Pragmatically, we want to define the volume of $\cR$ as $\vol(\cR')/\vol(\uU(1))$. For this purpose, we need volume forms on $\cR'$ and $\uU(1)$, or at least we need the
ratio of a volume-form on $\cR'$ and a volume form on $\uU(1)$.   We will see that the torsion gives precisely such a ratio.

§.§.§ Torsion of the Three-Holed Sphere

We are now ready to compute the torsion $\tau_Y$ of the three-holed sphere $Y$.    We will follow the general procedure described in section 3.4.2 of [14].   We will
review a few steps, emphasizing the points that are different for $\SU(1,1|1)$, but we will not give a self-contained explanation.

Since the torsion can be computed using any triangulation (or more generally any cell decomposition) of
a manifold, we can use the very simple description of a three-holed sphere shown in the right panel of fig. <ref>, with one two-cell, three boundary one-cells, and
one vertex $P$.   Let us discuss
the situation with an arbitrary gauge group $G$ before specializing to $G=\SU(1,1|1)$.
If we divide the space of flat connections on $Y$
not by all gauge transformations but only by gauge transformations that are trivial at $P$, we get a supermanifold $\h\cR\cong G\times G$ that is parametrized
by $U,V$ without any constraint or equivalence.   Then to get the moduli space $\cR$, we divide by $G$ acting on $\h\cR$ by conjugation:

The definition of the torsion requires a choice of a left- and right-invariant measure $\vol_G$ on the $G$ manifold; this is equivalent to a choice of invariant measure $\vol_\g$
on the Lie algebra $\g$ of $G$.   Such a choice determines a measure on any copy of the group manifold,
so in particular it determines measures $\vol_G(U)$ or $\vol_G(V)$ on the copies of $G$ parametrized by $U$ or $V$.
Similarly, if we define $U=R U_0 R^{-1}$, $V=R V_0 R^{-1}$, with some choices of $U_0$, $V_0$ (which in our application will be given by eqn.  (<ref>)), one
has a measure $\vol_G(R)$ on the copy of $G$ parametrized by $R$.
The torsion can be used to define
measures $\tau_Y$ and $\hat\tau_Y$ on $\cR$ and $\h\cR$, respectively.   One finds
τ_Y=_G(U)_G(V) and
τ_Y=τ_Y/_G(R) =_G(U)_G(V)/_G(R). It suffices to evaluate this formula for $\tau_Y$ at $R=1$, since every flat connection on $Y$ can be described by some pair $(U_0,V_0)$.

When the quotient $\cR =\h \cR/G$ is sufficiently nice, which happens for the examples $\SL(2,\R)$ and $\OSp(1|2)$ studied in [14], the quotient $\vol_G(U)\vol_G(V)/\vol_G(R)$
is a measure on $\cR$ and this is the appropriate moduli space measure.   For the example of $\SU(1,1|1)$ that we aim to understand here, it turns out instead
that $\vol_G(U)\vol_G(V)/{\vol_G(R)}$ is the ratio of a measure on $\cR'$, defined at the end of section <ref>, and a measure on the $\uU(1)$ $R$-symmetry group.
We interpret this to mean that, as essentially stated in section <ref>, integration over $\cR$ should be defined as integration over $\cR'$ divided by the volume of $\uU(1)$.

To evaluate the torsion, we first describe the volume form that we will use on $\mathfrak{su}(1,1|1)$.
Consider an arbitrary Lie algebra element $x \in \mathfrak{su}(1,1|1)$ with components $x= x_e {\sf e} + x_f {\sf f} + x_h {\sf h} + x_z \hq \,{\sf z} + \sum_{j=1}^4 x_j {\sf q}_j$. (Notice the factor of $\hq$ in the normalization in $x_z$, such that the range of $x_z$ is $2\pi$ in the $\hq$-fold of $\SU(1,1|1)$ we are working with.)
We choose the volume form on the Lie algebra to be
_ = 4 [ x̣_e x̣_f x̣_h x̣ _z | x̣_1  x̣_2  x̣_3 x̣_4],
and this also determines the volume form on the group.     As explained in [14], a symbol such as $[\d y|\d\eta]$ (where $y$ is bosonic and $\eta$ is fermionic)
represents the measure associated to a  basis
$\d y|\d\eta$ of 1-forms on a $\Z_2$-graded vector space.
A crucial fact is that in contrast to bosons, measures for fermions transform  oppositely to differential forms, in the sense that for[The reason
for the absolute value $|\lambda|$ in eqn (<ref>) in the bosonic case is that for a bosonic variable $y$, the measure $[\d y]$, as opposed to the 1-form $\d y$, is invariant under
$y\to -y$.]  $\lambda
\in\C$,
[λỵ|η̣]=|λ| [ỵ|η̣],    [ỵ|λη̣]=λ^-1[ỵ|η̣].

Evaluating ${\rm vol}_G(R)$ is immediate, because it suffices to evaluate this measure at $R=1$.  If we write $R=1+r$ then, at $R=1$,  $R^{-1} \d R = \d r$ and
$\vol_G(R)$ reduces to
vol_G(R)  =4 [ ṛ_e ṛ_f ṛ_h ṛ _z | ṛ_1  ṛ_2  ṛ_3 ṛ_4] .   Next we evaluate ${\rm vol}_G(U)$ for $U=RU_0 R^{-1}$, following the procedure outlined in  [14].
The quantity $U^{-1} \d U$ will get contributions both from the differential of $R$, resulting in terms proportional to components of $\d r$ which will cancel similar factors in
$\vol_G(R)$ when evaluating eqn. (<ref>), and contributions from the differential of $U_0$.   When
we repeat this for $V=R V_0 R^{-1}$,  we will be left finally with  $\tau=F[\d a \d b \d \kappa \d \phi_a \d \phi_b | \d\psi_1\d \psi_2 \d\psi_3\d \psi_4]/[\d r_z]$ for some function $F$.
As anticipated at the end of section
<ref>, we interpret the result as the ratio of a volume form on $\cR'$ and a volume form on $\uU(1)$.

The identity  (<ref>) can be greatly simplified by exploiting the fact that
the matrices $U_0$ and $V_0$ are respectively upper and lower triangular. Begin by evaluating the $q_3$ component of $U^{-1} \d U$ and the $q_2$ component of $V^{-1} \d V$:
(U^-1 Ụ )_3 = -(e^a/2-ıϕ_a+1)ṛ_3,    (V^-1 Ṿ )_2 = -(e^b/2+ıϕ_b+1)ṛ_2.
Since these expressions are proportional to $\d r_2$ and $\d r_3$, we can in evaluating the numerator of eqn. (<ref>) set to zero any other occurrences
of $\d r_2$ or $\d r_3$.  Then, we can cancel these factors of $\d r_2$ and $\d r_3$ with the factors
of $\d r_2$ and $\d r_3$ that appear in $\vol_G(R)$ in the denominator in eqn. (<ref>).   After this cancellation, we can forget about $\d r_2 $ and $\d r_3$ and set
them to zero.   We are left with a factor $(e^{\frac{a}{2}-\i\hq\phi_a}+1)^{-1}(e^{\frac{b}{2}+\i\hq\phi_b}+1)^{-1}$.   Note that these factors are inverted because for fermions,
measures transform oppositely to differential forms, as noted in eqn. (<ref>).
In the next step, evaluate the bosonic components
(U^-1 Ụ )_f = (e^a-1)ṛ_f ,    (V^-1 Ṿ )_e = (e^b-1)ṛ_e.
Because these expressions are proportional to $\d r_f$ and $\d r_e$, we can set to zero any other occurrences of $\d r_f$ and $\d r_e$ in the numerator
of eqn. (<ref>).   Then we can cancel $\d r_e$ and $\d r_f$  with similar occurrences in the denominator and set them to zero.
We are left this time with a factor $(e^a-1)(e^b-1)$.    Next evaluate
(U^-1 Ụ )_1 = -(e^-a/2-ıϕ_a+1)ṛ_1,
(V^-1 Ṿ )_4 = -(e^-b/2+ıϕ_b+1)ṛ_4.
We can set to zero any other occurrences of $\d r_1$ and $\d r_4$ in the numerator in eqn. (<ref>), and  cancel $\d r_1$ and $\d r_4$ with similar factors in the denominator.
We are left with a factor $(e^{-\frac{a}{2}-\i\hq\phi_a}+1)^{-1}(e^{-\frac{b}{2}+\i\hq\phi_b}+1)^{-1}$.
Next evaluate
(U^-1 Ụ )_h = ạ/2,    (V^-1 Ṿ )_h = -ḅ/2.
Since these expressions are proportional to $\d a$ and $\d b$, we can set to zero any occurrences of $\d a$ and $\d b$ that we find in the rest of the evaluation.  We are left with a factor
$-\frac{1}{4}\d a\d b$.
After dropping $\d a$ and $\d b$ where they appear elsewhere, we  obtain $(U^{-1} \d U )_z =\d\phi_a$ and $(V^{-1} \d V )_z =   \d\phi_b$, so we can set to zero any further occurrences of $\d\phi_a$ and $\d\phi_b$
that we may encounter, and we remain with a factor $\d\phi_a\d\phi_b$.   Then $(V^{-1} \d V )_f = -2 e^{-\frac{b}{2}} \d r_h$, allowing us to cancel $\d r_h$
with a similar factor in the denominator.
The final bosonic component is  $(U^{-1} \d U)_e =  e^{-\frac{a}{2}} \d\kappa$.
Now we can evaluate the remaining fermionic components

(U^-1  Ụ )_4 =ψ̣_4 - ıψ_4  ṛ_z ,    (V^-1 Ṿ )_1 =ψ̣_1 +  ıψ_1 ṛ_z.

(U^-1 Ụ )_2 =ψ̣_2 - ıψ_2 ṛ_z ,    (V^-1 Ṿ )_3 =ψ̣_3 +ıψ_3 ṛ_z.
The measure coming from (<ref>) and (<ref>) simplifies via $[\d\psi_1 +  \i \psi_1 \d r_z,\d\psi_2 - \i  \psi_2 \d r_z,\d\psi_3 +\i  \psi_3 \d r_z,\d\psi_4 - \i \psi_4  \d r_z] =[\d\psi_1\d\psi_2\d\psi_3\d\psi_4]$, since the Berezinian of the indicated change of basis
is 1.   We canceled almost all factors in the denominator of (<ref>), leaving a leftover factor of $[\d r_z]$ in the denominator.

After all these simplifications, keeping track of all the factors generated along the way, the identity (<ref>) becomes
8sinha/2 sinhb/2  [ạ   ḅ   κ̣  ϕ̣_a   ϕ̣_b | ^̣4ψ] /(e^a/2-ıϕ_a+1)(e^-a/2-ıϕ_a+1)(e^b/2+ıϕ_b+1)(e^-b/2+ıϕ_b+1)
where to shorten the notation we defined $[\d^4 \psi ] \equiv [\d \psi_1 \d \psi_2 \d \psi_3 \d \psi_4]$.
As promised, $\tau$ is the ratio of a measure on $\cR'$ and a measure on $\uU(1)$.

The result can be put in a more useful form by eliminating $\kappa$ in favor of the other moduli via eqn. (<ref>).
We have  $\d \kappa = \partial_c \kappa \, \d c + \ldots$, where the omitted terms involve derivatives with respect to other arguments and
do not contribute when this substitution is made for $\d\kappa$ in eqn. (<ref>).   So
8 (∂_c κ)sinha/2 sinhb/2 [ạ   ḅ   c̣  ϕ̣_a   ϕ̣_b | ^̣4ψ] /(e^a/2-ıϕ_a+1)(e^-a/2-ıϕ_a+1)(e^b/2+ıϕ_b+1)(e^-b/2+ıϕ_b+1) .
We are going to make two further adjustments of this formula to put it in its final form.

First, we have to decide what to do with the factor $1/[\d r_z]$.  The most
obvious thing to do, suggested by the discussion at the end of section <ref>, is to interpret the factor $1/[\d r_z]$ to mean that we should divide by
the volume of $\uU(1)$, computed using the measure $[\d r_z]$.  This will
replace $1/[\d r_z]$ with $1/2\pi$.  In
appendix <ref>, we consider the simpler case  of JT gravity coupled to a $\uU(1)$ gauge field, described by a $BF$ theory of
${\rm PSL}(2,\mathbb{R})\times \uU(1)$, and verify that this procedure is correct.

Second, the formulas that we have presented so far
have  been written in terms of $\phi_a$ and $\phi_b$, with no mention of $\phi_c$.
We can write the result  somewhat more symmetrically by “integrating in” $\phi_c$ with the identity,

\begin{equation}
1 = \int \d\phi_c~ \delta( \phi_a + \phi_b + \phi_c - F),
\end{equation}

where the coefficients in $F= f_{ij} \psi_{i} \psi_j + f_4 \psi_1 \psi_2 \psi_3 \psi_4$ were presented in eqn. (<ref>).
Making use of this, and replacing $1/[\d r_z]$ with $1/2\pi$, the final result for the torsion, which we will use from now on, is

τ_Y =-1/2π 2e^ıϕ_a-ıϕ_bsinha/2 sinhb/2 sinhc/2 (1+K)/(cosha/2 + cosϕ_a )(coshb/2 + cosϕ_b)δ(ϕ_a + ϕ_b + ϕ_c- F) [ạ ϕ̣_a  ḅ ϕ̣_b c̣ ϕ̣_c  | ^̣4ψ].
We remind the reader that to derive this expression we singled out a boundary with holonomy $W$. This was used in the definition of the fermionic moduli, which is why the formula
(<ref>) is not symmetrical in $a,b,c$.
Nevertheless, we show in sections <ref> and <ref>  that the permutation symmetry is restored, as it should be, after integrating over the fermionic moduli.
The formula (<ref>) is actually symmetrical in $a$ and $b$, up to a charge conjugation operation that reverses the signs of $\phi_a,\phi_b$,
because our definition of the fermionic moduli was symmetrical in $U$ and $V$, similarly
to what was explained in eqns. (3.96-7) of [14].

§.§.§ The Torsion of a Circle

Let us now discuss what happens when we glue together two manifolds with boundary  $Y_1$ and $Y_2$ along a common boundary circle $S$ to make a connected manifold $Y'$.
In quantum field theory in general, the gluing would be implemented by multiplying the path integrals on $Y_1$ and $Y_2$ and
summing over physical states propagating through $S$.   In standard approaches to $BF$ theory,
the sum can be carried out by integrating over the conjugacy class of the holonomy of a flat connection around $S$.   For gauge group $\SU(1,1|1)$, this
would mean an integration over a pair of parameters $a,\phi_a$ that appear as eigenvalues of the holonomy.

In the approach via the torsion, the appropriate gluing procedure is to multiply the torsions of $Y_1$ and $Y_2$ and divide by the torsion of $S$:
τ_Y'=τ_Y_11/τ_Sτ_Y_2. This procedure is explained heuristically in [14] (see eqn. (3.18)) and one can find a more detailed explanation in, for example, section 4 of [53].
The rough idea is that  $\tau_{Y_1}$ and $\tau_{Y_2}$ are both measures for the holonomy along $S$ (as well as other variables), and to avoid overcounting,
we have to divide by $\tau_S$, which is also such a measure.

The general procedure to compute $\tau_S$ for any gauge group is outlined in section 3.4.3 of [14].
For any $G$, let $U$ be the holonomy of a flat connection around $S$ and let $\h U$ be the linear transformation of the Lie algebra $\g$ of $G$ given
by $\h U(s)=U s U^{-1}$.  Then $\tau_S=\Ber'(\h U-1)$, where $\Ber'$ is a sort of superanalog of the determinant.   To make this more concrete in the case of $G=\SU(1,1|1)$,
up to conjugation we can assume that

U = diag (-e^a/2 + ıϕ,-e^-a/2 + ıϕ, e^2 ıϕ).
$\h U$ commutes with a decomposition of $\g=\mathfrak{su}(1,1|1)$ as $\g=\g^0\oplus \g^\perp$, where $\g^0$ is the subalgebra of diagonal matrices and $\g^\perp$ is the space
of strictly off-diagonal matrices.   The torsion has a corresponding factorization
τ_S=τ_S^⊥·τ_S^0.Let $\h U^0$, $\h U^\perp$ be the restrictions of $\h U$ to $\g^0$, $\g^\perp$.    Then $\h U^\perp$ is invertible and $\tau_S^\perp =\Ber'(\h U^\perp-1)$.   The definition
of $\Ber'$ is such that  if $M=\left(\begin{array}{c|c} M_b& 0 \\ \hline 0 & M_f\end{array}\right) $ is a linear transformation of a $\Z_2$-graded vector space with invertible bosonic
and fermionic blocks $M_b$, $M_f$, then $\Ber'(M)=|\det M_b|/\det M_f$.
In the present case, this leads to   τ_S^⊥=  Ber'(U^⊥-1) = sinh^2 a/2/(cosha/2+ cosϕ)^2.On the other hand, $\h U^0-1=0$, which leads to $\tau_S^0$ being a simple quotient of measures,
τ_S^0=ạ  ϕ̣  ϱ̣^-1   φ̣^-1. Here $a,\phi$ are the parameters of $U$ introduced in eqn. (<ref>), and $\varrho$, $\varphi$ are length-twist and $\uU(1)$-twist parameters that we already encountered in
discussing the double trumpet.    This formula generalizes eqn. (3.42) of [14].

The final answer for the torsion is

\begin{equation}\label{eq:torsioncirc}
\tau_S= \frac{\sinh^2 \frac{a}{2}}{(\cosh \frac{a}{2} + \cos \hq\phi)^2}\, \d a \, \d \phi \, (\d \varrho)^{-1} \, (\d \varphi)^{-1}.
\end{equation}

It is perhaps more to the point to write this as a formula for $1/\tau_S$, since that is what appears in the gluing law (<ref>):

1/τ_S =  (cosha/2 + cosϕ)^2/sinh^2 a/2 ạ^-1ϕ̣^-1ϱ̣  φ̣. This formula can be interpreted, in part, as follows.   Both $\tau_{Y_1}$ and $\tau_{Y_2}$ are proportional to $\d a\d\phi$, and $1/\tau_S$ removes an unwanted extra factor
of $\d a\d\phi$ that would be present in the product $\tau_{Y_1}\tau_{Y_2}$.   On the other hand, gluing introduces new parameters
$\varrho$ and $\varphi$, and $1/\tau_S$ generates a measure for integration over these parameters.

§.§.§ General Case: Measure on Moduli Space

Having computed the torsion of a three-holed sphere and a circle, we can assemble the pieces and compute the  moduli space measure of $\cN=2$ JT gravity on any oriented
NAdS$_2$ manifold.

For simplicity, we begin with a discussion of  surfaces without boundary. Any closed oriented surface $C$ of genus $g$ can be assembled by gluing together a set $T$  of $2g-2$ three-holed spheres $Y_t$, $t\in T$. These three-holed spheres have to be glued along a set $U$ of $3g-3$ circles $S_u$, $u\in U$.
Two three-holed spheres (or two boundaries of the same three-holed sphere) are glued along each $S_u$.

We denote as $\T_g$ the moduli space of flat connections modulo gauge transformations on a genus $g$ surface $C$ without boundary, and we write $\mu$
for the corresponding moduli space measure.
The measure $\mu$ is simply the tensor product of the $\tau_{Y_t}$ for all $t\in T$ times the tensor product of the $\frac{1}{\tau_{S_u}}$ for all $u\in U$.  This follows inductively
from the gluing formula.
μ_g,n=0=∏_t∈Tτ_Y_t∏_u∈U1/τ_S_u. Looking at eqn. (<ref>) for $\tau_{Y_t}$ and eqn. (<ref>) for $1/\tau_{S_u}$, we see that significant cancellation occurs in this product between factors
coming from three-holed spheres and factors coming from gluing along circles.

eqn. (<ref>) is actually only a formal expression.   As seen in eqn. (<ref>), $\tau_Y$ is proportional to a delta function of a sum $\phi_a+\phi_b+\phi_c$
of holonomy angles, modulo nilpotent corrections that lead to derivatives of the delta function.   If we actually evaluate $\mu_{g,n=0}$ using eqn. (<ref>), then,
since a closed surface without boundary has no external holonomies, after multiplying together all these delta functions and integrating over the $\phi$'s,  we will be left
over with a factor of $\delta(0)$ (or possibly a derivative thereof, $\delta^{[n]}(0)$ for some $n$).   It is quite likely that in $\N=2$ JT supergravity, the partition function of a closed
surface is not well-defined.   What are well-defined, though subtle, and what we need for understanding
holographic duality, are partition functions for surfaces with NAdS external boundaries.
Also well-defined, and closely related, are volumes for surfaces with geodesic boundaries and with specified holonomy parameters $b,\phi$ on each boundary.

To describe the appropriate measure in the case of an NAdS boundary, we have to know how to glue a trumpet onto a three-holed sphere.
We have described in section <ref>
the trumpet path integral $Z_{\rm tr}(b,\phi)$,
with the Schwarzian mode on the asymptotic boundary and standard boundary conditions on the interior geodesic;
we also have computed the path integral on a three-holed sphere using torsion boundary conditions.   We need to know how to compare or combine results computed with the two
different types of boundary condition.
It is slightly tricky to do this because  a convenient way to combine the Schwarzian path integral and the torsion in one calculation is not known.   Instead we proceed indirectly
as in [14].   In the standard approach, the measure for the double trumpet is[For brevity, we do not show the external parameters $\alpha_i,\beta_i$ characterizing the
Z_tr(b,ϕ)  ḅϕ̣ϱ̣φ̣ Z_tr(b,ϕ). If we knew how to calculate a trumpet path integral with the Schwarzian mode on the outer boundary and torsion boundary conditions on the inner boundary, we would
get a measure $\t Z_{\rm tr}(b,\phi)\,\d b\d\phi. $    Then to get the appropriate measure for the double trumpet, we would glue together two trumpets using the gluing
law for the torsion, meaning that we would multiply the torsions of the two trumpets and divide by $\tau_S$, the torsion of a circle where the gluing occurs.   This would give
Z_tr(b,ϕ)ḅϕ̣1/τ_SZ_tr(b,ϕ) ḅϕ̣. Using eqn. (<ref>)  for $1/\tau_S$, this becomes
Z_tr(b,ϕ) (coshb/2+cosq ϕ)^2/sinh^2b/2  ḅ ϕ̣ϱ̣φ̣  Z_tr(b,ϕ). Comparing these formulas, we get
Z_tr(b,ϕ) = Z_tr(b,ϕ) sinhb/2/coshb/2+cosq ϕ.

Now consider a Riemann surface $C$ of genus $g$ with $n$ boundaries, all of NAdS type.   Thus $C$ is made by gluing $n$ trumpets onto an internal manifold $C_\intt$ along
$n$ circles $S_u$, $u\in U_\ext$, where $U_\ext$ is the set of boundaries of $C_\intt$.   Write $\tau_{C_\intt}$ for the torsion measure of the moduli space $\cR_{g,n}$ of flat connections
on $C_\intt$.  The gluing law for the torsion tells us
that the $\N=2$ JT supergravity path integral on $C$ is the integral over $\cR_{g,n}$ of
∏_u∈U_(  Z_tr(b_u,ϕ_u) ḅ_uϕ̣_u 1/τ_S_u ) ·τ_C_.Using eqn. (<ref>) for $\t Z_{\rm tr}(b,\phi)$ and also eqn. (<ref>) for $1/\tau_{S}$, this becomes
∏_u∈U_ Z_tr(b_u,ϕ_u) coshb_u/2+cosϕ_u/sinhb_u/2 ϱ̣_u φ̣_u ·τ_C_. In the approach of [6], the expected answer is
∏_u∈U_ Z_tr(b_u,ϕ_u) ḅ_u ϕ̣_u ϱ̣_u φ̣_u ·μ_g,n(b_1,ϕ_1;⋯;b_n,ϕ_n) ,where $\mu_{g,n}(b_1,\phi_1;\cdots; b_n,\phi_n)$ is the Riemannian or symplectic measure on the moduli space $\T_{g,n}(b_1,\phi_1;\cdots;b_n,\phi_n)$
of flat connections on $C_\intt$ with the holonomy
around $S_u$ specified to be $b_u,\phi_u$.
Comparing the two formulas, we find a formula for $\mu_{g,n}(b_1,\phi_1;\cdots;b_n,\phi_n)$:
μ_g,n(b_1,ϕ_1;⋯;b_n,ϕ_n)=∏_u∈U_      coshb_u/2+cosϕ_u/sinhb_u/21/ḅ_uϕ̣_u
τ_C_. Here the factor of $1/\d b_u\d\phi_u$ is an instruction to remove the factor $\d b_u\d\phi_u$ that is present[This factor is present in $\tau_{C_\intt}$ because
$\tau_{C_\intt}$ is a measure on $\cR_{g,n}$,  the moduli space of flat connections on $C_\intt$ with no restriction on the boundary holonomies except that they are hyperbolic;
it is absent in $\mu_{g,n}$, because $\mu_{g,n}$ is a measure on the moduli space $\T_{g,n}$ of flat connections on $C_\intt$ with specified conjugacy classes of the boundary
in $\tau_{C_\intt}$.
$C_\intt$ can be built by gluing together $2g-2+n$ three-holed spheres $Y_t$, $t\in T$, along $3g-3+n $ circles $S_u$, $u\in U_\intt$.   Then we can write a formula
for $\tau_{C_\intt}$ that is precisely analogous to eqn. (<ref>):
τ_C_=∏_t∈T τ_Y_t∏_u∈U_1/τ_S_u. Finally, using (<ref>) for $\tau_{Y_t}$ and eqn. (<ref>) for $1/\tau_{S_u}$, and inserting eqn. (<ref>) in eqn. (<ref>),
we get a more explicit formula for $\mu_{g,n}$:

μ_g,n(b_1,ϕ_1;⋯;b_n,ϕ_n)    =    ∏_u∈U_int [ạ_u ϕ̣_u] ·[ϱ̣_u φ̣_u]∏_t∈T
-2e^ıϕ_a_t - ıϕ_b_t(coshc_t/2 + cosϕ_c_t) /(2π)          ×(1+K_t) δ(ϕ_a_t+ϕ_b_t+ϕ_c_t-F_t)  [^̣4 ψ_t] .
In writing this expression we used the same labels for each $t\in Y$ that we used in section <ref>. The three lengths are
$a_t$, $b_t$ and $c_t$, the three $\uU(1)$ holonomies are $\phi_{a_t}$, $\phi_{b_t}$ and $\phi_{c_t}$ and the fermionic moduli are  $\psi_t$.
The nilpotent terms $K_t$ and $F_t$ are implicitly given by the functions appearing in (<ref>) and (<ref>) involving only the parameters of $t\in Y$.

Since $C$ is assumed to be oriented, the measure (<ref>) will equal  the Weil-Petersson symplectic measure  of the moduli space $\mathcal{T}_{g,n}$ (assuming that the symplectic
structure is normalized properly).   $\T_{g,n}$ as we have defined it so far is a version of $\N=2$ Teichmüller space.[For an exploration
of  $\cN=2$ Teichmüller theory and the associated Weil-Petersson structure, see [27].]  To get the corresponding moduli space $\mM_{g,n}$ of $\N=2$
super Riemann surfaces, we have to divide $\T_{g,n}$ by the mapping class group.   Since $BF$ theory – and its partition function, the torsion – is diffeomorphism-invariant,
$\mu_{g,n}$ descends to a measure on $\mM_{g,n}$.
Finally, we can define formally the volume of $\mM_{g,n}$:
V_g,n = ∫__g,n μ_g,n.
Although we will be able to compute these integrals and in some sense they certainly deserve to be called volumes, actually their interpretation is subtle.
That is because $\mM_{g,n}$ is not really a supermanifold, partly because of the constraint on the external $R$-symmetry holonomies, and partly because dividing by global
$R$-symmetries is not a very nice operation, as
was explained at the end of section <ref>.
We will see in actual computations, for example eqn. (<ref>) and generalizations calculated later, that these “volumes” in their dependence on the $R$-symmetry
holonomies are derivatives of a delta function.   So for given external holonomies they are either zero or infinite; they cannot be interpreted literally as volumes in any usual sense.
To get actual numbers that would be volumes in a conventional sense, one has to integrate over one or more of the external holonomies to eliminate the delta function.
In fact, we will do just that in a particular way when we use these volumes as inputs to compute the partition function $Z_{g,n}$ via eqn. (<ref>) below.

Geometrically, integrating over external holonomies is similar to treating them as additional moduli.   When one does this, one replaces $\mM_{g,n}$ with a space $\X_{g,n}$
that is still not a supermanifold because the operation of dividing by global $R$-symmetries is not nice, but that at least locally can be interpreted as $\X'_{g,n}/\uU(1)$, where $\X'_{g,n}$ is
a supermanifold.   Then the “volume” of $\X_{g,n}$ is plausibly a quotient ${\rm vol}(\X'_{g,n})/{\rm vol}(\uU(1))$.

In this paper, it will not be important to understand the precise sense in which $V_{g,n}$ is a volume.   More relevant is that we can calculate $V_{g,n}$ (after learning in
section <ref> how to deal with the mapping class group), and that we can express the JT gravity partition function, which in turn can be compared to the matrix
model predictions, in terms of $V_{g,n}$.
To get the formula  for $Z_{g,n}$ that is analogous to that of [6] in the bosonic case,
we have to integrate the measure (<ref>) over ${\mM}_{g,n}$.   This replaces $\mu_{g,n}$ by $V_{g,n}$.
In addition, since $V_{g,n}$ does not depend on the gluing parameters,
we can explicitly integrate over $\varrho_u$ and $\varphi_u$, giving for each $u\in U_\ext$ a factor $2\pi b_u$.
Finally we get

Z_g,n= [ ∏_u=1^n ∫_0^∞b_uḅ_u ∫_0^2π2πϕ̣_u Z_tr (β_u,α_u;b_u)] V_g,n(b_1,…, b_n),
where ${\sf b}=(b,\phi)$ and $\overline{{\sf b}}=(b,-\phi)$.

will now do a direct calculation of $V_{0,3}$,  enabling us to justify the formula (<ref>)  that was used in section <ref> to match with the matrix model prediction for
To compute the volumes in general, we will in section <ref> use the preceding formulas as input to develop a Mirzakhani-style recursion relation.
This will give general results for the volumes and the partition functions that can be compared to random matrix theory.

§.§ Example: Computation of $V_{0,3}$

In bosonic JT gravity, $V_{0,3}=1$, because $\mM_{0,3}$, the moduli space of geodesic three-holed spheres with specified boundary lengths,  consists of a single point.  In $\N=1$ JT supergravity, $V_{0,3}=0$,
because of fermionic moduli.  The case of $\cN=2$ is more involved and more interesting.

$\mM_{0,3}$  has dimension $0|4$ with four fermionic moduli $\psi_1,\ldots, \psi_4$. We denote the three boundary lengths now by $b_1$, $b_2$ and $b_3$ and the corresponding $\uU(1)$ holonomies by $\phi_1$, $\phi_2$ and $\phi_3$. The measure $\mu_{0,3}$ on $\mM_{0,3}$ is given by eqns.  (<ref>) and (<ref>).   Explicitly
V_0,3    =    ∫__0,3 μ_0,3    =    -2e^ıϕ_1 - ıϕ_2(coshb_3/2 + cosϕ_3) /(2π)∫(1+K) δ(ϕ_1+ϕ_2+ϕ_3-F)  [^̣4 ψ_t].
This particular calculation is  relatively simple because the mapping class group of a three-holed sphere is trivial. If we did not have the nilpotent contributions $K= \sum_{i<j} k_{ij} \psi_i \psi_j  +k_4 \psi_1\psi_2\psi_3\psi_4$ and $F=\sum_{i<j} f_{ij} \psi_i \psi_j + f_4 \psi_1\psi_2\psi_3\psi_4$ that
were presented in eqns. (<ref>) and (<ref>), $V_{0,3}$ would vanish due to the rule of fermionic integration $\int \d \psi =0$. To obtain a non-zero answer requires a term proportional to the product $\psi_1\psi_2\psi_3\psi_4$.

We define the fermionic measure by $\int [\d^4 \psi] \psi_1 \psi_2 \psi_3 \psi_4 = 1$, and therefore

\begin{align}
\int [\d^4 \psi]  ~\big(1 + K\big)~\delta\Big(\sum_{i=1}^3\phi_i - F\Big) = & k_4 \,\delta\Big(\sum_{i=1}^3\phi_i \Big) \nonumber \\
&- (f_{12} k_{34} + f_{14} k_{23} + f_{23} k_{14} + f_{34} k_{12} + f_4  )  \delta'\Big(\sum_{i=1}^3\phi_i\Big)\nonumber\\
& + (f_{12} f_{34}+ f_{14} f_{23}) \delta''\Big(\sum_{i=1}^3\phi_i\Big).    \label{eqn:fermint2}
\end{align}

Using (<ref>) and (<ref>) for the coefficients, the formula for $V_{0,3}$  simplifies considerably:[We would like to thank D. Stanford for finding a mistake in an earlier version of this calculation.]

\begin{eqnarray}
V_{0,3} &=& -\frac{1}{2\pi}\frac{1}{4\hq^{\,2}} \Big[ G(\phi_1+\phi_2,\phi_3) \delta''(\phi_1 + \phi_2 + \phi_3) +2\partial_{\phi_1}G(\phi_1+\phi_2,\phi_3) \delta'(\phi_1 + \phi_2 + \phi_3) \nn
&&~~~+ \partial^2_{\phi_1}G(\phi_1+\phi_2,\phi_3) \delta(\phi_1 + \phi_2 + \phi_3)\Big],\label{eqn:V03final222}
\end{eqnarray}

where we define
G(ϕ,ϕ_3) =coshb_3/2 + cosϕ_3/coshb_3/2 + cosϕ.
Surprisingly, the three terms in eqn. (<ref>)   can be neatly combined to a single delta function:

\begin{equation}\label{surprising}
G(\phi,\phi_3) \delta''(\phi_3 + \phi) + 2 \partial_\phi  G(\phi,\phi_3) \delta'(\phi_3 + \phi)+\partial^2_\phi  G(\phi,\phi_3) \delta(\phi_3 + \phi)= \delta''(\phi_3 + \phi).
\end{equation}

One can prove this identity easily by multiplying by  an arbitrary test function and integrating, and then integrating by parts.  The only property of $G(\phi,\phi_3)$ that is required
is that $G(\phi_3,\phi_3)=1$.  By virtue of eqn. (<ref>), the Weil-Petersson volume of the $\cN=2$ hyperbolic three-holed sphere is equal to

V_0,3(b_1, b_2, b_3) = - 1/2π 1/4^  2 δ”(ϕ_1+ϕ_2+ϕ_3).
This precisely justifies eqn.  (<ref>) and therefore justifies claims made in section <ref> that follow from this formula.
It might be possible to define the fermionic moduli in a way that enables one to get this simple result without  the complicated intermediate steps, but it is not immediately clear how to do that.

In the next section we will find a recursion relation for $V_{g,n}$, and the formula for $V_{0,3}$  will be an initial condition in the recursion. This will extend the match with the random matrix ensemble from the three-holed sphere to an
arbitrary oriented surface.

§ TOPOLOGICAL RECURSION FROM GRAVITY AND RANDOM MATRICES

In the previous section, we determined  the path integral measure of $\cN=2$ JT supergravity on an oriented manifold with geodesic boundaries.   In order to calculate the partition
function, we need to integrate this measure over the appropriate moduli space.

Even though the measure is extremely simple, how to integrate it over  a fundamental domain of the mapping class group  is not obvious.   For ordinary Riemann surfaces, Mirzakhani   [54, 55] solved this problem by developing a recursion relation that can be used
to effectively compute the volumes.  This recursion relation was extended to $\N=1$ super Riemann surfaces in [14], and here we will extend this story to $\N=2$ super Riemann
surfaces. With the help of this result, we will be able to verify the duality between $\N=2$ JT supergravity and an $(\upalpha_0,\upbeta)=(1,2)$ AZ ensemble, to all orders in
perturbation theory.

§.§ Review of Mirzakhani's Approach

We begin with a brief
review of Mirzakhani's method, and then in section <ref>, we  explain the modifications needed to extend this approach to $\cN=2$ hyperbolic surfaces.  The
reader might find it helpful to consult appendix D of [14], which contains some more detailed explanations.   In this discussion, for brevity we refer to a three-holed
sphere with a hyperbolic metric and geodesic boundaries as a geodesic three-holed sphere.

Let $\Sigma$ be a hyperbolic surface of genus  $g$ with $n+1$ geodesic boundaries.  We single out one boundary of length $b$,
and denote the other $n$ boundary lengths as  $B=\{ b_1,\ldots,b_n\}$. It will be useful to label the boundary
geodesics  by their lengths as $\gamma_{b}, \gamma_{b_1},\ldots, \gamma_{b_n}$.

Generically, as explained in the text, a geodesic $\ell_p$ orthogonal to the boundary of $\Sigma$ at $p\in\gamma_b$ determines a hyperbolic three-holed
sphere $Y\subset \Sigma$ that has $\gamma_b$  as one of its boundaries.  Three possibilities are sketched here.

The first step is a simple decomposition of $\gamma_b$. Consider a point $p\in \gamma_b$, and let $\ell_p$ be the geodesic orthogonal to $\gamma_b$ at $p$.
The set of $p$ such that $\ell_p$ continues forever without intersecting itself or returning to the boundary has measure zero, according to a theorem of
Birman and Series.
After throwing away this set of measure zero, we can assume that $\ell_p$ eventually intersects itself or returns to the boundary of $\Sigma$.
When that happens, we terminate $\ell_p$.   We take the union of $\ell_p$ and the boundary component or
components it ends on, and thicken this set
slightly to get what topologically is a three-holed sphere whose boundaries are either boundaries of $\Sigma$ and therefore geodesics, or
else embedded circles in $\Sigma$.   The lengths of such embedded circles can be minimized in their homotopy classes to get geodesics.
In this way, the choice of a generic point $p\in \gamma_b$
determines an embedded geodesic three-holed sphere[The moduli space $\mM_{0,3}$ of a three-holed sphere is a point, of volume 1.  So we can assume
that $\Sigma$ is not itself a three-holed sphere, ensuring that
$Y$ is a proper subset of $\Sigma$ and always has at least one boundary in the interior of $\Sigma$.
In the extension to $\N=2$ super Riemann surfaces, the volume of $\mM_{0,3}$ is not simply 1, but we have already computed it in
section <ref>, so we will again assume that $\Sigma$ is not a three-holed sphere.]
$Y\subset \Sigma$.   All this is pictured in fig. <ref>.

We can now decompose $\gamma_b$ (with a set of measure zero discarded) as the union of three subsets:

* Set $A$: $\ell_p$  intersects itself or returns to the original boundary $\gamma_b$, and as depicted in figs.  <ref>(b,c), the three-holed sphere $Y$
has one boundary $\gamma_b$ and  two boundaries in the interior of $\Sigma$.     The complement of $Y$ is
connected and is a surface $\Sigma'$ of genus $g-1$ with $n+2$ boundaries, as schematically shown here:
[scale=0.5, baseline=([yshift=-0.1cm]current bounding box.center)]

at (-5.8,0) $b$;
[thick] (-5.2,0) ellipse (.15 and .5);
[thick] (-3.2,1.5) ellipse (.15 and .5);
[thick] (-3.2,-1.5) ellipse (.15 and .5);
[thick, bend right=45] (-5.2,.5) to (-4.2,1);
[thick, bend left=30] (-4.2,1) to (-3.2,2);
[thick, bend left=45] (-5.2,-.5) to (-4.2,-1);
[thick, bend right=30] (-4.2,-1) to (-3.2,-2);
[thick, bend right=95] (-3.2,1) to (-3.2,-1);
at (-2.7,1.5) $b'$;
at (-2.6,-1.5) $b''$;

[thick] (-2,1.5) ellipse (.15 and .5);
[thick] (-2,2) to (-.2,2);
[thick] (-2,1) to (-.2,1);
[thick] (-2,-1.5) ellipse (.15 and .5);
[thick] (-2,-2) to (-.2,-2);
[thick] (-2,-1) to (-.2,-1);
[gray!60,fill=gray!60] (1,0) ellipse (2 and 2.5);
[thick] (1,0) ellipse (2 and 2.5);
at (1,0) $g-1$;

We denote the lengths of the new boundaries as $b'$ and $b''$.

* Set $B$: again $\ell_p$ intersects itself or returns to $\gamma_b$, and $Y$ has two interior boundaries,
but now $\Sigma'$ is disconnected and is the union of surfaces $\Sigma_1,$ $\Sigma_2$ of genera
$h_1,$ $h_2$ such that $h_1+h_2=g$, and with boundary lengths $b',B_1$ and $b'',B_2$
such that $B_1 \cup B_2 = B$:
[scale=0.5, baseline=([yshift=-0.1cm]current bounding box.center)]

[thick] (-2,1.5) ellipse (.15 and .5);
[thick] (-2,2) to (-.2,2);
[thick] (-2,1) to (-.2,1);
[gray!60,fill=gray!60] (1,1.5) ellipse (1.2 and 1.2);
[thick] (1,1.5) ellipse (1.2 and 1.2);

[thick] (-2,-1.5) ellipse (.15 and .5);
[thick] (-2,-2) to (-.2,-2);
[thick] (-2,-1) to (-.2,-1);
[gray!60,fill=gray!60] (1,-1.5) ellipse (1.2 and 1.2);
[thick] (1,-1.5) ellipse (1.2 and 1.2);

at (-5.8,0) $b$;
[thick] (-5.2,0) ellipse (.15 and .5);
[thick] (-3.2,1.5) ellipse (.15 and .5);
[thick] (-3.2,-1.5) ellipse (.15 and .5);
[thick, bend right=45] (-5.2,.5) to (-4.2,1);
[thick, bend left=30] (-4.2,1) to (-3.2,2);
[thick, bend left=45] (-5.2,-.5) to (-4.2,-1);
[thick, bend right=30] (-4.2,-1) to (-3.2,-2);
[thick, bend right=95] (-3.2,1) to (-3.2,-1);
at (-2.7,1.5) $b'$;
at (-2.6,-1.5) $b''$;
at (1,1.5) $h_1$;
at (1,-1.5) $h_2$;

* Set $C$: The geodesic  reaches another boundary $\gamma_{b_k}$, or it self-intersects and $\gamma_{b_k}$ is one of the boundaries of $Y$.
Either way,  $Y$ has the two boundaries just specified and one internal boundary, say of length $b'$.
The remaining surface $\Sigma'$ is of genus $g$ with $n$ boundaries:
[scale=0.5, baseline=([yshift=-0.1cm]current bounding box.center)]

[thick] (-3.2,0) ellipse (.15 and .5);
[thick] (-5.2,1.5) ellipse (.15 and .5);
[thick] (-5.2,-1.5) ellipse (.15 and .5);
[thick, bend left=45] (-3.2,.5) to (-4.2,1);
[thick, bend right=30] (-4.2,1) to (-5.2,2);
[thick, bend right=45] (-3.2,-.5) to (-4.2,-1);
[thick, bend left=30] (-4.2,-1) to (-5.2,-2);
[thick, bend left=95] (-5.2,1) to (-5.2,-1);
at (-5.8,1.5) $b$;
at (-5.8,-1.5) $b_k$;
at (-2.6,0) $b'$;

[thick] (-2,0) ellipse (.15 and .5);
[thick] (-2,-.5) to (-.2,-.5);
[thick] (-2,.5) to (-.2,.5);
[gray!60,fill=gray!60] (1,0) ellipse (1.2 and 1.2);
[thick] (1,0) ellipse (1.2 and 1.2);
at (1,0) $g$;

Keeping $\Sigma$ fixed, the sum of the lengths of the three sets $A,B,C$ is $b$, since $A\cup B\cup C=\gamma_b$, except for a set of measure 0.  If we integrate over all moduli,
including the moduli of $\Sigma$ and also the choice of the point $p\in\gamma_b$, the result will be simply $b V_{g,n+1}(b,B)$.
However, we can make use of the fact that the choice of $p$ generically determines a decomposition of $\Sigma$ as the union of a three-holed sphere $Y$ with some other
surface $\Sigma'$ (possibly disconnected).  In general, $Y$ can be any geodesic three-holed sphere in $\Sigma$ one of whose boundaries is $\gamma_b$.
So we can organize the integral over the moduli of the pair $\Sigma,p$ as a sum over all choices of $Y$, and for each $Y$ an integral over $p$ and an integral over the moduli
of the remaining surface $\Sigma'$.   Since the Euler characteristic of $\Sigma'$ is always less negative than the Euler characteristic of $\Sigma$, this leads to a recursion
relation that ultimately completely determines the volumes.

To carry out this program, we need to know how to make the decomposition of $\gamma_b$ in the three sets $A,B,C$ if $Y$ is given.
Let $Y$ be a hyperbolic three-holed sphere
with geodesic boundaries $\gamma$, $\gamma'$ and $\gamma''$ of lengths $b$, $b'$ and $b''$ respectively. In the bosonic case, $Y$ has
no moduli apart from these lengths (in the generalization to super Riemann surfaces, $Y$ does have fermionic moduli and we will have to integrate over them).
We denote by ${\sf T}(b,b',b'')$ the length of the set of points $p\in\gamma$  such that $\ell_p$ ends on $\gamma'$. Of course the set of points such
that $\ell_p$ ends on $\gamma''$ then has length ${\sf T}(b,b'',b')$.   We also denote by ${\sf D}(b,b',b'')$ the length of the set of points $p\in\gamma$ such that $\ell_p$  self-intersects or
returns to $\gamma$. Since these are all the options (apart from a set of measure zero), we have

b = D(b,b',b”) + T(b,b',b”)+T(b,b”,b').
In terms of these functions, which we compute shortly, the recursion relation becomes
b V^bos._g,n+1 (b, B)    =    1/2∫_0^∞  b'db'  b” db” D_bos. (b,b',b”)  V^bos._g-1,n+2(b',b”,B)        + 1/2∫_0^∞  b'db'  b”db” D_bos. (b,b',b”) ∑_stable V^bos._h_1,|B_1|+1 (b',B_1) V^bos_h_2,|B_2|+1 (b”,B_2)      +∑_k=1^|B| ∫_0^∞ b'db'  (b -T_bos.(b,b',b_k) ) V^bos._g,n (b',B /b_k)
The three terms on each line of the right hand side correspond to each of the three possible behaviors of $\ell_p$. We have denoted  the bosonic volumes by $V_{g,n}^{\rm bos.}(b,B)$ to distinguish them from their supersymmetric counterpart which we analyze later, and similarly for the kernels appearing in the integrands. The qualification “stable” means that one
excludes the case that one of the components is of genus 0 with less than 3 boundaries, and hence does not have a hyperbolic metric.

In eqn. (<ref>), the reason for the coefficient ${\sf D}_{\rm bos.}(b,b',b'')$ in the first two lines is relatively easy to understand.
The outcomes that correspond to these lines arise if and only if  $\ell_p$, which is initially perpendicular to $\gamma_b$ at $p$, self-intersects or returns to $\gamma_b$
while remaining in a three-holed sphere $Y$ whose other boundaries are internal geodesics $\gamma_{b'}$, $\gamma_{b''}$ of $\Sigma$.
In that case, the thickening of $\gamma_b\cup\ell_p$ to a geodesic three-holed sphere  will always be $Y$.
The measure of the subset of $\gamma_b$ for which $\ell_p$ has this behavior with this particular $Y$ is ${\sf D}_{\rm bos.}(b,b',b'')$,
so that is the coefficient in the first two lines of eqn. (<ref>).
The coefficient in the last line is less obvious.   There are two ways that the thickening process can lead to a three-holed sphere $Y$ two of whose boundaries are $\gamma_b$
and another external boundary $\gamma_{b_k}$ of $\Sigma$.  First and perhaps more obvious, $\ell_p$ might terminate on $\gamma_{b_k}$ while remaining in $Y$.   This
corresponds to a set of measure ${\sf T}_{\rm bos.}(b,b_k,b')$.   Second and  possibly less obvious, $\ell_p$ might self-intersect or return to $\gamma_b$, in such a way that the
thickening process leads to a three-holed sphere with $\gamma_{b_k}$ as one of its boundaries.   This corresponds to a subset of $\gamma_b$ of measure ${\sf D}_{\rm bos.}(b,b',b_k)$.
The combined measure of the two subsets of $\gamma_b$ is ${\sf D}_{\rm bos.}(b,b',b_k)+{\sf T}_{\rm bos.}(b,b_k,b')=b-{\sf T}_{\rm bos.}(b,b',b_k)$, which is the
coefficient claimed in (<ref>).

To complete the derivation, we need an explicit expression for the functions ${\sf D}_{\rm bos.}(b,b',b'')$ and ${\sf T}_{\rm bos.}(b,b',b'')$. Thanks to (<ref>), we only need to determine ${\sf T}_{\rm bos.}(b,b',b'')$, the length of the segment $S$ of points along $\gamma$ that end up in $\gamma'$. We explain this calculation in a way that  straightforwardly  generalizes
to the $\cN=2$ case.   It is convenient to do the calculation by first “unwrapping” $Y$ in the upper half plane $H$,
which we parametrize by a complex coordinate $z = x + \i y $, $y>0$,  with metric  ṣ^2 = x̣^2+ỵ^2/y^2 .As suggested in the right panel of fig. <ref>(b), the fundamental group of $Y$  is generated by loops that wind once around $\gamma$ or once around $\gamma'$, starting and
ending at the base-point $P$.   Taking the bosonic limit of the formulas that we used in $\N=2$ supergravity, we can assume that the hyperbolic
flat connection on $Y$ has  holonomies $U,V$ around $\gamma$ and $\gamma'$, respectively:

U = ±(
e^b/2     κ

0    e^-b/2
),2cm V=±(
e^-b'/2     0

1    e^b'/2

We can view $Y$ as $\h Y/\Gamma$, where $\Gamma$ is the fundamental group of $Y$, and $\h Y$ is a region in $H$ that is bounded by geodesics that are lifts to $H$ of the boundaries
$\gamma,\gamma',\gamma''$ of $Y$.   In fact, $\h Y$ has infinitely many
geodesic boundaries that project in $Y$ to $\gamma$, $\gamma'$, or $\gamma''$.   For example, $\gamma$ has a lift that is invariant under the action of $U$ on $H$; its other
lifts in $\h Y$ are invariant under elements of $\Gamma$ that are conjugate to $U$.   There is a unique $U$-invariant geodesic $\h\gamma$ in $H$; it connects the two fixed points of $U$
on the conformal boundary of $H$.   The conformal boundary of $H$ is at $y=0$; it is a copy of $\RP^1$, which can be parametrized by  homogeneous coordinates $u,v$
with $x=u/v$.
As a $2\times 2$ matrix acting on $\begin{pmatrix}u\\[-4pt] v\end{pmatrix}$, $U$ has two eigenvectors, which correspond to the
two fixed points of $U$ on $ \RP^1$. The eigenvectors of $U$ are  $\Big(\begin{array}{c}
1 \\[-4pt]
0\end{array}\Big)$ and $\Big(\begin{array}{c}
-\frac{\kappa}{2\sinh \frac{b}{2}} \\[-4pt] 1\end{array}\Big)$, so the end-points of the geodesic $\widehat{\gamma}$ are located at $x_0= - \kappa/(2\sinh \frac{b}{2})$ and $x_3=\infty$.
Likewise, $\gamma'$ has a $V$-invariant lift $\h \gamma'$ to $\h Y$.   The endpoints of $\h\gamma'$ correspond to the eigenvectors of $V$, namely
0 \\[-4pt]
\end{array}\Big)$ and $\Big(\begin{array}{c}
-2\sinh \frac{b'}{2} \\[-4pt]
\end{array}\Big)$, corresponding to $x_1=0$ and $x_2=-2\sinh \frac{b'}{2}$.

Let $\ell$ be a non-self-intersecting geodesic  in $Y$ orthogonal to $\gamma$ at a point $p$ and ending at   $p'\in\gamma'$.
Keeping $p'$ fixed, $\ell$ is the shortest path from $p'$ to $\gamma$ in its homotopy class.   A homotopy class of non-self-intersecting paths
from $p'$ to $\gamma$ is unique except for the possibility of adjoining at the beginning of the path a loop that winds around $\gamma'$ any integer number  of times.
As we vary $p'$, $\ell$ will continue to exist, being always the shortest path from $\gamma'$ to $\gamma$ in its homotopy class; $p$ will change, of course.  This continues
as $p'$ winds around $\ell'$ any number of times, and since in that way we explore all possible homotopy classes of non-intersecting paths from $\gamma'$ to $\gamma$,
it follows that the space of non-self-intersecting geodesics  in $H$ that are orthogonal to $\gamma$ and end at $\gamma'$ is connected.   Since that is the case, in lifting to $H$,
we can pick a specific
pair of lifts $\h\gamma$ and $\h\gamma'$ to $H$ and consider geodesics in $\h Y$ orthogonal to $\h\gamma$ and ending on $\h\gamma'$.  There is no need to consider
different lifts of $\gamma$ and $\gamma'$.    We can simply consider the lifts described in the last paragraph, such that $\h\gamma$ connects boundary points $x_0$ and $x_3$,
while $\h\gamma'$ connects $x_1$ and $x_2$.

[scale=1, baseline=([yshift=-0.1cm]current bounding box.center)]
[ultra thick] (-5,0) – (3,0);
[ thick ] (0,0) – (0,6);
[thick] (-4,0) to [out=90,in=180] (-3,1) to [out =0,in=90] (-2,0);
[dashed] (-2,0) to [out = 90, in =180] (0,2);
[dashed] (-4,0) to [out = 90, in =180] (0,4);
at (0,-0.4) $x_0$;
at (-2,-0.4) $x_1$;
at (-4,-0.4) $x_2$;
at (1.2,2) $(x_0,x_0-x_1)$;
at (1.2,4) $(x_0,x_0-x_2)$;
at (.4,3) $\mathcal{S}$;
at (0.4,5.5) $\widehat{\gamma}$;
at (-3,1.3) $\widehat{\gamma}'$;

The picture in the upper half-plane $H$ with its conformal boundary shown as a horizontal line.    $\widehat{\gamma}$ is the solid vertical line and $\widehat{\gamma}'$ is the solid semicircle.  The dashed lines, which are portions of circles that are orthogonal to both $\h\gamma$ and the conformal boundary, denote the  limiting cases of geodesics that are orthogonal to $\widehat{\gamma}$ and reach $\widehat{\gamma}'$. The interval between them represents the region $\mathcal{S}$ of $\gamma$.  Its length can be straightforwardly determined from
the picture.

To work out the length of the desired subset of $\gamma$ is now straightforward.   We just need to know that geodesics in $H$ are either vertical straight lines
or semicircles (in the Euclidean metric $\d x^2+\d y^2$) that are orthogonal to the line $x=0$.  So the geodesic $\h\gamma$ from $x_0$ to $x_3$ is
the vertical line $x=x_0$ and the geodesic $\h\gamma'$ from $x_1$ to $x_2$ is the semi-circle shown in
fig. <ref>. We denote by $x_1$ and $x_2<x_1$ the end-points of $\widehat{\gamma}'$.
Points $p\in\h\gamma$ such that the geodesic $\ell_p$ orthogonal to $\h\gamma$ at $p$ reaches $\h\gamma'$ make up an interval $\mathcal S$,
and the desired function ${\sf T}_{\rm bos.}(b,b',b'')$ is the length of $\mathcal S$.   The endpoints of $\mathcal S$ are easily determined as in the figure,
and computing the length of $\mathcal S$ in the hyperbolic metric (<ref>), we get $\log\frac{(x_0-x_2)}{(x_0-x_1)}$.    Equivalently (as $x_3=\infty$),

T_bos.(b,b',b”) = log(x_0-x_2)(x_1-x_3)/(x_0-x_1)(x_2-x_3).
In order words, it is the logarithm of a cross-ratio between the four endpoints of the two geodesics $\widehat{\gamma}$ and $\widehat{\gamma}'$.

The final step is to find an expression for $x_0,\ldots, x_3$ in terms of the lengths $b$, $b'$ and $b''$.  All we need to know is that $\kappa=- 2 \cosh \frac{b-b'}{2} - 2 \cosh \frac{b''}{2}$
(eqn. <ref>).  With this information, we can compute
T_bos.(b,b',b”) = log[coshb”/2 + coshb+b'/2/coshb”/2 + coshb-b'/2].
This completes the derivation of Mirzakhani's recursion relation, since we can easily determine ${\sf D}_{\rm bos.}$ from (<ref>) and replace it in (<ref>).

§.§ Geometry of $\cN=2$ Hyperbolic Surfaces

In this section we present some important facts about $\cN=2$ hyperbolic surfaces, study the geometry of the $\cN=2$ three-holed sphere, and derive the kernels that will appear in the recursion, which is written in its final form in the next section <ref>.

§.§.§ Chiral vs. Non-Chiral Formalism

Up to this point, we have not had to be very specific about what we mean by an $\N=2$ super Riemann surface.  We only cared about the moduli of $\N=2$ super Riemann
surfaces.   However, to obtain the recursion relation, we will have to use some geometric facts about $\N=2$ super Riemann surfaces themselves, as opposed to their moduli.

There are actually two essentially different kinds of $\N=2$ super Riemann surface, which share the same moduli space.  The difference shows up when one writes
an action for a theory with $\N=2$ supersymmetry.  There are “$D$-terms,” which are written as an integral $\int \d^2z \d^4\theta K$ over a supermanifold of  complex
dimension $1|2$ or real dimension $2|4$, with some generic operator $K$.   And there are “$F$-terms,” which are written as an integral $\int \d^2z \d^2\theta W$,
over a supermanifold  of complex dimension $1|1$ or real
dimension $2|2$,
with some chiral operator $W$.

To illustrate the difference between the two, we can describe what is a genus 0 super Riemann surface of one type or the other.
The chiral version of a genus 0 super Riemann surface is $\CP^{1|1}$, with homogeneous coordinates $u,v|\theta$.  We observe that $\CP^{1|1}$ does admit
an action of the supergroup $\SL(2|1)$, acting on $\left(\begin{array}{c}u\\[-4pt] v    \\[-1pt] \hline \theta\end{array}\right)$.  We can introduce a hermitian metric
on $\CP^{1|1}$ and reduce to a real form $\SU(1,1|1)$ of $\SL(2|1)$.
The non-chiral version of a genus 0 super Riemann surface is actually $\CP^{1|2}$, with homogeneous coordinates $u,v|\uptheta,\uptheta'$, endowed with a symplectic
form that is homogeneous of degree 2:

ω= ụ ṿ +1/2(^2+'^2).
The group of linear transformations of these homogeneous coordinates that preserves the symplectic form is the complex Lie group
$\OSp(2|2)_\C$.   Again, we can introduce a hermitian metric
and a real form which we will simply call $\OSp(2|2)$.

In fact, the group $\OSp(2|2)_\C$ is equivalent to a double cover[See footnote <ref> for an explanation of this double cover, which however
will not play a major role in the following.] of $\SL(2|1)$, and, with the right choice of the real forms, $\OSp(2|2)$ is similarly a double cover of $\SU(1,1|1)$.
There is a natural one-to-one correspondence between super Riemann surfaces of the two types; see, for example,  section 9 of [56].    So they have the
same moduli spaces and that is why up to this point, there was no need to make a choice.

In this article, we will need very little systematic theory of super Riemann surfaces.   For either of the two types of super Riemann surface, we just define
$z=u/v$, define the analog $\h H$ of the upper half plane by ${\mathrm{Im}}\,z>0$, and define a hyperbolic super Riemann surface to be a quotient
$\h H/\Gamma$, where $\Gamma$ is a discrete subgroup of $\SU(1,1|1)$ or $\OSp(2|2)$, with the property that if we reduce modulo the fermionic variables
and forget the $R$-symmetries, then $H/\Gamma$ is a hyperbolic Riemann surface in the ordinary sense.

But should we use the chiral or non-chiral version of a super Riemann surface?   In fact, it turns out that if we work with non-chiral $\N=2$ super Riemann surfaces,
then Mirzakhani's derivation in the bosonic case and the derivation in [14] for $\N=1$ have straightforward generalizations to $\N=2$.   It appears to be less straightforward
to make similar arguments in terms of chiral $\N=2$ super Riemann surfaces.  An indication that the generalization to $\N=2$ is more direct if the arguments are expressed
in terms of non-chiral super Riemann
surfaces is the following.   Mirzakhani's derivation in the bosonic case leads to a cross-ratio of four points (eqn. (<ref>)), and similarly, the $\N=1$ derivation in [14]
leads to an $\N=1$ version of the cross-ratio.   So we can guess that we will need an $\N=2$ version of the cross ratio.
In the non-chiral formalism, there is no problem with this.
If  a point $P_1\in \CP^{1|2}$ has homogeneous coordinates $u_1, v_1|\uptheta_1,\uptheta'_1$  (where we denote the odd variables as
$\uptheta$ and $\uptheta'$ and use the numerical subscript to label the point) and similarly for $P_2, P_3, P_4$, then one defines

⟨1,2⟩= u_1 v_2- u_2 v_1 - _1 _2- '_1 '_2,
and the cross ratio is as usual $\langle1,2\rangle \langle3,4\rangle /\langle1,3\rangle \langle2,4\rangle $.   (In addition to this complex cross ratio,
four points in $\CP^{1|2}$ have four odd complex moduli, which will not play an important role.)  By contrast, there is no simple definition of a cross ratio for four points in $\CP^{1|1}$,
because there is no appropriate invariant in the $2|1$-dimensional representation of $\SL(2|1)$.

Other facts about $\N=1$ super Riemann surfaces that were important in  [14], appendix D, also have simple analogs for $\N=2$ in the non-chiral formulation but not in the
chiral formulation.   In particular, the notions of geodesics, perpendicular geodesics, intersecting geodesics, and the like will be spelled out in section <ref>.

It is not clear that these notions have equally satisfactory analogs if we work with chiral super Riemann surfaces.
So we will use the non-chiral approach, though we cannot rule out the possibility that a simple proof might somehow be possible in the chiral approach.
A drawback of working with non-chiral super Riemann surfaces is that we will have to convert some $\SU(1,1|1)$ formulas into $\OSp(2|2)$ formulas.
This gives formulas that are equivalent but generally more complicated.   That is actually why we did not choose to write the whole article in terms of $\OSp(2|2)$.

As already explained, in either picture based on $\SU(1,1|1)$ or $\OSp(2|2)$, the super upper half plane $\h H$ is defined by ${\mathrm {Im}}\,z>0$, with $z=u/v$.
In section <ref>, we will want to know in the $\OSp(2|2)$ picture what should be meant by the conformal boundary of $\h H$.   In fact, the natural conformal boundary is defined
by taking $u,v|\uptheta,\uptheta'$ to all be real.   Note that the real supergroup $\OSp(2|2)$ does act on the locus of real $u,v|\uptheta,\uptheta'$, so the boundary in
this sense is $\OSp(2|2)$-invariant.   Equivalently, after scaling to set $v=1$, the boundary is parametrized by odd and even real variables $x|\uptheta,\uptheta'$.
We will call this boundary $\partial\h H$.   The reader might be slightly surprised that $\partial\h H$ is of codimension $1|2$, not $1|0$, relative to $\h H$.   This
is actually a common state of affairs in supergeometry: the most natural conformal boundary of a supermanifold
is not necessarily of codimension $1|0$.   For example, in the AdS/CFT correspondence,
in a superspace description, the conformal boundary will frequently have fewer fermionic coordinates than the bulk.

§.§.§ Converting to  $\OSp(2|2)$

A first step is to write the $\SU(1,1|1)$ generators of eqns. (<ref>) and (<ref>) as $\OSp(2|2)$ generators, that is, as linear transformations
of $u,v|\uptheta,\uptheta'$.
For bosonic generators, we have
0    1   0    0

0   0     0   0

0    0     0     0

0     0     0     0

0     0   0    0

1    0     0    0

0    0    0     0

0    0    0     0

1   0   0    0

0   -1     0   0

0   0     0    0

0   0    0    0

0   0   0    0

0   0     0   0

0   0     0    1

0   0    -1   0

and for fermionic generators, we have
q_1   =   (
0   0   -1/2√(2)   -ı/2√(2)

0   0     0   0

0   1/2√(2)     0    0

0   ı/2√(2)    0    0

0   0   1/√(2)   -ı/√(2)

0   0     0   0

0   - 1/√(2)     0    0

0   ı/√(2)    0    0

q_3   =   (
0   0   0    0

0   0     1/2√(2)   ı/2√(2)

1/2√(2)   0     0    0

ı/2√(2)   0    0    0

0   0   0    0

0   0     1/√(2)   - ı/√(2)

1/√(2)   0     0    0

-ı/√(2)   0    0    0

To obtain the kernel that appears  in the recursion relation  requires a knowledge of the measure on the space of flat connections on a three-holed sphere $Y$ .
We determined this in section <ref>  in the $\SU(1,1|1)$ language, and these results can be translated to $\OSp(2|2)$.
As a first step, we can rewrite eqn.  (<ref>) for the monodromies by which we parametrized a flat connection on $Y$ in terms of
algebra generators:
U_0= e^a/2 h + a/2sinha/2 κe + (π+ϕ_1)z e^ψ_2 q_2+ψ_4 q_4,   V_0 =e^-b/2 h + b/2sinhb/2f + (π+ ϕ_2) z e^ψ_1 q_1+ψ_3 q_3.
To write the holonomies as operators on $u,v|\uptheta,\uptheta'$, we now just have to substitute into these formulas the expressions
(<ref>) and (<ref>) for the generators.  The expressions for $\kappa$ and the $\uU(1)$ constraint in terms of $a,b|\psi_1,\psi_2,\psi_3,\psi_4$
were derived in section  <ref>.

In the rest of this derivation, we use conventions similar to those that we used in reviewing Mirzakhani's bosonic derivation.   Thus the boundary lengths of a geodesic three-holed
sphere will be $b,b',b''$ (not $a,b,c$ as hitherto) and the $R$-symmetry twist parameters will be $\phi,\phi',\phi''$ (not $\phi_a,\phi_b,\phi_c$).

§.§.§ The Supersymmetric Setup

We now consider the kernel that appears in the Mirzakhani-style recursion relation for the $\cN=2$ volumes.   We consider an $\cN=2$  three-holed sphere
$Y$ with geodesic boundaries $\gamma,\gamma',\gamma''$, and boundary   data $\b=(b,\phi)$, $\b'=(b',\phi')$, $\b''=(b'',\phi'')$, and $\OSp(2|2)$ holonomies $U,V,W$.
We want to find the measure $\h {\sf T}(\b,\b',\b'')$ of
the set $\mathcal S$ consisting of points $p\in \gamma$ such that the geodesic $\ell_p$ orthogonal to $\gamma$ at $p$ reaches $\gamma'$.
The answer is actually  the obvious superanalog of eqn. (<ref>).    If $P_0,P_3$ are the endpoints of the $U$-invariant geodesic $\h\gamma\subset \h H$,
and  $P_1,P_2$ are the endpoints of a $V$-invariant geodesic $\h\gamma'\subset \h H$,
T=log⟨0,2⟩⟨1,3⟩/⟨0,1⟩⟨2,3⟩,similarly to eqn. (<ref>).   The $\N=2$ version of the symbol $\la~,~\ra$ was defined in eqn. (<ref>).

To justify eqn. (<ref>), we first need to explain what is meant by the terms “geodesics,” “orthogonal,” and so on.   For $\N=1$ super Riemann surfaces, such
a discussion was given in appendix D of [14], leading to the obvious $\N=1$ superanalog of eqns. (<ref>) and (<ref>).   A similar discussion applies
for $\N=2$, assuming that one works with non-chiral $\N=2$ super Riemann surfaces, as we will now  explain.

First,  what is meant by a “geodesic” in $\h H$?   We recall that the $\N=2$ super upper half plane $\h H$ contains the ordinary upper half plane $H$, embedded
at $\uptheta=\uptheta'=0$.   We want to regard an ordinary geodesic in $H$ as a geodesic in $\h H$.  We also want the notion of a geodesic to be $\OSp(2|2)$-invariant.
So we declare a geodesic in $\h H$  to be any curve – that is, any submanifold of dimension $1|0$  – that is equivalent under $\OSp(2|2)$ to a geodesic in $H$.
Note that this definition implies that there is a well-defined length element along any geodesic $\gamma\subset \h H$: we pick an $\OSp(2|2)$ transformation that maps
$\gamma$ to $H$ and then we use the ordinary length element for a geodesic in $H$.   Another consequence of the definition is that a geodesic in $\h H$ has
endpoints in what we defined in section <ref> as $\partial\h H$, the conformal boundary of $\h H$, parametrized by real $x|\uptheta,\uptheta'$.   Indeed, if $\gamma\subset H$,
then it has endpoints in the conformal boundary of $H$.   These are points of the form $x|0,0$, with real $x$, so they are in $\partial \h H$.  Applying an $\OSp(2|2)$ transformation
and using the fact that $\partial \h H$ is $\OSp(2|2)$-invariant, we learn that any geodesic in $\h H$ has endpoints in $\partial \h H$.  We can also describe this by saying that
any geodesic in $\h H$ has real endpoints (that is, endpoints whose coordinates $x|\uptheta,\uptheta'$ are all real).
Now, if $\gamma$ is a geodesic in $\h H$ and $p\in \gamma$, we want to define the geodesic $\ell_p$ that is orthogonal to $\gamma$ at $p$.  If $\gamma\in H$,
we just use the classical definition of the orthogonal geodesic $\ell_p$. Any case can be reduced by this one by an $\OSp(2|2)$ transformation.   We also need to know
what it means to say that a geodesic $\ell_p$ is self-intersecting or intersects another geodesic $\gamma'$.   Since a geodesic is a curve of dimension $1|0$,
two geodesics in a super Riemann surface of dimension $2|4$ generically do not have a point in common, and indeed asking for two geodesics to have a point in common
is not the right condition.
One way to explain the right condition (slightly different from but equivalent to the explanation in [14] for $\N=1$)
is to use the fact that we are interested in hyperbolic super Riemann surfaces.  This means
that in discussing whether two geodesics intersect, we can lift the situation to $\h H$ and give a definition there.
We  say that an intersection of two geodesics $\gamma_1,\gamma_2\subset \h H$ consists of a pair of points
$P_1\in\gamma_1, $ $P_2\in\gamma_2$ with $\la 1,2\ra=0$.  Since  there are two
real parameters in the choices of $P_1$ and $P_2$, the complex condition $\la 1,2\ra=0$, equivalent to two real conditions, is a good criterion for intersection of geodesics.

The special case of this that we most need is  the condition that two geodesics $\gamma_1$ and $\gamma_2$ have an endpoint in common.   Suppose that $\gamma_1$ has an endpoint with coordinates $x_1|\uptheta_1,\uptheta'_1$, and $\gamma_2$ has an endpoint  with coordinates $x_2|\uptheta_2,\uptheta'_2$. These coordinates are all real, since the endpoints of any geodesic are real.  The condition under which we say that $\gamma_1$ and $\gamma_2$
have a common endpoint  is that $\la 1,2\ra=0$, again using the pairing (<ref>).
Since the coordinates are real, this is a single real condition, just as it would be classically.

Now, going back to the $U$- and $V$-invariant geodesics $\h\gamma,\h\gamma'\subset \h U$, we pick a point
$p\in \h\gamma$ and ask if the orthogonal geodesic $\ell_p$ to $\gamma$ at $p$ has a common endpoint with $\h\gamma'$.   In the bosonic case, the picture corresponding to
this question is drawn
in fig. <ref>; two points $p$ satisfy the desired condition, one for each endpoint of $\h\gamma'$.   We have included the fermionic variables in such a way
that we did not change the number of equations (the condition $\la 1,2\ra=0$ on the endpoints  is a single real condition, reducing to $x_1=x_2$ in the classical limit)
or unknowns (we defined a geodesic to have dimension $1|0$).    So also supersymmetrically the condition that $\ell_p$ and $\h\gamma'$ have an endpoint in common
determines two points in
$\h\gamma$.   These points are the endpoints of the set $\mathcal S\subset \h\gamma$ whose length is supposed to be $\h {\sf T}$.

At this point, then, to compute $\h {\sf T}$ we have to analyze the situation of fig. <ref> in the supersymmetric case.    It might seem that we have to do a genuinely
new supersymmetric calculation, but actually one can justify the  obvious answer of eqn. (<ref>) without any really new calculation.
First, we can make an $\OSp(2|2)$ transformation that maps $\h\gamma$ to $H$ and, of course, also acts on $\h\gamma'$.   (We can even pick the $\OSp(2|2)$ transformation
so that $\h\gamma$ is a vertical geodesic in $H$, as pictured in  fig. <ref>.   But this is not necessary for the argument.)   Then the endpoints $P_0$, $P_3$
of $\h \gamma$ are at $x_0|0,0$ and $x_3|0,0$.   Since $\h\gamma\subset H$, by definition all geodesics orthogonal to $\h\gamma$ are in $H$ also, so for any $p\in\h\gamma$,
an endpoint $Q$ of $\ell_p$ is in $H$, say at $\t x|0,0$. The endpoints $P_1,P_2$ of $\h\gamma'$ are generically not in $H$; they are at, say, $x_1|\uptheta_1,\uptheta'_1$
and $x_2|\uptheta_2,\uptheta_2'$.   But since $Q$ is at $\uptheta=\uptheta'=0$, the conditions $\la 1,Q\ra=0$ and $\la 2,Q\ra=0$ do not depend on the fermionic coordinates of
$P_1,P_2$; those conditions reduce to $y=x_1$ and $y=x_2$.  At this point, the calculation of the length of the set $\mathcal S$ reduces precisely to the calculation that we already
explained in the purely bosonic case, leading to  T=log(x_0-x_2)(x_1-x_3)/(x_0-x_1)(x_2-x_3),as in eqn. (<ref>).

This is the answer in a coordinate system with $\h\gamma\in H$, implying that the points $P_0$, $P_3$ are at $x_0|0,0$ and $x_3|0,0$.   We would like a formula
for $\h{\sf T}$ that does not depend on this condition.   To find such a formula, we just note that the fact that $P_0,P_3$ have vanishing fermionic coordinates implies
that $x_0-x_2=\la 0,2\ra$, $x_1-x_3=\la 1,3\ra$, etc.   So the covariant version of the answer is just eqn. (<ref>).

One point that might puzzle the reader is that in analyzing the function $\h {\sf T}$, we made use of the real structure of $\OSp(2|2)\cong \SU(1,1|1)$, but
in computing the measure on moduli space
we ignored this
real structure (at least as regards the fermionic moduli).  There is actually no contradiction.   $\SU(1,1|1)$ does have a real structure,
and we are free to use it, if convenient, in computing the function $\h {\sf T}$.   On the other hand, the measure can be analytically continued to complex values of the moduli,
and in computing the measure, we can disregard the real structure if that is convenient.

§.§.§ Evaluation of the Kernel Appearing in the Recursion

It remains to reduce eqn. (<ref>) to an explicit formula.   Unfortunately, this will lead us through some rather complicated formulas, though the simplicity of the final
results suggests that there might be a simpler route to those results.

The first step is to find explicit formulas for the endpoints $P_0,\cdots,P_3$ that appear in this derivation,   In doing so, we will describe these points by homogeneous coordinates
$u_0,v_0|\uptheta_0,\uptheta_0'$, $u_1,v_1|\uptheta_1,\uptheta_1'$, $u_2,v_2|\uptheta_2,\uptheta_2'$
and $u_3,v_3|\uptheta_3,\uptheta_3'$.
We view the holonomies $U,V$ as $2|2\times 2|2$ matrices
acting on $u,v | \uptheta,\uptheta'$.
These matrices have two bosonic and two fermionic eigenvectors.   The bosonic eigenvectors of $U$ are the points $P_0,P_3$, and the bosonic eigenvectors of $V$ are the points
After a detailed calculation, we find the explicit eigenvectors:
u_0,v_0|_0,'_0   =   -κ/2sinhb/2,1 |ψ_2+κ/ 2 sinhb/2 ψ_4/√(2)(1-e^ıϕ-b/2) , -ıψ_2 + κ/2 sinhb/2 ψ_4 /√(2)(1-e^ıϕ-b/2),
u_1,v_1|_1,'_1   =    0,1 | ψ_1/2√(2)(e^b'/2-ıϕ'-1) ,ıψ_1/2√(2)(e^b'/2-ıϕ'-1)  ,
u_2,v_2|_2,'_2   =    -2sinhb'/2,1|ψ_1 - 2 sinhb'/2 ψ_3/2√(2)(e^-b'/2-ıϕ'-1) ,ıψ_1 - 2 sinhb'/2 ψ_3/2√(2)(e^-b'/2-ıϕ'-1),

u_3,v_3|_3,'_3    =    1,0 | ψ_4/√(2)(-1+e^b/2+ıϕ),-ıψ_4/√(2)(-1+e^b/2+ıϕ).
As explained in section <ref>, we now have to evaluate
T = log0,21,3/ 0,12,3=log[ u_0v_2-u_2v_0-_0_2-'_0'_2/u_0v_1-u_1 v_0-_0_1-'_0'_1u_1v_3-u_3v_1-_1_3-'_1'_3/u_2v_3-u_3 v_2-_2_3-'_2'_3].
This function depends on the three geodesic lengths, the two independent $\uU(1)$ holonomies, and the four fermionic moduli.  An explicit formula, using eqn. (<ref>)
for the eigenvectors, is
T =log[-κ+2sinhb/2  2 sinhb'/2+(κψ_4 + 2 sinhb/2 ψ_2)(ψ_1 - 2 sinhb'/2 ψ_3)/2(1-e^-b/2+ıϕ)(1-e^-b'/2-ıϕ')/- κ+(κψ_4 + 2 sinhb/2 ψ_2)ψ_1/2(1-e^-b/2+ıϕ)(1-e^b'/2-ıϕ')1-ψ_1 ψ_4/2(e^b/2+ıϕ-1)(1-e^b'/2-ıϕ')/1+(ψ_1-2sinhb'/2 ψ_3)ψ_4/2(e^b/2+ıϕ-1)(e^-b'/2-ıϕ'-1)]To make the dependence on fermionic moduli explicit, we can use the expansion   $\kappa = \kappa_0 + \sum_{i<j} \kappa_{ij} \psi_i \psi_j + \kappa_4 \psi_1\ldots \psi_4$ presented in (<ref>). Then we expand  $\widehat{{\sf T}}$
T=T_0 +∑_i<j T_ij ψ_iψ_j + T_4 ψ_1 ψ_2 ψ_3 ψ_4,
Finding the explicit formulas for $\widehat{{\sf T}}_{ij}$ and $\widehat{{\sf T}}_4$ is straightforward but tedious. The function $\widehat{{\sf T}}_0$ is simpler and, importantly, is equal to the bosonic length appearing in the original Mirzakhani recursion
T_0 = T_bos. = log[coshb”/2+ coshb+b'/2 /coshb”/2+ coshb-b'/2 ].
We will not attempt to write down the answers for $\widehat{{\sf T}}_{ij}$ and $\widehat{{\sf T}}_4$, partly because they are long, and also because we ultimately only need the simpler functions that arise after  fermionic integration.

We are now ready to complete the calculation, and integrate $\h{\sf T}$ over $\mathcal{M}_{0,3}$, the $0|4$ dimensional moduli space of the three-holed sphere.   We define
T(b,b',b”) = 2π  ∫_ℳ_0,3 μ_0,3 ·T .
We included a factor of $2\pi$ on the right hand side to simplify the final form of the recursion. The measure obtained in the previous section is
μ_0,3 = - 2 e^ıϕ- ıϕ' (coshb”/2 + cosϕ”)/(2π) (1+K)δ(ϕ+ϕ'+ϕ”-F) [^̣4 ψ]
When integrating $\mu_{0,3} \cdot \widehat{\sf T}$ over fermionic moduli there are three terms, depending on whether we pick $\widehat{{\sf T}}_0$, $\widehat{{\sf T}}_{ij}$ or $\widehat{{\sf T}}_4$ to absorb the fermionic integrations. The three terms  are given schematically by

\begin{align}\notag
\int [\d^4 \psi]~  \delta(\phi+\phi'+\phi''-F)\,(1+K) \, \widehat{{\sf T}}_0 & \propto \delta''(\sum\phi) \widehat{{\sf T}}_0,\\ \notag
\int [\d^4 \psi] \delta(\phi+\phi'+\phi''-F)\,(1+K) \,\widehat{{\sf T}}_{ij} \psi_i\psi_j& = \delta'(\sum\phi)\, f_{ij}  \widehat{{\sf T}}_{ij}+ \delta(\sum \phi) ~k_{ij}  \widehat{{\sf T}}_{ij},
\\
\int [\d^4 \psi] \delta(\phi+\phi'+\phi''-F)\,(1+K) \,\widehat{{\sf T}}_4 &= \delta(\sum \phi ) \widehat{{\sf T}}_{4}.
\end{align}

The first term involving $\widehat{{\sf T}}_0$ was essentially already computed in section <ref>. Since $\widehat{{\sf T}}_0$ is already independent of fermionic moduli, the answer is proportional to the quantity $V_{0,3}$, computed in that section. The second term involves the new function $\widehat{{\sf T}}_{ij}$. Surprisingly, we find that $f_{ij}\widehat{\sf T}_{ij}=0$. Finally, we are left with two contributions proportional to $\delta(\phi+\phi'+\phi'')$, from $k_{ij} \widehat{{\sf T}}_{ij}$ and from $\widehat{{\sf T}}_4$. The sum of these  is surprisingly simple:
k_ij T_ij + T_4 =e^ıϕ' - ıϕ/2(coshb”/2 + cosϕ”)[  1/16 cosh^2 b+b'+b”/4-1/16 cosh^2 b-b'+b”/4+(b”→-b”)].There might be a more direct route to this relatively simple formula, but this is not obvious.   The final result for $\sf T$ is presented in eqn. (<ref>) below.

To formulate the recursion relation, we need one more function ${\sf D}({\sf b},{\sf b}',{\sf b}'')$.
This is the integral over the fermionic moduli  of the segment of $\gamma_b$ from which orthogonal geodesics reach neither $\gamma_{b'}$ nor $\gamma_{b''}$. It
can be computed from the following sum rule:

2π∫__0,3 μ_0,3 ·b =D(b,b',b”) + T(b,b',b”) + T(b,b”,b').
It is important to formulate this sum rule after integration over moduli, since the way we parametrize the fermionic moduli spoils the symmetry between the $b'$ and $b''$ boundaries. The extra factor of $2\pi$ was introduced for future convenience, like the rescaling of ${\sf T}$. As opposed to the case of $\cN=1$ super Riemann surfaces, the left hand side of
eqn. (<ref>) does  not vanish and produces an answer proportional to $V_{0,3}$.  The result for  ${\sf D}({\sf b},{\sf b}',{\sf b}'')$ is stated in eqn. (<ref>).

We are now ready to write down the final version of the $\N=2$ recursion relation.

§.§ Recursion Relation for $\cN=2$ Volumes

§.§.§ Final Evaluation

In this section we will summarize the results so far and write down the recursion relation for the Weil-Petersson volumes for $\cN=2$ hyperbolic
surfaces in a way that is self-contained and can be read independently from the details of the derivation outlined so far.

The computation in  section <ref> leads to explicit formulas for the two kernels ${\sf T}({\sf b},{\sf b}',{\sf b}'')$ and ${\sf D}({\sf b},{\sf b}',{\sf b}'')$ that appear in the recursion relation.
Here  we remind the reader that ${\sf b} = (b,\phi)$.  The two functions can be expanded

T(b,b',b”)    =   -1/4^ 2δ”(ϕ+ϕ'+ϕ”)T_2(b,b',b”)-δ(ϕ+ϕ'+ϕ”)  T_0(b,b',b”)

D(b,b',b”)    =   -1/4^ 2δ”(ϕ+ϕ'+ϕ”) D_2(b,b',b”)-δ(ϕ+ϕ'+ϕ”)  D_0(b,b',b”).
The functions ${\sf T}_{0}$, ${\sf T}_2$ and ${\sf D}_{0}$, ${\sf D}_2$ appearing on the right hand side depend  only on the lengths; the subscript indicates the number of derivatives acting on the delta function they multiply. ${\sf T}_{0}$, ${\sf T}_2$ are
T_2(b,b',b”)    =   log[coshb”/2+coshb+b'/2 /coshb”/2+ coshb-b'/2 ],

T_0(b,b',b”)    =    1/ 16cosh^2 b+b'+b”/4-1/16cosh^2 b-b'+b”/4+(b”→- b”).
${\sf D}_{0}$, ${\sf D}_2$  are obtained through the sum rule
D_2(b,b',b”)    =   b-T_2(b,b',b”) - T_2(b,b”,b')

D_0(b,b',b”)    =   -T_0(b,b',b”) - T_0(b,b”,b')
that comes from eqn. (<ref>) by integration over $\mM_{0,3}$.
One can check explicitly that

\begin{align}{\sf T}_0(b,b',b'')& = \partial_{b'}^2 {\sf T}_2(b,b',b'')\cr {\sf D}_0(b,b',b'')&= \partial_{b'}^2 {\sf D}_2 (b,b',b''). \label{goodrel}\end{align}
These relations are  not  accidents and will be crucial to match the $\cN=2$ generalization of the Mirzakhani recursion relation with the matrix ensemble.

We have now all the ingredients we need to obtain the final form of the recursion relation that computes the volumes of moduli spaces of $\mathcal{N}=2$ hyperbolic surfaces

b V_g,n+1 (b, B)    =    1/2(2π) ∫_0^∞b' ḅ' ∫_0^2πϕ̣' ∫_0^∞b” ḅ” ∫_0^2πϕ̣”   D (b,b',b”)      ×(V_g-1,n+2(b̅',b̅”,B) + ∑_stable V_h_1,|B_1|+1 (b̅',B_1) V_h_2,|B_2|+1 (b̅”,B_2) )      +∑_k=1^|B| ∫_0^∞b' ḅ'∫_0^2π ϕ̣'   [ -b/4^ 2 δ”(ϕ+ϕ'+ϕ_k) -T(b,b',b_k) ] V_g,n (b̅',B /b_k) .

Here ${\sf B}=\{ {\sf b}_1,\ldots, {\sf b}_n\}$ is the set of boundary parameters, and following our notation an overline denote orientation reversal for a given boundary, so that if ${\sf b}=(b,\phi)$ then $\bar{{\sf b}} = (b,-\phi)$.  The coefficients in this relation have precisely the same interpretation
as in the bosonic recursion relation (<ref>), except that they have been integrated over the fermionic moduli of a three-holed sphere.
The qualification “stable” under the sum has the same meaning it had in the bosonic case reviewed earlier. Integration over the $\uU(1)$ twist parameters will produce a factor of $(2\pi)^2=\int \d\varphi' \d \varphi'' $ in the first line, and of $(2\pi)= \int \d \varphi'$ in the last. We will explain below the relevance of these seemingly innocent factors and the role they play. One of those factors cancels with a similar one in the normalization of $\mu_{0,3}$ and the normalization of the functions ${\sf T}$ and ${\sf D}$.

As a final ingredient, we need  two initial conditions for this recursion relation.  One initial condition is familiar:

V_0,3(b_1, b_2, b_3) = - 1/2π 1/4^ 2 δ”(ϕ_1+ϕ_2+ϕ_3).
We also need $V_{1,1}$, which can be easily derived by imitating the derivation of the recursion with a slight modification:[A Riemann surface of genus 1 with one
hole or puncture has a $\Z_2$ symmetry.   As in [14], by $V_{1,1}$, we mean the volume of $\mM_{1,1}$ in the orbifold sense, including a factor of $1/2$ because
every point in $\mM_{1,1}$ has this $\Z_2$ symmetry.   If we do not do this, the statement of the recursion relation (<ref>) would be more awkward.]
V_1,1 (b)    =    1/2b ∫_0^∞b' ḅ' ∫_0^2π ϕ̣'  D(b,b',b̅'),   =   - (2π) 1/4^ 2δ”(ϕ) b^2+4π^2/48 - (2π) δ(ϕ) 1/8
The recursion relation determines all $V_{g,n}$ in terms of these two.

Eqn. (<ref>) is the main result of this section.   As we show in section <ref>, it  implies the duality between $\cN=2$ JT gravity and the
random matrix ensemble.    First, we will present some general properties that follow from the recursion relation.

§.§.§ $\cN=2$ Volumes at 	“Fixed Charge”

The volume of the three-holed sphere with geodesic boundaries of fixed $\uU(1)$ holonomies $\phi,\phi',\phi''$ is proportional to $\delta''(\phi+\phi'+\phi'')$. Similarly the kernels appearing in the recursion are sums of delta functions, possibly with derivatives, of a sum of $\uU(1)$ holonomies.  This implies, by induction, that the same is true for the volume of every $\mM_{g,n}$. More precisely, from the recursion relation, one can deduce an expansion
V_g,n(B) =(2π)^2g-1 ∑^2g-2+n_m=0 (-1)^m/(4^ 2)^m δ^(2m)(ϕ_B)   v_g,n,m(B),
where we introduced some notation to avoid clutter here and later. On the right hand side, $2m$ is the number of derivatives acting on the delta function.
For  ${\sf B}=\{ {\sf b}_1,\ldots, {\sf b}_n\}$,
we define $\phi_{\sf B} = \phi_1 +\ldots +\phi_n$. The function $v_{g,n,m}(B)$ depends only on the lengths $B=\{ b_1,\ldots, b_n\}$. The choice of overall normalization in (<ref>) is for later convenience.

It is convenient to introduce  “fixed charge” volumes.
The simple dependence of $V_{g,n}(\sf B)$ on the holonomies implies the following condition for the Fourier transforms:

∫^̣nϕ/(2π)^n  e^ı∑_i=1^n q_i ϕ_i   V_g,n(B) = 0,   unless  q_1 = q_2 …= q_n.
Note that because $V_{g,n}({\sf B})$ is supported at $\phi_1+\cdots+\phi_n=0$, this integral makes sense and has the stated property as long as all differences
$q_i-q_j$ are integers; it is not necessary for the individual $q_i$ to be integers.
Vanishing of eqn. (<ref>) when the $q_i$ are not all equal is equivalent to the statement that different supermultiplets are statistically independent, to all orders
in the topological expansion.
To study any one supermultiplet,  we can set $q_1 = \ldots =q_n = q$ and define the “fixed charge” volumes

V^(q)_g,n(B) =(2π)^2-2g ∫^̣nϕ/(2π)^n  e^ıq ϕ_B   V_g,n(B).
The normalization will simplify the recursion relation that these volumes satisfy. Here $q$ is not the $\uU(1)$ $R$-charge of a state, but a label of a supermultiplet.
One can verify (see eqn. (<ref>)) that when these volumes are glued to trumpets in order to compute the path integral with NAdS boundary conditions, the $R$-charges of the supermultiplet states are $q \pm \hq /2$; therefore $q$ has the same interpretation as in section <ref>.

In Fourier space,  the “fixed charge” volumes are simply polynomials in $q^2$ of the form
V^(q)_g,n(B) = ∑^2g-2+n_m=1 (q^2/4^ 2)^m v_g,n,m(b_1,…, b_n).
One can easily prove this property by induction on the recursion relation, without going through (<ref>), since the Fourier transforms of the kernels have
this  property. One can also prove, imitating the proof by Mirzakhani, that the coefficient of a given power of $q^2$ in the fixed charge volumes is a polynomial in the variables $b_i^2$, of a degree we will specify below.

For future reference in proving the match with the  matrix model, it is useful to translate the recursion for the volumes $V_{g,n}({\sf B})$ into a recursion for the “fixed charge” volumes $V_{g,n}^{(q)}(B)$. One can simply apply a Fourier transform on (<ref>). The result is
b V^(q)_g,n+1 (b, B)    =    1/2∫_0^∞b'db'  b”db” (q^2/4^ 2  D_2 (b,b',b”) - D_0(b,b',b”))

×(V^(q)_g-1,n+2(b',b”,B)+ ∑_stable V^(q)_h_1,|B_1|+1 (b',B_1) V^(q)_h_2,|B_2|+1 (b”,B_2) )      +∑_k=1^|B| ∫_0^∞b'db' [ q^2/4^ 2( b -T_2(b,b',b_k)) +T_0(b,b',b_k) ] V^(q)_g,n (b',B /b_k)
This way of presenting the recursion also avoids the cluttering associated to keeping track of the holonomies. The purpose of normalizing the “fixed charge” volumes with a factor of $(2\pi)^{2-2g}$ in (<ref>) was to obtain (<ref>) in this form. The initial conditions for this recursion can be read off from the previous results:
V_0,3^(q) =q^2/4^ 2 1_v_0,3,1,    V_1,1^(q) = q^2/4^ 2 b^2+4π^2/48_v_1,1,1 +(- 1/8)_v_1,1,0.

§.§.§ Relation to Bosonic Volumes

There is a simple property we can notice easily starting from (<ref>). Consider the coefficient $v_{g,n,m}$ with the highest number of derivatives, or equivalently the highest power of $q$, with $m=2g-2+n$. This coefficient satisfies a simplified version of the  recursion (<ref>) in which we keep only the terms proportional to $q^2/4\hq^{\,2}$ in the recursion kernels. Up to a simple rescaling, this
is precisely the same as the bosonic volume recursion of Mirzakhani, given in (<ref>). A careful analysis of the normalization therefore yields the relation
v_g,n,2g-2+n(B) = V^bos._g,n(B).
This property holds  for the initial conditions given in eqn. (<ref>), and the recursion relation implies that it is true for all $g,n$.
We can rewrite the same property in terms of the volumes at fixed holonomies as
V_g,n(B) = (2π)^2g-2   V^bos_g,n(B)   (-1)^n 2πδ^(4g-4+2n)(ϕ_B)/(4^ 2)^2g-2+n +….
In bosonic JT gravity coupled to a $\uU(1)$ gauge theory, one would get a similar result, but without the factors $(-1)^n/(4 \hq^{\, 2})^{2g-2+n}$
and without the derivatives acting on the delta function. The derivatives of the delta function have their origin in the fermionic moduli, and
the other extra factors are due to choices of normalization in $\cN=2$ JT gravity.

This relation makes it clear that $v_{g,n,2g-2+n}$ is a polynomial in $b_i^2$ of degree $3g-3+n$, since it follows from Mirzakhani's results. Moreover, one can show using similar techniques that $v_{g,n,2g-2+n-s}$ is a polynomial in $b_i^2$ of degree $3g-3+n-s$. This implies that $v_{g,n,0}$ is a polynomial of degree $g-1$ in $b_i^2$. This is  true even for $g=0$, as $v_{0,n,0}=0$.

§.§.§ Some Simple Examples

Next we give some concrete simple examples of $\cN=2$ volumes obtained from the recursion relation.

For any genus $g\geq 1$, there can be terms in the volumes with no derivatives acting on the delta function, since this is true for the seed $V_{1,1}({\sf b})$. That is not so for the case of genus zero, since the seed $V_{0,3}$ has two derivatives. Therefore all volumes at genus zero are sums of expressions with at least two derivatives. We can analytically compute the two-derivative coefficients and verify that  they are non-zero. After performing the integrals over holonomies, the recursion  for $v_{0,n,1}$ reduces to

b v_0,n+1,1(b,B) =∑_k=1^|B| ∫_0^∞b'ḅ'  T_0(b,b',b_k)  v_0,n,1(b',B/b_k).
One can prove by direct evaluation that the following ansatz for the term with two derivatives solves this recursion:
V_0,n(B) =(2π)^-1(-1)^n(n-1)!/2 1/4^ 2δ”(ϕ_1 + …+ ϕ_n) + (higher derivatives).
or equivalently $v_{0,n,1} =(-1)^{n+1}\frac{(n-1)!}{2}$.

To illustrate some of the preceding  statements, we can compute $V_{0,4}^{(q)}$ and $V_{0,5}^{(q)}$:
V_0,4^(q)(b_1,b_2,b_3,b_4)    =    q^4/4^2^ 4 4π^2 +b_1^2 +b_2^2+b_3^2+b_4^2/2_v_0,4,2=V^bos._0,4(B) +q^2/4^ 2 (-3)_v_0,4,1 ,

V_0,5^(q)(b_1,b_2,b_3,b_4,b_5)    =     q^6/4^3^ 3 V^bos._0,5(B) - q^4/4^2^ 452π^2 + 9 ∑_j=1^5 b_j^2/2 + q^2/4^ 2 12.

From these expressions, we can recover
the bosonic volumes and  the two derivative term predicted in eqn. (<ref>).  We see as well that in these examples, the volumes
are polynomials in the variables $b_i^2$ of the expected degree.

§.§ Equivalence with Random Matrix Recursion

Now we will finally complete the circle of ideas and show that the $\cN=2$ JT gravity path integral is equal to the result from the random matrix ensemble that we have described.
We recall eqn. (<ref>) expressing partition functions in terms of volumes:
Z_g,n= [ ∏_j=1^n ∫_0^∞(b_j ḅ_j) ∫_0^2π(2πϕ̣_j) Z_tr (β_j,α_j;b_j)] V_g,n(b_1,…, b_n).We need a more explicit version of this formula. For this, we use the explicit form  (<ref>) of the $\cN=2$ JT gravity trumpet to rewrite this as
Z_g,n   =    (2π)^n∑_q_1,…,q_n∈ℤ+δ-1/2  ∏_i=1^n [(e^ıα_i (q_i-/2)+e^ıα_i (q_i +/2)) e^-β_i E_0(q_i)]

×∫_0^∞∏_j=1^n b_j ḅ_j   Z_tr^bos.(β,b) ∫_0^2π ∏_j=1^nϕ̣_j/2π e^ıϕ_j q_j  V_g,n(b_1,…,b_n),
where the bosonic trumpet is $Z_{\rm tr}^{\rm bos.}(\beta,b) =e^{-\frac{b^2}{4\beta}}/(2\sqrt{\pi \beta} )$.
As in the discussion of eqn. (<ref>), the integration over the $\phi_i$ has simple properties because the integrand is supported at $\phi_1+\cdots+\phi_n=0$ and
the differences of the exponents $q_i-q_j$ are integers.
Then we obtain a partition function consistent,  to all orders in $e^{-S_0}$, with statistically independent supermultiplets

Z_g,n= (2π)^2g+n-2∑_q∈ℤ+δ-1/2 ∏_i=1^n (e^ıα_i (q-/2)+e^ıα_i (q+/2))  Z_g,n^(q)(β_1,…,β_n),
where the contribution from each supermultiplet $Z_{g,n}^{(q)}$ is given by[We can now complete the proof that the number of BPS states does not fluctuate in gravity. Since, as we argued above, $V^{(q)}_{g,n}$ are polynomials in $b_i^2$, if $E_0(q)>0$, the integral over the $b_i$ cannot produce a dependence with $\beta$ that cancels the exponential decay $e^{-\beta E_0(q)}$ at large $\beta$. For the multiplet with $E_0(0)=0$ it would require a pole at $b=0$ to modify the number of BPS states, also ruled out by the fact that $V_{g,n}^{(q)}$ are polynomials. ]
Z_g,n^(q) (β_1,…,β_n)=  [∏_j=1^n  ∫_0^∞b_j ḅ_j    Z_tr^bos.(β_i,b_i)e^-β_i E_0(q)]V^(q)_g,n(b_1,…,b_n).
The factor of $(2\pi)^{2g+n-2}$ can be absorbed by a shift of $S_0$, since the exponent is minus the Euler characteristic of the surface.
For this reason, we start by focusing on $Z_{g,n}^{(q)}$.

To show the equivalence with the matrix model recursion, we will proceed in the following way.   Let  $Z_{g,n,\M}^{(q)}$  be the solution of the matrix model recursion of an $(\upalpha_0,\upbeta)=(1,2)$ AZ ensemble with spectral curve $y_q(x)$ and threshold energy $E_0(q)$, both for now generic.[$\N=2$ JT supergravity has a number of BPS states proportional to $e^{S_0}$. We show in appendix <ref> that the loop equations of an AZ ensemble with measure $(\upalpha,\upbeta)=(1+2\nu,2)$ are independent of $\nu$ when $\nu$ is proportional to $L$ (apart for allowing a pole in the spectral curve at $x=0$). Therefore the loop equations only depend on $\upalpha_0 =\upalpha- \upbeta \nu$ which we take to be one.]
Similarly, define matrix model “volumes”:
Z_g,n,^(q) (β_1,…,β_n)=  [∏_j=1^n  ∫_0^∞b_j ḅ_j Z_tr^bos.(β_i,b_i)e^-β_i E_0(q)]V^(q)_g,n,(b_1,…,b_n).
We want to show
that $Z_{g,n,\M}^{(q)}$, with the right choice of spectral curve, obeys the same recursion relation as $Z_{g,n}^{(q)}$.
A small generalization of a calculation in [57], which we explain in appendix <ref>, shows that one can rewrite the matrix model loop equations (presented in eqn. (<ref>)) in terms of the matrix model volumes, as
b V^(q)_g,n+1, (b, B)    =    1/2 ∫_0^∞ b'db' ∫_0^∞ b”db” D_q(b'+b”,b)      ×(V^(q)_g-1,n+2,(b',b”,B) + ∑_stable V^(q)_h_1,|B_1|+1, (b',B_1) V^(q)_h_2,|B_2|+1, (b”,B_2) )      + 1/2∑_k=1^|B| ∫_0^∞ b'db' (D_q(b'+b_k,b) + D_q(b'-b_k,b) ) V^(q)_g,n, (b',B /b_k)
where the function $D_q(x,y)$ depends on the spectral curve $y_q(x)$ through the following integral transform

D_q(x,y) = -∫_ε+ıℝ ẓ/2πı e^-xz sinh(yz) /z y_q(E_0(q)-z^2).
The contour of integration is along the imaginary axis with a small positive real part $\varepsilon>0$. This result was deduced in  [57]  for a Dyson ensemble with $\upbeta=2$, but is also valid for an $(\upalpha_0,\upbeta)=(1,2)$ AZ ensemble,
since the matrix model recursion is the same.[The matrix model recursion is different  for ensembles describing theories with time-reversal symmetries.]
The first obvious observation is that the threshold energy has to match the one computed in section <ref> for a supermultiplet with average charge $q$, which is  $E_0(q)=q^2/4\hq^{\,2}$.  We can derive the following identities by summing over residues
-∫_ε+ıℝ ẓ/2πı 4π/z sin2πz e^-x z sinh(y z) = y + log(cosh^2x-y/4/cosh^2x+y/4 ),

-∫_ε+ıℝ ẓ/2πı 4πz^2/z sin2πz e^-x z sinh(y z) = - sinhx/2 sinhy/2/2(coshx/2 + coshy/2)^2
The first and second lines are also equal to
-∫_ε+ıℝ ẓ/2πı 4π/z sin2πz e^-x z sinh(y z) = y - T_2(y,b',x-b') - T_2(y,x-b',b'),

- ∫_ε+ıℝ ẓ/2πı 4πz^2/z sin2πz e^-x z sinh(y z) =-    T_0(y,b',x-b')-T_0(y,x-b',b')
These relations imply that the recursion relation we derived for the volumes of moduli spaces of $\cN=2$ hyperbolic surfaces, at “fixed charge” $q$, is equivalent to the matrix model loop equation for spectral curve $1/y_q(q^2/4\hq^{\,2} - z^2) \sim (q^2/4\hq^{\,2} -z^2)/\sin(2\pi z)$. To verify the claim, insert this spectral curve in (<ref>), then write (<ref>) explicitly and compare with (<ref>). Working out the prefactors, and including the contribution from the overall $(2\pi)^{2g+n-2}$ in (<ref>), which further multiplies the spectral curve by a power of $1/2\pi$, we obtain the final version of the spectral curve associated to the matrix model dual to pure $\cN=2$ JT gravity
y_q(x) =1/2π sin(2π√(-x + q^2/4^ 2))/4πx.
This is precisely the spectral curve we obtained in section <ref> by performing the gravitational path integral on the disk topology. If we write $x=q^2/4\hq^{\,2}-z^2$ and multiply the spectral curve by $2\pi$, which produces an overall factor of $(2\pi)^{2g+n-2}$, then we obtain $y_q(z) \sim \frac{4\hq^{\,2}}{q^2}\frac{\sin 2\pi z}{4\pi} $ up to a correction of
relative order $1/q$.   This is the spectral curve of bosonic JT gravity, up to a rescaling, giving another explanation of the fact that the “fixed charge” volumes are proportional to the bosonic ones at large $q$: $V_{g,n}^{(q)}(B) \sim (q^2/4\hq^{\,2})^{2g+n-2} V^{\rm bos}_{g,n}(B)$.

§.§ Deformations of $\N=2$ JT gravity

We have successfully proven, to all orders in the topological expansion, that $\N=2$ JT gravity without additional symmetries is dual to the type of random
matrix ensemble proposed in section <ref> for $\N=2$ quantum mechanics, namely the supermultiplets of different $R$-charge are statistically independent, and each is
governed by a $(1,2)$ AZ ensemble.

Such ensembles depend on two choices. The first is the spectrum of BPS states for each $R$-charge. This can be extracted from the disk partition function and, for the ensembles relevant to black holes, the BPS spectrum does not fluctuate (this is explained in section <ref>). The second is the choice of matrix potential that appears in (<ref>), or
equivalently the choice of spectral curve.  This is specified by the leading order density of states of non-BPS multiplets in the topological expansion, also computed in section <ref> for pure $\N=2$ JT gravity. In this section, we want to address the question of how to give a gravity interpretation to other choices of BPS spectrum and other matrix potentials in the random matrix ensemble.

The recipe in gravity to modify the matrix potential was provided in [58] and [59] for bosonic JT gravity, and generalized to $\N=1$ JT gravity in [60]. Introduce a gas of defects in the gravitational path integral of pure JT gravity. In the $\N=2$ case, the defect species are parametrized by the opening angle $\gamma$, and also a $\uU(1)$$R$-symmetry holonomy $e^{\i \hq \phi}$ around the defect.[This was referred to as $\alpha$ in [58] and [59], but we avoid this notation here to avoid confusion with the $\uU(1)$$R$-symmetry chemical potential at the NAdS boundaries.] These two parameters can be combined into the following $\SU(1,1|1)$ holonomy matrix around the defect

e^ıϕ cosγ/2   -e^ıϕ sinγ/2   0

e^ıϕ sinγ/2    e^ıϕ cosγ/2   0

0   0   e^2ıϕ
In the upper left block, we have simply written the product of a rotation and an $R$-symmetry transformation.
This deformation of $\N=2$ JT gravity involves an integral over the location of the defect and a sum over the possible number of them, besides the usual degrees of freedom of pure gravity. For each defect insertion, we add a weighting factor $w$, which is another parameter of the deformation. The monodromy $U$ is conjugate to a diagonal matrix of the form (<ref>) with imaginary geodesic length $b= \i \gamma$. Therefore the result for the path integral with such defects can be obtained by analytic continuation of the path integral with geodesic boundaries. More precisely, this is true  when $0<\gamma\leq \pi$; for attempts to extend the results beyond this range, see [61, 62].

For generic values of $\gamma$ and $\phi$,  the one-defect correction to the disk partition function is given by
Z_1-def   =   w ∑_q∈ℤ + δ-1/2 ( e^ıα(q-/2)  + e^ıα(q+/2))  e^- ıϕq e^-βE_0(q) 1/2π  e^γ^2/4β/2√(πβ)   =   ∑_q∈ℤ + δ-1/2 ( e^ıα(q-/2)  + e^ıα(q+/2)) ∫_E_0(q)^∞Ẹ  e^-βE  w e^-ıϕqcosh(γ√(E-E_0(q)))/8 π^2 √(E-E_0(q)) .The first line is the analytic continuation of (<ref>) upon $b\to \i \gamma$. In the second line we wrote the defect partition function in a way that makes evident what is the correction to the non-BPS density of states. A first observation is that the threshold energy continues to be $E=E_0(q)$, for each $q$, in the presence of the defect.  To be more exact,
this is true order by order in the defect expansion, but the $1/\sqrt{E-E_0(q)}$ singularity indicates that there is actually a shift of order $w$ in the threshold energy, assuming that
the theory can be described by a matrix integral. A second observation is that the result is complex. It is therefore natural to consider $w$ to be a complex parameter as well, and to introduce for each defect species a  conjugate defect with holonomy $U^{-1}$ and complex conjugate weight. Therefore we include
defect species parametrized by $(\gamma,\phi)$ in pairs, such that we preserve unitarity, and to linear order the correction to $\rho_q(E)$ for $E>E_0(q)$ is

ρ_q(E) = sinh( 2π√(E-E_0(q)))/8π^3 E + Re [ w  e^-ıϕqcosh(γ√(E-E_0(q)))/4 π^2 √(E-E_0(q))] + 𝒪(w^2).
We conclude that adding such generic defects modifies the spectral curve, and therefore the matrix potential, without modifying the number of BPS states. We leave for future work to derive the finite $w$ expression for the non-BPS density of states, analogous to the ones derived in [58] and [59]. In particular it is important to verify that the resummation over the number of defects can be interpreted in terms of a shift of the threshold energy $E_0(q)$, since otherwise the square root singularity of (<ref>) near threshold is not consistent with a matrix integral.

Instead of considering further the non-BPS sector, in this article we focus on the new element not present in the previous literature: how to modify the number of BPS
states, or in other words  the choice of component of solutions to the constraint $Q^2=0$. It turns out this can also be done with a gas of defects. Even though a generic defect does not contribute to the number of BPS states, supersymmetric defects do.[An analog for $\N=1$ is that the number of supersymmetric states, which vanishes in the absence
of defects, becomes nonzero in the presence of a gas of Ramond punctures [14].]  Locally, a defect preserves supersymmetry whenever the deficit angle is related to the $R$-symmetry holonomy by

γ= 2π±2 ϕ.
When this relation is satisfied, the holonomy (<ref>) commutes with two out of the four fermionic generators. The sign in eqn. (<ref>) determines
precisely which fermionic symmetry  is preserved by the defect. The partition function on the disk with such a defect inserted is analyzed in appendix <ref>. It is not a simple analytic continuation of the  usual trumpet partition function and is instead
Z_susy-defect(β,α)   =    w∑_n∈ℤ e^2πınδ  cos((α+2πn)/2) /2π^3 (1∓(α+2πn)/π) e^1/4β(2π±ϕ)^2- ^  2/β(α-ϕ+ 2πn)^2   =   w∑_n∈ℤ e^2πınδ  cos((α+2πn)/2) /2π^3 (1∓(α+2πn)/π)  (β→∞)
In the second line we take $\beta \to \infty$ and therefore project onto BPS states. Using Poisson resummation one can derive
Z_susy-defect(β→∞,α) = w ∑_k ∈ℤ+δ e^ıαk  e^∓ıπk//4π^2  Θ(k),
where we defined the function

Θ(k)= 1     |k|</2

1/2     |k|=/2

0     |k|>/2.
Again we see that this correction to the spectrum is complex and therefore for each defect we need to insert an orientation reversed one. The correction to the BPS spectrum is then

N_BPS(k) = e^S_0cos(πk/)/4π^2 + Re[ w  e^ ıπk//2π^2  ] Θ(k)
This equation is exact in both $S_0$ and $w$ as long as $0<\gamma\leq\pi$. For $\gamma$ in this range, configurations either with handles or with more than one defect do not contribute to the BPS spectrum. The reason is that for $\gamma$ in this range, as soon as we have two or more defects or any number of handles there will be a geodesic homotopic to each NAdS boundary. The form of the trumpet partition function will guarantee that there will be no contribution to the number of BPS states.

We can consider multiple supersymmetric defect species, with different values of $\phi$. The contribution to the number of BPS states comes only from either the disk, or a disk with one defect (of any species). Since the one defect contribution to the number of BPS states computed in eqn. (<ref>) is independent of $\phi$, a sum over species will only modify the effective value of $w$.   The answer will still take the form in (<ref>), at least as long as all defects satisfy $0<\gamma \leq \pi$.

For the reasons given above, the most general shift in the spectrum of BPS states is a linear combination of $\cos (\frac{\pi k}{\hq})$, associated to the real part of $w$, or $\sin (\frac{\pi k}{\hq})$, associated to the imaginary part of $w$. Only for $w\in \mathbb{R}$ can we preserve charge conjugation invariance, since $\sin(\frac{\pi k}{\hq})$ changes sign under $k\to-k$. Moreover, a contribution $\sin (\frac{\pi k}{\hq})$ produces BPS states at charges $k=\pm \hq/2$. Since the disk predicts a vanishing number of BPS states at these charges, a model with non-real $w$ would seemingly predict a negative number of BPS states for either $k=+\hq/2$ or $-\hq/2$. Such a theory of gravity would then be apparently non-unitary if $\delta=1/2$, so that the spectrum of charges is half-integral.[A similar issue arises in other examples.  For example, adding a gas of defects to bosonic JT gravity can lead to a negative density of states. On the matrix model side, it has been proposed in [60] that this  signals a phase transition.]

Similar to their non-supersymmetric counterpart, the spectrum of non-BPS states is affected by contributions with an arbitrary number of supersymmetric defects, but we will not analyze this problem in this article.

The spectrum of BPS states that we can achieve with a gas of supersymmetric defects is very restrictive. For example if $w\in \mathbb{R}$, then the correction from a defect has the same dependence on $k$ as the leading answer from the disk. Interestingly, this change in the spectrum does not allow us to construct a theory of gravity with a number of BPS states reproducing the $\N=2$ SYK answer (with the only exception being $\hq=3$.) It would be nice to find a gravitational calculation that reproduces the BPS spectrum of SYK for any[We
thank J. Maldacena for raising this question.] $\hq$.

Finally, it is possible to interpret the deformations by a gas of defects as a modification of the dilaton potential of two-dimensional $\N=2$ dilaton-gravity, with a modification that
depends on both the dilaton and the scalar field coming from the $R$-symmetry $\uU(1)$ $BF$ theory. This approach actually explains why we need to consider defects coming in pairs with opposite orientation: in this way the gravity action is real. Since we did not analyze the explicit form of the $\N=2$ JT gravity action, it is outside of the scope of this article to describe this perspective.

§ INCORPORATING TIME-REVERSAL SYMMETRY

§.§ Boundary Viewpoint

In this section, we will incorporate time-reversal symmetry in $\N=2$ matrix models and JT supergravity.

Time-reversal is an anti-unitary transformation that commutes with the Hamiltonian.   But we have to choose whether it will commute or anticommute with the $R$-charge generator
$J$.   The standard convention in particle physics[The
reason for this convention is that historically, $\tT$ was defined to be an approximate symmetry of ordinary matter (originally, it was believed to be an exact symmetry).  For this
to be the case, additive conserved charges such as baryon and lepton number must be $\tT$-invariant.]
is that an anti-unitary symmetry that commutes with additive conserved charges as well as with the Hamiltonian is called $\tT$, while an anti-unitary symmetry that commutes with the Hamiltonian but anticommutes with additive conserved charges is called $\CT$.
So we will consider two cases:

(A)   The boundary quantum mechanics system  (or ensemble) may have an antiunitary symmetry $\tT$ that commutes with both $H$ and $J$.  To be a symmetry, $\tT$ will have to conjugate
$Q$ and $Q^\dagger$ to linear combinations of themselves.  Since $[J,Q]=\h q \,Q$, $[J,Q^\dagger]=-\h q \,Q^\dagger$, $\tT$ must conjugate $Q$ to a multiple of itself and likewise
$Q^\dagger$.   After possibly replacing $\tT$ with $\tT e^{\i\gamma J}$ for some $\gamma$, we can assume that $\tT$ commutes with $Q$ and $Q^\dagger$.
So $\tT$ simply commutes with the whole supersymmetry algebra of the boundary theory.

In theories with fermions, one can often define a time-reversal symmetry such that $\tT^2=(-1)^\fF$, the operator that distinguishes bosons and fermions.   In the present context, however,
we have a fermionic operator $Q$ that commutes with $\tT$, and therefore with $\tT^2$.   So necessarily $\tT^2$ commutes with fermionic operators, rather than anticommuting with them.
(This leads in Euclidean signature to what is called a ${\mathrm{pin}}^-$ structure, rather than ${\mathrm{pin}}^+$.)

The operator $\tT^2$ is a unitary operator that commutes with the whole supersymmetry algebra.
The unitary operators that we know about that commute with the supersymmetry algebra are $c$-numbers $e^{\i a}$, $a\in \R$ and also operators $e^{2\pi \i n J/\h q}$, $n\in\Z$.
We could consider a boundary theory with an additional unitary symmetry
beyond  the supersymmetry algebra; then there would also be an additional gauge symmetry in bulk.   However, we will avoid introducing additional symmetries.
Given this, $\tT^2$ must be of the form $e^{\i a}e^{2\pi \i n J/\h q}$ for some $a,n$.
However, $\tT^2$ is further constrained by the fact that it commutes with $\tT$.   The only operators $e^{\i a}e^{2\pi \i n J/\h q}$ that commute with $\tT$ are $\pm 1$.
So the only options are $\tT^2=\pm 1$.

Since $\tT$ commutes with $J$, $\tT$ symmetry places no constraint on the charge defect defined by $J=\delta~{\mathrm{mod}}~\Z$.

(B) The boundary system may have an antiunitary symmetry $\CT$ that commutes with $H$ but anticommutes with $J$.   In this case, $\CT$ must conjugate $Q$ to a multiple of
$Q^\dagger$, and vice-versa.  So  $\CT Q\CT^{-1}= e^{\i\alpha} Q^\dagger$, $\CT Q^\dagger \CT^{-1} =e^{-\i\alpha}Q$,
for some real $\alpha$.   By  replacing $\CT $ with $\CT e^{-\i \alpha J/\h q}$, we can reduce to $\alpha=0$:
Q ^-1=Q^†,  Q^†^-1=Q.

Again $\CT^2$ commutes with the whole supersymmetry algebra, so if we do not wish to introduce additional symmetries, we must have $\CT^2=e^{\i a} e^{2\pi\i n J/\hq}$
for some $a,n$.   The condition that $\CT^2$ commutes with $\CT$ implies that actually $\CT^2=\pm e^{2\pi \i n J/\h q}$ for some $n$.  But we are still free to substitute
$\CT\to\CT e^{-2\pi\i k J/\h q}$, $k\in\Z$, without disturbing eqn. (<ref>).   By choosing $k$ so that either $2k=n$ or $2k=n+\h q$, we can reduce to $\CT^2=\pm 1$.

Since $J$ is odd under $\CT$, the possible values of $\delta$ in a $\CT$-invariant theory are $0,1/2$.  (In the reduction in the last paragraph, one has to take into account
that if $\delta=1/2$, then $\exp(2\pi\i J)=-1$.)

When the boundary theory has a $\tT$ or $\CT$ symmetry, we expect the two-dimensional bulk supergravity theory to have such a symmetry as well.   In the present article,
we study the bulk theory primarily in Euclidean signature.   In that context, it is more natural to consider an ordinary (linear) reflection symmetry rather than an antilinear time-reversal
symmetry.   So we make use of the $\CPT$ theorem, which tells us that a relativistic theory has a spatial reflection symmetry $\rR$ that commutes with additive conserved
charges such as $J$ if and only if it has a $\CT$ symmetry, and it has a spatial reflection symmetry $\CR$ that anticommutes with additive conserved charges such as $J$
if and only if it has a $\tT$ symmetry.

We will next discuss the  random matrix ensembles that are natural in $\N=2$ supersymmetry extended by $\tT$ or $\CT$ symmetry.
The bulk picture with $\CR$ or $\rR$ symmetry is discussed in section <ref>.

(A) Since $\tT$ commutes with $J$, it maps the space of BPS states for any given $J$ to itself and places no constraint on the dimensions of these spaces.[There is one
small exception:
if $\tT^2=-1$, the dimensions of spaces of BPS states must be even, by what is usually called Kramers doubling.    However, this factor of 2
is essentially invisible in a double-scaled model, in which dimensions are proportional to $e^{S_0}$.  There is a similar doubling for the non-BPS supermultiplets.   A random matrix
description of a theory with $\tT^2=-1$ uses an ensemble in which this doubling is built in.]   Likewise, the space of $(k,k+\h q \,)$ representations is $\tT$-invariant,
for each $k$.

It is straightforward to repeat the analysis of sections <ref> and <ref> to incorporate the $\tT$ symmetry.[The following amounts mostly to repeating the analysis
in [14] for the cases that the anomaly coefficient vanishes (no anomaly, $\tT^2=1$) or equals 4 (the anomaly is captured by saying $\tT^2=-1$).]
First, consider a single $(k,k+\h q\,)$ multiplet
assuming that the others do not exist.   We have to modify the analysis of section <ref> by assuming that the matrix $\qQ$ is $\tT$-invariant and similarly
by restricting the symmetry group $G=\uU(L_k)\times \uU(L_{k+\h q})$ to its $\tT$-invariant subgroup.    For $\tT^2=1$, this means that we should assume $\qQ$ to be a real
matrix and take $G={\rm O}(L_k)\times {\rm O}(L_{k+\h q})$.  In other words, $\qQ$ is now a bifundamental of a product of orthogonal, rather than unitary, groups.
The derivation of eqn. (<ref>) is modified as follows. There is no continuous symmetry rotating the
phase of a given eigenvalue, so $\upalpha=0$ in eqn. (<ref>); and $\w$ is real in eqn. (<ref>), as a result of which the factors $(\lambda_i\pm \lambda_j)^2$
in eqn. (<ref>) are replaced by $|\lambda_i\pm\lambda_j|$.   The resulting measure is $\prod_i \d\lambda_i \prod_{i<j}|\lambda_i^2-\lambda_j^2|$,
an AZ ensemble with $(\upalpha,\upbeta)=(0,1)$.    In the subsequent derivation
assuming the presence of $\nu$ zero-modes of charge $k$ or $k+\h q$, there are now $\nu$ instead of $2\nu$ additional
symmetry generators for each $\lambda_i$, so the factor $\lambda_i^{2\nu}$ in eqn. (<ref>) is replaced by $\lambda_i^\nu$.   The final result is then
$\prod_i \d\lambda_i\, \lambda_i^\nu\prod_{i<j}|\lambda_i^2-\lambda_j^2|$, an AZ ensemble with $(\upalpha,\upbeta)=(\nu,1)$.
We can write this as in eqn. (<ref>), but now with $\upbeta=1$:
=_0+ν,1cm (_0,)=(0,1). The derivation in section <ref> to allow
for potential couplings between the different multiplets requires similar small corrections.
The argument $\t\lambda u+\lambda v$ of the delta function
in eqn. (<ref>) is now real, so the delta function is an ordinary delta function of a real variable, and also $\w$ is
now real in eqn. (<ref>).  These effects cancel as before and the multiplets remain statistically independent.
Upon allowing for the presence of BPS states,  there are now half as many symmetries mapping a non-vanishing row
or column to a row or column of zeros, so the contribution to $\upalpha$ is half as big as before.  The final prediction is that the $(k,k+\h q\,)$ multiplets are statistically independent
and each one is described by an AZ multiplet with parameters satisfying (<ref>).

In case $\tT^2=-1$, $\qQ$ is a matrix of quaternions and the unitary groups $\uU(L_k)$ and $\uU(L_{k+\h q})$ are replaced by symplectic groups.   In other words,
$\qQ$ is now a bifundamental of a product of symplectic groups.   The group that rotates a single eigenvalue is now $\SU(2)\cong
\Sp(1)$, with dimension 3, leading to $\upalpha_0=3$,
and the parameter $\w$ in eqn. (<ref>) is now quaternionic, depending on four real parameters and leading to $\upbeta=4$.   After also taking into account that all
parameters are quaternionic and there
are now four generators mapping a nonzero row or column of a bifundamental to a vanishing row or column, one finds that the $(k,k+\h q \,)$ multiplets are statistically
independent and described by an AZ multiplet with $(\upalpha,\upbeta)=(3+4\nu,4)$.
=_0+ν,1cm (_0,)=(3,4).

In a bulk description, the difference between a bifundamental of a pair of orthogonal group and a bifundamental of a pair of symplectic groups is a factor of $(-1)^\chi$, that
is a factor of $-1$ for each crosscap.   This is proved in section 4.2.4 of [14] using the loop equations.

(B) Since $\CT$ anticommutes with $J$, it maps the space of BPS states with $J=k$ to the corresponding space with $J=-k$, placing no constraint on the dimensions $L_k^0$
of the spaces of BPS states except that $L_k^0=L_{-k}^0$ (and in case $\delta=0$ and $\tT^2=-1$, $L_0^0$ must be even by Kramers doubling).   Similarly, the space
of $(k,k+\h q\,)$ representations is mapped by $\CT$ to the space of $(-k-\h q,-k)$ representations.   Thus for generic values of $k$, $\CT$ merely implies that the random matrix
description of the $(k,k+\h q)$ states is the same as the random matrix description of the $(-k-\h q,-k)$ states. In terms of the bulk charge $q=k+\hq/2$ introduced in section <ref>, $\CT$ acts as $q\to -q$.

We learn more only in the case of a supermultiplet with $R$-charges $(-\h q/2,\h q/2)$, or equivalently $q=0$.   As $\h q$ is odd, this
multiplet only exists if $\delta=1/2$.  For $\delta=1/2$, $\CT$ maps the $(-\h q/2,\h q/2)$
states to themselves, and therefore for this multiplet, the random matrix description will be different.
Since $J$ has eigenvalues $\pm \h q/2$ for these states, it can be put in a block diagonal form
J=q/2( 1     0
0     -1  ). The blocks are of equal size, as they are exchanged by $\CT$. Since $\CT$ anticommutes with $J$, and $\CT^2=\varepsilon$ where $\varepsilon=\pm 1$,   we can assume  $\CT$ to take the form
=( 0     1
ε    0  )⋆.The symmetry group is reduced to the group of unitary transformations that commute with both $J$ and $\CT$.   These are unitary
transformations of the form
=( U     0
0     U̅  ),with $U$ unitary (the adoint, complex conjugate, and transpose of a matrix $M$ will be denoted $M^\dagger, \,\bar M,\, M^\tr$).
Thus the symmetry group is $\uU(L_{\h q/2})$.
Because $Q$ and $Q^\dagger$ carry $R$-charges $\h q$ and $-\h q$, they take the general form
Q=( 0    M
0     0  ),1cm Q^†=( 0     0
M^†    0  ).The condition $\CT Q^\dagger\CT^{-1}= Q$ tells us  that $M=\varepsilon M^\tr$ is  symmetric or antisymmetric depending on whether $\varepsilon=+1$ or $\varepsilon=-1$,
and the symmetry $Q\to \U Q\U^{-1}$ becomes
$M\to U M U^\tr$.   In other words, $M$ is a complex  second rank tensor of the unitary group $\uU(L_{\h q/2})$, a symmetric tensor if $\varepsilon=1$ and an antisymmetric
one if $\varepsilon=-1$.

Ensembles of such symmetric or antisymmetric matrices with unitary symmetry also arise in $\N=1$ JT supergravity with anomaly coefficient 2 or 6 mod 8 [14].
The symmetric tensors correspond to an AZ ensemble with $(\upalpha_0,\upbeta)=(1,1)$ and the antisymmetric tensors correspond to an AZ ensemble with $(\upalpha_0,\upbeta)=(1,4)$.
This can be shown by arguments similar to those in section <ref>. We consider in turn the two cases:

(1) The canonical form of a symmetric second rank tensor $M$ is $\diag(\lambda_1,\lambda_2,\cdots)$,
with all $\lambda_j\geq 0$.  For each $j$, the symmetry group ${\rm U}(L_{\h q/2})$ has one generator rotating the phase of $\lambda_j$; this leads to $\upalpha=1$.
For each pair $\lambda_i, \lambda_j$, the symmetry group has two generators rotating the corresponding eigenvectors of $M$ among themselves, as opposed to the four
in eqns. (<ref>), (<ref>).   This leads to $\upbeta=1$ rather than 2.  So the ensemble is an AZ multiplet with $(\upalpha,\upbeta)=(1,1)$.

(2) The canonical form of an antisymmetric tensor $M$ is a direct sum of
$2\times 2$ blocks of the form $\begin{pmatrix}0 & \lambda_j\cr -\lambda_j & 0\end{pmatrix}$, with all $\lambda_j\geq 0$.   The unitary group $\uU(L_{\h q/2})$ has for each $j$
a $\uU(2)$ subgroup acting on this block, of which an ${\mathrm{SU}}(2)$ subgroup leaves $M$ invariant and the one remaining generator rotates the phase of $\lambda_j$, leading to $\upalpha=1$.  By looking at generators that act between a pair of these $2\times 2$ blocks of $M$, one finds
$\upbeta=4$.  So $(\upalpha,\upbeta)=(1,4)$ for antisymmetric tensors.

There are no BPS states of charge $\pm \h q/2$, so for these multiplets, there is no
distinction betweeen $\upalpha$ and $\upalpha_0$.

In a bulk description, the difference between a random ensemble of symmetric complex tensors with unitary symmetry and
a random ensemble of antisymmetric complex tensors with that symmetry consists of a factor $(-1)^\chi$, just like the difference between the two cases in (A).  This is proved using the
loop equations in section 4.2.4 of [14], and it also has a simple proof based on Feynman diagrams explained in appendix B of that paper.

§.§ The Cross-Cap With $\rR$ or $\CR$ Symmetry

In Lorentz signature with coordinates $t=x^0, x=x^1$, consider gamma matrices $\gamma_\mu$, $\mu=0,1$ satisfying $\{\gamma_\mu,\gamma_\nu\}=\pm 2\eta_{\mu\nu}$ with
$\eta_{\mu\nu}=\diag(-1,1)$.   The massless Dirac equation $(\gamma^0\partial_0+\gamma^1\partial_1)\psi=0$ has a time-reversal symmetry
$\tT\psi(t,x)=\gamma_0\psi(-t,x)$ and a reflection symmetry $\rR\psi(t,x) =\gamma_1\psi(t,-x)$.   We observe that $\tT^2=\pm 1$ acting on $\psi$ if and only if $\rR^2=\mp 1$ acting on
$\psi$; the
signs are  determined by the sign in the Clifford algebra.   An operator version of this statement is that[Here we are considering $\tT^2$ and $\rR^2$ as automorphisms of
the operator algebra. Even if $\tT^2$ or $\rR^2$ is the identity as an automorphism of the operator algebra, it may still act as $-1$ on a quantum Hilbert space, under some
conditions.]       $\tT^2=1$ or $(-1)^\fF$ if and only if $\rR^2=(-1)^\fF$ or 1.
This relationship is actually a general prediction of the $\CPT$ theorem.

Here we did not incorporate any additive conserved charge.   If one does so, for example by taking $\psi$ to be a complex fermion with a $\uU(1)$ symmetry, one
finds that because time-reversal acts by an antilinear operator while a spatial reflection  acts by a linear one, time-reversal  will commute with the conserved charges if and only if the
reflection anticommutes with them,
and vice-versa.  To see this, expand $\psi $ in Majorana fermions by $\psi=\psi_1+\i \psi_2$.   A time-reversal symmetry can be defined to act by, say,
$\psi_1(t,x)\to \gamma_0 \psi_1(-t,x)$, $\psi_2(t,x)\to \varepsilon \gamma_0\psi_2(-t,x)$, with $\varepsilon=\pm 1$.  Regardless of what signs we pick in the definition of time-reversal,
the reflection symmetry that is related to it by the $\CPT$ theorem will act with the same signs: $\psi_1(t,x)\to \gamma_1 \psi(t,-x)$, $\psi_2(t,x)\to \varepsilon \gamma_1\psi_2(t,-x)$.
Then, as time-reversal is anti-unitary, we see that it maps $\psi$ to $\psi^\dagger=\psi_1-\i\psi_2$ if $\varepsilon=1$, but to $\psi$ if $\varepsilon=-1$.  For a unitary reflection
symmetry, these statements are reversed.

So a $\tT$ symmetry satisfying $\tT^2=1$ (or $(-1)^\fF$)  is always accompanied by a $\CR$ symmetry satisfying $\CR^2=(-1)^\fF$ (or 1), and a $\CT$ symmetry satisfying
$\CT^2=1$ (or $(-1)^\fF$) is always accompanied by an $\rR$ symmetry satisfying $\rR^2=(-1)^\fF$ (or 1).
The boundary ensembles discussed in section <ref> have a $\tT$ or $\CT$ symmetry satisfying $\tT^2=1$ or $\CT^2=1$, so they will have a $\CR$ or $\rR$ symmetry
satisfying $\CR^2=(-1)^\fF$ or $\rR^2=(-1)^\fF$.

One more important detail that can be understood from the preceding derivation is the following.   To get $\tT^2=1$ acting on $\psi$, the sign in the Clifford algebra was
$\{\gamma_\mu,\gamma_\nu\}=-2\eta_{\mu\nu}$.   This Clifford algebra has a two-dimensional real representation, so Majorana fermions with $\tT^2=1$ can exist in two
spacetime dimensions.    However, when we Wick rotate to Euclidean signature, where we will perform computations, the Clifford
algebra becomes $\{\gamma_\mu,\gamma_\nu\}=-2\delta_{\mu\nu}$.   This algebra does not have a two-dimensional real representation,
but it does have a representation by imaginary $2\times 2$ matrices.   The formula
$\rR\psi(t,x)=\gamma_1\psi(t,-x)$ shows that in acting on fermions $\rR$ (or $\CR$ if conserved charges are incorporated) is proportional to a gamma matrix, so we have to expect
that as a $2\times 2$ matrix acting on fermions, $\rR$ or $\CR$ can be chosen to be imaginary but cannot chosen to be real.
Since $\gamma_1^2=-1$, $\gamma_1 $ has eigenvalues $\pm \i$
and determinant 1; hence we expect $\rR$ to have determinant 1 as an operator on fermions.

(a) $C$ is a cylinder with opposite points on the right boundary identified (as schematically indicated by the $\Large\times$) to make
an unorientable “crosscap” manifold.   After this identification, the boundary on the right becomes a circle of half the original circumference,
and it is no longer a boundary but an interior circle, ultimately an interior geodesic.
A flat connection on $C$ with holonomy $U$ around the boundary and holonomy $V$ around the interior geodesic can exist
if and only if $V^2=U$. (b) To calculate the torsion, one can use the very simple cell decomposition of $C$ (or more precisely of a topological space that for these
purposes is equivalent to $C$)
indicated here, with a single vertex $P$, two one-cells labeled $V$ and $U$ (the two one-cells labeled  $V$ that appear in the diagram are identified to make the crosscap), and
one two-cell.

With these preliminaries out of the way, our goal now is to compute the “torsion” (or moduli space integration measure) for a basic unorientable two-manifold
from which all others can be built.
This is the “crosscap” manifold (fig. <ref>(a)), which is obtained from a cylinder by identifying opposite points on one boundary, to build an unorientable manifold $C$
with a single boundary circle.
Gluing this manifold onto a trumpet gives an unorientable asymptotically NAdS spacetime $C'$ with one asymptotic boundary; the path integral
on $C'$ is expected to reproduce the leading correction to the random matrix partition function due to time-reversal symmetry.   The main novel step
in computing this correction is to compute the torsion of a flat connection on $C$.  Once this is known, relatively standard steps similar to those that we have already explained
lead to the prediction for the correction to the partition function.

In computing the torsion, we consider a flat connection on $C$ in which the holonomy around the boundary is $U$ and the holonomy around the “interior” circle $\gamma$ where the
identification is made  is $V$.   ($\gamma$ will ultimately be a geodesic in a hyperbolic metric.)
Flatness of the connection implies that $U=V^2$, because the boundary on the left of fig. <ref>(a) is homotopic to a double cover of  $\gamma$.
We can satisfy these conditions with

-e^ıq ϕe^b/2   0   0

0   -e^ıq ϕe^-b/2   0

0   0   e^2ıq ϕ
),1cm V=(
ıe^ıq ϕ/2e^b/4   0   0

0   -ıe^ıq ϕ/2e^-b/4   0

0   0   e^ıq ϕ
Not coincidentally, this is the same $U$ that we used in eqn. (<ref>)  to describe a trumpet with an inner geodesic of
length $b$ and an $R$-symmetry parameter $\phi$.  Clearly, $V^2=U$,
and if we set $b=\phi=0$ so that $U$ reduces to $\diag(-1,-1|1)$, then $V$ has the properties described earlier for a reflection $\rR$ (acting on the fermions, it is imaginary
with determinant 1).  Replacing $\i$ by $-\i$ in the formula for $V$ would give an equivalent formula, up to a shift of $\phi$ by $2\pi$.

In eqn. (<ref>), $V$ is an element of the supergroup $\SU(1,1|1)$, or more exactly its complexification,
as is appropriate for a model with reflection symmetry $\rR$.  To get $\CR$ symmetry, we have to combine the expression for $V$ with a charge conjugation operation;
to postpone this slight  complication, we consider $\rR$ symmetry first, case (B) in the language of section <ref>.

In section 3.4.5 of [14], using the existence of a very simple cell decomposition of
the crosscap manifold (fig. <ref>(b)), a simple general recipe was described to compute $\tau_C$, the torsion of a flat bundle on $C$.  Here we will just use this
recipe and refer to the paper for an explanation.

For $x\in \mathfrak g$, let $\h V(x)= V x V^{-1}$. Then the torsion of the crosscap manifold $C$ is $\Ber' (\h V-1)$.   Decompose the Lie algebra $\mathfrak g$ of
$\SU(1,1|1)$, as $\mathfrak g^0\oplus \mathfrak g^\perp$, where $\g^0$ is the kernel of $\h V-1$ and $\g^\perp$ is its orthocomplement.   For the case of $\rR$ symmetry, $\mathfrak g^0$ consists of the subalgebra of diagonal matrices
and $\mathfrak g^\perp$ consists of the strictly off-diagonal matrices.   Because this decomposition is $\h V$-invariant, one has a factorization
τ_C=   τ_C^0τ_C^⊥,  with
τ_C^⊥= ' (V^⊥-1), 1cm τ_C^0= '( V^0-1).Here $\h V^0$ and $\h V^\perp$ are the restrictions of $\h V$ to  to $\mathfrak g^0$ and $\mathfrak g^\perp$.
For the case of $\CR$ symmetry that we consider later, the decomposition $\g=\g^0\oplus \g^\perp$ is slightly different, but it is still $\h V$-invariant, so eqn. (<ref>) still holds.

$\h V^\perp-1$ is invertible and preserves the $\Z_2$-grading of the Lie algebra, with separate blocks $\h V^\perp_+-1$ and $\h V^\perp_{-}-1$
acting on the bosonic and fermionic parts of $\g_\perp$.  So   '(V^⊥-1)=|( V^⊥_+-1)|/( V^⊥_--1).(See the paragraph after (<ref>) for a definition of $\Ber'$.)
In the present example, the eigenvalues of $\h V^\perp_{+}-1$  are  $-e^{\pm b/2}-1$, leading to  $|\det\,(\h V^\perp_{+}-1)|=(1+e^{b/2})(1+e^{-b/2})=4\cosh^2 \frac{b}{4}.$ The action on the fermionic generators is
V: q_1 q_2 q_3 q_4→ ı  e^ıϕ/2 e^b/4 q_1 ı  e^-ıϕ/2e^b/4 q_2-ı  e^ıϕ/2 e^-b/4 q_3-ı  e^-ıϕ/2e^-b/4 q_4. Therefore the eigenvalues of $\h V^\perp_{-}-1$  are $-1\pm \i e^{\pm b/4}e^{-\i \h q \phi/2}$ and $-1\pm \i e^{\pm b/4} e^{\i\h q \phi/2}$, so
$\det \,(\h V^\perp_{-}-1) = 4(\cos \frac{\h q\phi}{2}-\i \sinh \frac{b}{4})^2$.  Combining these results, we get
τ_C^⊥=cosh^2 b/4/(cosqϕ/2-ısinhb/4)^2. $\tau_C^0$ can be computed precisely as in the derivation of
eqn. (3.49) of [14].  It   is a simple ratio of measures for parameters $b,\phi$ that appear in $U$ and $V$ and potential gluing parameters $\varrho,\varphi$
for gluing the left boundary in fig. <ref>(a) onto something else:
τ_C^0=1/4 ḅϕ̣(ϱ̣φ̣)^-1. The factor of $1/4$ here has the same origin as the $1/2$ in eqn. (3.49) of [14]: one can write $\tau_C^0=\d \bar b\d\bar\phi (\d\varrho\d\varphi)^{-1}$, where
$\bar b$ is the length of the interior geodesic in the figure and $\bar\phi$ is the $R$-symmetry monodromy around this geodesic; as $b=2\bar b$, $\phi=2\bar\phi$,
this leads to  eqn. (<ref>).
So the crosscap torsion is
τ_C=τ_C^⊥τ_C^0 =cosh^2 b/4/4(cosqϕ/2-ısinhb/4)^2 ḅϕ̣(ϱ̣φ̣)^-1.

To turn this into a prediction that can be compared to random matrix theory, we have to combine this result with the boundary path integral of the Schwarzian mode.
As in eqn. (<ref>),  the trumpet path integral with boundary conditions natural for the torsion calculation is given by
Z_tr(b,ϕ) = Z_tr(b,ϕ) sinhb/2/coshb/2+cosq ϕ, where $Z_{\rm tr}(b,\phi)$ was computed in (<ref>). A trumpet that ends on a crosscap can be obtained by gluing an ordinary trumpet onto the crosscap manifold of fig. <ref>(a).   The gluing law for the torsion
gives a formula for the path integral that results from this gluing:
Z_xcap(b,ϕ)= Z_tr(b,ϕ)ḅϕ̣1/τ_Sτ_C, Using eqn. (<ref>) for $\t Z_{\rm tr}(b,\phi)$,  eqn. (<ref>) for $\tau_S$, and eqn. (<ref>) for $\tau_C$, we get finally the measure
for a trumpet that ends on a crosscap:
Z_xcap,(b,ϕ) =Z_tr(b,ϕ) ( coshb/2+ cosq ϕ)b/4 /8(cosqϕ/2-ısinhb/4)^2 ḅϕ̣, where we make explicit the fact that this is the prediction in a model
with $\CT $ or $\rR$ symmetry.
Since we have already computed $Z_{\rm tr}(b,\phi)$, this result can be compared to the matrix model prediction, as we will do in section <ref>.

Now we consider the case of $\tT$ or $\CR$ symmetry, case (A) in the language of section <ref>.    This means that we have to include  charge conjugation, which we will
call $\cC$, as part of the monodromy.   As we explain in detail momentarily, $\cC$ is an outer automorphism of $\SU(1,1|1)$, satisfying $\cC^2=1$. It acts trivially on $\SU(1,1)$ but reverses the
sign of the $R$-charge generator, which
was called $\sf z$ in eqn.  (<ref>).  Roughly, including $\cC$ means replacing the monodromy $V$ by $V_\cC=\cC V$.  However, because the $R$-charge is odd under $\cC$, the conjugacy
class of $V_\cC$ is independent of $\phi$.   To see this, write more explicitly $V_\phi$ for the $\phi$-dependent matrix  $V$ defined in eqn. (<ref>), and observe that,
because ${\sf z}$ is odd under $\cC$, we have
$e^{\lambda {\sf z}} \cC V_\phi e^{-\lambda {\sf z}}=\cC e^{-2\lambda {\sf z}} V_\phi=\cC V_{\phi- \lambda/\hq}$.
Hence by conjugation we can set $\phi=0$ in $V$, in which case also $\phi=0$ in $U=V^2$  (concretely, this happens because as the $R$-charge is odd under $\cC$,
$U=\cC V_\phi \cC V_\phi$ is actually independent of $\phi$).   So we consider the monodromies
-e^b/2   0   0

0   -e^-b/2   0

0   0    1
),1cm V_=V= (
ıe^b/4   0   0

0   -ıe^-b/4   0

0   0   1
$\cC$ cannot be written as a $3\times 3$ matrix, because it is an outer automorphism of the Lie algebra of $\SU(1,1|1)$ that
conjugates the $2|1$-dimensional representation  that we use in writing
most formulas
into an inequivalent representation, with reversed $R$-charges.   (The formula  $V_\cC =\cC V$ should really be interpreted as a formula in the group or in a $\cC$-invariant
representation such as the adjoint representation, not a formula in $3\times 3$ matrices as written in eqn. (<ref>).)
$\cC$ acts on the Lie algebra of $\SU(1,1|1)$, in the notation of eqn.     (<ref>), by

\begin{align}\label{actionc} {\sf q}_1& \leftrightarrow - {\sf q}_2 \cr
{\sf q}_3&\leftrightarrow {\sf q}_4 \cr
{\sf z}&\leftrightarrow -{\sf z} ,\end{align}

with ${\sf e},{\sf f},{\sf h}$ invariant.   This defines an automorphism of the Lie algebra, and clearly $\cC^2=1$.

To compute the torsion, we have to replace $\h V(x)=V x V^{-1}$ by $\h V_\cC(x)=\cC(V x V^{-1})   $ with  $V$ as defined in eqn. (<ref>).
One now has $\tau_C^\perp=\Ber'(\h V^\perp_{\cC}-1)$.
In the previous example, $\mathfrak g^\perp_+$ was spanned by the $\SU(1,1)$ generators ${\sf e,f}$. These generators are $\cC$-invariant,
so acting on them,
there is no difference between $\h V$ and $\h V_\cC$, and they contribute a factor $4\cosh^2\frac{b}{4}$ to $|\det(\h V_{\cC \perp,+}-1)|$, as before.
However, as the $R$-charge generator $\sf z$ is $\cC$-odd, it is not in the kernel of  $\h V_\cC-1$ and instead has eigenvalue $-2$.
Hence it contributes a factor of 2 to $|\det\,(\h V^\perp_{\cC,+}-1)|$, which now equals
Acting on  $\mathfrak g^\perp_-$, $\h V_\cC$ is quite different from $\h V$.   It acts by
V_: q_1 q_2 q_3 q_4→ -ıe^b/4 q_2 -ıe^b/4 q_1-ıe^-b/4 q_4-ıe^-b/4q_3. From this, one finds $\det(\h V^\perp_{\cC,-}-1)=4\cosh^2\frac{b}{4}=\frac{1}{2}\det(\h V^\perp_{\cC,+}-1)$.  So eqn. (<ref>) is replaced by
τ_C^⊥=2. What about $\tau_C^0$?     The kernel of $\h V_\cC$ is generated by $\sf h$, and the same derivation as in eqn. (3.49) of [14] leads to
$\tau_C^0=\frac{1}{2}\d b\d\varrho^{-1}$.    Putting these factors together,
τ_C=τ_C^⊥τ_C^0 =ḅ ϱ̣^-1. This is a measure for a flat connection on the crosscap manifold, where the monodromy $U$ around the outer circle automatically has $\phi=0$.

The prediction for the path integral measure for a trumpet ending on a crosscap in a theory with $\tT$ symmetry can be deduced from eqn. (<ref>) using
this formula for $\tau_C$:
Z_xcap,(b)=Z_tr(b,0) coshb/4/sinhb/4 ḅ φ̣. We have written the left hand side only as a function of $b$, with no $\phi$; and on the right hand side, we have set $\phi=0$ in the trumpet partition function $Z_{\rm tr}(b,\phi)$.

Compared to eqn. (<ref>), we see a notable difference: there is no $\d\phi$, and instead there is a $\d\varphi$.   Technically, the reason that this happened
is that because the $R$-charge generator $\sf z$ is not annihilated by $\h V_\cC$, $\tau_C^0$ did not contain a factor $\d\phi (\d\varphi)^{-1}$ that was present in the previous
analysis.   Eqn. (<ref>) is telling us that instead of integrating over an $R$-symmetry angle $\phi$ in the monodromy $V$, we have to integrate over an $R$-symmetry
parameter $\varphi$ that appears in gluing the crosscap to the trumpet.

The fact that there is no $R$-symmetry angle $\phi$ to integrate over is indeed something we already know, since the monodromies
(<ref>) do not depend on any such parameter.