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diff --git a/2-point.tex b/2-point.tex index 9be13e9..ce2d1bc 100644 --- a/2-point.tex +++ b/2-point.tex @@ -612,314 +612,7 @@ $\pmb\sigma_1$ is special among the set of $\pmb\sigma$ replicas, since it alone is constrained to lie a given overlap from the $\mathbf s$ replicas. This replica asymmetry will be important later. -\subsection{The Hessian factors} - -The double partial derivatives of the energy are Gaussian with the variance -\begin{equation} - \overline{(\partial_i\partial_jH(\mathbf s))^2}=\frac1Nf''(1) -\end{equation} -which means that the matrix of partial derivatives belongs to the GOE class. Its spectrum is given by the Wigner semicircle -\begin{equation} - \rho(\lambda)=\begin{cases} - \frac2{\pi}\sqrt{1-\big(\frac{\lambda}{\mu_\text m}\big)^2} & \lambda^2\leq\mu_\text m^2 \\ - 0 & \text{otherwise} - \end{cases} -\end{equation} -with radius $\mu_\text m=\sqrt{4f''(1)}$. Since the Hessian differs from the -matrix of partial derivatives by adding the constant diagonal matrix $\omega -I$, it follows that the spectrum of the Hessian is a Wigner semicircle shifted -by $\omega$, or $\rho(\lambda+\omega)$. - -The average over factors depending on the Hessian alone can be made separately -from those depending on the gradient or energy, since for random Gaussian -fields the Hessian is independent of these \cite{Bray_2007_Statistics}. In -principle the fact that we have conditioned the Hessian to belong to stationary -points of certain energy, stability, and proximity to another stationary point -will modify its statistics, but these changes will only appear at subleading -order in $N$ \cite{Ros_2019_Complexity}. At leading order, the various expectations factorize, each yielding -\begin{equation} - \overline{\big|\det\operatorname{Hess}H(\mathbf s,\omega)\big|\,\delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s,\omega)\big)} - =e^{N\int d\lambda\,\rho(\lambda+\mu)\log|\lambda|}\delta(N\mu-N\omega) -\end{equation} -Therefore, all of the Lagrange multipliers are fixed to the stabilities $\mu$. We define the function -\begin{equation} - \begin{aligned} - \mathcal D(\mu) - &=\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\ - &=\begin{cases} - \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2} - & \mu^2\leq\mu_\text m^2 \\ - \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2} - -\left|\frac{\mu}{\mu_\text m}\right|\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1} - -\log\left(\left|\frac{\mu}{\mu_\text m}\right|-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu^2>\mu_\text m^2 - \end{cases} - \end{aligned} -\end{equation} -and the full factor due to the Hessians is -\begin{equation} - e^{Nm\mathcal D(\mu_0)+Nn\mathcal D(\mu_1)}\left[\prod_a^m\delta(N\mu_0-N\varsigma_a)\right]\left[\prod_a^n\delta(N\mu_1-N\omega_a)\right] -\end{equation} - -\subsection{The other factors} - -Having integrated over the Lagrange multipliers using the $\delta$-functions -resulting from the average of the Hessians, any $\delta$-functions in the -remaining integrand we Fourier transform into their integral representation -over auxiliary fields. The resulting integrand has the form -\begin{equation} - e^{ - Nm\hat\beta_0E_0+Nn\hat\beta_1E_1 - -\sum_a^m\left[(\pmb\sigma_a\cdot\hat{\pmb\sigma}_a)\mu_0 - -\frac12\hat\mu_0(N-\pmb\sigma_a\cdot\pmb\sigma_a) - \right] - -\sum_a^n\left[(\mathbf s_a\cdot\hat{\mathbf s}_a)\mu_1 - -\frac12\hat\mu_1(N-\mathbf s_a\cdot\mathbf s_a) - -\frac12\hat\mu_{12}(Nq-\pmb\sigma_1\cdot\mathbf s_a) - \right] - +\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t) - } -\end{equation} -where we have introduced the linear operator -\begin{equation} - \mathcal O(\mathbf t) - =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left( - i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0 - \right) - + - \sum_a^n\delta(\mathbf t-\mathbf s_a)\left( - i\hat{\mathbf s}_a\cdot\partial_{\mathbf t}-\hat\beta_1 - \right) -\end{equation} -Here the $\hat\beta$s are the fields auxiliary to the energy constraints, the -$\hat\mu$s are auxiliary to the spherical and overlap constraints, and the -$\hat{\pmb\sigma}$s and $\hat{\mathbf s}$s are auxiliary to the constraint that -the gradient be zero. -We have written the $H$-dependent terms in this strange form for the ease of taking the average over $H$: since it is Gaussian-correlated, it follows that -\begin{equation} - \overline{e^{\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t)}} - =e^{\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')\overline{H(\mathbf t)H(\mathbf t')}} - =e^{N\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')f\big(\frac{\mathbf t\cdot\mathbf t'}N\big)} -\end{equation} -It remains only to apply the doubled operators to $f$ and then evaluate the simple integrals over the $\delta$ measures. We do not include these details, which are standard. - -\subsection{Hubbard--Stratonovich} - -Having expanded this expression, we are left with an argument in the exponential which is a function of scalar products between the fields $\mathbf s$, $\hat{\mathbf s}$, $\pmb\sigma$, and $\hat{\pmb\sigma}$. We will change integration coordinates from these fields to matrix fields given by their scalar products, defined as -\begin{equation} \label{eq:fields} - \begin{aligned} - C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b && - R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b && - D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\ - C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b && - R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b && - R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b && - D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\ - C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b && - R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b && - D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b - \end{aligned} -\end{equation} -We insert into the integral the product of $\delta$-functions enforcing these -definitions, integrated over the new matrix fields, which is equivalent to -multiplying by one. Once this is done, the many scalar products appearing -throughout can be replaced by the matrix fields, and the original vector fields -can be integrated over. Conjugate matrix field integrals created when the -$\delta$-functions are promoted to exponentials can be evaluated by saddle -point in the standard way, yielding an effective action depending on the above -matrix fields alone. - -\subsection{Saddle point} - -We will always assume that the square matrices $C^{00}$, $R^{00}$, $D^{00}$, -$C^{11}$, $R^{11}$, and $D^{11}$ are hierarchical matrices, with each set of -three sharing the same hierarchical structure. In particular, we immediately -define $c_\mathrm d^{00}$, $r_\mathrm d^{00}$, $d_\mathrm d^{00}$, $c_\mathrm d^{11}$, $r_\mathrm d^{11}$, and -$d_\mathrm d^{11}$ as the value of the diagonal elements of these matrices, -respectively. Note that $c_\mathrm d^{00}=c_\mathrm d^{11}=1$ due to the spherical constraint. - -Defining the `block' fields $\mathcal Q_{00}=(\hat\beta_0, \hat\mu_0, C^{00}, -R^{00}, D^{00})$, $\mathcal Q_{11}=(\hat\beta_1, \hat\mu_1, C^{11}, R^{11}, -D^{11})$, and $\mathcal Q_{01}=(\hat\mu_{01},C^{01},R^{01},R^{10},D^{01})$ -the resulting complexity is -\begin{equation} - \Sigma_{01} - =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\int d\mathcal Q_{00}\,d\mathcal Q_{11}\,d\mathcal Q_{01}\,e^{Nm\mathcal S_0(\mathcal Q_{00})+Nn\mathcal S_1(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})} -\end{equation} -where -\begin{equation} \label{eq:one-point.action} - \begin{aligned} - &\mathcal S_0(\mathcal Q_{00}) - =\hat\beta_0E_0-r^{00}_\mathrm d\mu_0-\frac12\hat\mu_0(1-c^{00}_\mathrm d)+\mathcal D(\mu_0)\\ - &\quad+\frac1m\bigg\{ - \frac12\sum_{ab}^m\left[ - \hat\beta_1^2f(C^{00}_{ab})+(2\hat\beta_1R^{00}_{ab}-D^{00}_{ab})f'(C^{00}_{ab})+(R_{ab}^{00})^2f''(C_{ab}^{00}) - \right]+\frac12\log\det\begin{bmatrix}C^{00}&R^{00}\\R^{00}&D^{00}\end{bmatrix} -\bigg\} -\end{aligned} -\end{equation} -is the action for the ordinary, one-point complexity, and remainder is given by -\begin{equation} \label{eq:two-point.action} - \begin{aligned} - &\mathcal S(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00}) - =\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\ - &\quad+\frac1n\sum_b^n\left\{-\frac12\hat\mu_{12}(q-C^{01}_{1b})+\sum_a^m\left[ - \hat\beta_0\hat\beta_1f(C^{01}_{ab})+(\hat\beta_0R^{01}_{ab}+\hat\beta_1R^{10}_{ab}-D^{01}_{ab})f'(C^{01}_{ab})+R^{01}_{ab}R^{10}_{ab}f''(C^{01}_{ab}) - \right]\right\} - \\ - &\quad+\frac1n\bigg\{ - \frac12\sum_{ab}^n\left[ - \hat\beta_1^2f(C^{11}_{ab})+(2\hat\beta_1R^{11}_{ab}-D^{11}_{ab})f'(C^{11}_{ab})+(R^{11}_{ab})^2f''(C^{11}_{ab}) - \right]\\ - &\quad+\frac12\log\det\left( - \begin{bmatrix} - C^{11}&iR^{11}\\iR^{11}&D^{11} - \end{bmatrix}- - \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix}^T - \begin{bmatrix} - C^{00}&iR^{00}\\iR^{00}&D^{00} - \end{bmatrix}^{-1} - \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix} - \right) - \bigg\} - \end{aligned} -\end{equation} -Because of the structure of this problem in the twin limits of $m$ and $n$ to -zero, the parameters $\mathcal Q_{00}$ can be evaluated at a saddle point of -$\mathcal S_0$ alone. This means that these parameters will take the same value -they take when the ordinary, 1-point complexity is calculated. For a replica -symmetric complexity of the reference point, this results in -\begin{align} - \hat\beta_0 - &=-\frac{\mu_0f'(1)+E_0\big(f'(1)+f''(1)\big)}{u_f}\\ - r_\mathrm d^{00} - &=\frac{\mu_0f(1)+E_0f'(1)}{u_f} \\ - d_\mathrm d^{00} - &=\frac1{f'(1)} - -\left( - \frac{\mu_0f(1)+E_0f'(1)}{u_f} - \right)^2 -\end{align} -where we define for brevity (here and elsewhere) the constants -\begin{align} - u_f=f(1)\big(f'(1)+f''(1)\big)-f'(1)^2 - && - v_f=f'(1)\big(f''(1)+f'''(1)\big)-f''(1)^2 -\end{align} -Note that because the coefficients of $f$ must be nonnegative for $f$ to -be a sensible covariance, both $u_f$ and $v_f$ are strictly positive. Note also -that $u_f=v_f=0$ if $f$ is a homogeneous polynomial as in the pure models. -These expressions are invalid for the pure models because $\mu_0$ and $E_0$ -cannot be fixed independently; we would have done the equivalent of inserting -two identical $\delta$-functions. For the pure models, the terms $\hat\beta_0$ and -$\hat\beta_1$ must be set to zero in our prior formulae (as if the energy was -not constrained) and then the saddle point taken. - - -In general, we except the $m\times n$ matrices $C^{01}$, $R^{01}$, $R^{10}$, -and $D^{01}$ to have constant \emph{rows} of length $n$, with blocks of rows -corresponding to the \textsc{rsb} structure of the single-point complexity. For -the scope of this paper, where we restrict ourselves to replica symmetric -complexities, they have the following form at the saddle point: -\begin{align} \label{eq:01.ansatz} - C^{01}= - \begin{subarray}{l} - \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\ - \left[ - \begin{array}{ccc} - q&\cdots&q\\ - 0&\cdots&0\\ - \vdots&\ddots&\vdots\\ - 0&\cdots&0 - \end{array} - \right]\begin{array}{c} - \\\uparrow\\m-1\\\downarrow - \end{array} -\end{subarray} - && - R^{01} - =\begin{bmatrix} - r_{01}&\cdots&r_{01}\\ - 0&\cdots&0\\ - \vdots&\ddots&\vdots\\ - 0&\cdots&0 - \end{bmatrix} - && - R^{10} - =\begin{bmatrix} - r_{10}&\cdots&r_{10}\\ - 0&\cdots&0\\ - \vdots&\ddots&\vdots\\ - 0&\cdots&0 - \end{bmatrix} - && - D^{01} - =\begin{bmatrix} - d_{01}&\cdots&d_{01}\\ - 0&\cdots&0\\ - \vdots&\ddots&\vdots\\ - 0&\cdots&0 - \end{bmatrix} -\end{align} -where only the first row is nonzero. The other entries, which correspond to the -completely uncorrelated replicas in an \textsc{rsb} picture, are all zero -because uncorrelated vectors on the sphere are orthogonal. - -The inverse of block hierarchical matrix is still a block hierarchical matrix, since -\begin{equation} - \begin{bmatrix} - C^{00}&iR^{00}\\iR^{00}&D^{00} - \end{bmatrix}^{-1} - = - \begin{bmatrix} - (C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00} & -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} \\ - -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} & (C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00} - \end{bmatrix} -\end{equation} -Because of the structure of the 01 matrices, the volume element will depend -only on the diagonals of the matrices in this inverse block matrix. If we define -\begin{align} - \tilde c_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}]_{\text d} \\ - \tilde r_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00}]_{\text d} \\ - \tilde d_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00}]_{\text d} -\end{align} -as the diagonals of the blocks of the inverse matrix, then the result of the product is -\begin{equation} - \begin{aligned} - & \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix}^T - \begin{bmatrix} - C^{00}&iR^{00}\\iR^{00}&D^{00} - \end{bmatrix}^{-1} - \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix} \\ - &\qquad=\begin{bmatrix} - q^2\tilde d_\mathrm d^{00}+2qr_{10}\tilde r^{00}_\mathrm d-r_{10}^2\tilde d^{00}_\mathrm d - & - i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right] - \\ - i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right] - & - d_{01}^2\tilde c^{00}_\mathrm d+2r_{01}d_{01}\tilde r^{00}_\mathrm d-r_{01}^2\tilde d^{00}_\mathrm d - \end{bmatrix} - \end{aligned} -\end{equation} -where each block is a constant $n\times n$ matrix. Because the matrices -$C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in the replica symmetric case, -the diagonals of the blocks above take a simple form: -\begin{align} - \tilde c_\mathrm d^{00}=f'(1) && - \tilde r_\mathrm d^{00}=r^{00}_\mathrm df'(1) && - \tilde d_\mathrm d^{00}=d^{00}_\mathrm df'(1) -\end{align} -Once these expressions are inserted into the complexity, the limits of $n$ and -$m$ to zero can be taken, and the parameters from $D^{01}$ and $D^{11}$ can be -extremized explicitly. The resulting expression for the complexity, which must +The resulting expression for the complexity, which must still be extremized over the parameters $\hat\beta_1$, $r^{01}$, $r^{11}_\mathrm d$, $r^{11}_0$, and $q^{11}_0$, is \begin{equation} @@ -958,8 +651,6 @@ this paper was produced. Second, the complexity can be calculated in the near neighborhood of a reference point by expanding in small $1-q$. This is what we describe in the next subsection. -\subsection{Expansion in the near neighborhood} - If there is no overlap gap between the reference point and its nearest neighbors, their complexity can be calculated by an expansion in $1-q$. First, we'll use this method to describe the most common type of stationary point in @@ -1324,6 +1015,317 @@ INFN. \appendix +\section{Details of calculation for the two-point complexity} +\label{sec:complexity-details} + +\subsection{The Hessian factors} + +The double partial derivatives of the energy are Gaussian with the variance +\begin{equation} + \overline{(\partial_i\partial_jH(\mathbf s))^2}=\frac1Nf''(1) +\end{equation} +which means that the matrix of partial derivatives belongs to the GOE class. Its spectrum is given by the Wigner semicircle +\begin{equation} + \rho(\lambda)=\begin{cases} + \frac2{\pi}\sqrt{1-\big(\frac{\lambda}{\mu_\text m}\big)^2} & \lambda^2\leq\mu_\text m^2 \\ + 0 & \text{otherwise} + \end{cases} +\end{equation} +with radius $\mu_\text m=\sqrt{4f''(1)}$. Since the Hessian differs from the +matrix of partial derivatives by adding the constant diagonal matrix $\omega +I$, it follows that the spectrum of the Hessian is a Wigner semicircle shifted +by $\omega$, or $\rho(\lambda+\omega)$. + +The average over factors depending on the Hessian alone can be made separately +from those depending on the gradient or energy, since for random Gaussian +fields the Hessian is independent of these \cite{Bray_2007_Statistics}. In +principle the fact that we have conditioned the Hessian to belong to stationary +points of certain energy, stability, and proximity to another stationary point +will modify its statistics, but these changes will only appear at subleading +order in $N$ \cite{Ros_2019_Complexity}. At leading order, the various expectations factorize, each yielding +\begin{equation} + \overline{\big|\det\operatorname{Hess}H(\mathbf s,\omega)\big|\,\delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s,\omega)\big)} + =e^{N\int d\lambda\,\rho(\lambda+\mu)\log|\lambda|}\delta(N\mu-N\omega) +\end{equation} +Therefore, all of the Lagrange multipliers are fixed to the stabilities $\mu$. We define the function +\begin{equation} + \begin{aligned} + \mathcal D(\mu) + &=\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\ + &=\begin{cases} + \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2} + & \mu^2\leq\mu_\text m^2 \\ + \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2} + -\left|\frac{\mu}{\mu_\text m}\right|\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1} + -\log\left(\left|\frac{\mu}{\mu_\text m}\right|-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu^2>\mu_\text m^2 + \end{cases} + \end{aligned} +\end{equation} +and the full factor due to the Hessians is +\begin{equation} + e^{Nm\mathcal D(\mu_0)+Nn\mathcal D(\mu_1)}\left[\prod_a^m\delta(N\mu_0-N\varsigma_a)\right]\left[\prod_a^n\delta(N\mu_1-N\omega_a)\right] +\end{equation} + +\subsection{The other factors} + +Having integrated over the Lagrange multipliers using the $\delta$-functions +resulting from the average of the Hessians, any $\delta$-functions in the +remaining integrand we Fourier transform into their integral representation +over auxiliary fields. The resulting integrand has the form +\begin{equation} + e^{ + Nm\hat\beta_0E_0+Nn\hat\beta_1E_1 + -\sum_a^m\left[(\pmb\sigma_a\cdot\hat{\pmb\sigma}_a)\mu_0 + -\frac12\hat\mu_0(N-\pmb\sigma_a\cdot\pmb\sigma_a) + \right] + -\sum_a^n\left[(\mathbf s_a\cdot\hat{\mathbf s}_a)\mu_1 + -\frac12\hat\mu_1(N-\mathbf s_a\cdot\mathbf s_a) + -\frac12\hat\mu_{12}(Nq-\pmb\sigma_1\cdot\mathbf s_a) + \right] + +\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t) + } +\end{equation} +where we have introduced the linear operator +\begin{equation} + \mathcal O(\mathbf t) + =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left( + i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0 + \right) + + + \sum_a^n\delta(\mathbf t-\mathbf s_a)\left( + i\hat{\mathbf s}_a\cdot\partial_{\mathbf t}-\hat\beta_1 + \right) +\end{equation} +Here the $\hat\beta$s are the fields auxiliary to the energy constraints, the +$\hat\mu$s are auxiliary to the spherical and overlap constraints, and the +$\hat{\pmb\sigma}$s and $\hat{\mathbf s}$s are auxiliary to the constraint that +the gradient be zero. +We have written the $H$-dependent terms in this strange form for the ease of taking the average over $H$: since it is Gaussian-correlated, it follows that +\begin{equation} + \overline{e^{\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t)}} + =e^{\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')\overline{H(\mathbf t)H(\mathbf t')}} + =e^{N\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')f\big(\frac{\mathbf t\cdot\mathbf t'}N\big)} +\end{equation} +It remains only to apply the doubled operators to $f$ and then evaluate the simple integrals over the $\delta$ measures. We do not include these details, which are standard. + +\subsection{Hubbard--Stratonovich} + +Having expanded this expression, we are left with an argument in the exponential which is a function of scalar products between the fields $\mathbf s$, $\hat{\mathbf s}$, $\pmb\sigma$, and $\hat{\pmb\sigma}$. We will change integration coordinates from these fields to matrix fields given by their scalar products, defined as +\begin{equation} \label{eq:fields} + \begin{aligned} + C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b && + R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b && + D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\ + C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b && + R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b && + R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b && + D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\ + C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b && + R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b && + D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b + \end{aligned} +\end{equation} +We insert into the integral the product of $\delta$-functions enforcing these +definitions, integrated over the new matrix fields, which is equivalent to +multiplying by one. Once this is done, the many scalar products appearing +throughout can be replaced by the matrix fields, and the original vector fields +can be integrated over. Conjugate matrix field integrals created when the +$\delta$-functions are promoted to exponentials can be evaluated by saddle +point in the standard way, yielding an effective action depending on the above +matrix fields alone. + +\subsection{Saddle point} + +We will always assume that the square matrices $C^{00}$, $R^{00}$, $D^{00}$, +$C^{11}$, $R^{11}$, and $D^{11}$ are hierarchical matrices, with each set of +three sharing the same hierarchical structure. In particular, we immediately +define $c_\mathrm d^{00}$, $r_\mathrm d^{00}$, $d_\mathrm d^{00}$, $c_\mathrm d^{11}$, $r_\mathrm d^{11}$, and +$d_\mathrm d^{11}$ as the value of the diagonal elements of these matrices, +respectively. Note that $c_\mathrm d^{00}=c_\mathrm d^{11}=1$ due to the spherical constraint. + +Defining the `block' fields $\mathcal Q_{00}=(\hat\beta_0, \hat\mu_0, C^{00}, +R^{00}, D^{00})$, $\mathcal Q_{11}=(\hat\beta_1, \hat\mu_1, C^{11}, R^{11}, +D^{11})$, and $\mathcal Q_{01}=(\hat\mu_{01},C^{01},R^{01},R^{10},D^{01})$ +the resulting complexity is +\begin{equation} + \Sigma_{01} + =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\int d\mathcal Q_{00}\,d\mathcal Q_{11}\,d\mathcal Q_{01}\,e^{Nm\mathcal S_0(\mathcal Q_{00})+Nn\mathcal S_1(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})} +\end{equation} +where +\begin{equation} \label{eq:one-point.action} + \begin{aligned} + &\mathcal S_0(\mathcal Q_{00}) + =\hat\beta_0E_0-r^{00}_\mathrm d\mu_0-\frac12\hat\mu_0(1-c^{00}_\mathrm d)+\mathcal D(\mu_0)\\ + &\quad+\frac1m\bigg\{ + \frac12\sum_{ab}^m\left[ + \hat\beta_1^2f(C^{00}_{ab})+(2\hat\beta_1R^{00}_{ab}-D^{00}_{ab})f'(C^{00}_{ab})+(R_{ab}^{00})^2f''(C_{ab}^{00}) + \right]+\frac12\log\det\begin{bmatrix}C^{00}&R^{00}\\R^{00}&D^{00}\end{bmatrix} +\bigg\} +\end{aligned} +\end{equation} +is the action for the ordinary, one-point complexity, and remainder is given by +\begin{equation} \label{eq:two-point.action} + \begin{aligned} + &\mathcal S(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00}) + =\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\ + &\quad+\frac1n\sum_b^n\left\{-\frac12\hat\mu_{12}(q-C^{01}_{1b})+\sum_a^m\left[ + \hat\beta_0\hat\beta_1f(C^{01}_{ab})+(\hat\beta_0R^{01}_{ab}+\hat\beta_1R^{10}_{ab}-D^{01}_{ab})f'(C^{01}_{ab})+R^{01}_{ab}R^{10}_{ab}f''(C^{01}_{ab}) + \right]\right\} + \\ + &\quad+\frac1n\bigg\{ + \frac12\sum_{ab}^n\left[ + \hat\beta_1^2f(C^{11}_{ab})+(2\hat\beta_1R^{11}_{ab}-D^{11}_{ab})f'(C^{11}_{ab})+(R^{11}_{ab})^2f''(C^{11}_{ab}) + \right]\\ + &\quad+\frac12\log\det\left( + \begin{bmatrix} + C^{11}&iR^{11}\\iR^{11}&D^{11} + \end{bmatrix}- + \begin{bmatrix} + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix}^T + \begin{bmatrix} + C^{00}&iR^{00}\\iR^{00}&D^{00} + \end{bmatrix}^{-1} + \begin{bmatrix} + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix} + \right) + \bigg\} + \end{aligned} +\end{equation} +Because of the structure of this problem in the twin limits of $m$ and $n$ to +zero, the parameters $\mathcal Q_{00}$ can be evaluated at a saddle point of +$\mathcal S_0$ alone. This means that these parameters will take the same value +they take when the ordinary, 1-point complexity is calculated. For a replica +symmetric complexity of the reference point, this results in +\begin{align} + \hat\beta_0 + &=-\frac{\mu_0f'(1)+E_0\big(f'(1)+f''(1)\big)}{u_f}\\ + r_\mathrm d^{00} + &=\frac{\mu_0f(1)+E_0f'(1)}{u_f} \\ + d_\mathrm d^{00} + &=\frac1{f'(1)} + -\left( + \frac{\mu_0f(1)+E_0f'(1)}{u_f} + \right)^2 +\end{align} +where we define for brevity (here and elsewhere) the constants +\begin{align} + u_f=f(1)\big(f'(1)+f''(1)\big)-f'(1)^2 + && + v_f=f'(1)\big(f''(1)+f'''(1)\big)-f''(1)^2 +\end{align} +Note that because the coefficients of $f$ must be nonnegative for $f$ to +be a sensible covariance, both $u_f$ and $v_f$ are strictly positive. Note also +that $u_f=v_f=0$ if $f$ is a homogeneous polynomial as in the pure models. +These expressions are invalid for the pure models because $\mu_0$ and $E_0$ +cannot be fixed independently; we would have done the equivalent of inserting +two identical $\delta$-functions. For the pure models, the terms $\hat\beta_0$ and +$\hat\beta_1$ must be set to zero in our prior formulae (as if the energy was +not constrained) and then the saddle point taken. + + +In general, we except the $m\times n$ matrices $C^{01}$, $R^{01}$, $R^{10}$, +and $D^{01}$ to have constant \emph{rows} of length $n$, with blocks of rows +corresponding to the \textsc{rsb} structure of the single-point complexity. For +the scope of this paper, where we restrict ourselves to replica symmetric +complexities, they have the following form at the saddle point: +\begin{align} \label{eq:01.ansatz} + C^{01}= + \begin{subarray}{l} + \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\ + \left[ + \begin{array}{ccc} + q&\cdots&q\\ + 0&\cdots&0\\ + \vdots&\ddots&\vdots\\ + 0&\cdots&0 + \end{array} + \right]\begin{array}{c} + \\\uparrow\\m-1\\\downarrow + \end{array} +\end{subarray} + && + R^{01} + =\begin{bmatrix} + r_{01}&\cdots&r_{01}\\ + 0&\cdots&0\\ + \vdots&\ddots&\vdots\\ + 0&\cdots&0 + \end{bmatrix} + && + R^{10} + =\begin{bmatrix} + r_{10}&\cdots&r_{10}\\ + 0&\cdots&0\\ + \vdots&\ddots&\vdots\\ + 0&\cdots&0 + \end{bmatrix} + && + D^{01} + =\begin{bmatrix} + d_{01}&\cdots&d_{01}\\ + 0&\cdots&0\\ + \vdots&\ddots&\vdots\\ + 0&\cdots&0 + \end{bmatrix} +\end{align} +where only the first row is nonzero. The other entries, which correspond to the +completely uncorrelated replicas in an \textsc{rsb} picture, are all zero +because uncorrelated vectors on the sphere are orthogonal. + +The inverse of block hierarchical matrix is still a block hierarchical matrix, since +\begin{equation} + \begin{bmatrix} + C^{00}&iR^{00}\\iR^{00}&D^{00} + \end{bmatrix}^{-1} + = + \begin{bmatrix} + (C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00} & -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} \\ + -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} & (C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00} + \end{bmatrix} +\end{equation} +Because of the structure of the 01 matrices, the volume element will depend +only on the diagonals of the matrices in this inverse block matrix. If we define +\begin{align} + \tilde c_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}]_{\text d} \\ + \tilde r_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00}]_{\text d} \\ + \tilde d_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00}]_{\text d} +\end{align} +as the diagonals of the blocks of the inverse matrix, then the result of the product is +\begin{equation} + \begin{aligned} + & \begin{bmatrix} + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix}^T + \begin{bmatrix} + C^{00}&iR^{00}\\iR^{00}&D^{00} + \end{bmatrix}^{-1} + \begin{bmatrix} + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix} \\ + &\qquad=\begin{bmatrix} + q^2\tilde d_\mathrm d^{00}+2qr_{10}\tilde r^{00}_\mathrm d-r_{10}^2\tilde d^{00}_\mathrm d + & + i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right] + \\ + i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right] + & + d_{01}^2\tilde c^{00}_\mathrm d+2r_{01}d_{01}\tilde r^{00}_\mathrm d-r_{01}^2\tilde d^{00}_\mathrm d + \end{bmatrix} + \end{aligned} +\end{equation} +where each block is a constant $n\times n$ matrix. Because the matrices +$C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in the replica symmetric case, +the diagonals of the blocks above take a simple form: +\begin{align} + \tilde c_\mathrm d^{00}=f'(1) && + \tilde r_\mathrm d^{00}=r^{00}_\mathrm df'(1) && + \tilde d_\mathrm d^{00}=d^{00}_\mathrm df'(1) +\end{align} +Once these expressions are inserted into the complexity, the limits of $n$ and +$m$ to zero can be taken, and the parameters from $D^{01}$ and $D^{11}$ can be +extremized explicitly. \section{Details of calculation for the isolated eigenvalue} \label{sec:eigenvalue-details} |