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authorJaron Kent-Dobias <jaron@kent-dobias.com>2024-06-11 16:32:54 +0200
committerJaron Kent-Dobias <jaron@kent-dobias.com>2024-06-11 16:32:54 +0200
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Added appendix on BRST.
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@@ -448,6 +448,9 @@ Finally, the marginal complexity is defined by evaluating the complexity conditi
\subsection{General features of saddle point computation}
+Several elements of the computation of the marginal complexity, and indeed the
+ordinary dominant complexity, follow from the formulae of the above section in
+the same way.
\begin{align}
\label{eq:delta.grad}
&\delta\big(\nabla H(\mathbf x_a,\pmb\omega_a)\big)
@@ -1104,9 +1107,15 @@ absolute value sign) that make up the basic Kac--Rice measure, so that we can wr
\end{aligned}
\end{equation}
where we have written $d1=d\theta_1\,d\bar\theta_1$ and $d\pmb\phi=d\mathbf
-x\,d\bar{\pmb\eta}\,d\pmb\eta\,\frac{d\hat{\mathbf x}}{(2\pi)^N}$. Besides some deep connections
-to the physics of BRST, this compact notation dramatically simplifies the
-analytical treatment of the problem. The reason why this simplification is
+x\,d\bar{\pmb\eta}\,d\pmb\eta\,\frac{d\hat{\mathbf x}}{(2\pi)^N}$. Besides some
+deep connections to the physics of BRST, this compact notation dramatically
+simplifies the analytical treatment of the problem. The energy of stationary points can also be fixed using this notation, by writing
+\begin{equation}
+ \int d\pmb\phi\,\frac{d\hat\beta}{2\pi}\,e^{\hat\beta E+\int d1\,(1-\hat\beta\bar\theta_1\theta_1)H(\pmb\phi(1))}
+\end{equation}
+which a small calculation confirms results in the same expression as \eqref{eq:delta.energy}.
+
+The reason why this simplification is
possible is because there are a large variety of superspace algebraic and
integral operations with direct corollaries to their ordinary real
counterparts. For instance, consider a super linear operator $M(1,2)$, which
@@ -1155,6 +1164,72 @@ save for the inverse of $\det D$. The same method can be used to calculate the
superdeterminant in arbitrary superspaces, where for $\mathbb R^{N|2D}$ each
basis has $2^{2D-1}$ elements. For instance, for $\mathbb R^{N|4}$ we have $\mathbf e(1,2)=\{1,\bar\theta_1\theta_1,\bar\theta_2\theta_2,\bar\theta_1\theta_2,\bar\theta_2\theta_1,\bar\theta_1\bar\theta_2,\theta_1\theta_2,\bar\theta_1\theta_1\bar\theta_2\theta_2\}$ and $\mathbf f(1,2)=\{\bar\theta_1,\theta_1,\bar\theta_2,\theta_2,\bar\theta_1\theta_1\bar\theta_2,\bar\theta_2\theta_2\theta_1,\bar\theta_1\theta_1\theta_2,\bar\theta_2\theta_2\theta_1\}$.
+\section{BRST symmetry}
+\label{sec:brst}
+
+The superspace representation is also helpful because it can make manifest an
+unusual symmetry in the dominant complexity of minima that would otherwise be
+obfuscated. This arises from considering the Kac--Rice formula as a kind of
+gauge fixing procedure \cite{Zinn-Justin_2002_Quantum}. Around each stationary
+point consider making the coordinate transformation $\mathbf u=\nabla H(\mathbf
+x)$. Then in the absence of fixing the trace, the Kac--Rice measure becomes
+\begin{equation}
+ \int d\nu(\mathbf x,\pmb\omega\mid E)
+ =\int\sum_\sigma d\mathbf u\,\delta(\mathbf u)\,
+ \delta\big(NE-H(\mathbf x_\sigma)\big)
+\end{equation}
+where the sum is over stationary points. This integral has a symmetry of its
+measure of the form $\mathbf u\mapsto\mathbf u+\delta\mathbf u$. Under the
+nonlinear transformation that connects $\mathbf u$ and $\mathbf x$, this
+implies a symmetry of the measure in the Kac--Rice integral of $\mathbf
+x\mapsto\mathbf x+(\operatorname{Hess}H)^{-1}\delta\mathbf u$. This symmetry, while exact, is
+nonlinear and difficult to work with.
+
+When the absolute value sign has been dropped and Grassmann vectors introduced,
+this symmetry can be simplified considerably. Due to the expansion properties
+of Grassmann integrals, any appearance of $-\bar{\pmb\eta}\pmb\eta^T$ in the
+integrand resolves to $(\operatorname{Hess}H)^{-1}$. The
+symmetry of the measure can then be written
+\begin{equation}
+ \mathbf x\mapsto \mathbf x-\bar{\pmb\eta}\pmb\eta^T\delta\mathbf u
+ =\mathbf x+\bar{\pmb\eta}\delta\epsilon
+\end{equation}
+where $\delta\epsilon=-\pmb\eta^T\delta\mathbf u$ is a Grassmann number. This
+establishes that $\delta\mathbf x=\bar{\pmb\eta}\delta\epsilon$, now linear. The rest of
+the transformation can be built by requiring that the action is invariant after
+expansion in $\delta\epsilon$. Ignoring for a moment the piece of the measure
+fixing the trace of the Hessian, this gives
+\begin{align}
+ \delta\mathbf x=\bar{\pmb\eta}\,\delta\epsilon &&
+ \delta\hat{\mathbf x}=-i\hat\beta\bar{\pmb\eta}\,\delta\epsilon &&
+ \delta\pmb\eta=-i\hat{\mathbf x}\,\delta\epsilon &&
+ \delta\bar{\pmb\eta}=0
+\end{align}
+so that the differential form of the symmetry is
+\begin{equation}
+ \mathcal D=\bar{\pmb\eta}\frac\partial{\partial\mathbf x}
+ -i\hat\beta\bar{\pmb\eta}\frac\partial{\partial\hat{\mathbf x}}
+ -i\hat{\mathbf x}\frac\partial{\partial\pmb\eta}
+\end{equation}
+The Ward identities associated with this symmetry give rise to relationships among the order parameters. These identities are
+\begin{align}
+ 0=\frac1N\mathcal D\langle\mathbf x_a^T\pmb\eta_b\rangle
+ =\frac1N\left[
+ \langle\bar{\pmb\eta}_a^T\pmb\eta_b\rangle-
+ i\langle\mathbf x_a^T\hat{\mathbf x}_b\rangle
+ \right]
+ =G_{ab}+R_{ab} \\
+ 0=\frac iN\mathcal D\langle\hat{\mathbf x}_a^T\pmb\eta_b\rangle
+ =\frac1N\left[
+ \hat\beta\langle\bar{\pmb\eta}_a^T\pmb\eta_b\rangle
+ +\langle\hat{\mathbf x}_a^T\hat{\mathbf x}_b\rangle
+ \right]
+ =\hat\beta G_{ab}+D_{ab}
+\end{align}
+These identities establish $G_{ab}=-R_{ab}$ and $D_{ab}=\hat\beta R_{ab}$,
+allowing elimination of the matrices $G$ and $D$ in favor of $R$. Fixing the
+trace to $\mu$ explicitly breaks this symmetry, and the simplification is lost.
+
\section{Complexity of dominant optima in the least-squares problem}
\label{sec:dominant.complexity}