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| -rw-r--r-- | marginal.tex | 70 |
1 files changed, 45 insertions, 25 deletions
diff --git a/marginal.tex b/marginal.tex index b9fabc9..442e7e2 100644 --- a/marginal.tex +++ b/marginal.tex @@ -641,16 +641,31 @@ treat the determinant keeping the absolute value signs, as in previous works \cite{Folena_2020_Rethinking, Kent-Dobias_2023_How}. However, other of our examples are for models where the same techniques are impossible. -For the cases studied here, fixing the trace results in a relationship -between $\mu$ and the Lagrange multipliers enforcing the constraints. This is -because the trace of $\partial\partial H$ is typically an order of $N$ smaller -than the trace of $\partial\partial g_i$. The result is that +Finally, the $\delta$-function fixing the trace of the Hessian to $\mu$ in +\eqref{eq:kac-rice.measure.2} must be addressed. One could treat it using a +Fourier representation as in (\ref{eq:delta.grad}--\ref{eq:delta.eigen}), but +this is inconvenient because a term of the form +$\operatorname{Tr}\partial\partial H(\mathbf x)$ in the exponential integrand +cannot be neatly captured in superspace representation introduced in the next +section. However, in the cases we study in this paper a simplification can be made: the trace of $\partial\partial H$ can be separated into two pieces, one +that is spatially independent and one that is typically small, or +\begin{equation} \label{eq:mu.star} + \operatorname{Tr}\partial\partial H(\mathbf x)=N\mu^*_H+\Delta_H(\mathbf x) +\end{equation} +where $\overline{\mu^*_H}=\mu^*$ and $\overline{\Delta_H(\mathbf x)}=O(N^0)$. +Then fixing the trace of the Hessian to $\mu$ implies that \begin{equation} - \mu - =\frac1N\operatorname{Tr}\operatorname{Hess}H(\mathbf x) - =\frac1N\sum_{i=1}^r\omega_i\operatorname{Tr}\partial\partial g_i(\mathbf x) - +O(N^{-1}) + \begin{aligned} + \mu + &=\frac1N\operatorname{Tr}\operatorname{Hess}H(\mathbf x) + =\frac1N\left(\partial\partial H(\mathbf x)+ + \sum_{i=1}^r\omega_i\operatorname{Tr}\partial\partial g_i(\mathbf x)\right) + \\ + &=\mu^*+\frac1N\sum_{i=1}^r\omega_i\operatorname{Tr}\partial\partial g_i(\mathbf x) + +O(N^{-1}) + \end{aligned} \end{equation} +for typical samples $H$. In particular, here we study only cases with quadratic $g_i$, which results in a linear expression relating $\mu$ and the $\omega_i$ that is independent of $\mathbf x$. Since $H$ contains the disorder of the problem, this simplification means @@ -720,7 +735,7 @@ functions and the determinant made. The new measures \delta\big((\mathbf s_a^\alpha)^T\partial\mathbf g(\mathbf x_a)\big) \\ d\pmb\omega&=\bigg(\prod_{i=1}^rd\omega_i\bigg) - \,\delta\bigg(N\mu-\sum_i^r\omega_i\operatorname{Tr}\partial\partial g_i\bigg) + \,\delta\bigg(N\mu-\mu^*-\sum_i^r\omega_i\operatorname{Tr}\partial\partial g_i\bigg) \end{align} collect the individual measures of the various fields embedded in the superfield, along with their constraints. \end{widetext} @@ -765,7 +780,8 @@ unit normal distribution \cite{Crisanti_1993_The}. We focus on marginal minima in models with $f'(0)=0$, which corresponds to models without a random external field. Such a random field would correspond in each individual sample $H$ to a signal, and therefore complicate the analysis by correlating the positions of -stationary points and the eigenvectors of their Hessians. +stationary points and the eigenvectors of their Hessians. Here, $\mu^*$ of +\eqref{eq:mu.star} is zero. The marginal optima of these models can be studied without the methods introduced in this paper, and have been in the past \cite{Folena_2020_Rethinking, @@ -1030,7 +1046,7 @@ for $\mathbf x,\mathbf x'\in\mathbb R^N$ by \overline{H_i(\mathbf x)H_j(\mathbf x')} =N\delta_{ij}f_i\left(\frac{\mathbf x\cdot\mathbf x'}N\right) \end{equation} -with the functions $f_1$ and $f_2$ not necessarily the same. +with the functions $f_1$ and $f_2$ not necessarily the same. As for the spherical spin glasses, $\mu^*$ of \eqref{eq:mu.star} is zero. In this problem, there is an energetic competition between the independent spin glass energies on each sphere and their tendency to align or anti-align through @@ -1374,9 +1390,12 @@ As in the previous sections, we used the method of Lagrange multipliers to analy -V_k(\mathbf x)\partial\partial V_k(\mathbf x)\right]+\omega I \end{aligned} \end{align} -As in the spherical and multispherical spin glasses, fixing the trace of the Hessian -is equivalent to constraining the value of the Lagrange -multiplier $\omega=\mu$. +Unlike in the spherical and multispherical spin glasses, the value $\mu^*$ +defined in \eqref{eq:mu.star} giving the typical value of +$\frac1N\operatorname{Tr}\partial\partial H$ is not always zero. Instead +$\mu^*=-f'(0)$, nonzero where there is a linear term in $V$. Fixing the trace +of the Hessian is therefore equivalent to setting $\omega=\mu+f'(0)$. + The derivation of the marginal complexity for this model is complicated, but can be made schematically like that of the derivation of the equilibrium free @@ -1389,10 +1408,11 @@ $\lambda^*$ is given by \begin{aligned} \mathcal N(E,\mu,\lambda^*)^n &=\int d\hat\beta\,d\hat\lambda\prod_{a=1}^n\lim_{m_a\to0}\prod_{\alpha=1}^{m_a}d\pmb\phi_a^\alpha - \exp\left\{ + \\ + &\qquad\times\exp\left\{ \delta^{\alpha1}N(\hat\beta E+\hat\lambda\lambda^*) -\frac12\int d1\,d2\,\left[B^\alpha(1,2)\sum_{k=1}^MV_k(\pmb\phi_a^\alpha(1,2))^2 - -\mu\|\pmb\phi_a^\alpha(1,2)\|^2\right] + -\big(\mu+f'(0)\big)\|\pmb\phi_a^\alpha(1,2)\|^2\right] \right\} \end{aligned} \end{equation} @@ -1509,7 +1529,7 @@ with an effective action \begin{equation} \begin{aligned} &\mathcal S_\mathrm{RSS}(\hat\beta,\hat\lambda,r,d,g,q_0,\tilde q_0\mid\lambda^*,E,\mu,\beta) - =\hat\beta E-\mu(r+g+\hat\lambda) + =\hat\beta E-\big(\mu+f'(0)\big)(r+g+\hat\lambda) +\hat\lambda\lambda^* +\frac12\log\left(\frac{d+r^2}{g^2} \times\frac{1-2q_0+\tilde q_0^2}{(1-q_0)^2}\right) \\ @@ -1532,7 +1552,7 @@ taking the zero-temperature limit, we find \begin{equation} \begin{aligned} &\mathcal S_\mathrm{RSS}(\hat\beta,\hat\lambda,r,d,g,y,\Delta z\mid\lambda^*,E,\mu,\infty) - =\hat\beta E-\mu(r+g+\hat\lambda) + =\hat\beta E-\big(\mu+f'(0)\big)(r+g+\hat\lambda) +\hat\lambda\lambda^* +\frac12\log\left(\frac{d+r^2}{g^2}\times\frac{y^2-2\Delta z}{y^2}\right) \\ @@ -2012,9 +2032,9 @@ the replicated count of stationary points can be written =\int d\hat\beta\prod_{a=1}^n\,d\pmb\phi_a\, \exp\bigg[ N\hat\beta E \\ - &\qquad-\frac12\int d1\,\left( + &-\frac12\int d1\,\left( B(1)\sum_{k=1}^MV_k(\pmb\phi_a(1))^2 - -\mu\|\pmb\phi_a(1)\|^2 + -\big(\mu+f'(0)\big)\|\pmb\phi_a(1)\|^2 \right) \bigg] \end{aligned} @@ -2057,9 +2077,9 @@ Making the $M$ independent Gaussian integrals, we find \begin{equation} \begin{aligned} &\mathcal N(E,\mu)^n - =\int d\hat\beta\left(\prod_{a=1}^nd\pmb\phi_a\right) - \exp\bigg[ - nN\hat\beta E+\frac\mu2\sum_a^n\int d1\,\|\pmb\phi_a\|^2 \\ + =\int d\hat\beta\left(\prod_{a=1}^nd\pmb\phi_a\right) \\ + &\times\exp\bigg[ + nN\hat\beta E+\frac{\mu+f'(0)}2\sum_a^n\int d1\,\|\pmb\phi_a\|^2 \\ &\quad-\frac M2\log\operatorname{sdet}\left( \delta_{ab}\delta(1,2)+B(1)f\left(\frac{\pmb\phi_a(1)\cdot\pmb\phi_b(2)}N\right) \right) @@ -2080,7 +2100,7 @@ We therefore have &\mathcal N(E,\mu)^n =\int d\hat\beta\,d\mathbb Q\, \exp\bigg\{ - nN\hat\beta E+N\frac\mu2\operatorname{sTr}\mathbb Q + nN\hat\beta E+N\frac{\mu+f'(0)}2\operatorname{sTr}\mathbb Q +\frac N2\log\operatorname{sdet}\mathbb Q -\frac M2\log\operatorname{sdet}\left[ \delta_{ab}\delta(1,2)+B(1)f(\mathbb Q_{ab}(1,2)) @@ -2110,7 +2130,7 @@ where the effective action is given by \begin{equation} \begin{aligned} \mathcal S_\mathrm{KR}(\hat\beta,C,R,D,G) - &=\hat\beta E+\lim_{n\to0}\frac1n\Bigg(-\mu\operatorname{Tr}(G+R) + &=\hat\beta E+\lim_{n\to0}\frac1n\Bigg(-\big(\mu+f'(0)\big)\operatorname{Tr}(G+R) +\frac12\log\det\big[G^{-2}(CD+R^2)\big] +\alpha\log\det\big[I+G\odot f'(C)\big] \\ |