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@@ -1,4 +1,4 @@ -\documentclass[]{iopart} +\documentclass[12pt]{iopart} \usepackage[utf8]{inputenc} % why not type "Stokes" with unicode? \usepackage[T1]{fontenc} % vector fonts @@ -407,15 +407,28 @@ Writing $\partial=\partial/\partial z$, this gives \end{equation} which is nothing but the projection of $(\partial\mathcal S)^*$ into the tangent space of the manifold, with the projection operator -$P=(Dz)^*[(Dz)^\dagger(Dz)]^{-1}(Dz)^T$. Note that $P$ is hermitian. For the spherical models, where $\tilde\Omega$ is the complex phase spaced defined by all points $z\in\mathbb C^N$ such that $z^Tz=1$, the projection operator is given by +$P=(Dz)^*[(Dz)^\dagger(Dz)]^{-1}(Dz)^T$. Note that $P$ is hermitian. + +All one needs to work out the projection operator is a coordinate system. Here we sketch the derivation for the spherical models. +Take $z_0=e^N$. In a neighborhood of $z_0$, the map $u^\alpha=z^\alpha$ is a coordinate system. +Its inverse is $z^i=u^i$ for $1\leq i\leq N-1$ and +\begin{equation} + z^N=\sqrt{N-u^2}. +\end{equation} +The Jacobian is +\begin{equation} + D_\alpha z^i=\frac{\partial z^i}{\partial u^\alpha}=\delta^i_\alpha-\delta_N^{i}\frac{u^\alpha}{\sqrt{N-u^2}} +\end{equation} +Taking the appropriate combination of $Dz$ yields \begin{equation} P=I-\frac{zz^\dagger}{|z|^2} \end{equation} -something that we be worked out in detail in a following section. One can +One can quickly verify that this operator indeed projects the dynamics onto the -manifold: its tangent at any point $z$ is given by $\partial(z^Tz)=z$, and -$Pz=z-z|z|^2/|z|^2=0$. For any vector $u$ perpendicular to $z$, i.e., -$z^\dagger u=0$, $Pu=u$. +manifold: the vector perpendicular to the manifold at any point $z$ is given by +$\partial(z^Tz)=z$, and $Pz=z-z|z|^2/|z|^2=0$. For any vector $u$ perpendicular +to $z$, i.e., $z^\dagger u=0$, $Pu=u$, the identity. + \begin{figure} \includegraphics{figs/thimble_flow.pdf} @@ -437,19 +450,21 @@ coordinate. This is because $(\tilde\Omega, g)$ is Kähler and therefore admits a symplectic structure, but that the flow conserves $\operatorname{Im}\beta\mathcal S$ can be shown using \eref{eq:flow} and the holomorphic property of $\mathcal S$: -\begin{eqnarray} - \frac d{dt}\operatorname{Im}\beta\mathcal S - &=\dot z\partial\operatorname{Im}\beta\mathcal S+\dot z^*\partial^*\operatorname{Im}\beta\mathcal S \\ - &=\frac i4\left( - (\beta\partial \mathcal S)^\dagger P\beta\partial\mathcal S-(\beta\partial\mathcal S)^TP^*(\beta\partial\mathcal S)^* - \right) \\ - &=\frac{i|\beta|^2}4\left( - (\partial\mathcal S)^\dagger P\partial\mathcal S-[(\partial\mathcal S)^\dagger P\partial\mathcal S]^* - \right) \\ - &=\frac{i|\beta|^2}4\left( - \|\partial\mathcal S\|^2-(\|\partial\mathcal S\|^*)^2 - \right)=0. -\end{eqnarray} +\begin{equation} + \eqalign{ + \frac d{dt}\operatorname{Im}\beta\mathcal S + &=\dot z\partial\operatorname{Im}\beta\mathcal S+\dot z^*\partial^*\operatorname{Im}\beta\mathcal S \\ + &=\frac i4\left( + (\beta\partial \mathcal S)^\dagger P\beta\partial\mathcal S-(\beta\partial\mathcal S)^TP^*(\beta\partial\mathcal S)^* + \right) \\ + &=\frac{i|\beta|^2}4\left( + (\partial\mathcal S)^\dagger P\partial\mathcal S-[(\partial\mathcal S)^\dagger P\partial\mathcal S]^* + \right) \\ + &=\frac{i|\beta|^2}4\left( + \|\partial\mathcal S\|^2-(\|\partial\mathcal S\|^*)^2 + \right)=0. + } +\end{equation} A consequence of this conservation is that the flow in the action takes a simple form: \begin{equation} @@ -524,31 +539,35 @@ means that it obeys the Cauchy--Riemann equations Using these relationships alongside the Wirtinger derivative $\partial\equiv\frac12(\partial_x-i\partial_y)$ allows the order of the derivatives and the real or imaginary parts to be commuted, with -\begin{eqnarray} - \partial_x\operatorname{Re}\tilde\mathcal S=\operatorname{Re}\partial\tilde\mathcal S - \qquad - \partial_y\operatorname{Re}\tilde\mathcal S=-\operatorname{Im}\partial\tilde\mathcal S \\ - \partial_x\operatorname{Im}\tilde\mathcal S=\operatorname{Im}\partial\tilde\mathcal S - \qquad - \partial_y\operatorname{Im}\tilde\mathcal S=\operatorname{Re}\partial\tilde\mathcal S -\end{eqnarray} +\begin{equation} + \eqalign{ + \partial_x\operatorname{Re}\tilde\mathcal S=\operatorname{Re}\partial\tilde\mathcal S + \qquad + \partial_y\operatorname{Re}\tilde\mathcal S=-\operatorname{Im}\partial\tilde\mathcal S \\ + \partial_x\operatorname{Im}\tilde\mathcal S=\operatorname{Im}\partial\tilde\mathcal S + \qquad + \partial_y\operatorname{Im}\tilde\mathcal S=\operatorname{Re}\partial\tilde\mathcal S + } +\end{equation} Using these relationships, the hessian \eref{eq:real.hessian} can be written in the more manifestly complex way -\begin{eqnarray} - \operatorname{Hess}_{x,y}\operatorname{Re}\beta\mathcal S - &=\left[\matrix{ - \hphantom{-}\operatorname{Re}\beta\partial\partial\tilde\mathcal S & - -\operatorname{Im}\beta\partial\partial\tilde\mathcal S \cr - -\operatorname{Im}\beta\partial\partial\tilde\mathcal S & - -\operatorname{Re}\beta\partial\partial\tilde\mathcal S - }\right] \\ - &=\left[\matrix{ - \hphantom{-}\operatorname{Re}\beta\operatorname{Hess}\mathcal S & - -\operatorname{Im}\beta\operatorname{Hess}\mathcal S \cr - -\operatorname{Im}\beta\operatorname{Hess}\mathcal S & - -\operatorname{Re}\beta\operatorname{Hess}\mathcal S - }\right] -\end{eqnarray} +\begin{equation} + \eqalign{ + \operatorname{Hess}_{x,y}\operatorname{Re}\beta\mathcal S + &=\left[\matrix{ + \hphantom{-}\operatorname{Re}\beta\partial\partial\tilde\mathcal S & + -\operatorname{Im}\beta\partial\partial\tilde\mathcal S \cr + -\operatorname{Im}\beta\partial\partial\tilde\mathcal S & + -\operatorname{Re}\beta\partial\partial\tilde\mathcal S + }\right] \\ + &=\left[\matrix{ + \hphantom{-}\operatorname{Re}\beta\operatorname{Hess}\mathcal S & + -\operatorname{Im}\beta\operatorname{Hess}\mathcal S \cr + -\operatorname{Im}\beta\operatorname{Hess}\mathcal S & + -\operatorname{Re}\beta\operatorname{Hess}\mathcal S + }\right] + } +\end{equation} where $\operatorname{Hess}\mathcal S$ is the hessian with respect to $z$ given in \eqref{eq:complex.hessian}. @@ -561,17 +580,19 @@ R^N$ are such that =\lambda\left[\matrix{v_x \cr v_y}\right] \end{equation} where the eigenvalue $\lambda$ must be real because the hessian is real symmetric. The problem can be put into a more obviously complex form by a change of basis. Writing $v=v_x+iv_y$, we find -\begin{eqnarray} - &\left[\matrix{0&(i\beta\operatorname{Hess}\mathcal S)^*\cr i\beta\operatorname{Hess}\mathcal S&0}\right] - \left[\matrix{v \cr iv^*}\right]\\ - &\qquad=\left[\matrix{1&i\cr i&1}\right] - (\operatorname{Hess}_{x,y}\operatorname{Re}\beta\mathcal S) - \left[\matrix{1&i\cr i&1}\right]^{-1} - \left[\matrix{1&i\cr i&1}\right] - \left[\matrix{v_x \cr v_y}\right] \\ - &\qquad=\lambda\left[\matrix{1&i\cr i&1}\right]\left[\matrix{v_x \cr v_y}\right] - =\lambda\left[\matrix{v \cr iv^*}\right] -\end{eqnarray} +\begin{equation} + \eqalign{ + &\left[\matrix{0&(i\beta\operatorname{Hess}\mathcal S)^*\cr i\beta\operatorname{Hess}\mathcal S&0}\right] + \left[\matrix{v \cr iv^*}\right]\\ + &\qquad=\left[\matrix{1&i\cr i&1}\right] + (\operatorname{Hess}_{x,y}\operatorname{Re}\beta\mathcal S) + \left[\matrix{1&i\cr i&1}\right]^{-1} + \left[\matrix{1&i\cr i&1}\right] + \left[\matrix{v_x \cr v_y}\right] \\ + &\qquad=\lambda\left[\matrix{1&i\cr i&1}\right]\left[\matrix{v_x \cr v_y}\right] + =\lambda\left[\matrix{v \cr iv^*}\right] + } +\end{equation} It therefore follows that the eigenvalues and vectors of the real hessian satisfy the equation \begin{equation} \label{eq:generalized.eigenproblem} \beta\operatorname{Hess}\mathcal S v=\lambda v^* @@ -628,14 +649,16 @@ $\beta\operatorname{Hess}\mathcal S$. A direct relationship between these singul values and the eigenvalues of the hessian immediately follows by taking an eigenvector $v\in\mathbb C$ that satisfies \eref{eq:generalized.eigenproblem}, and writing -\begin{eqnarray} - \sigma v^\dagger u - &=v^\dagger(\beta\operatorname{Hess} S)^\dagger(\beta\operatorname{Hess} S)u - =(\beta\operatorname{Hess} Sv)^\dagger(\beta\operatorname{Hess} S)u\\ - &=(\lambda v^*)^\dagger(\beta\operatorname{Hess} S)u - =\lambda v^T(\beta\operatorname{Hess} S)u - =\lambda^2 v^\dagger u -\end{eqnarray} +\begin{equation} + \eqalign{ + \sigma v^\dagger u + &=v^\dagger(\beta\operatorname{Hess} S)^\dagger(\beta\operatorname{Hess} S)u + =(\beta\operatorname{Hess} Sv)^\dagger(\beta\operatorname{Hess} S)u\\ + &=(\lambda v^*)^\dagger(\beta\operatorname{Hess} S)u + =\lambda v^T(\beta\operatorname{Hess} S)u + =\lambda^2 v^\dagger u + } +\end{equation} Thus if $v^\dagger u\neq0$, $\lambda^2=\sigma$. It follows that the eigenvalues of the real hessian are the singular values of the complex matrix $\beta\operatorname{Hess}\mathcal S$, and their eigenvectors coincide up to a @@ -719,19 +742,21 @@ The coordinates $u$ can be constructed implicitly in the close vicinity of the s where the sum is over pairs $(\lambda, v)$ which satisfy \eqref{eq:generalized.eigenproblem} and have $\lambda>0$. It is straightforward to confirm that these coordinates satisfy \eqref{eq:thimble.integration.def} asymptotically close to the stationary point, as -\begin{eqnarray} - \beta\mathcal S(s(u)) - &=\beta\mathcal S(s_\sigma) - +\frac12(s(u)-s_\sigma)^T(\beta\operatorname{Hess}\mathcal S)(s(u)-s_\sigma)+\cdots \\ - &=\beta\mathcal S(s_\sigma) - +\frac{|\beta|}2\sum_{ij}\frac{v^{(i)}_k}{\sqrt{\lambda^{(i)}}}(\beta[\operatorname{Hess}\mathcal S]_{k\ell})\frac{v^{(j)}_\ell}{\sqrt{\lambda^{(j)}}}u_iu_j+\cdots \\ - &=\beta\mathcal S(s_\sigma) - +\frac{|\beta|}2\sum_{ij}\frac{v^{(i)}_k}{\sqrt{\lambda^{(i)}}}\frac{\lambda^{(j)}(v^{(j)}_k)^*}{\sqrt{\lambda^{(j)}}}u_iu_j+\cdots \\ - &=\beta\mathcal S(s_\sigma) - +\frac{|\beta|}2\sum_{ij}\frac{\sqrt{\lambda^{(j)}}}{\sqrt{\lambda^{(i)}}}\delta_{ij}u_iu_j+\cdots \\ - &=\beta\mathcal S(s_\sigma) - +\frac{|\beta|}2\sum_iu_i^2+\cdots -\end{eqnarray} +\begin{equation} + \eqalign{ + \beta\mathcal S(s(u)) + &=\beta\mathcal S(s_\sigma) + +\frac12(s(u)-s_\sigma)^T(\beta\operatorname{Hess}\mathcal S)(s(u)-s_\sigma)+\cdots \\ + &=\beta\mathcal S(s_\sigma) + +\frac{|\beta|}2\sum_{ij}\frac{v^{(i)}_k}{\sqrt{\lambda^{(i)}}}(\beta[\operatorname{Hess}\mathcal S]_{k\ell})\frac{v^{(j)}_\ell}{\sqrt{\lambda^{(j)}}}u_iu_j+\cdots \\ + &=\beta\mathcal S(s_\sigma) + +\frac{|\beta|}2\sum_{ij}\frac{v^{(i)}_k}{\sqrt{\lambda^{(i)}}}\frac{\lambda^{(j)}(v^{(j)}_k)^*}{\sqrt{\lambda^{(j)}}}u_iu_j+\cdots \\ + &=\beta\mathcal S(s_\sigma) + +\frac{|\beta|}2\sum_{ij}\frac{\sqrt{\lambda^{(j)}}}{\sqrt{\lambda^{(i)}}}\delta_{ij}u_iu_j+\cdots \\ + &=\beta\mathcal S(s_\sigma) + +\frac{|\beta|}2\sum_iu_i^2+\cdots + } +\end{equation} The Jacobian of this transformation is \begin{equation} \frac{\partial s_i}{\partial u_j}=\sqrt{\frac{|\beta|}{\lambda^{(j)}}}v^{(j)}_i @@ -751,11 +776,13 @@ Z_\sigma(\beta)=e^{-\beta\mathcal S(s_\sigma)}\int du\,\det\frac{ds}{du}e^{-\fra Now we take the saddle point approximation, assuming the integral is dominated by its value at the stationary point such that the determinant can be approximated by its value at the stationary point. This gives -\begin{eqnarray} - Z_\sigma(\beta) - &\simeq e^{-\beta\mathcal S(s_\sigma)}\left.\det\frac{ds}{du}\right|_{s=s_\sigma}\int du\,e^{-\frac{|\beta|}2u^Tu} \\ - &=e^{-\beta\mathcal S(s_\sigma)}\left(\prod_i^D\sqrt{\frac1{\lambda_0^{(i)}}}\right)\det U\left(\frac{2\pi}{|\beta|}\right)^{D/2} -\end{eqnarray} +\begin{equation} + \eqalign{ + Z_\sigma(\beta) + &\simeq e^{-\beta\mathcal S(s_\sigma)}\left.\det\frac{ds}{du}\right|_{s=s_\sigma}\int du\,e^{-\frac{|\beta|}2u^Tu} \\ + &=e^{-\beta\mathcal S(s_\sigma)}\left(\prod_i^D\sqrt{\frac1{\lambda_0^{(i)}}}\right)\det U\left(\frac{2\pi}{|\beta|}\right)^{D/2} + } +\end{equation} We are left with evaluating the determinant of the unitary part of the coordinate transformation. In circumstances you may be used to, only the absolute value of the determinant from the coordinate transformation is relevant, and since the determinant of a @@ -776,12 +803,12 @@ $\det U=i^k$. As the argument of $\beta$ is changed, we know how the eigenvector Z_\sigma(\beta)\simeq\left(\frac{2\pi}\beta\right)^{D/2}i^{k_\sigma}|\det\operatorname{Hess}\mathcal S(s_\sigma)|^{-\frac12}e^{-\beta\mathcal S(s_\sigma)} \end{equation} -\begin{eqnarray} +\begin{equation} Z(\beta)^* =\sum_{\sigma\in\Sigma_0}n_\sigma Z_\sigma(\beta)^* =\sum_{\sigma\in\Sigma_0}n_\sigma(-1)^{k_\sigma}Z_\sigma(\beta^*) =Z(\beta^*) -\end{eqnarray} +\end{equation} \section{The ensemble of symmetric complex-normal matrices} @@ -844,7 +871,7 @@ eigenvalues of the real $2N\times2N$ block matrix \end{equation} as we saw in \S\ref{sec:stationary.hessian}. The eigenvalue spectrum of this block matrix can be studied by ordinary means. Defining the `partition function' -\begin{equation} +\begin{equation} \fl \qquad Z(\sigma)=\int dx\,dy\,\exp\left\{ -\frac12\left[\matrix{x&y}\right] \left(\sigma I- @@ -858,24 +885,26 @@ implies a Green function G(\sigma)=\frac\partial{\partial\sigma}\log Z(\sigma) \end{equation} This can be put into a manifestly complex form in the same way it was done in \S\ref{sec:stationary.hessian}, using the same linear transformation of $x$ and $y$ into $z$ and $z^*$. This gives -\begin{eqnarray} - Z(\sigma) - &=\int dz\,dz^*\,\exp\left\{ - -\frac12\left[\matrix{z^*&-iz}\right] - \left(\sigma I- - \left[\matrix{0&(iB)^*\cr iB&0}\right] - \right) - \left[\matrix{z\cr iz^*}\right] - \right\} \\ - &=\int dz\,dz^*\,\exp\left\{ - -\frac12\left( - 2z^\dagger z\sigma-z^\dagger B^*z^*-z^TBz +\begin{equation} + \eqalign{ + Z(\sigma) + &=\int dz\,dz^*\,\exp\left\{ + -\frac12\left[\matrix{z^*&-iz}\right] + \left(\sigma I- + \left[\matrix{0&(iB)^*\cr iB&0}\right] \right) - \right\} \\ - &=\int dz\,dz^*\,\exp\left\{ - -z^\dagger z\sigma+\operatorname{Re}(z^TBz) - \right\} -\end{eqnarray} + \left[\matrix{z\cr iz^*}\right] + \right\} \\ + &=\int dz\,dz^*\,\exp\left\{ + -\frac12\left( + 2z^\dagger z\sigma-z^\dagger B^*z^*-z^TBz + \right) + \right\} \\ + &=\int dz\,dz^*\,\exp\left\{ + -z^\dagger z\sigma+\operatorname{Re}(z^TBz) + \right\} + } +\end{equation} which is a general expression for the singular values $\sigma$ of a symmetric complex matrix $B$. @@ -989,12 +1018,11 @@ using a Lagrange multiplier $\mu$ by writing \tilde\mathcal S(z)=\mathcal S(z)-\frac\mu2(z^Tz-N) \end{equation} The gradient of the constraint is simple with $\partial g=z$, and \eqref{eq:multiplier} implies that -\begin{eqnarray} +\begin{equation} \mu - &=\frac1Nz^T\partial\mathcal S - =\frac1Nz_{i_p}\sum_{p=2}^\infty a_p\frac p{p!}\sum_{i_1\cdots i_p}J_{i_1\cdots i_p}z_{i_1}\cdots z_{i_{p-1}} \\ - &=\sum_{p=2}^\infty a_pp\frac{\mathcal S_p(z)}N -\end{eqnarray} + =\frac1Nz^T\partial\mathcal S + =\sum_{p=2}^\infty a_pp\frac{\mathcal S_p(z)}N +\end{equation} which for the pure $p$-spin in particular implies that $\mu=p\epsilon$ for specific energy $\epsilon$. \subsection{2-spin} @@ -1028,11 +1056,11 @@ of generality, assume these are associated with the first and second cardinal directions. Since the gradient is proportional to $z$, any components that are zero at some time will be zero at all times. The dynamics for the components of interest assuming all others are zero are -\begin{eqnarray} +\begin{equation} \dot z_1 - &=-z_1^*\left(d_1^*-\frac{d_1^*z_1^*z_1+d_2^*z_2^*z_2}{|z_1|^2+|z_2|^2}\right) \\ - &=-(d_1-d_2)^*z_1^*\frac{|z_2|^2}{|z_1|^2+|z_2|^2} -\end{eqnarray} + =-z_1^*\left(d_1^*-\frac{d_1^*z_1^*z_1+d_2^*z_2^*z_2}{|z_1|^2+|z_2|^2}\right) + =-(d_1-d_2)^*z_1^*\frac{|z_2|^2}{|z_1|^2+|z_2|^2} +\end{equation} and the same for $z_2$ with all indices swapped. Since $\Delta=d_1-d_2$ is real, if $z_1$ begins real it remains real, with the same for $z_2$. Since the stationary points are at real $z$, we make this restriction, and find @@ -1069,21 +1097,23 @@ separatrix of a third. This means that when the imaginary energies of two stationary points are brought to the same value, their surfaces of constant imaginary energy join. -\begin{eqnarray} - Z(\beta) - &=\int_{S^{N-1}}ds\,e^{-\beta H_2(s)} - =\sum_{\sigma\in\Sigma_0}n_\sigma\int_{\mathcal J_\sigma}ds\,e^{-\beta H_2(s)} \\ - &\simeq\sum_{k=0}^D2i^k\left(\frac{2\pi}\beta\right)^{D/2}e^{-\beta H_2(s_\sigma)}\prod_{\lambda_0>0}\lambda_0^{-1/2} \\ - &=2\sum_{k=0}^D\exp\left\{ - i\frac\pi2k+\frac D2\log\frac{2\pi}\beta-N\beta\epsilon_k-\frac12\sum_{\ell\neq k}\log2|\epsilon_k-\epsilon_\ell| - \right\} -\end{eqnarray} -\begin{eqnarray} +\begin{equation} + \eqalign{ + Z(\beta) + &=\int_{S^{N-1}}ds\,e^{-\beta H_2(s)} + =\sum_{\sigma\in\Sigma_0}n_\sigma\int_{\mathcal J_\sigma}ds\,e^{-\beta H_2(s)} \\ + &\simeq\sum_{k=0}^D2i^k\left(\frac{2\pi}\beta\right)^{D/2}e^{-\beta H_2(s_\sigma)}\prod_{\lambda_0>0}\lambda_0^{-1/2} \\ + &=2\sum_{k=0}^D\exp\left\{ + i\frac\pi2k+\frac D2\log\frac{2\pi}\beta-N\beta\epsilon_k-\frac12\sum_{\ell\neq k}\log2|\epsilon_k-\epsilon_\ell| + \right\} + } +\end{equation} +\begin{equation} \fl Z(\beta) - &=2\int d\epsilon\,\rho(\epsilon)\exp\left\{ + =2\int d\epsilon\,\rho(\epsilon)\exp\left\{ i\frac\pi2k_\epsilon+\frac D2\log\frac{2\pi}\beta-N\beta\epsilon-\frac D2\int d\epsilon'\,\rho(\epsilon')\log2|\epsilon-\epsilon'| \right\} -\end{eqnarray} +\end{equation} Since the $J$ of the 2-spin model is a symmetric real matrix with variance $1/N$, its eigenvalues are distributed by a semicircle distribution of radius 2, @@ -1184,7 +1214,7 @@ trick. Based on the experience from similar problems \cite{Castellani_2005_Spin- expected to be exact wherever the complexity is positive. As in \S\ref{sec:stationary.hessian}, these can be bright into a manifestly complex form using Cauchy--Riemann relations. This gives -\begin{equation} \label{eq:real.kac-rice} +\begin{equation} \mathcal N =\int dz^*dz\,d\hat z^*d\hat z\,d\eta^*d\eta\,d\gamma^*d\gamma\exp\left\{ \operatorname{Re}\left( @@ -1256,13 +1286,15 @@ where the inverse supermatrix is defined by I\delta(1,2)=\int d3\,Q^{-1}(1,3)Q(3,2) \end{equation} Making such a transformation, we arrive at the saddle point equations -\begin{eqnarray} - 0 - &=\int d3\,\frac{\partial S_\mathrm{eff}}{\partial Q(1,3)}Q(3,2) \\ - &=\frac p{16}\int d3\,Q^{(p-1)}(1,3)Q(3,2)-\frac p2\left[ - \matrix{\frac\epsilon2&0\cr0&\frac{\epsilon^*}2} - \right]Q(1,2)+\frac12I\delta(1,2) -\end{eqnarray} +\begin{equation} + \eqalign{ + 0 + &=\int d3\,\frac{\partial S_\mathrm{eff}}{\partial Q(1,3)}Q(3,2) \\ + &=\frac p{16}\int d3\,Q^{(p-1)}(1,3)Q(3,2)-\frac p2\left[ + \matrix{\frac\epsilon2&0\cr0&\frac{\epsilon^*}2} + \right]Q(1,2)+\frac12I\delta(1,2) + } +\end{equation} When expanded, the supermatrix $Q$ contains nine independent bilinear combinations of the original variables: $z^\dagger z$, $\hat z^T z$, $\hat z^\dagger z$, $\hat z^T\hat z$, $\hat z^\dagger\hat z$, $\eta^\dagger\eta$, @@ -1290,21 +1322,21 @@ real sphere. The complexity can then be written } \end{equation} $I_p(u)=0$ if $|\epsilon|^2<|\epsilon_\mathrm{th}|^2$. -\begin{eqnarray} - I_p(u) - &= - \left(\frac12+\frac1{r^{p-2}-1}\right)^{-1}(\operatorname{Re}u)^2 - - - \left(\frac12-\frac1{r^{p-2}+1}\right)^{-1}(\operatorname{Im}u)^2\\ - &\qquad-\log\left( - r^{p-2}\left| - u+\sqrt{u^2-1} - \right|^2 - \right)+2\operatorname{Re} - \left( - u\sqrt{u^2-1} - \right) -\end{eqnarray} +\begin{equation} + \eqalign{ + I_p(u) + &=\left(\frac12+\frac1{r^{p-2}-1}\right)^{-1}(\operatorname{Re}u)^2 + -\left(\frac12-\frac1{r^{p-2}+1}\right)^{-1}(\operatorname{Im}u)^2 \\ + &\qquad-\log\left( + r^{p-2}\left| + u+\sqrt{u^2-1} + \right|^2 + \right)+2\operatorname{Re} + \left( + u\sqrt{u^2-1} + \right) + } +\end{equation} where the branch of the square roots are chosen such that the real part of the root has the opposite sign as the real part of $u$, e.g., if $\operatorname{Re}u<0$ then $\operatorname{Re}\sqrt{u^2-1}>0$. If the real part @@ -1331,14 +1363,14 @@ gives a sense of whether many Stokes lines should be expected, and when. To determine this, we perform the same Kac--Rice procedure as in the previous section, but now with two probe points, or replicas of the system. The number of stationary points with given energies $\epsilon_1$ and $\epsilon_2$ are -\begin{eqnarray} +\begin{equation} \mathcal N - &=\int d\phi_1\,d\phi_2^*\,d\phi_2\,\exp\left\{ + =\int d\phi_1\,d\phi_2^*\,d\phi_2\,\exp\left\{ \int d1 \left[ \tilde\mathcal S_p(\phi_1(1))+\operatorname{Re}\tilde\mathcal S_p(\phi_2(1)) \right] \right\} -\end{eqnarray} +\end{equation} \begin{equation} Q(1,2) =\left[ @@ -1369,15 +1401,17 @@ stationary points with given energies $\epsilon_1$ and $\epsilon_2$ are +\frac12Q^{-1}(1,2) \end{equation} where $\odot$ denotes element-wise multiplication. -\begin{eqnarray} - 0 - &=\int d3\,\frac{\partial S_\mathrm{eff}}{\partial Q(1,3)}Q(3,2) \\ - &=\frac p4\int d3\, - \left\{\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot Q^{(p-1)}(1,3)\right\}Q(3,2) - -\frac p2 \left[ - \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}\right]Q(1,2) - +\frac12I\delta(1,2) -\end{eqnarray} +\begin{equation} + \eqalign{ + 0 + &=\int d3\,\frac{\partial S_\mathrm{eff}}{\partial Q(1,3)}Q(3,2) \\ + &=\frac p4\int d3\, + \left\{\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot Q^{(p-1)}(1,3)\right\}Q(3,2) + -\frac p2 \left[ + \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}\right]Q(1,2) + +\frac12I\delta(1,2) + } +\end{equation} Despite being able to pose the saddle point problem in a compact way, a great deal of complexity lies within. The supermatrix $Q$ depends on 35 independent bilinear products, and when the superfields are expanded produces 48 (not entirely independent) equations. @@ -1417,10 +1451,12 @@ $x_1^Tx_1=1$, and $x_2^Tx_2=1+y_2^Ty_2$. Then, by their definitions, \begin{equation} \Delta=2(1+|y_2|^2-\sqrt{1-|y_2|^2}\cos\theta_{xx}) \end{equation} -\begin{eqnarray} - \delta\Delta&=2-2x_1^Tx_2-2ix_1^Ty_2=2(1-|x_2|\cos\theta_{xx}-i|y_2|\cos\theta_{xy}) \\ - &=2(1-\sqrt{1-|y_2|^2}\cos\theta_{xx}-i|y_2|\cos\theta_{xy}) -\end{eqnarray} +\begin{equation} + \eqalign{ + \delta\Delta&=2-2x_1^Tx_2-2ix_1^Ty_2=2(1-|x_2|\cos\theta_{xx}-i|y_2|\cos\theta_{xy}) \\ + &=2(1-\sqrt{1-|y_2|^2}\cos\theta_{xx}-i|y_2|\cos\theta_{xy}) + } +\end{equation} There is also an inequality between the angles $\theta_{xx}$ and $\theta_{xy}$ between $x_1$ and $x_2$ and $y_2$, respectively, which takes that form $\cos^2\theta_{xy}+\cos^2\theta_{xx}\leq1$. This results from the fact that @@ -1444,16 +1480,18 @@ function of $\Delta$ and $\arg\delta$. \subsection{Pure $p$-spin: is analytic continuation possible?} -\begin{eqnarray} - Z(\beta) - &=\sum_{\sigma\in\Sigma_0}n_\sigma Z_\sigma(\beta) \\ - &=\sum_{\sigma\in\Sigma_0}\left(\frac{2\pi}\beta\right)^{D/2}i^{k_\sigma} - |\det\operatorname{Hess}\mathcal S(s_\sigma)|^{-\frac12} - e^{-\beta\mathcal S(s_\sigma)} \\ - &\simeq\sum_{k=0}^D\int d\epsilon\,\mathcal N_\mathrm{typ}(\epsilon,k) - \left(\frac{2\pi}\beta\right)^{D/2}i^k - \left(|\det\operatorname{Hess}\mathcal S(s_\sigma)|^{-\frac12}\bigm|\mathcal S(s_\sigma)=N\epsilon,k_\sigma=k\right) e^{-\beta N\epsilon} -\end{eqnarray} +\begin{equation} + \eqalign{ + Z(\beta) + &=\sum_{\sigma\in\Sigma_0}n_\sigma Z_\sigma(\beta) \\ + &=\sum_{\sigma\in\Sigma_0}\left(\frac{2\pi}\beta\right)^{D/2}i^{k_\sigma} + |\det\operatorname{Hess}\mathcal S(s_\sigma)|^{-\frac12} + e^{-\beta\mathcal S(s_\sigma)} \\ + &\simeq\sum_{k=0}^D\int d\epsilon\,\mathcal N_\mathrm{typ}(\epsilon,k) + \left(\frac{2\pi}\beta\right)^{D/2}i^k + \left(|\det\operatorname{Hess}\mathcal S(s_\sigma)|^{-\frac12}\bigm|\mathcal S(s_\sigma)=N\epsilon,k_\sigma=k\right) e^{-\beta N\epsilon} + } +\end{equation} Following Derrida \cite{Derrida_1991_The}, \begin{equation} \mathcal N_\mathrm{typ}(\epsilon,k)=\overline\mathcal N(\epsilon,k)+\eta(\epsilon,k)\overline\mathcal N(\epsilon,k)^{1/2} @@ -1529,74 +1567,18 @@ Jacobian over the thimble. } \label{fig:obuchi_3-spin} \end{figure} +This zeroth-order analysis for the $p$-spin suggests that analytic continuation +can be sometimes done despite the presence of a great many complex stationary +points. In particular, when weight is concentrated in certain minima Stokes +lines do not appear to interrupt the proceedings. How bad the situation in is +other regimes, like for smaller $|\beta|$, remains to be seen: our analysis +cannot tell between the effects of Stokes points changing the contour and the +large-$|\beta|$ saddle-point used to evaluate the thimble integrals. Taking the +thimbles to the next order in $\beta$ may reveal more explicitly where Stokes +points become important. \section{The $p$-spin spherical models: numerics} -To confirm the presence of Stokes lines under certain processes in the $p$-spin, we studied the problem numerically. - -\bibliographystyle{unsrt} -\bibliography{stokes} - -\appendix - -\section{Geometry} - -The surface $M\subset\mathbb C^N$ defined by $N=f(z)=z^2$ is an $N-1$ dimensional -\emph{Stein manifold}, a type of complex manifold defined by the level set of a -holomorphic function \cite{Forstneric_2017_Stein}. One can define a Hermitian -metric $h$ on $M$ by taking the restriction of the standard metric of $\mathbb -C^N$ to vectors tangent along $M$. For any smooth function $\phi:M\to\mathbb -R$, its gradient is a holomorphic vector field given by -\begin{equation} - \operatorname{grad}\phi=\bar\partial^\sharp\phi -\end{equation} -Suppose $u:M\to\mathbb C^{N-1}$ is a coordinate system. Then -\begin{equation} - \operatorname{grad}\phi=h^{\bar\beta\alpha}\bar\partial_{\bar\beta}\phi\frac{\partial}{\partial u^\alpha} -\end{equation} -Let $z=u^{-1}$. -\begin{equation} - \frac\partial{\partial u^\alpha}=\frac{\partial z^i}{\partial u^\alpha}\frac\partial{\partial z^i} -\end{equation} -\begin{equation} - \bar\partial_{\bar\beta}\phi=\frac{\partial\bar z^i}{\partial\bar u^{\bar\beta}}\frac{\partial\phi}{\partial\bar z^i} -\end{equation} -\begin{equation} - \operatorname{grad}\phi - =\frac{\partial\bar z^{\bar\jmath}}{\partial\bar u^{\bar\beta}}h^{\bar\beta\alpha}\frac{\partial z^i}{\partial u^\alpha}\frac{\partial\phi}{\partial\bar z^{\bar\jmath}}\frac\partial{\partial z^i} -\end{equation} -At any point $z_0\in M$, $z_0$ is parallel to the normal to $M$. Without loss of generality, take $z_0=e^N$. In a neighborhood of $z_0$, the map $u^\alpha=z^\alpha$ is a coordinate system. -its inverse is $z^i=u^i$ for $1\leq i\leq N-1$ and -\begin{equation} - z^N=\sqrt{N-u^2}. -\end{equation} -The Jacobian is -\begin{equation} - \frac{\partial z^i}{\partial u^\alpha}=\delta^i_\alpha-\delta_N^{i}\frac{u^\alpha}{\sqrt{N-u^2}} -\end{equation} -and therefore the Hermitian metric induced by the map is -\begin{equation} - h_{\alpha\bar\beta}=\frac{\partial\bar z^i}{\partial\bar u^\alpha}\frac{\partial z^{\bar\jmath}}{\partial u^{\bar\beta}}\delta_{i\bar\jmath} - =\delta_{\alpha\bar\beta}+\frac{\bar u^{\alpha}u^{\bar\beta}}{|N-u^2|} -\end{equation} -The metric can be inverted explicitly: -\begin{equation} - h^{\bar\beta\alpha} - =\delta^{\bar\beta\alpha}-\frac{\bar u^{\bar\beta}u^\alpha}{|N-u^2|+|u|^2}. -\end{equation} -Putting these pieces together, we find -\begin{equation} - \frac{\partial\bar z^{\bar\jmath}}{\partial\bar u^{\bar\beta}}h^{\bar\beta\alpha}\frac{\partial z^i}{\partial u^\alpha} - =\delta^{\bar\jmath i}-\frac{z^{\bar\jmath}\bar z^i}{|z|^2} -\end{equation} -\begin{equation} - \operatorname{grad}\phi - =\left(\delta^{\bar\jmath i}-\frac{z^{\bar\jmath}\bar z^i}{|z|^2}\right) - \frac{\partial\phi}{\partial\bar z^{\bar\jmath}}\frac\partial{\partial z^i} -\end{equation} - -\section{Numerics} - To study Stokes lines numerically, we approximated them by parametric curves. If $z_0$ and $z_1$ are two stationary points of the action with $\operatorname{Re}\mathcal S(z_0)>\operatorname{Re}\mathcal S(z_1)$, then we @@ -1625,4 +1607,8 @@ the parameter vectors $g$. Stokes lines are found or not between points by using the Levenberg--Marquardt algorithm starting from $g^{(i)}=0$ for all $i$, and approximating the cost integral by a finite sum. +\section*{References} +\bibliographystyle{unsrt} +\bibliography{stokes} + \end{document} |