Converted more vectors to bold, and clarified response function computation.

1 files changed, 22 insertions, 9 deletions

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index e918176..a7a03cc 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -1006,16 +1006,26 @@ At $\hat \beta>\hat \beta_f$ there is a further transition.
 
 \subsection{\textit{R} and \textit{D}: response functions}
 
-The matrix field $R$ is related to responses of the stationary points to perturbations of the tensors $J$:
+The matrix field $R$ is related to responses of the stationary points to
+perturbations of the tensors $J$. Since the only dependence on $J$ lies in the
+measure, once the normalization $\mathcal N$ is replicated one finds
+\begin{equation}
+  \begin{aligned}
+    \frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}
+    &=\lim_{n\to0}\frac1{N^p}\sum_{i_1\cdots i_p}\frac\partial{\partial J^{(p)}_{i_1\cdots i_p}}
+    \int\left(\prod_a^nd\nu(\mathbf s_a)\right)\,s^1_{i_1}\cdots s^1_{i_p} \\
+    & =\lim_{n\to0}\int\left(\prod_a^nd\nu(\mathbf s_a)\right)\sum_b^n\left[
+      \hat\beta\left(\frac{\mathbf s_1\cdot\mathbf s_b}N\right)^p+
+        p\left(-i\frac{\mathbf s_1\cdot\hat{\mathbf s}_b}N\right)\left(\frac{\mathbf s_1\cdot\mathbf s_b}N\right)^{p-1}
+      \right]
+  \end{aligned}
+\end{equation}
+Taking the average of this expression over disorder and averaging over the equivalent replicas in the integral gives, similar to before,
 \begin{equation}
   \begin{aligned}
     \overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}}
-    =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1n\sum_{ab}\left[
-      \hat\beta\left(\frac{s_a\cdot s_b}N\right)^p+
-      p\left(-i\frac{\hat s_a\cdot s_b}N\right)\left(\frac{s_a\cdot s_b}N\right)^{p-1}
-    \right]} \\
-    =\lim_{n\to0}\frac1n\sum_{ab}(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1})
-    =\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x))
+    &=\lim_{n\to0}\int D[C,R,D]\,\frac1n\sum_{ab}^n(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1})e^{nN\Sigma[C,R,D]}\\
+    &=\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x))
   \end{aligned}
 \end{equation}
 In particular, when the energy is unconstrained ($\hat\beta=0$) and there is replica symmetry,
@@ -1029,7 +1039,10 @@ tend to align with a field. The energy constraint has a significant
 contribution due to the perturbation causing stationary points to move up or
 down in energy.
 
-The matrix field $D$ is related to the response of the complexity to such perturbations:
+The matrix field $D$ is related to the response of the complexity to
+perturbations to the variance of the tensors $J$. This can be found by taking
+the expression for the complexity and inserting the dependence of $f$ on the
+coefficients $a_p$, then differentiating:
 \begin{equation}
   \begin{aligned}
     \frac{\partial\Sigma}{\partial a_p}
@@ -1051,7 +1064,7 @@ When the saddle point of the Kac--Rice problem is supersymmetric,
 \end{equation}
 and in particular for $p=1$
 \begin{equation}
-  \frac{\partial\Sigma}{a_1}
+  \frac{\partial\Sigma}{\partial a_1}
   =\frac{\hat\beta}4\overline{\frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}}}
 \end{equation}
 i.e., the change in complexity due to a linear field is directly related to the