From 28cfd58481308e5d670be280e91ab5d458b86c9b Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Mon, 9 Sep 2024 10:09:21 +0200
Subject: Some more retooling

---
 topology.tex | 50 ++++++++++++++++++++++++++++++++++++++++----------
 1 file changed, 40 insertions(+), 10 deletions(-)

(limited to 'topology.tex')

diff --git a/topology.tex b/topology.tex
index ee658a0..4b48e3f 100644
--- a/topology.tex
+++ b/topology.tex
@@ -382,6 +382,15 @@ and $\mathcal S_\chi(m^*)=0$. If this solution were always well-defined, it woul
   V_0^2>V_{\text{\textsc{sat}}\ast}^2\equiv\frac{f'(1)}\alpha-f(1)
 \end{equation}
 This corresponds precisely to the satisfiability transition found in previous work by a replica symmetric analysis of the cost function \eqref{eq:cost} \cite{Fyodorov_2019_A, Fyodorov_2020_Counting, Fyodorov_2022_Optimization, Tublin_2022_A, Vivo_2024_Random}.
+However, the action becomes complex in the region $m^2<m_\text{min}^2$ for
+\begin{equation}
+  m_\text{min}^2
+  =1-\frac{f(1)^2}{f'(1)}\frac{V_0^2[2(1+\sqrt{1-\alpha})-\alpha]-\alpha f(1)}{
+    2V_0^2f(1)(2-\alpha)-\alpha(V_0^4+f(1)^2)
+  }
+\end{equation}
+When $m_*^2<m_\text{min}^2$, this solution is no longer valid. Likewise, when $m_\text{min}^2>0$, the solution at $m=0$ is also not valid. In fact, it is not clear what the average value of the Euler characteristic should be at all
+
 However, when the magnitude of $V_0$ is sufficiently large, with
 \begin{equation}
   V_0^2>V_\text{on}^2\equiv\frac{1-\alpha+\sqrt{1-\alpha}}\alpha f(1)
@@ -498,13 +507,7 @@ will clarify the meaning of our effective action for the average Euler
 characteristic in the range of overlaps $m$ where it takes a complex value. This is because the average number of stationary points is a
 nonnegative number. If the region of complex $\mathcal S_\chi$ has a
 well-defined number of stationary points, it indicates that we are looking at a
-situation with a negative average Euler characteristic. On the other hand, if
-the average number of stationary points yields a complex value at some latitude
-$m$, it must be because it is either too large or small in $N$ to be captured by
-the calculation, e.g., that it behaves like $e^{N^2\Sigma}$ or
-$e^{-N^2\Sigma}$. The following calculation indicates this second situation:
-the region of complex action is due to a lack of stationary points to
-contribute to the Euler characteristic at those latitudes.
+situation with a negative average Euler characteristic. In fact, this is what we find below.
 
 To compute the complexity, we follow a similar procedure to the Euler
 characteristic. The main difference lies in how we treat the absolute value
@@ -528,9 +531,36 @@ $\mathbf b$ are regular vectors. This introduces many new order parameters into
 the problem, but this is a difficulty of scale rather than principle. With this
 identity substituted for the usual determinant one, the problem can be solved
 much as before. The details of this solution are relegated to
-Appendix~\ref{sec:complexity.details}. The result is that, to largest order in
-$N$, the logarithm of the average number of stationary points is the same as
-the logarithm of the average Euler characteristic.
+Appendix~\ref{sec:complexity.details}. We again reduce the result to a single integral over the overlap $m$ with the height axis, of the form
+\begin{equation}
+  \overline{\mathcal N_H(\Omega)}
+  \propto\int dm\,e^{N\mathcal S_\mathcal N(m)}
+\end{equation}
+For $m^2>m_c$ the effective action $\mathcal S_\mathcal N(m)$ is the same as
+$\mathcal S_\chi(m)$ for the Euler characteristic, where we have defined
+\begin{equation}
+  m_c^2
+  =1-\frac{2(1-\alpha)f(1)f'(1)}{
+    (2-\alpha)f(1)f''(1)+2(1-\alpha)f'(1)^2
+    -\sqrt{\alpha f''(1)}\sqrt{4V_0^2(1-\alpha)f'(1)^2+\alpha f(1)^2f''(1)}
+  }
+\end{equation}
+For $m^2\leq m_c$ it instead takes the value
+\begin{align}
+  &\mathcal S_\mathcal N^*(m)
+  =
+  -\frac12\alpha\left(
+    1+\frac{V_0^2\big[f''(1)(1-m^2)-f'(1)\big]}{(1-m^2)[f''(1)f(1)+f'(1)^2]-f(1)f'(1)}
+  \right)
+  \\
+  &\quad+\frac12(1-\alpha)\log\left(
+    \frac{f''(1)(1-m^2)}{f'(1)(1-\alpha)}
+  \right)
+  -\frac12\alpha\log\left(
+    \frac{\alpha f'(1)V_0^2}{(1-m^2)[f''(1)f(1)+f'(1)^2]-f(1)f'(1)}
+  \right)
+  \notag
+\end{align}
 
 \section{Implications for the topology of solutions}
 
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