

$B(\Pi\mathcal{L}|_{\operatorname{div}t})=B(\mathcal{L}|_{\operatorname{div}t})^{-1};\quad
B(\mathcal{L}|_{\operatorname{div}t})=B(\mathcal{M}|_{\operatorname{div}t})\,,$

where the latter identification is given by multiplying by any generating
holomorphic section of $\mathcal{L}^{-1}\otimes\mathcal{M}$ in a neighborhood
of $\operatorname{div}t$. Applying this to all the restrictions in the formula
above, we compute

$B_{2}(t)\otimes
B_{1}(t)^{-1}=\frac{1}{[1\,|\;]^{3}\otimes\left[\frac{\zeta}{[dz|d\zeta]}\;\left|\;\frac{1}{[dz|d\zeta]},\frac{z}{[dz|d\zeta]}\right.\right]}\in
B(\mathcal{O})^{-3}\otimes B(\omega^{2})^{-1}.$

By a similar simplification of the expression for $B_{3}(t)$ from the previous
section, we obtain

$B_{3}(t)\otimes
B_{1}(t)^{-1}=\frac{\left[\frac{\zeta}{[dz|d\zeta]}\;\left|\;\frac{1}{[dz|d\zeta]},\frac{z}{[dz|d\zeta]}\right.\right]\otimes\left[\frac{\zeta[dz|d\zeta]^{3}}{z},\frac{\zeta[dz|d\zeta]^{3}}{z^{2}},\frac{\zeta[dz|d\zeta]^{3}}{z^{3}}\;\left|\;\frac{[dz|d\zeta]^{3}}{z},\frac{[dz|d\zeta]^{3}}{z^{2}}\right.\right]}{[1\,|\;]^{2}}\\\
\in B(\omega^{2})\otimes B(\omega^{3})\otimes B(\mathcal{O})^{-2}.$

Finally, noticing that the canonical section $[1\;|\;]$ defines a global
trivialization of $B(\mathcal{O})$, and can thus be omitted in the formulas
above, we obtain the following expression for the super Mumford form

(11) $\mu=B_{3}(t)\otimes B_{2}(t)\otimes
B_{1}(t)^{-2}=\left[\tfrac{\zeta[dz|d\zeta]^{3}}{z},\tfrac{\zeta[dz|d\zeta]^{3}}{z^{2}},\tfrac{\zeta[dz|d\zeta]^{3}}{z^{3}}\;\left|\;\tfrac{[dz|d\zeta]^{3}}{z},\tfrac{[dz|d\zeta]^{3}}{z^{2}}\right.\right]$

as a global section of the line bundle $B(\omega^{3})\otimes
B(\mathcal{O})^{-5}=B(\omega^{3})$ over $\mathcal{M}_{0,n,0}$.

##  4\. The super Kodaira-Spencer map over $\mathcal{M}_{0,n,0}$

We now proceed to work out the super Mumford isomorphism for the case of genus
0 SRS with NS markings, by computing the natural isomorphism given by the
super Kodaira-Spencer map.

The superanalog of the Kodaira-Spencer map in the case of deformation theory
of supermanifolds is well-known [11, 28, 9]. In our case this will be a map
from $\mathcal{T}_{X}\mathcal{M}_{0,n,0}$, which is a super vector space of
dimension $n-3|n-2$, to a suitable first homology. For our purposes it will be
more natural to work with the dual map, as it is more explicit to write bases
for spaces of sections $H^{0}$ rather than for the first homology $H^{1}$. We
first recall the setup, and then compute the super Kodaira-Spencer map
explicitly, by using an explicit cover to get the class in Čech cohomology.

###  4.1. The construction

For a SRS $X$, let $\mathcal{T}^{sc}_{X}\subset\mathcal{T}_{X}$ be the sheaf
of _superconformal vector fields_ , i.e. vector fields that preserve the
distribution $\mathcal{D}$, i.e. of $v\in\mathcal{T}_{X}$ such that
$[v,\mathcal{D}]\subset\mathcal{D}$. coordinates, see formula (12) in
Katherine Maxwell’s paper https://arxiv.org/pdf/2002.06625.pdf. Note that
$\mathcal{T}^{sc}_{X}$ is not an $\mathcal{O}_{X}$-module: the product $f\cdot
v$ of $v\in\mathcal{T}_{X}^{sc}$ and a (local holomorphic) function
$f\in\mathcal{O}_{X}$ may not lie in $\mathcal{T}^{sc}_{X}$. However, as a
sheaf of abelian groups, $\mathcal{T}^{sc}_{X}$ is isomorphic to
$\mathcal{D}^{\otimes 2}$, and in fact this natural isomorphism can be used to
induce an $\mathcal{O}_{X}$-module structure on $\mathcal{T}^{sc}_{X}$, see
[29]. We thus consider
$\mathcal{T}^{sc}_{X}(-\operatorname{NS})\subset\mathcal{T}^{sc}_{X}$ as the
subsheaf of those vector fields that preserve $\mathcal{D}$, and which also do
not move the NS divisor $\operatorname{NS}$.

The previous discussion can be globalized to consider $\mathcal{D}$ as a
subsheaf of the relative tangent sheaf
$\mathcal{T}_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}\subset\mathcal{T}_{\mathcal{C}_{0,n,0}}$
and then define the subsheaves
$\mathcal{T}^{sc}_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}(-\operatorname{NS})\subset\mathcal{T}_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}$
and
$\mathcal{T}^{sc}_{\mathcal{C}_{0,n,0}}(-\operatorname{NS})\linebreak[0]\subset\mathcal{T}_{\mathcal{C}_{0,n,0}}$
of vector fields that preserve the global $\mathcal{D}$ and fix the global
divisor $\operatorname{NS}$, which we think of as an unramified effective
Cartier divisor.

Altogether, this yields the exact sequence of sheaves of vector spaces

$0\to\mathcal{T}^{sc}_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}(-\operatorname{NS})\to\mathcal{T}^{sc}_{\mathcal{C}_{0,n,0}}(-\operatorname{NS})\to
p^{*}\mathcal{T}_{\mathcal{M}_{0,n,0}}\to 0\,.$

Then the _super Kodaira-Spencer isomorphism_ is the connecting homomorphism

$\mathcal{T}_{\mathcal{M}_{0,n,0}}\to
R^{1}p_{*}\mathcal{T}^{sc}_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}(-\operatorname{NS})=R^{1}p_{*}\mathcal{D}^{\otimes
2}(-\operatorname{NS})$

induced by the isomorphism

(12) $\begin{split}\mathcal{D}^{\otimes
2}&\xrightarrow{\sim}\mathcal{T}^{sc}_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}},\\\
fD_{\zeta}\otimes D_{\zeta}&\longmapsto
f\partial_{z}+\tfrac{(-1)^{\tilde{f}}}{2}D_{\zeta}(f)D_{\zeta},\end{split}$

see [30, 31, 32, 29]. Taking global sections, we globalize the super Kodaira-
Spencer isomorphism to the isomorphism

$\operatorname{KS}:H^{0}(\mathcal{M}_{0,n,0},\mathcal{T}_{\mathcal{M}_{0,n,0}})\rightarrow
H^{1}(\mathcal{C}_{0,n,0},\mathcal{D}^{\otimes 2}(-\operatorname{NS})).$

The _super Kodaira-Spencer isomorphism over a given SRS_
$X\in\mathcal{M}_{0,n,0}$ is

$\mathcal{T}_{\mathcal{M}_{0,n,0}}|_{X}\to H^{1}(X,\mathcal{D}|_{X}^{\otimes
2}(-\operatorname{NS})).$

To identify $H^{1}(\mathcal{C}_{0,n,0},\mathcal{D}^{\otimes
2}(-\operatorname{NS}))$ further, we apply _Serre duality_ :

$R^{1}p_{*}\mathcal{D}^{\otimes
2}(-\operatorname{NS})=p_{*}\omega_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}^{\otimes
3}(\operatorname{NS})^{*},$

which leads to the isomorphism

$\sigma\colon\thinspace H^{1}(\mathcal{C}_{0,n,0},\mathcal{D}^{\otimes
2}(-\operatorname{NS}))=H^{0}(\mathcal{C}_{0,n,0},\omega^{3}(\operatorname{NS}))^{*}.$

Combined with the super Kodaira-Spencer isomorphism, this gives the dual
isomorphism:

$p_{*}\omega_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}^{\otimes
3}(\operatorname{NS})\to\mathcal{T}_{\mathcal{M}_{0,n,0}}^{*},$

which globalizes to the _dual super Kodaira-Spencer isomorphism_

$\operatorname{KS}^{*}\colon\thinspace
H^{0}(\mathcal{C}_{0,n,0},\omega_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}^{\otimes
3}(\operatorname{NS}))\to
H^{0}(\mathcal{M}_{0,n,0},\mathcal{T}_{\mathcal{M}_{0,n,0}}^{*}).$

###  4.2. Proof of the main theorem

The rest of the paper is largely dedicated to the explicit computation of the
super Kodaira-Spencer map and identification of bases for the spaces involved.
Before we get into that, we would like to outline the proof of 1. Consider the
following short exact sequence (SES) of sheaves over $\mathcal{C}_{0,n,0}$, by
restricting to the NS divisor:

(13)
$0\to\omega^{3}{\rightarrow}\omega^{3}(\operatorname{NS}){\rightarrow}\omega^{3}(\operatorname{NS})|_{\operatorname{NS}}\to
0\,.$

This is the analog for a divisor with multiple components of the case $k=3$ of
the SES described in Section 3. This SES gives a canonical isomorphism of the
Berezinian line bundles:

$B(\omega^{3})\otimes
B(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})\xrightarrow{\sim}B(\omega^{3}(\operatorname{NS}))\,,$

which we are going to identify explicitly. Since $H^{0}(\omega^{3})=0$, and
$H^{1}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})=0$, as the relative
(topological) dimension of the $\operatorname{NS}$ divisor over the moduli
space $\mathcal{M}_{0,n,0}$ is 0 and $\mathcal{M}_{0,n,0}$ is affine, the
associated exact sequence in cohomology is simply

(14) $0\rightarrow
H^{0}(\omega^{3}(\operatorname{NS})){\rightarrow}H^{0}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}}){\rightarrow}H^{1}(\omega^{3})\rightarrow
0\,.$

The choice of any even generating section of the line bundle
$\omega^{2}(\operatorname{NS})$ in an affine open neighborhood of the divisor
$\operatorname{NS}$ yields an isomorphism

(15)
$B(\omega|_{\operatorname{NS}})\xrightarrow{\sim}B(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})\,,$

which does not depend on the choice of the section and is thereby canonical,
see [1, 12]. provided none of the NS punctures is at $\infty$, and the
Berezinian line bundle
$B(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})$. Thus, we have a
canonical isomorphism

$\varphi:B(\omega^{3})\otimes
B(\omega|_{\operatorname{NS}})\xrightarrow{\sim}B(\omega^{3}(\operatorname{NS}))\,.$

In Appendix A we provide further details and compute the image of the super
Mumford form $\mu\in\Gamma(\mathcal{M}_{0,n,0},B(\omega^{3}))$, given by
$\eqref{superMum}$, tensored with a global section of
$B(\omega|_{\operatorname{NS}})$ arising from vertex-operator insertions,
under the isomorphism $\varphi$, see 6:

$\varphi\left(\mu\otimes\left\langle\bigotimes_{j=1}^{n}V_{j}(z_{j}|\zeta_{j})\Big{[}\,\zeta[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{|}[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{]}\right\rangle\right)\\\
=\left\langle\prod_{j=1}^{n}V_{j}(z_{j}|\zeta_{j})\right\rangle\operatorname{Ber}\left(\\{\zeta,\zeta
z,\dots,\zeta
z^{n-4}\,|\,1,\dots,z^{n-3}\\}\cdot\frac{[dz|d\zeta]^{3}}{\prod_{j=1}^{n}(z-z_{j}-\zeta\zeta_{j})}\right)\,,$

where $(z_{1}|\zeta_{1})$, …, $(z_{1}|\zeta_{1})$ are the NS punctures. In
Section 5 we explain how vertex-operator insertions make up a section of
$B(\omega|_{\operatorname{NS}})$. The Berezinian of the dual super Kodaira-
Spencer isomorphism, computed in the remaining part of Section 4, with
detailed linear algebra computations given in Appendix B, naturally identifies
$B(\omega^{3}(\operatorname{NS}))$ with the space
$H^{0}(\mathcal{M}_{0,n,0},\omega_{\mathcal{M}_{0,n,0}})=H^{0}(\mathcal{M}_{0,n,0},\operatorname{Ber}(\mathcal{T}_{\mathcal{M}_{0,n,0}}^{*}))$
of holomorphic volume forms on $\mathcal{M}_{0,n,0}$:

$\operatorname{Ber}(\operatorname{KS}^{*})\colon\thinspace
B(\omega^{3}(\operatorname{NS}))\xrightarrow{\sim}H^{0}(\mathcal{M}_{0,n,0},\omega_{\mathcal{M}_{0,n,0}}).$

4 below gives an explicit computation,

$d\nu:=-\frac{1}{2^{n-2}}(z_{3}-z_{1}-\tfrac{1}{2}\zeta_{3}\zeta_{1})(z_{3}-z_{2}-\tfrac{1}{2}\zeta_{3}\zeta_{2})[dz_{4}\dots
dz_{n}\,|\,d\zeta_{3}\dots d\zeta_{n}]\,,$

of the image

$\operatorname{Ber}(\operatorname{KS}^{*})\left(\operatorname{Ber}\left(\\{\zeta,\zeta
z,\dots,\zeta
z^{n-4}\,|\,1,\dots,z^{n-3}\\}\cdot\frac{[dz|d\zeta]^{3}}{\prod_{j=1}^{n}(z-z_{j}-\zeta\zeta_{j})}\right)\right)$

under the isomorphism $\operatorname{Ber}(\operatorname{KS}^{*})$, when all
the NS punctures are finite. and yields 1. The case $z_{1}=\infty$ is treated
in Appendix C. ∎

###  4.3. The Čech cohomology computation

First, note that for computations with supermanifolds, it is enough to take an
acyclic (Leray) cover of the underlying bosonic manifold. Below is a precise
statement, which is surely known to seasoned supergeometeres, though we could
not find it in the literature.

For a topological space $X$ and a sheaf $\mathcal{F}$ on it, recall that an
open cover $\\{U_{i}\\}$ is called _Leray with respect to $\mathcal{F}$_ if
for the derived-functor cohomology, we have $H^{q}(U_{i_{1}}\cap\dots\cap
U_{i_{p}},\mathcal{F})=0$ for all finite sets $\\{i_{1},\dots,i_{p}\\}$ of
indices and all $q>0$. In this case, the derived-functor cohomology
$H^{q}(X,\mathcal{F})$ will agree with the Čech coholomogy of $\mathcal{F}$
with respect to this cover, justifying using the same notation for both.

######  Lemma 2 (Folklore).

  1. $(1)$

If $X$ is a separated superscheme of finite type and $\mathcal{F}$ is a quasi-
coherent sheaf on $X$, then any cover of the underlying scheme
$X_{\operatorname{red}}$ by affine open schemes is Leray with respect to
$\mathcal{F}$.

  2. $(2)$

If $X$ is a complex supermanifold and $\mathcal{F}$ is a coherent sheaf on
$X$, then any cover of the underlying manifold $X_{\operatorname{red}}$ by
open Stein manifolds is Leray with respect to $\mathcal{F}$.

This lemma follows from the following lemma.

######  Lemma 3 (Folklore).

  1. $(1)$

If $X$ is a superscheme of finite type whose reduction
$X_{\operatorname{red}}$ is affine and $\mathcal{F}$ is a quasi-coherent sheaf
on $X$, then $H^{q}(X,\mathcal{F})=0$ for all $q>0$.

  2. $(2)$

If $X$ is a complex supermanifold whose reduction $X_{\operatorname{red}}$ is
Stein and $\mathcal{F}$ is a coherent sheaf on $X$, then
$H^{q}(X,\mathcal{F})=0$ for all $q>0$.

###### Proof.

(1) Let $J_{X}=((\mathcal{O}_{X})_{1})$ be the ideal sheaf of odd nilpotents
on the superscheme $X$. If $X$ is of finite type and $\mathcal{F}$ is a quasi-
coherent sheaf on it, then $\mathcal{F}$ is a finite iterated extension by
quasi-coherent sheaves $J_{X}^{n}\mathcal{F}/J_{X}^{n+1}\mathcal{F}$ of
$\mathcal{O}_{X_{\operatorname{red}}}=\mathcal{O}_{X}/J_{X}$-modules, whose
higher cohomology vanishes by the algebraic version of Cartan’s theorem B, see
[33, Theorem III.3.7]. Therefore, by downward induction on $n$, the higher
cohomology of $\mathcal{F}$ on $X$ will also vanish.

(2) The same argument works, except that Cartan’s theorem B itself is applied
to the coherent sheaves $J_{X}^{n}\mathcal{F}/J_{X}^{n+1}\mathcal{F}$ on
$X_{\operatorname{red}}$. ∎

######  Remark 2.

By a straightforward super generalization of Serre’s theorem on affineness, it
follows that if $X$ is a superscheme of finite type whose reduction
$X_{\operatorname{red}}$ is affine, then $X$ is affine as a superscheme.

To make formulas more symmetric, we will perform the computation below while
setting the points $z_{1}^{0},z_{2}^{0},z_{3}^{0}$ to be finite. (The case
$z_{1}^{0}=\infty$ is treated in Appendix C.) Recall that while all these 3
bosonic coordinates along with the corresponding fermionic coordinates
$\zeta_{1}^{0}$ and $\zeta_{2}^{0}$ can be fixed by the action of
$\operatorname{OSp}(1|2)^{0}$, the $\zeta_{3}$ remains free, and in
anticipation of this we will be taking one “unnecessary” open set, $W_{3}$, to
handle this.

We will compute the super Kodaira-Spencer map

$\operatorname{KS}:H^{0}(\mathcal{M}_{0,n,0},\mathcal{T}_{\mathcal{M}_{0,n,0}})\rightarrow
H^{1}(\mathcal{C}_{0,n,0},\mathcal{D}^{\otimes 2}(-\operatorname{NS}))$

in Čech cohomology. Since the super Kodaira-Spencer map is a morphism of
sheaves over $(\mathcal{M}_{0,n,0})_{\operatorname{red}}=M_{0,n}$, we may
compute it locally on $M_{0,n}$, which may be identified with an open subset
of $\mathbb{C}^{n-3}$. For any point $(z_{4}^{0},\dots,z_{n}^{0})\in M_{0,n}$,
we can choose $\varepsilon>0$ sufficiently small so that the ball
$B_{\tfrac{\varepsilon}{2}}(z_{4}^{0},\dots,z_{n}^{0})$ in $\mathbb{C}^{n-3}$
of radius $\varepsilon/2$ around this point is contained in $M_{0,n}$, which
is to say that within this ball all the $n-4$ bosonic points stay distinct,
and also distinct from $z_{1}=z^{0}_{1}$, $z_{2}=z^{0}_{2}$ and
$z_{3}=z^{0}_{3}$. Then, working over this ball in the universal family
$p\colon\thinspace\mathcal{C}_{0,n,0}\to\mathcal{M}_{0,n,0}$ of SRSs, we cover
the preimage
$p_{\operatorname{red}}^{-1}(B_{\tfrac{\varepsilon}{2}}(z_{4}^{0},\ldots,z_{n}^{0}))\subset(\mathcal{C}_{0,n,0})_{\operatorname{red}}$
by the following open subsets.

First recall that the fibers of $p_{\operatorname{red}}$ are simply copies of
${\mathbb{P}}^{1}$, since in the universal curve the point is allowed to
coincide with one of $z_{1}^{0},\dots,z_{n}^{0}$. We denote by
$U_{j}\subset\mathbb{C}$, $j=3,\ldots,n$, the open radius-$\varepsilon$ disk
around $z_{j}^{0}$ (adjusting $\varepsilon$, if necessary, to make sure that
the disks are all disjoint), and denote by $U_{0}$ the complement in
${\mathbb{P}}^{1}$ of the union of closed radius-$\varepsilon/2$ disks around
these points. Then ${\mathbb{P}}^{1}=U_{0}\cup U_{3}\cup\dots\cup U_{n}$, and
we now construct a suitable cover of
$p_{\operatorname{red}}^{-1}(B_{\tfrac{\varepsilon}{2}}(z_{4}^{0},\ldots,z_{n}^{0}))$
by the open sets $W_{\bullet}:=U_{\bullet}\times
B_{\tfrac{\varepsilon}{2}}(z_{4}^{0},\ldots,z_{n}^{0})$. We see that the open
cover $\\{W_{0},W_{3},\dots,W_{n}\\}$ is such that the only nonempty
intersections are the pairwise intersections between $W_{0}$ and $W_{j}$,
since $\varepsilon$ was sufficiently small ($\varepsilon<{\rm
min}\\{|z_{j}-z_{k}|,\ j<k\\}$). Moreover, these pairwise intersections are
simply open annuli, and thus Stein, and thereby acyclic for coherent sheaves.
We further take the open subsupermanifolds of $\mathcal{C}_{0,n,0}$ based on
each of these open subsets.

We utilize the same coordinates
$(z_{4},\ldots,z_{n}|\zeta_{3},\ldots,\zeta_{n})$ on the superdomain in
$\mathcal{M}_{0,n,0}$ over the common factor
$B_{\frac{\epsilon}{2}}(z_{4}^{0},\ldots,z_{n}^{0})$ of all $W_{j}$’s and
augment them with coordinates built on the standard affine coordinates
$(z|\zeta)$ on the superdomains in ${\mathbb{P}}^{1|1}$ over the factors
$U_{j}$, $3\leq j\leq n$. More precisely, we put over $W_{j}$ the coordinates

$\displaystyle(z_{4},\ldots,z_{n}|\zeta_{3},\ldots,\zeta_{n};w_{j}|\eta_{j})\,,$

where

(16)
$\begin{split}(w_{3}|\eta_{3})=&(z-z_{3}^{0}-\zeta\zeta_{3}|\zeta-\zeta_{3}),\\\
(w_{j}|\eta_{j})=&(z-z_{j}-\zeta\zeta_{j}|\zeta-\zeta_{j}),\quad j\geq
4.\end{split}$

Since our open subsets and their intersections (pairwise, as triple are all
empty) are Stein manifolds, they are acyclic for cohomology with coefficients
in coherent sheaves of $\mathcal{O}$-modules on the corresponding open
subsupermanifolds. Therefore this atlas can be used to compute Čech
cohomology. We thus compute the super Kodaira-Spencer map from (16) in this
atlas as

$\begin{split}\partial_{z_{j}}\stackrel{{\scriptstyle\operatorname{KS}}}{{\longmapsto}}&\Big{\\{}-\delta_{kj}\partial_{w_{j}}\in\Gamma(W_{0}\cap
W_{k},\mathcal{T}^{sc}_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}(-\operatorname{NS}))\Big{\\}}_{k=3}^{n},\qquad
j=4,\ldots,n,\\\
\partial_{\zeta_{a}}\stackrel{{\scriptstyle\operatorname{KS}}}{{\longmapsto}}&\Big{\\{}-\delta_{ka}(\partial_{\eta_{a}}-\zeta\partial_{w_{a}})=\delta_{ka}((2\eta_{a}+\zeta_{a})\partial_{w_{a}}-D_{\eta_{a}})\in\Gamma(W_{0}\cap
W_{k},\mathcal{T}^{sc}_{\mathcal{C}_{0,n,0}/\mathcal{M}_{0,n,0}}(-\operatorname{NS}))\Big{\\}}_{k=3}^{n},\\\
&\qquad a=3,\dots,n.\end{split}$

Composing with the isomorphism (12), we get

(17)
$\begin{split}\partial_{z_{j}}\stackrel{{\scriptstyle\operatorname{KS}}}{{\longmapsto}}&\Big{\\{}-\delta_{kj}D_{\eta_{j}}\otimes
D_{\eta_{j}}\in\Gamma(W_{0}\cap W_{k},\mathcal{D}^{\otimes
2}(-\operatorname{NS}))\Big{\\}}_{k=3}^{n},\qquad j=4,\ldots,n,\\\
\partial_{\zeta_{a}}\stackrel{{\scriptstyle\operatorname{KS}}}{{\longmapsto}}&\Big{\\{}\delta_{ka}((2\eta_{a}+\zeta_{a})D_{\eta_{a}}\otimes
D_{\eta_{a}})\in\Gamma(W_{0}\cap W_{k},\mathcal{D}^{\otimes
2}(-\operatorname{NS}))\Big{\\}}_{k=3}^{n},\qquad a=3,\dots,n.\end{split}$

###  4.4. The dual map

While we have an explicit basis for $H^{0}(\omega^{3}(\operatorname{NS}))$, we
want to compute the super Kodaira-Spencer map in coordinates
$(z_{4},\ldots,z_{n}|\zeta_{3},\ldots,\zeta_{n})$ on $\mathcal{M}_{0,n,0}$.
Equivalently, we can make explicit the dual map isomorphism

$\operatorname{KS}^{*}\colon\thinspace H^{0}(\omega^{3}(\operatorname{NS}))\to
H^{0}(\mathcal{M}_{0,n,0},\mathcal{T}_{\mathcal{M}_{0,n,0}}^{*})$

obtained by combining the super Kodaira-Spencer map with Serre duality and
applying linear duality. The Serre duality isomorphism

$\sigma:H^{1}(X,\mathcal{D}^{\otimes 2}(-\operatorname{NS}))\longrightarrow
H^{0}(\omega^{\otimes 3}(\operatorname{NS}))^{*}$

is given by

$\displaystyle\sigma(fD_{\zeta}\otimes
D_{\zeta})(\lambda)=\operatorname{Res}\left[\iota_{fD_{\zeta}\otimes
D_{\zeta}}\lambda\right].$

The contraction can be easily computed once we notice that

(18) $\iota_{D_{\zeta}}[dz|d\zeta]=-1.$

A nice way to prove (18) is by using the exact sequence

$0\to\mathcal{D}\to\mathcal{T}\to\mathcal{T}/\mathcal{D}\to 0$

and the isomorphism $\mathcal{D}^{2}=\mathcal{T}/\mathcal{D}$. Indeed, note
that $\left[\tfrac{\partial}{\partial
z}\,|\,\tfrac{\partial}{\partial\zeta}\right]$ is a local basis of
$\operatorname{Ber}\mathcal{T}$, and compute

$\left[\tfrac{\partial}{\partial
z}\,|\,\tfrac{\partial}{\partial\zeta}\right]=\left[\tfrac{\partial}{\partial
z}\,|\,D_{\zeta}\right]=[D_{\zeta}^{2}\,|\,D_{\zeta}]=D_{\zeta}\in\operatorname{Ber}\mathcal{T}=\mathcal{D}.$

Therefore,
$[dz|d\zeta]=D_{\zeta}^{*}\in\mathcal{D}^{*}=\operatorname{Ber}\mathcal{T}^{*}=\omega_{X}$,
which means $[dz|d\zeta](D_{\zeta})=1$ and implies (18).

So, the composition of Serre duality with the super Kodaira-Spencer map (17)
gives

(19)
$\begin{split}\sigma\circ\operatorname{KS}\colon\thinspace\partial_{z_{j}}&\longmapsto\sigma(-D_{\zeta}\otimes
D_{\zeta}|_{W_{0}\cap W_{j}}),\quad j=4,\ldots,n,\\\
\sigma\circ\operatorname{KS}\colon\thinspace\partial_{\zeta_{a}}&\longmapsto\sigma((2\eta_{a}+\zeta_{a})D_{\zeta}\otimes
D_{\zeta}|_{W_{0}\cap W_{a}}),\quad a=3,\ldots,n,\end{split}$

where we take into account that $D_{\eta_{j}}=D_{\zeta}|_{W_{0}\cap W_{j}}$,
which follows from (16). We want to evaluate $\sigma$ explicitly on the basis
(28) for $H^{0}(\omega^{3}(\operatorname{NS}))$, described in 6. For
convenience, we rewrite that basis in the form

(20) $\left\\{\zeta
z^{a}\Bigg{|}\,z^{b}\left(1-\sum_{k=1}^{n}\frac{\zeta_{k}\zeta}{z-z_{k}}\right)\right\\}\cdot\frac{[dz|d\zeta]^{3}}{\prod_{j=1}^{n}(z-z_{j})},$

where $b=0,\ldots,n-3$ and $a=0,\ldots,n-4$. In this formula, and until the
end of this section, to simplify notation, we will simply write
$z_{1},z_{2},z_{3}$ instead of $z_{1}^{0},z_{2}^{0},z_{3}^{0}$ and
$\zeta_{1},\zeta_{2}$ instead of $\zeta_{1}^{0},\zeta_{2}^{0}$. We stress that
these are fixed points in $\mathbb{C}$, not free bosonic coordinates on
$\mathcal{M}_{0,n,0}$ like $z_{4},\dots,z_{n}$ (and we still do not allow them
to take infinite value, except in Appendix C). However, as they appear
symmetrically in the formulas, we will avoid using the superscript $0$ here
and in the Appendices.

We compute

$\sigma\left(-D_{\zeta}\otimes D_{\zeta}|_{W_{0}\cap
W_{i}}\right)\left(\frac{z^{a}\zeta[dz|d\zeta]^{3}}{\prod_{j=1}^{n}(z-z_{j})}\right)=\operatorname{Res}|_{z=z_{i}}\left[\frac{-z^{a}\zeta[dz|d\zeta]}{\prod_{j=1}^{n}(z-z_{j})}\right]=\frac{-z_{i}^{a}}{\prod_{k\neq
i}(z_{i}-z_{k})},$ $\displaystyle\sigma\left(-D_{\zeta}\otimes
D_{\zeta}|_{W_{0}\cap
W_{i}}\right)\left(\frac{z^{b}[dz|d\zeta]^{3}}{\prod_{j=1}^{n}(z-z_{j})}\left(1-\sum_{k=1}^{n}\frac{\zeta_{k}\zeta}{z-z_{k}}\right)\right)=$
$\displaystyle\operatorname{Res}|_{z=z_{i}}\left[\frac{z^{b}[dz|d\zeta]}{\prod_{j=1}^{n}(z-z_{j})}\left(\sum_{k=1}^{n}\frac{\zeta_{k}\zeta}{z-z_{k}}\right)\right]$
$\displaystyle=\frac{z_{i}^{b}}{\prod_{j\neq i}(z_{i}-z_{j})}\sum_{k\neq
i}\frac{\zeta_{k}}{z_{i}-z_{k}}$
$\displaystyle+\zeta_{i}\frac{d}{dz}_{z=z_{i}}\frac{z^{b}}{\prod_{j\neq
i}(z-z_{j})}$ $\displaystyle=\frac{z_{i}^{b}}{\prod_{j\neq
i}(z_{i}-z_{j})}\sum_{k\neq i}\frac{\zeta_{k}-\zeta_{i}}{z_{i}-z_{k}}$
$\displaystyle+\frac{b\zeta_{i}z_{i}^{b-1}}{\prod_{j\neq i}(z_{i}-z_{j})}$

for $i=4,\ldots,n$ and, similarly,

$\displaystyle\sigma\left((2\eta_{c}+\zeta_{c})D_{\zeta}\otimes
D_{\zeta}|_{W_{0}\cap
W_{c}}\right)\left(\frac{z^{a}\zeta[dz|d\zeta]^{3}}{\prod_{j=1}^{n}(z-z_{j})}\right)$
$\displaystyle=\frac{-z_{c}^{a}\zeta_{c}}{\prod_{k\neq c}(z_{c}-z_{k})},$
$\displaystyle\sigma\left((2\eta_{c}+\zeta_{c})D_{\zeta}\otimes
D_{\zeta}|_{W_{0}\cap
W_{c}}\right)\left(\frac{z^{b}[dz|d\zeta]^{3}}{\prod_{j=1}^{n}(z-z_{j})}\left(1-\sum_{k=1}^{n}\frac{\zeta_{k}\zeta}{z-z_{k}}\right)\right)$
$\displaystyle=\frac{2z_{c}^{b}}{\prod_{j\neq c}(z_{c}-z_{j})}$
$\displaystyle\quad+\frac{z_{c}^{b}}{\prod_{j\neq c}(z_{c}-z_{j})}\sum_{k\neq
c}\frac{\zeta_{c}\zeta_{k}}{z_{c}-z_{k}}$

for $c=3,\ldots,n$.

###  4.5. The Berezinian of the dual map

######  Proposition 4.

The isomorphism $\operatorname{Ber}(\operatorname{KS}^{*})$ maps
$\operatorname{Ber}\left(\\{\zeta,\zeta z,\dots,\zeta
z^{n-4}\,|\,1,\dots,z^{n-3}\\}\cdot\frac{[dz|d\zeta]^{3}}{\prod_{j=1}^{n}(z-z_{j}-\zeta\zeta_{j})}\right)$
to
$-\frac{1}{2^{n-2}}(z_{3}-z_{1}-\tfrac{1}{2}\zeta_{3}\zeta_{1})(z_{3}-z_{2}-\tfrac{1}{2}\zeta_{3}\zeta_{2})[dz_{4},\dots,dz_{n}\mid
d\zeta_{3},\dots,d\zeta_{n}]$.

###### Proof.

We compute the dual super Kodaira-Spencer map $\operatorname{KS}^{*}$
explicitly to be given by the following matrix in the dual bases (28) of
$H^{0}(\omega^{3}(\operatorname{NS}))$ and $\\{dz_{4},\dots,dz_{n}\mid
d\zeta_{3},\dots,d\zeta_{n}\\}$ of
$H^{0}(\mathcal{M}_{0,n,0},\mathcal{T}_{\mathcal{M}_{0,n,0}}^{*})$:

$\displaystyle M_{n}=\begin{pmatrix}A&B\\\ C&D\end{pmatrix}\,,$

where $A$ is an $(n-3)\times(n-3)$ matrix with entries

$\displaystyle A_{jk}=\frac{-z_{j+3}^{k-1}}{\prod_{l\neq j+3}(z_{j+3}-z_{l})}$

with $1\leq l\leq n$ and $1\leq j,k\leq n-3$; $B$ is an $(n-3)\times(n-2)$
matrix with entries

$\displaystyle B_{jb}=\frac{z_{j+3}^{b-1}}{\prod_{l\neq
j+3}(z_{j+3}-z_{l})}\left(-\sum_{i\neq
j+3}\frac{\zeta_{j+3}-\zeta_{i}}{z_{j+3}-z_{i}}+(b-1)\frac{\zeta_{j+3}}{z_{j+3}}\right)$

with $1\leq l\leq n$ and $1\leq j\leq n-3$, $1\leq b\leq n-2$; $C$ is an
$(n-2)\times(n-3)$ matrix with entries

$\displaystyle C_{ak}=\frac{\zeta_{a+2}z_{a+2}^{k-1}}{\prod_{l\neq
a+2}(z_{a+2}-z_{l})}$

with $1\leq l\leq n$ and $1\leq a\leq n-2$, $1\leq k\leq n-3$; and $D$ is an
$(n-2)\times(n-2)$ matrix with entries

$\displaystyle D_{ab}=-\frac{z_{a+2}^{b-1}}{\prod_{l\neq
a+2}(z_{a+2}-z_{l})}\left(2+\sum_{i\neq
a+2}\frac{\zeta_{a+2}\zeta_{i}}{z_{a+2}-z_{i}}\right)$

with $1\leq l\leq n$ and $1\leq a\leq n-2$, $1\leq b\leq n-2$. In the end, it
is the Berezinian of $M_{n}$ what we need, and we determine it by a direct
computation.

######  Lemma 5.

The Berezinian of the map $M_{n}$ is given by

$\operatorname{Ber}M_{n}=\frac{\det(A)}{\det(D-CA^{-1}B)}=-\frac{1}{2^{n-2}}(z_{3}-z_{1}-\tfrac{1}{2}\zeta_{3}\zeta_{1})(z_{3}-z_{2}-\tfrac{1}{2}\zeta_{3}\zeta_{2}).$

Appendix B is dedicated to the proof of this lemma. The proposition follows. ∎

##  5\. Vertex-operator insertions

Here we discuss vertex-operator insertions and argue that they define a
section of the line bundle $B(\omega|_{\operatorname{NS}})$. Our construction
in principle follows the lines suggested by Witten [34, Appendix A] for the
moduli space $M_{g,1}$ of ordinary Riemann surfaces with one puncture, but the
case of Neveu-Schwarz punctures on a super Riemann surfaces seems to be
considerably subtler. A vertex operator is usually understood as representing
the asymptotic state of an incoming (at time $t=-\infty$), or outgoing (at
time $t=+\infty$) particle, with given spacetime momentum, spin, and other
possible quantum numbers. After integration over spacetime, the vacuum
expectation value (VEV) of the vertex operator becomes a section of the
contangent bundle of the worldsheet (a super Riemann surface) of the
propagating superstring, which presents itself as a boundary component of the
SRS. The assumed superconformal invariance of the vertex operator means that
the vertex operator is a cotangent vector to the super Riemann surface $X$ at
the puncture replacing the boundary component. This puncture can be of Ramond
or Neveu-Schwarz type, but in this paper, we focus on NS punctures. Let us
also consider only _conformal_ , i.e., _left-moving_ , vertex operators, as a
start, to keep doing complex analytic geometry as long as possible. The
holomorphic cotangent bundle $\mathcal{T}^{*}_{X}$ on $X$ is a rank-$(1|1)$
super vector bundle, which fits into the short exact sequence

$0\to(\mathcal{D}^{*})^{\otimes 2}\to\mathcal{T}^{*}_{X}\to\mathcal{D}^{*}\to
0,$

dual to (1). Recall that $\mathcal{D}^{*}=\omega_{X}$ and restrict this
sequence to the NS puncture $p\in X$:

$0\to\omega_{X}^{\otimes
2}\bigm{|}_{p}\to\mathcal{T}^{*}_{X}|_{p}\to\omega_{X}|_{p}\to 0.$

If two cotangent vectors $v_{1},v_{2}\in\mathcal{T}^{*}_{X}|_{p}$ form a basis
such that $v_{1}$ generates $\omega_{X}^{\otimes 2}|_{p}$ and $v_{2}$ projects
to a generator of $\omega_{X}|_{p}$, then their Berezinian $[v_{1}|v_{2}]$ is
a vector in $\operatorname{Ber}\mathcal{T}^{*}_{X}|_{p}=\omega_{X}|_{p}$. This
vector is a _vertex-operator insertion_ , which is meant to have been
integrated over the field configurations (to produce a VEV) and is usually
denoted with extra brackets: $\langle[v_{1}|v_{2}]\rangle$.

We claim the vertex-operator insertion $[v_{1}|v_{2}]\in\omega_{X}|_{p}$ gives
rise canonically to a vector in the odd line
$\operatorname{Ber}H^{0}(\operatorname{NS}_{p},\omega_{X}|_{\operatorname{NS}_{p}})$.
Let $\operatorname{NS}_{p}$ be the NS divisor corresponding to the NS puncture
$p$. The NS divisor is determined by its algebra of functions
$\mathcal{O}_{X}|_{\operatorname{NS}_{p}}=S(\mathcal{D}^{*}|_{p})=\mathcal{O}_{X}|_{p}\oplus\mathcal{D}^{*}|_{p}$,
see Section 2. Tensoring this algebra with
$\omega_{X}|_{\operatorname{NS}_{p}}$ over
$\mathcal{O}_{X}|_{\operatorname{NS}_{p}}=\mathcal{O}_{\operatorname{NS}_{p}}$,
we get

$\omega_{X}|_{\operatorname{NS}_{p}}=\omega_{X}|_{p}\oplus\omega_{X}^{\otimes
2}|_{p},$

whence we see that
$\operatorname{Ber}H^{0}(\operatorname{NS}_{p},\omega_{X}|_{\operatorname{NS}_{p}})=\omega_{X}^{\otimes
2}|_{p}\otimes\omega^{-1}_{X}|_{p}=\omega_{X}|_{p}$. Thus, the vertex-operator
insertion $[v_{1}|v_{2}]\in\omega_{X}|_{p}$ can naturally be regarded as a
vector in
$B(\omega_{X}|_{\operatorname{NS}_{p}})=\operatorname{Ber}H^{0}(\operatorname{NS}_{p},\omega_{X}|_{\operatorname{NS}_{p}})$.
In the above, we have made sure to avoid splitting super vector spaces and
sheaves into even and odd components, which allows the argument to easily
generalize to families of super Riemann surfaces. For the universal family
$\mathcal{C}_{g,1,0}\to\mathcal{M}_{g,1,0}$, a vertex-operator insertion
becomes a section $V(p)=[v_{1}|v_{2}]\in
H^{0}(\mathcal{M}_{g,1,0},B(\omega|_{\operatorname{NS}_{p}}))$, where $p$ is
understood as a section $p:\mathcal{M}_{g,1,0}\to\mathcal{C}_{g,1,0}$ of the
universal family.

When we have $n$ NS punctures $p_{1},\dots p_{n}$, the corresponding total NS
divisor is the sum
$\operatorname{NS}=\operatorname{NS}_{p_{1}}+\dots\linebreak[0]+\operatorname{NS}_{p_{n}}$,
and the vertex-operator insertions $V_{1}(p_{1}),\dots,V_{n}(p_{n})$ multiply
to produce a section $\bigotimes_{j=1}^{n}V_{j}(p_{j})$ of
$\operatorname{Ber}(\omega|_{\operatorname{NS}})=\operatorname{Ber}(\omega|_{\operatorname{NS}_{p_{1}}})\otimes\dots\otimes\operatorname{Ber}(\omega|_{\operatorname{NS}_{p_{1}}})$.
If $(z|\zeta)$ are holomorphic coordinates near the $j$’th NS puncture
$p_{j}=(z_{j}|\zeta_{j})$, where
$\Big{[}\,\zeta[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{|}[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{]}$
gives a basis of $\operatorname{Ber}(\omega|_{\operatorname{NS}_{j}})$, then
$V_{j}(p_{j})$ may be expressed as
$V_{j}(z_{j}|\zeta_{j})\Big{[}\,\zeta[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{|}[dz|d\zeta]\\\
vert_{\operatorname{NS}_{j}}\Big{]}$, whence the product of these gives a
section

$\bigotimes_{j=1}^{n}V_{j}(z_{j}|\zeta_{j})\Big{[}\,\zeta[dz|d\zeta]|_{\operatorname{NS}_{j}}\mid[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{]}\in
H^{0}(\mathcal{M}_{0,n,0},\operatorname{Ber}(\omega|_{\operatorname{NS}})),$

as claimed in (6), where we used the brackets $\langle\;\rangle$ there to
stress that it includes integration over the spacetime configurations.

##  6\. Tree-level scattering amplitudes in the physics literature

To better understand the physics context of our results and match them with
the well-known computations, we review how the holomorphic measure $d\nu$
arises in the determination the superstring scattering amplitudes in the
physics literature for the type IIB superstring theory. The matter superfields
in terms of the holomorphic coordinates $(z|\zeta)$ on a SRS and the
antiholomorphic coordinates $(\bar{z}|\bar{\zeta})$ on the complex conjugate
SRS are given by

(21)
$\displaystyle{\boldsymbol{X}}^{\mu}(z|\zeta;\bar{z}|\bar{\zeta})=X_{L}^{\mu}(z)+X_{R}^{\mu}(\bar{z})+\zeta\psi^{\mu}(z)+\bar{\zeta}\bar{\psi}^{\mu}(\bar{z}).$

In the usual approach to superstring scattering amplitudes one also introduces
the superghost fields

$\displaystyle B(z|\zeta)=$ $\displaystyle b(z)+\zeta\beta(z),\qquad
C(z|\zeta)=c(z)+\zeta\gamma(z),$ $\displaystyle\bar{B}(\bar{z}|\bar{\zeta})=$
$\displaystyle\bar{b}(\bar{z})+\bar{\zeta}\bar{\beta}(\bar{z}),\qquad\bar{C}(\bar{z}|\bar{\zeta})=\bar{c}(\bar{z})+\bar{\zeta}\bar{\gamma}(\bar{z}).$

In order to compute the scattering amplitude among $n$ physical states, one
inserts and then integrates superconformal vertex operators of conformal
weight $(1|1)$ at the punctures. However, to eliminate the zero modes of the
superghosts and get sensible path integrals, one must then also introduce
picture changing operators (PCO), necessary to get a sensible nontrivial
measure (allowing for nonvanishing result). It turns out that the vertex
operators can then be taken in the form

${\mathcal{V}}(z|\zeta;\bar{z}|\bar{\zeta})=c(z)\bar{c}(\bar{z})\delta(\gamma(z))\delta(\bar{\gamma}(\bar{z}))V(z|\zeta;\bar{z}|\bar{\zeta})\,,$

where $V$ is a superconformal field of conformal weight $(1|1)$, and the delta
distributions are meant to ensure that the $\gamma$ components vanish at the
puncture. When computing the amplitudes, for $n\geq 3$, one has also to
introduce the insertion of $\delta(\beta)\delta(\bar{\beta})$, conventionally
chosen at $(z_{3}|\zeta_{3},\bar{z}_{3}|\bar{\zeta}_{3})$, which can be used
to eliminate $\delta(\gamma(z_{3}))\delta(\bar{\gamma}(\bar{z}_{3}))$. For
$n\geq 4$, $n-3$ vertices can be replaced by integrated vertices

$K_{V_{j}}=\int_{\Sigma}d\zeta_{j}d\bar{\zeta}_{j}dz_{j}d\bar{z}_{j}\
V_{j}(z_{j}|\zeta_{j};\bar{z}_{j}|\bar{\zeta}_{j})\,,$

where $\Sigma$ is the genus zero super Riemann surface underlying the
scattering representation, so that we can write for the amplitude

${\mathcal{A}}_{n}=\left\langle
c\bar{c}\delta(\gamma)V_{1}(z_{1}^{0}|0;\bar{z}_{1}^{0}|0)c\bar{c}\delta(\gamma)V_{2}(z_{2}^{0}|0;\bar{z}_{2}^{0}|0)c\bar{c}\delta(\gamma)D_{\zeta_{3}}D_{\bar{\zeta}_{3}}V_{3}(z_{3}^{0}|0;\bar{z}_{3}^{0}|0)\prod_{j=4}^{n}K_{V_{j}}\right\rangle\,,$

where $\langle\dots\rangle$ means the VEV. In particular, the first three
bosonic coordinates are fixed at the arbitrary values
$z^{0}_{1},z^{0}_{2},z^{0}_{3}$, while the first two odd variables are
conventionally fixed at $\zeta_{a}=0$, $a=1,2$. Notice that the ghosts
disappeared from the integrated vertices so they factor in the VEV as

$k_{gh}=\left\langle
c\bar{c}\delta(\gamma)(z_{1}^{0};\bar{z}_{1}^{0})c\bar{c}\delta(\gamma)(z_{2}^{0};\bar{z}_{2}^{0})c\bar{c}\delta(\gamma)(z_{3}^{0};\bar{z}_{3}^{0})\right\rangle\,.$

A simple physics calculation using bosonization and operator product expansion
(OPE) then yields

$\displaystyle
k_{gh}=(z_{3}^{0}-z_{1}^{0})(z_{3}^{0}-z_{2}^{0})(\bar{z}_{3}^{0}-\bar{z}_{1}^{0})(\bar{z}_{3}^{0}-\bar{z}_{2}^{0})\,,$

and we remark that the amplitude can then be rewritten as

(22) ${\mathcal{A}}_{n}=g_{c}^{n-2}\int_{\mathcal{M}_{0,n,0}}\left\langle
V_{1}(z_{1}^{0}|0;\bar{z}_{1}^{0}|0)V_{2}(z_{2}^{0}|0;\bar{z}_{2}^{0}|0)V_{3}(z_{3}^{0}|\zeta_{3};\bar{z}_{3}^{0}|\bar{\zeta}_{3})\prod_{j=4}^{n}V_{j}(z_{j}|\zeta_{j};\bar{z}_{j}|\bar{\zeta}_{j})\right\rangle
Dm\,,$

where $g_{c}$ denotes the closed string coupling and the integration measure
is

$Dm=k_{gh}[dz_{4}\dots dz_{n}|d\zeta_{3}\dots d\zeta_{n}][d\bar{z}_{4}\dots
d\bar{z}_{n}|d\bar{\zeta}_{3}\dots d\bar{\zeta}_{n}]\,.$

Direct comparison with (9) allows us to identify $g_{c}=1/4$ (when neglecting
the dilaton field).

##  7\. Conclusion

We can now compare our results with the standard derivation of tree-level NS
amplitudes available in the literature. Instead of deducing the amplitude
formula (22) by constructing a measure through the introduction of ghosts and
PCOs, we have suggested the slightly more general formula (9), containing only
the physical fields and the canonical measure, as determined by the super
Mumford isomorphism over $\mathcal{M}_{0,n,0}$. By comparing the two formulas,
we see that they coincide after replacing $g_{c}$ by $\tfrac{\langle
e^{\phi}\rangle}{4}$, if we include the dilaton contribution in (9).

##  Appendix A The super Mumford form over $\mathcal{M}_{0,n,0}$

Here we compute the image of the super Mumford form

(23)
$\mu=\left[\frac{\zeta[dz|d\zeta]^{3}}{z},\frac{\zeta[dz|d\zeta]^{3}}{z^{2}},\frac{\zeta[dz|d\zeta]^{3}}{z^{3}}\;\left|\;\frac{[dz|d\zeta]^{3}}{z},\frac{[dz|d\zeta]^{3}}{z^{2}}\right.\right]\in
B(\omega^{3}),$

see (11), tensored with a global section

$\left\langle\bigotimes_{j=1}^{n}V_{j}(z_{j}|\zeta_{j})\Big{[}\,\zeta[dz|d\zeta]\\\
vert_{\operatorname{NS}_{j}}\Big{|}[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{]}\right\rangle\in
B(\omega|_{\operatorname{NS}})$

arising from vertex-operator insertions, under the isomorphism

(24) $\varphi\colon\thinspace B(\omega^{3})\otimes
B(\omega|_{\operatorname{NS}})\to B(\omega^{3}(\operatorname{NS})),$

which comes from the short exact sequence

(25) $0\to\omega^{3}\stackrel{{\scriptstyle
f}}{{\longrightarrow}}\omega^{3}(\operatorname{NS})\stackrel{{\scriptstyle
g}}{{\longrightarrow}}\omega^{3}(\operatorname{NS})|_{\operatorname{NS}}\to
0\,$

and the canonical isomorphism (15):

(26)
$B(\omega|_{\operatorname{NS}})\xrightarrow{\sim}B(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})\,.$

Even though all the isomorphisms are defined in a canonical, coordinate-
independent way, since we want to perform explicit computations, we work with
the standard affine coordinates $(z|\zeta)$ on ${\mathbb{P}}^{1|1}$. Note that
the isomorphisms are linear over holomorphic functions, such as
$\left\langle\prod_{j=1}^{n}V_{j}(z_{j}|\zeta_{j})\right\rangle$, on the
supermoduli space and, therefore, it is enough to compute the image of

(27)
$\mu\otimes\bigotimes_{j=1}^{n}\Big{[}\,\zeta[dz|d\zeta]|_{\operatorname{NS}_{j}}\mid[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{]}$

under $\varphi$.

######  Proposition 6.

The image of (27) under the isomorphism $\varphi$ of (24) is the Berezinian of
the following basis of $H^{0}(\omega^{3}(\operatorname{NS}))$:

(28) $\displaystyle\\{\zeta,\zeta z,\dots,\zeta
z^{n-4}\,|\,1,\dots,z^{n-3}\\}\cdot\frac{[dz|d\zeta]^{3}}{\prod_{j=1}^{n}(z-z_{j}-\zeta\zeta_{j})}\,.$

###### Proof.

For each $j=1,\dots,n$, multiplication by the generator
$\frac{[dz|d\zeta]^{2}}{(z-z_{j}-\zeta\zeta_{j})}$ of $\omega^{2}(NS)$ (in a
small enough open neighborhood of the $j$th NS puncture) maps the basis

$\left\\{\zeta[dz|d\zeta]\biggm{|}_{\operatorname{NS}_{j}}\;\Biggm{|}\;[dz|d\zeta]\biggm{|}_{\operatorname{NS}_{j}}\right\\}$

of $H^{0}(\omega|_{\operatorname{NS}_{j}})$ to the basis

(29)
$\left\\{\frac{\zeta[dz|d\zeta]^{3}}{z-z_{j}-\zeta\zeta_{j}}\biggm{|}_{\operatorname{NS}_{j}}\;\Biggm{|}\;\frac{[dz|d\zeta]^{3}}{z-z_{j}-\zeta\zeta_{j}}\biggm{|}_{\operatorname{NS}_{j}}\right\\}$

of $H^{0}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}_{j}})$, where the
“restriction” $|_{\operatorname{NS}_{j}}$ means considering a local section of
$\omega^{3}(\operatorname{NS})$ modulo the subsheaf
$(z-z_{j}-\zeta\zeta_{j})\,\omega^{3}(\operatorname{NS})$. This maps the
Berezinian

(30)
$\bigotimes_{j=1}^{n}\Big{[}\,\zeta[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{|}[dz|d\zeta]|_{\operatorname{NS}_{j}}\Big{]}$

of the basis of $H^{0}(\omega_{\operatorname{NS}})$ to the Berezinian

(31)
$\bigotimes_{j=1}^{n}\left[\frac{\zeta[dz|d\zeta]^{3}}{z-z_{j}-\zeta\zeta_{j}}\biggm{|}_{\operatorname{NS}_{j}}\;\Biggm{|}\;\frac{[dz|d\zeta]^{3}}{z-z_{j}-\zeta\zeta_{j}}\biggm{|}_{\operatorname{NS}_{j}}\right]$

of the basis of $H^{0}(\omega^{3}(\operatorname{NS})_{\operatorname{NS}})$ and
induces the canonical isomorphism (26) on the Berezinians of the cohomology.
It remains to show that the canonical isomorphism

(32) $B(\omega^{3})\otimes
B(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})\xrightarrow{\sim}B(\omega^{3}(\operatorname{NS}))$

induced by (25) matches the super Mumford form $\mu$, the Berezinian (31), and
the Berezinian of the basis (28):

(33) $\mu\otimes\eqref{NS|NS}\mapsto[\eqref{basisA}].$

We will work with the long exact sequence in cohomology

(34) $0\rightarrow H^{0}(\omega^{3}(\operatorname{NS}))\stackrel{{\scriptstyle
g}}{{\longrightarrow}}H^{0}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})\stackrel{{\scriptstyle\delta}}{{\longrightarrow}}H^{1}(\omega^{3})\rightarrow
0\,$

associated with the short exact sequence (25) of sheaves. In terms of the
sequence (34), the isomorphism (32) rewrites as

$\operatorname{Ber}(H^{0}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}}))\xrightarrow{\sim}\operatorname{Ber}(H^{1}(\omega^{3}))\otimes\operatorname{Ber}(H^{0}(\omega^{3}(\operatorname{NS}))).$

We will describe the relation between the bases of the three terms of (34) and
the resulting relation between their Berezinians.

The map $g$ in (34) is obtained by restricting the elements of the basis (28)
to the divisor $\operatorname{NS}$. This gives the following elements of
$H^{0}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})$:

$\displaystyle\bigoplus_{j}\frac{\zeta z_{j}^{a}}{\prod_{k\neq
j}(z_{j}-z_{k})}\,\frac{[dz|d\zeta]^{3}}{z-z_{j}-\zeta\zeta_{j}}\biggm{|}_{\operatorname{NS}_{j}},\qquad
0\leq a\leq n-4,$
$\displaystyle\bigoplus_{j}\left[\frac{z_{j}^{a}}{\prod_{k\neq
j}(z_{j}-z_{k})}-\frac{\zeta z_{j}^{a}}{\prod_{k\neq
j}(z_{j}-z_{k})}\left(\sum_{k\neq
j}\frac{\zeta_{j}-\zeta_{k}}{z_{j}-z_{k}}-a\frac{\zeta_{j}}{z_{j}}\right)\right]\frac{[dz|d\zeta]^{3}}{z-z_{j}-\zeta\zeta_{j}}\biggm{|}_{\operatorname{NS}_{j}},\quad
0\leq a\leq n-3.$

To determine the map $\delta$ in (34) explicitly, we first note that

(35) $\left\\{\zeta z^{2},\zeta
z,\zeta\biggm{|}z^{2},z\right\\}\cdot\frac{[dz|d\zeta]^{3}}{z^{3}}$

is a basis of $H^{1}(\omega^{3})$ whose Berezinian is the super Mumford form
$\mu$, see (23). On the other hand, by Serre duality
$H^{1}(\omega^{3})=H^{0}(\omega^{-2})^{*}$, and

$\left\\{1,z,z^{2}\,|\,-\zeta,-\zeta z\right\\}\cdot[dz|d\zeta]^{-2}$

is the dual basis of $H^{0}(\omega^{-2})$. The map $\delta$ sends $\lambda\in
H^{0}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})$ to
$\delta(\lambda)\in H^{1}(\omega^{3})=H^{0}(\omega^{-2})^{*}$ such that for
any $\alpha\in H^{0}(\omega^{-2})$, we have

$\displaystyle\delta(\lambda)(\alpha)=\operatorname{Res}_{\operatorname{NS}}(\lambda\alpha).$

This map $\delta$ is given explicitly on the basis (29) by

$\displaystyle\frac{\zeta[dz|d\zeta]^{3}}{z-z_{j}-\zeta\zeta_{j}}\biggm{|}_{\operatorname{NS}_{j}}$
$\displaystyle\longmapsto\left(\frac{\zeta}{z}+z_{j}\frac{\zeta}{z^{2}}+z_{j}^{2}\frac{\zeta}{z^{2}}\right)[dz|d\zeta]^{3},$
$\displaystyle\frac{[dz|d\zeta]^{3}}{z-z_{j}-\zeta\zeta_{j}}\biggm{|}_{\operatorname{NS}_{j}}$
$\displaystyle\longmapsto\left(\frac{1}{z}+z_{j}\frac{1}{z^{2}}\right)[dz|d\zeta]^{3}$

for $j=1,\dots,n$.

Thus, in matrix form with respect to the chosen bases, we can write

$\displaystyle g=\begin{pmatrix}A&B\\\ 0&C\end{pmatrix},$

with

$\displaystyle A=$ $\displaystyle\begin{pmatrix}\frac{1}{\prod_{k\neq
1}(z_{1}-z_{k})}(1,z_{1},\ldots,z_{1}^{n-4})\\\ \frac{1}{\prod_{k\neq
2}(z_{2}-z_{k})}(1,z_{2},\ldots,z_{2}^{n-4})\\\ \ldots\ldots\ldots\\\
\frac{1}{\prod_{k\neq
n}(z_{n}-z_{k})}(1,z_{n},\ldots,z_{n}^{n-4})\end{pmatrix},$ $\displaystyle B=$
$\displaystyle\begin{pmatrix}\frac{1}{\prod_{k\neq
1}(z_{1}-z_{k})}\left(\sum_{k\neq 1}\frac{\zeta_{1}-\zeta_{k}}{z_{1}-z_{k}}\
;\ \sum_{k\neq 1}\frac{(\zeta_{1}-\zeta_{k})z_{1}}{z_{1}-z_{k}}-\zeta_{1}\ ;\
\ldots\ ;\ \sum_{k\neq
1}\frac{(\zeta_{1}-\zeta_{k})z_{1}^{n-3}}{z_{1}-z_{k}}-(n-3)z_{1}^{n-4}\zeta_{1}\right)\\\
\frac{1}{\prod_{k\neq 2}(z_{2}-z_{k})}\left(\sum_{k\neq
2}\frac{\zeta_{2}-\zeta_{k}}{z_{2}-z_{k}}\ ;\ \sum_{k\neq
2}\frac{(\zeta_{2}-\zeta_{k})z_{2}}{z_{2}-z_{k}}-\zeta_{2}\ ;\ \ldots\ ;\
\sum_{k\neq
2}\frac{(\zeta_{2}-\zeta_{k})z_{2}^{n-3}}{z_{2}-z_{k}}-(n-3)z_{2}^{n-4}\zeta_{2}\right)\\\
\ldots\ldots\ldots\\\ \frac{1}{\prod_{k\neq n}(z_{n}-z_{k})}\left(\sum_{k\neq
n}\frac{\zeta_{n}-\zeta_{k}}{z_{n}-z_{k}}\ ;\ \sum_{k\neq
n}\frac{(\zeta_{n}-\zeta_{k})z_{n}}{z_{n}-z_{k}}-\zeta_{n}\ ;\ \ldots\ ;\
\sum_{k\neq
n}\frac{(\zeta_{n}-\zeta_{k})z_{n}^{n-3}}{z_{n}-z_{k}}-(n-3)z_{n}^{n-4}\zeta_{n}\right)\end{pmatrix},$
$\displaystyle C=$ $\displaystyle\begin{pmatrix}\frac{1}{\prod_{k\neq
1}(z_{1}-z_{k})}(1,z_{1},\ldots,z_{1}^{n-3})\\\ \frac{1}{\prod_{k\neq
2}(z_{2}-z_{k})}(1,z_{2},\ldots,z_{2}^{n-3})\\\ \ldots\ldots\ldots\\\
\frac{1}{\prod_{k\neq
n}(z_{n}-z_{k})}(1,z_{n},\ldots,z_{n}^{n-3})\end{pmatrix},$

and

$\displaystyle\delta=\begin{pmatrix}1&1&\ldots&1&0&0&\ldots&0\\\
z_{1}&z_{2}&\ldots&z_{n}&0&0&\ldots&0\\\
z_{1}^{2}&z_{2}^{2}&\ldots&z_{n}^{2}&0&0&\ldots&0\\\
0&0&\ldots&0&1&1&\ldots&1\\\
0&0&\ldots&0&z_{1}&z_{2}&\ldots&z_{n}\end{pmatrix}.$

Let us pick a splitting of the short exact sequence (34), that is to say, a
map $\delta^{\prime}:H^{1}(\omega^{3})\to
H^{0}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})$ such that
$\delta\delta^{\prime}=\operatorname{id}$. A priori there are many choices,
but the relation between the Berezinians will not depend on the choice. Since
we have fixed the coordinates $\\{z_{4},\ldots
z_{n}\,|\,\zeta_{3},\ldots,\zeta_{n}\\}$, we are left to work with the matrix
elements associated to the remaining ones, which renders an invertible
$5\times 5$ matrix. We get

$\displaystyle\delta^{\prime}=\begin{pmatrix}\frac{z_{2}z_{3}}{(z_{2}-z_{1})(z_{3}-z_{1})}&\frac{-(z_{2}+z_{3})}{(z_{2}-z_{1})(z_{3}-z_{1})}&\frac{1}{(z_{2}-z_{1})(z_{3}-z_{1})}&0&0\\\
\frac{z_{1}z_{3}}{(z_{1}-z_{2})(z_{3}-z_{2})}&\frac{-(z_{1}+z_{3})}{(z_{1}-z_{2})(z_{3}-z_{2})}&\frac{1}{(z_{1}-z_{2})(z_{3}-z_{2})}&0&0\\\
\frac{z_{2}z_{1}}{(z_{2}-z_{3})(z_{1}-z_{3})}&\frac{-(z_{2}+z_{1})}{(z_{2}-z_{3})(z_{1}-z_{3})}&\frac{1}{(z_{2}-z_{3})(z_{1}-z_{3})}&0&0\\\
0&0&0&0&0\\\ \vdots&\vdots&\vdots&\vdots&\vdots\\\ 0&0&0&0&0\\\
0&0&0&\frac{z_{2}}{z_{2}-z_{1}}&\frac{-1}{z_{2}-z_{1}}\\\
0&0&0&\frac{z_{1}}{z_{1}-z_{2}}&\frac{-1}{z_{1}-z_{2}}\\\ 0&0&0&0&0\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\ 0&0&0&0&0\end{pmatrix}.$

The transition matrix $M$ from the basis of
$H^{0}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})$ formed by the
image of the basis (28) under $g$ and the image of the basis (35) under
$\delta^{\prime}$ to the basis (29) over all $j=1,\dots,n$ is thus represented
by a $2n\times 2n$ block matrix

$\displaystyle M=\begin{pmatrix}\tilde{A}&\tilde{B}\\\
0&\tilde{C}\end{pmatrix},$

where

$\displaystyle\tilde{A}=$ $\displaystyle\begin{pmatrix}\frac{1}{\prod_{k\neq
1}(z_{1}-z_{k})}(1,z_{1},\ldots,z_{1}^{n-4}),&\frac{(z_{2}z_{3},-z_{2}-z_{3},1)}{(z_{2}-z_{1})(z_{3}-z_{1})}\\\
\frac{1}{\prod_{k\neq
2}(z_{2}-z_{k})}(1,z_{2},\ldots,z_{2}^{n-4}),&\frac{(z_{1}z_{3},-z_{1}-z_{3},1)}{(z_{1}-z_{2})(z_{3}-z_{2})}\\\
\frac{1}{\prod_{k\neq
3}(z_{3}-z_{k})}(1,z_{3},\ldots,z_{3}^{n-4}),&\frac{(z_{1}z_{2},-z_{1}-z_{2},1)}{(z_{1}-z_{3})(z_{2}-z_{3})}\\\
\frac{1}{\prod_{k\neq 4}(z_{4}-z_{k})}(1,z_{4},\ldots,z_{4}^{n-4}),&(0,0,0)\\\
\vdots&\vdots\\\ \frac{1}{\prod_{k\neq
n}(z_{n}-z_{k})}(1,z_{n},\ldots,z_{n}^{n-4}),&(0,0,0)\end{pmatrix},$
$\displaystyle\tilde{B}=$ $\displaystyle\begin{pmatrix}\frac{1}{\prod_{k\neq
1}(z_{1}-z_{k})}\left(\sum_{k\neq 1}\frac{\zeta_{1}-\zeta_{k}}{z_{1}-z_{k}};\
\sum_{k\neq 1}\frac{(\zeta_{1}-\zeta_{k})z_{1}}{z_{1}-z_{k}}-\zeta_{1};\
\ldots;\ \sum_{k\neq
1}\frac{(\zeta_{1}-\zeta_{k})z_{1}^{n-3}}{z_{1}-z_{k}}-(n-3)z_{1}^{n-4}\zeta_{1};0;0\right)\\\
\frac{1}{\prod_{k\neq 2}(z_{2}-z_{k})}\left(\sum_{k\neq
2}\frac{\zeta_{2}-\zeta_{k}}{z_{2}-z_{k}};\ \sum_{k\neq
2}\frac{(\zeta_{2}-\zeta_{k})z_{2}}{z_{2}-z_{k}}-\zeta_{2};\ \ldots;\
\sum_{k\neq
2}\frac{(\zeta_{2}-\zeta_{k})z_{2}^{n-3}}{z_{2}-z_{k}}-(n-3)z_{2}^{n-4}\zeta_{2};0;0\right)\\\
\ldots\ldots\ldots\\\ \frac{1}{\prod_{k\neq n}(z_{n}-z_{k})}\left(\sum_{k\neq
n}\frac{\zeta_{n}-\zeta_{k}}{z_{n}-z_{k}};\ \sum_{k\neq
n}\frac{(\zeta_{n}-\zeta_{k})z_{n}}{z_{n}-z_{k}}-\zeta_{n};\ \ldots;\
\sum_{k\neq
n}\frac{(\zeta_{n}-\zeta_{k})z_{n}^{n-3}}{z_{n}-z_{k}}-(n-3)z_{n}^{n-4}\zeta_{n};0;0\right)\end{pmatrix},$
$\displaystyle\tilde{C}=$ $\displaystyle\begin{pmatrix}\frac{1}{\prod_{k\neq
1}(z_{1}-z_{k})}(1,z_{1},\ldots,z_{1}^{n-3}),&\frac{(z_{2},-1)}{z_{2}-z_{1}}\\\
\frac{1}{\prod_{k\neq
2}(z_{2}-z_{k})}(1,z_{2},\ldots,z_{2}^{n-3}),&\frac{(z_{1},-1)}{z_{1}-z_{2}}\\\
\frac{1}{\prod_{k\neq 3}(z_{3}-z_{k})}(1,z_{3},\ldots,z_{3}^{n-3}),&(0,0,0)\\\
\vdots&\vdots\\\ \frac{1}{\prod_{k\neq
n}(z_{n}-z_{k})}(1,z_{n},\ldots,z_{n}^{n-3}),&(0,0,0)\end{pmatrix}.$

Accordingly, the Berezinians of the two bases of
$H^{0}(\omega^{3}(\operatorname{NS})|_{\operatorname{NS}})$ are related by
$\operatorname{Ber}M$, which the following lemma claims to be simply equal to
1, proving (33) and thereby the proposition. ∎

######  Lemma 7.

$\operatorname{Ber}M=1$.

###### Proof.

We have $\operatorname{Ber}M=\det\tilde{A}\det\tilde{C}^{-1}$. In order to
compute these determinants, let us notice that

$\displaystyle\tilde{A}=\begin{pmatrix}A_{1}&A_{2}\\\ A_{3}&0\end{pmatrix},$

where $A_{2}$ is the $3\times 3$ matrix

$\displaystyle A_{2}=\begin{pmatrix}1&1&1\\\ z_{1}&z_{2}&z_{3}\\\
z_{1}^{2}&z_{2}^{2}&z_{3}^{2}\end{pmatrix}^{-1},$

and $A_{3}$ is the $(n-3)\times(n-3)$ matrix

$\displaystyle A_{3}=\begin{pmatrix}\frac{1}{\prod_{k\neq
4}(z_{4}-z_{k})}(1,z_{4},\ldots,z_{4}^{n-4})\\\ \vdots\\\
\frac{1}{\prod_{k\neq
n}(z_{n}-z_{k})}(1,z_{n},\ldots,z_{n}^{n-4})\end{pmatrix}.$

Therefore,

$\displaystyle\det A_{2}=$ $\displaystyle\prod_{1\leq j<k\leq
3}\frac{1}{z_{k}-z_{j}},$ $\displaystyle\det A_{3}=$
$\displaystyle\prod_{l=4}^{n}\left(\frac{1}{\prod_{k\neq
l}(z_{l}-z_{k})}\right)\det\begin{pmatrix}1,z_{4},\ldots,z_{4}^{n-4}\\\
\vdots\\\
1,z_{n},\ldots,z_{n}^{n-4}\end{pmatrix}=\prod_{l=4}^{n}\left(\frac{1}{\prod_{k\neq
l}(z_{l}-z_{k})}\right)\prod_{4\leq r<m\leq n}(z_{m}-z_{r}),$

and222In bringing each of the far right columns to the left, in order to get a
block-diagonal matrix, the determinant changes sign $3(n-3)$ times.

$\displaystyle\det\tilde{A}=(-1)^{n-3}\det A_{2}\det
A_{3}=(-1)^{n-3}\prod_{1\leq j<k\leq
3}\frac{1}{z_{k}-z_{j}}\prod_{l=4}^{n}\left(\frac{1}{\prod_{k\neq
l}(z_{l}-z_{k})}\right)\prod_{4\leq r<m\leq n}(z_{m}-z_{r}).$

In the same way, we have

$\displaystyle\tilde{C}=\begin{pmatrix}C_{1}&C_{2}\\\ C_{3}&0\end{pmatrix},$

where $C_{2}$ is the $2\times 2$ matrix

$\displaystyle C_{2}=\begin{pmatrix}1&1\\\ z_{1}&z_{2}\end{pmatrix}^{-1},$

and $C_{3}$ is the $(n-2)\times(n-2)$ matrix

$\displaystyle C_{3}=\begin{pmatrix}\frac{1}{\prod_{k\neq
3}(z_{3}-z_{k})}(1,z_{3},\ldots,z_{3}^{n-3})\\\ \vdots\\\
\frac{1}{\prod_{k\neq
n}(z_{n}-z_{k})}(1,z_{n},\ldots,z_{n}^{n-3})\end{pmatrix}.$

Therefore,

$\displaystyle\det C_{2}=$ $\displaystyle\frac{1}{z_{2}-z_{1}},$
$\displaystyle\det C_{3}=$
$\displaystyle\prod_{l=3}^{n}\left(\frac{1}{\prod_{k\neq
l}(z_{l}-z_{k})}\right)\det\begin{pmatrix}1,z_{3},\ldots,z_{3}^{n-3}\\\
\vdots\\\
1,z_{n},\ldots,z_{n}^{n-3}\end{pmatrix}=\prod_{l=3}^{n}\left(\frac{1}{\prod_{k\neq
l}(z_{l}-z_{k})}\right)\prod_{3\leq r<m\leq n}(z_{m}-z_{r}),$

and

$\displaystyle\det\tilde{C}=\det C_{2}\det
C_{3}=\frac{1}{z_{2}-z_{1}}\prod_{l=3}^{n}\left(\frac{1}{\prod_{k\neq
l}(z_{l}-z_{k})}\right)\prod_{3\leq r<m\leq n}(z_{m}-z_{r}).$

By taking the quotient we get

$\displaystyle\operatorname{Ber}M=$
$\displaystyle\frac{\det\tilde{A}}{\det\tilde{C}}=\frac{(-1)^{n-3}}{(z_{3}-z_{2})(z_{3}-z_{1})(z_{2}-z_{1})}\prod_{l=4}^{n}\left(\frac{1}{\prod_{k\neq
l}(z_{l}-z_{k})}\right)\prod_{4\leq r<m\leq n}(z_{m}-z_{r})$
$\displaystyle\times(z_{2}-z_{1})\prod_{l=3}^{n}\left(\prod_{k\neq
l}(z_{l}-z_{k})\right)\prod_{3\leq r<m\leq n}\frac{1}{z_{m}-z_{r}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{n-3}}{(z_{3}-z_{2})(z_{3}-z_{1})}\left(\prod_{k\neq
3}(z_{3}-z_{k})\right)\prod_{3<m\leq n}\frac{1}{z_{m}-z_{3}}.$

The first factor times the last $n-3$ give $\prod_{k\neq
3}\frac{1}{z_{3}-z_{k}}$, so we get $\operatorname{Ber}M=1$. ∎

##  Appendix B Proof of 5: the Berezinian of the super Kodaira-Spencer map

We first notice that, for $N=D-CA^{-1}B$, in terms of rows, we can write

$\displaystyle N=\begin{pmatrix}-\frac{2}{\prod_{j\neq
3}(z_{3}-z_{j})}\left(\left(1+\frac{1}{2}\sum_{k\neq
3}\frac{\zeta_{3}\zeta_{k}}{z_{3}-z_{k}}\right)(1,z_{3},\dots,z_{3}^{n-3})+\frac{\zeta_{3}}{2}(1,z_{3},\dots,z_{3}^{n-4})A^{-1}B\right)\\\
-\frac{2}{\prod_{j\neq
4}(z_{4}-z_{j})}\left(1,z_{4},\ldots,z_{4}^{n-3}\right)\\\ \ldots\\\
-\frac{2}{\prod_{j\neq
n}(z_{n}-z_{j})}\left(1,z_{n},\ldots,z_{n}^{n-3}\right)\end{pmatrix}$

so that writing the first row as a sum, we can write $\det N=\det
N^{\prime}+\det N^{\prime\prime}$, where

$\displaystyle N^{\prime}=\begin{pmatrix}-\frac{2}{\prod_{j\neq
3}(z_{3}-z_{j})}\left(\left(1+\frac{1}{2}\sum_{k=1}^{2}\frac{\zeta_{3}\zeta_{k}}{z_{3}-z_{k}}\right)(1,z_{3},\dots,z_{3}^{n-3})\right)\\\
-\frac{2}{\prod_{j\neq
4}(z_{4}-z_{j})}\left(1,z_{4},\ldots,z_{4}^{n-3}\right)\\\ \ldots\\\
-\frac{2}{\prod_{j\neq
n}(z_{n}-z_{j})}\left(1,z_{n},\ldots,z_{n}^{n-3}\right)\end{pmatrix}$

and

$\displaystyle N^{\prime\prime}=\begin{pmatrix}-\frac{1}{\prod_{j\neq
3}(z_{3}-z_{j})}\left(\left(\sum_{k=4}^{n}\frac{\zeta_{3}\zeta_{k}}{z_{3}-z_{k}}\right)(1,z_{3},\dots,z_{3}^{n-3})+\zeta_{3}(1,z_{3},\dots,z_{3}^{n-4})A^{-1}B\right)\\\
-\frac{2}{\prod_{j\neq
4}(z_{4}-z_{j})}\left(1,z_{4},\ldots,z_{4}^{n-3}\right)\\\ \ldots\\\
-\frac{2}{\prod_{j\neq
n}(z_{n}-z_{j})}\left(1,z_{n},\ldots,z_{n}^{n-3}\right)\end{pmatrix}.$

It is immediate to see that $\det A/\det N^{\prime}$ gives the expression (5).
So we just need to prove that $\det N^{\prime\prime}=0$. This is equivalent to
say that $\det\tilde{N}=0$, where

$\displaystyle\tilde{N}=\begin{pmatrix}\left(\sum_{k=4}^{n}\frac{\zeta_{3}\zeta_{k}}{z_{3}-z_{k}}\right)(1,z_{3},\dots,z_{3}^{n-3})+\zeta_{3}(1,z_{3},\dots,z_{3}^{n-4})A^{-1}B\\\
(1,z_{4},\ldots,z_{4}^{n-3})\\\ \ldots\\\
(1,z_{n},\ldots,z_{n}^{n-3})\end{pmatrix}.$

Now, we observe that the first line has the form
$\sum_{k=4}^{n}\chi_{k}\zeta_{3}\zeta_{k}$, where $\chi_{k}$ is a row. So, it
is sufficient to prove that $\det\tilde{N}_{k}=0$ for each $k$, where
$\tilde{N}_{k}$ is obtained by replacing $\chi_{k}$ in place of the first row
of $\tilde{N}$. By the structure of the element of the matrix $\tilde{N}$, we
see that its determinant is cyclically symmetric in the $z_{k}$, $k>3$, so it
is sufficient to prove the vanishing result just for one given $k$, say $k=4$.
We also notice that in $A^{-1}B$ the common denominators in $A$ and $B$ cancel
out. It is also convenient to notice that of the rows of $B$ only the last
terms in the parenthesis of $B_{jb}$ contribute to the determinant, since the
first terms give rows that are combinations of the remaining rows. Therefore,
we can replace the matrix $A^{-1}B$ with $\tilde{A}^{-1}\tilde{B}_{4}$, where

$\displaystyle\tilde{A}=\begin{pmatrix}1&z_{4}&z_{4}^{2}&\ldots&z_{4}^{n-4}\\\
1&z_{5}&z_{5}^{2}&\ldots&z_{5}^{n-4}\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
1&z_{n}&z_{n}^{2}&\ldots&z_{n}^{n-4}\\\ \end{pmatrix}$

is a Vandermonde matrix (up to the change of sign in the second column), and

$\displaystyle\tilde{B}_{4}=\begin{pmatrix}0&-\zeta_{4}&-2\zeta_{4}z_{4}&\ldots&-(n-3)\zeta_{4}z_{4}^{n-4}\\\
0&0&0&\ldots&0\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
0&0&0&\ldots&0\end{pmatrix}.$

Thus, we see that

$\displaystyle\det\tilde{N}_{4}=\det\tilde{N}_{4}^{\prime}+\det\tilde{N}_{4}^{\prime\prime}\,,$

where

$\displaystyle\tilde{N}_{4}^{\prime}=\begin{pmatrix}\frac{\zeta_{3}\zeta_{4}}{z_{3}-z_{4}}(1,z_{3},\dots,z_{3}^{n-3})\\\
1,z_{4},\ldots,z_{4}^{n-3}\\\ \ldots\\\
1,z_{n},\ldots,z_{n}^{n-3}\end{pmatrix}$

and

$\displaystyle\tilde{N}_{4}^{\prime\prime}=\begin{pmatrix}\zeta_{3}(1,z_{3},\dots,z_{3}^{n-4})\tilde{A}^{-1}\tilde{B}_{4}\\\
1,z_{4},\ldots,z_{4}^{n-3}\\\ \ldots\\\
1,z_{n},\ldots,z_{n}^{n-3}\end{pmatrix}.$

We now prove that
$\det\tilde{N}_{4}^{\prime\prime}=-\det\tilde{N}_{4}^{\prime}$. Factoring out
$\tfrac{\zeta_{3}\zeta_{4}}{z_{3}-z_{4}}$ from the first row of
$\tilde{N}_{4}^{\prime}$ gives a Vandermonde matrix, so that the determinant
is equal to

(36)
$\displaystyle\det\tilde{N}_{4}^{\prime}=\frac{\zeta_{3}\zeta_{4}}{z_{3}-z_{4}}\prod_{3\leq
j<k\leq
n}(z_{k}-z_{j})=-\zeta_{3}\zeta_{4}\prod_{l=5}^{n}(z_{l}-z_{3})\prod_{4\leq
j<k\leq n}(z_{k}-z_{j}).$

We can now compute $\det\tilde{N}_{4}^{\prime\prime}$. To this end, we notice
that the entry number $k$ of the first row of $\tilde{N}_{4}^{\prime\prime}$
is $\sum_{a=1}^{n-3}z_{3}^{a-1}\tilde{A}^{-1}_{a1}(\tilde{B}_{4})_{1k}$. This
can be computed as follows. Since $\tilde{A}^{-1}_{a1}\det\tilde{A}$ is the
co-factor of the matrix $\tilde{A}$ in position $a1$, we see that
$\sum_{a=1}^{n-3}z_{3}^{a-1}\tilde{A}^{-1}_{a1}\det\tilde{A}$ is the
determinant of the matrix $\bar{A}$ obtained by replacing $z_{4}$ with $z_{3}$
in the first row of $\tilde{A}$. Since both $\tilde{A}$ and
$\tilde{A}^{\prime}$ are Vandermonde matrices, we get

$\displaystyle\sum_{a=1}^{n-3}z_{3}^{a-1}\tilde{A}^{-1}_{a1}=\frac{\det\tilde{A}^{\prime}}{\det\tilde{A}}=\prod_{j=5}^{n}\frac{z_{j}-z_{3}}{z_{j}-z_{4}},$

and so

(37) $\displaystyle\det\tilde{N}_{4}^{\prime\prime}=$
$\displaystyle-\zeta_{3}\zeta_{4}\frac{\prod_{l=5}^{n}(z_{l}-z_{3})}{\prod_{j=5}^{n}(z_{j}-z_{4})}\det\begin{pmatrix}(0,1,2z_{4},\dots,(n-3)z_{4}^{n-4})\\\
(1,z_{4},\ldots,z_{4}^{n-3})\\\ \ldots\\\
(1,z_{n},\ldots,z_{n}^{n-3})\end{pmatrix}.$

Now, all co-factors of the first line of the last matrix are of Vandermonde
type and, in particular, are thus divisible by $\prod_{j=5}^{n}(z_{j}-z_{4})$.
It follows that $\det\tilde{N}_{4}^{\prime\prime}$ is a polynomial in the
$z_{j}$ of the same degree as $\det\tilde{N}_{4}^{\prime}$, which vanishes
when $z_{j}=z_{3}$ (for some $j\geq 5$) or when $z_{j}=z_{k}$ (for some $5\leq
j<k\leq n$). Since the determinant $\det\tilde{N}_{4}^{\prime}$ is simply the
product of these linear factors, and $\det\tilde{N}_{4}^{\prime\prime}$ is a
polynomial of the same degree divisible by the same degree, it follows that
$\det\tilde{N}_{4}^{\prime\prime}/\det\tilde{N}_{4}^{\prime}$ is just a
constant, which we can compute by evaluating for any specific values of
$z_{j}$. For $z_{4}=0$ we get

(38) $\displaystyle\det\tilde{N}_{4}^{\prime\prime}=$
$\displaystyle-\zeta_{3}\zeta_{4}\frac{\prod_{l=5}^{n}(z_{l}-z_{3})}{\prod_{j=5}^{n}z_{j}}\det\begin{pmatrix}(0,1,0,\dots,0)\\\
(1,0,\ldots,0)\\\ (1,z_{5},\ldots,z_{5}^{n-3})\\\ \ldots\\\
(1,z_{n},\ldots,z_{n}^{n-3})\end{pmatrix}=\zeta_{3}\zeta_{4}\frac{\prod_{l=5}^{n}(z_{l}-z_{3})}{\prod_{j=5}^{n}z_{j}}\det\begin{pmatrix}(z_{n}^{2},\ldots,z_{5}^{n-3})\\\
\ldots\\\ (z_{n}^{2},\ldots,z_{n}^{n-3})\end{pmatrix}$ (39) $\displaystyle=$
$\displaystyle\zeta_{3}\zeta_{4}\prod_{l=5}^{n}(z_{l}-z_{3})\prod_{j=5}^{n}z_{j}\prod_{5\leq
j<k\leq n}(z_{k}-z_{j}),$

which coincides with the evaluation of $-\det\tilde{N}_{4}^{\prime}$, computed
in (36), for $z_{4}=0$.

##  Appendix C The $z_{1}=\infty$ case

Here we consider the $z_{1}=\infty$ case, or, equivalently, of
$(w\,|\,\eta)=(0\,|\,\eta_{1})$ in the coordinates
$(w\,|\,\eta)=\left(\tfrac{1}{z}\,,|\,\tfrac{\zeta}{z}\right)$ in the chart on
${\mathbb{P}}^{1|1}$ centered at $z=\infty$. We will deduce the resulting
expression (8) in that case of 1 by letting $z_{1}\to\infty$ in formula (7),
which gives the expression of the measure for a finite value of $z_{1}$. More
precisely, we will let $(w_{1}\,|\,\eta_{1})\to(0\,|\,\eta_{1})$ (or simply
$w_{1}\to 0$) for
$(w_{1}\,|\,\eta_{1})=\left(\tfrac{1}{z_{1}}\,|\,\tfrac{\zeta_{1}}{z_{1}}\right)$.

There is one subtlety, though. We want the vertex-operator insertion

(40)
$V_{1}(z_{1}|\zeta_{1})\left[\,\zeta[dz|d\zeta]\\!\bigm{|}_{\operatorname{NS}_{1}}\Bigm{|}\,[dz|d\zeta]\\!\bigm{|}_{\operatorname{NS}_{1}}\,\right]=V_{1}\left(\tfrac{1}{w_{1}}\Big{|}\tfrac{\zeta_{1}}{w_{1}}\right)\tfrac{1}{w_{1}}\left[\eta[dw|d\eta]\\!\bigm{|}_{\operatorname{NS}_{1}}\;\Bigm{|}\;[dw|d\eta]\\!\bigm{|}_{\operatorname{NS}_{1}}\right]$

at $\operatorname{NS}_{1}=\\{w-w_{1}+\eta\eta_{1}=0\\}$ in (6) to converge to
a finite vertex operator

$\widetilde{V_{1}}(0|\eta_{1})\Big{[}\,\eta[dw|d\eta]|_{\operatorname{NS}_{1}}\Big{|}[dw|d\eta]|_{\operatorname{NS}_{1}}\Big{]}$

as $w_{1}\to 0$, i.e., we want

$\lim_{w_{1}\to
0}V_{1}\left(\tfrac{1}{w_{1}}\Big{|}\tfrac{\zeta_{1}}{w_{1}}\right)\tfrac{1}{w_{1}}=\widetilde{V_{1}}(0|\eta_{1})\,.$

Here we again drop the vacuum expectation value brackets $\langle\,\rangle$
from the notation.

On the other hand,

$d\nu=-\frac{1}{2^{n-2}w_{1}}\left(z_{3}w_{1}-1-\tfrac{1}{2}\zeta_{3}\eta_{1}\right)\left(z_{3}-z_{2}-\tfrac{1}{2}\zeta_{3}\zeta_{2}\right)[dz_{4}\dots
dz_{n}\,|\,d\zeta_{3}\dots d\zeta_{n}]$

and hence

$\displaystyle V_{1}(z_{1}|\zeta_{1})d\nu$
$\displaystyle=-\frac{V_{1}\big{(}\frac{1}{w_{1}}\mid\frac{\zeta_{1}}{w_{1}}\big{)}}{2^{n-2}w_{1}}\left(z_{3}w_{1}-1-\tfrac{1}{2}\zeta_{3}\eta_{1}\right)\left(z_{3}-z_{2}-\tfrac{1}{2}\zeta_{3}\zeta_{2}\right)[dz_{4}\dots
dz_{n}\,|\,d\zeta_{3}\dots d\zeta_{n}]$
$\displaystyle\to\quad\widetilde{V_{1}}(0|\eta_{1})\tfrac{1}{2^{n-2}}\left(1+\tfrac{1}{2}\zeta_{3}\eta_{1}\right)\left(z_{3}-z_{2}-\tfrac{1}{2}\zeta_{3}\zeta_{2}\right)[dz_{4}\dots
dz_{n}\,|\,d\zeta_{3}\dots d\zeta_{n}]$

as $w_{1}\to 0$. In 1, we use $V_{1}$ in lieu of $\widetilde{V_{1}}$ as well
as slightly abuse notation and denote the above limit by
$V_{1}(0|\eta_{1})d\nu$.

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