Some work.

1 files changed, 39 insertions, 22 deletions

diff --git a/topology.tex b/topology.tex
index a7b68dc..d0b986e 100644
--- a/topology.tex
+++ b/topology.tex
@@ -199,16 +199,6 @@ is the vector of partial derivatives with respect to all $N+M+1$ variables.
 This integral is now in a form where standard techniques from mean-field theory
 can be applied to calculate it.
 
-In order for certain Gaussian integrals in the following calculation to be
-well-defined, it is necessary to treat instead the Lagrangian problem above
-with $\pmb\omega\mapsto i\pmb\omega$. This transformation does not effect the
-Dirac $\delta$ functions of the gradient, but it does change the determinant by
-a factor of $i^{N+M+1}$. We will see that the result of the rest of the
-calculation neglecting this factor is real. Since the Euler characteristic is
-also necessarily real, this indicates an inconsistency with this transformation
-when $N+M+1$ is odd. In fact, the Euler characteristic is always zero for
-odd-dimensional manifolds. This is the signature of it in this problem.
-
 \subsubsection{Calculation of the average Euler characteristic}
 
 To evaluate the average of $\chi$ over the constraints, we first translate the $\delta$ functions and determinant to integral form, with
@@ -574,7 +564,7 @@ To make the calculation compact, we introduce superspace coordinates. Define the
 \begin{equation}
   \pmb\phi(1)=\mathbf x+\bar\theta_1\pmb\eta+\bar{\pmb\eta}\theta_1+\bar\theta_1\theta_1i\hat{\mathbf x}
   \qquad
-  \sigma_k(1)=\omega_k+\bar\theta_1\gamma_k+\bar\gamma_k\theta_1+\bar\theta_1\theta_1\hat\omega_k
+  \sigma_k(1)=\omega_k+\bar\theta_1\gamma_k+\bar\gamma_k\theta_1+\bar\theta_1\theta_1i\hat\omega_k
 \end{equation}
 The Euler characteristic can be expressed using these supervectors as
 \begin{equation}
@@ -584,8 +574,8 @@ The Euler characteristic can be expressed using these supervectors as
     &=\int d\pmb\phi\,d\pmb\sigma\,\exp\left\{
       \int d1\left[
         H\big(\pmb\phi(1)\big)
-        +\frac i2\sigma_0(1)\left(\|\pmb\phi(1)\|^2-N\right)
-        +i\sum_{k=1}^M\sigma_k(1)\Big(V_k\big(\pmb\phi(1)\big)-V_0\Big)
+        +\frac12\sigma_0(1)\left(\|\pmb\phi(1)\|^2-N\right)
+        +\sum_{k=1}^M\sigma_k(1)\Big(V_k\big(\pmb\phi(1)\big)-V_0\Big)
       \right]
     \right\}
   \end{aligned}
@@ -597,14 +587,14 @@ Since this is an exponential integrand linear in the functions $V_k$, we can ave
     =\int d\pmb\phi\,d\pmb\sigma\,\exp\Bigg\{
       \int d1\left[
         H(\pmb\phi(1))
-        +\frac{i}2\sigma_0(1)\big(\|\pmb\phi(1)\|^2-N\big)
-        -iV_0\sum_{k=1}^M\sigma_k(1)
+        +\frac12\sigma_0(1)\big(\|\pmb\phi(1)\|^2-N\big)
+        -V_0\sum_{k=1}^M\sigma_k(1)
       \right] \\
-      -\frac12\int d1\,d2\,\sum_{k=1}^M\sigma_k(1)\sigma_k(2)f\left(\frac{\pmb\phi(1)\cdot\pmb\phi(2)}N\right)
+      +\frac12\int d1\,d2\,\sum_{k=1}^M\sigma_k(1)\sigma_k(2)f\left(\frac{\pmb\phi(1)\cdot\pmb\phi(2)}N\right)
     \Bigg\}
   \end{aligned}
 \end{equation}
-This is a Gaussian integral in the Lagrange multipliers with $1\leq k\leq M$.
+This is a super-Gaussian integral in the Lagrange multipliers with $1\leq k\leq M$.
 Performing that integral yields
 \begin{equation}
   \begin{aligned}
@@ -612,7 +602,7 @@ Performing that integral yields
     &=\int d\pmb\phi\,d\sigma_0\,\exp\Bigg\{
       \int d1\left[
         H(\pmb\phi(1))
-        +\frac{i}2\sigma_0(1)\big(\|\pmb\phi(1)\|^2-N\big)
+        +\frac12\sigma_0(1)\big(\|\pmb\phi(1)\|^2-N\big)
       \right] \\
     &\hspace{5em}-\frac M2V_0^2\int d1\,d2\,f\left(\frac{\pmb\phi(1)\cdot\pmb\phi(2)}N\right)^{-1}
       -\frac M2\log\operatorname{sdet}f\left(\frac{\pmb\phi(1)\cdot\pmb\phi(2)}N\right)
@@ -636,7 +626,7 @@ These new variables can replace $\pmb\phi$ in the integral using a generalized H
     \,\exp\Bigg\{
       N\int d1\left[
         \mathbb M(1)
-        +\frac{i}2\sigma_0(1)\big(\mathbb Q(1,1)-1\big)
+        +\frac12\sigma_0(1)\big(\mathbb Q(1,1)-1\big)
       \right] \\
     &\hspace{5em}-\frac M2V_0^2\int d1\,d2\,f(\mathbb Q)^{-1}(1,2)
       -\frac M2\log\operatorname{sdet}f(\mathbb Q)
@@ -695,8 +685,8 @@ while $\bar H$, $H$, $\bar{\hat H}$, $\hat H$, $\bar H_0$ and $H_0$ are Grassman
 \end{align}
 We can treat the integral over $\sigma_0$ immediately. It gives
 \begin{equation}
-  \int d\sigma_0\,e^{N\int d1\,\frac i2\sigma_0(1)(\mathbb Q(1,1)-1)}
-  =2\pi\,\delta(C-1)\,\delta(G+R)\,\bar HH
+  \int d\sigma_0\,e^{N\int d1\,\frac12\sigma_0(1)(\mathbb Q(1,1)-1)}
+  =2\pi i\,\delta(C-1)\,\delta(G+R)\,\bar HH
 \end{equation}
 This therefore sets $C=1$ and $G=-R$ in the remainder of the integrand, as well
 as setting everything depending on $\bar H$ and $H$ to zero. With these solutions inserted, the remaining remaining terms in the exponential give
@@ -730,10 +720,37 @@ as setting everything depending on $\bar H$ and $H$ to zero. With these solution
 \section{Details of the calculation of the average number of stationary points}
 \label{sec:complexity.details}
 
+\begin{align}
+  \pmb\phi(1,2)
+  &=\mathbf x
+  +\bar\theta_1\pmb\eta_1+\bar{\pmb\eta}_1\theta_1\bar\theta_2\theta_2
+  +\bar\theta_2\pmb\eta_2+\bar{\pmb\eta}_2\theta_2\bar\theta_1\theta_1 \\
+  &\qquad+\frac1{\sqrt2}(\bar\theta_1\theta_2+\bar\theta_2\theta_1)\mathbf a
+  +\frac i{\sqrt2}(\bar\theta_1\theta_1+\bar\theta_2\theta_2)\mathbf b
+  +\bar\theta_1\theta_1\bar\theta_2\theta_2i\hat{\mathbf x}
+  \notag \\
+  \sigma_k(1,2)
+  &=\omega_k
+  +\bar\theta_1\gamma_{1k}+\bar{\gamma}_{1k}\theta_1\bar\theta_2\theta_2
+  +\bar\theta_2\gamma_{2k}+\bar{\gamma}_{2k}\theta_2\bar\theta_1\theta_1 \\
+  &\qquad+\frac1{\sqrt2}(\bar\theta_1\theta_2+\bar\theta_2\theta_1)c_k
+  +\frac i{\sqrt2}(\bar\theta_1\theta_1+\bar\theta_2\theta_2)d_k
+  +\bar\theta_1\theta_1\bar\theta_2\theta_2\hat\omega_k
+  \notag
+\end{align}
+\begin{equation}
+  \mathcal N_H(\Omega)
+  =\lim_{\epsilon\to0}\int d\pmb\phi\,d\pmb\sigma\,e^{
+   \int d1\,d2\,L(\pmb\phi(1,2),\pmb\sigma(1,2))
+   -\frac{i\epsilon}2
+   (\|\mathbf a\|^2-\|\mathbf b\|^2+\|\mathbf c\|^2-\|\mathbf d\|^2)
+ }
+\end{equation}
+
 \section{The quenched shattering energy in {\oldstylenums 1}\textsc{frsb} models}
 \label{sec:1frsb}
 
-\cite{Kent-DObias_2023_How}
+\cite{Kent-Dobias_2023_How}
 \[
   \chi_0(q)=\frac1{\hat\omega_1}f''(q)^{-1/2}-\frac{r_d^2}{d_d}
 \]