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| -rw-r--r-- | bezout.bib | 14 | ||||
| -rw-r--r-- | bezout.tex | 34 |
2 files changed, 47 insertions, 1 deletions
@@ -7,4 +7,18 @@ address = {rue S. Jacques, Paris} } +@article{Nguyen_2014_The, + author = {Nguyen, Hoi H. and O'Rourke, Sean}, + title = {The Elliptic Law}, + journal = {International Mathematics Research Notices}, + publisher = {Oxford University Press (OUP)}, + year = {2014}, + month = {10}, + number = {17}, + volume = {2015}, + pages = {7620--7689}, + url = {https://doi.org/10.1093%2Fimrn%2Frnu174}, + doi = {10.1093/imrn/rnu174} +} + @@ -64,7 +64,7 @@ It is easy to see that {\em for a pure $p$ spin}, at any critical point $\epsil For the constraint we shall choose here $z^2=N$, rather than $|z^2|=N$, in order to preserve the holomorphic nature of the functions.. In addition, the nonholomorphic spherical constraint has a disturbing lack of critical points -nearly everywhere, since $0=\partial H/\partial z^*=-p\epsilon z$ is only +nearly everywhere, since $0=\partial^* H=-p\epsilon z$ is only satisfied for $\epsilon=0$, as $z=0$ is forbidden by the constraint. Since $H$ is holomorphic, a point is a critical point of its real part if and @@ -106,6 +106,38 @@ form |\det\partial\partial H|^2. \end{equation} +The Hessian of \eqref{eq:constrained.hamiltonian} is $\partial_i\partial_j +H=\partial_i\partial_j H_0-p\epsilon\delta_{ij}$, or the Hessian of +\eqref{eq:bare.hamiltonian} with a constant added to its diagonal. The +eigenvalue distribution $\rho$ of the constrained Hessian is therefore related +to the eigenvalue distribution $\rho_0$ of the unconstrained one by a similar +shift, or $\rho(\lambda)=\rho_0(\lambda+p\epsilon)$. The Hessian of +\eqref{eq:bare.hamiltonian} is +\begin{equation} \label{eq:bare.hessian} + \partial_i\partial_jH_0 + =\frac{p(p-1)}{p!}\sum_{k_1\cdots k_{p-2}}^NJ_{ijk_1\cdots k_{p-2}}z_{k_1}\cdots z_{k_{p-2}}, +\end{equation} +which makes its ensemble that of Gaussian complex symmetric matrices. Given its variances +$\langle|\partial_i\partial_j H_0|^2\rangle=p(p-1)a^{p-2}/2N$ and +$\langle(\partial_i\partial_j H_0)^2\rangle=p(p-1)\kappa/2N$, $\rho_0(\lambda)$ is constant inside the ellipse +\begin{equation} \label{eq:ellipse} + \left(\frac{\mathop{\mathrm{Re}}(\lambda e^{i\theta})}{a^{p-2}+|\kappa|}\right)^2+ + \left(\frac{\mathop{\mathrm{Im}}(\lambda e^{i\theta})}{a^{p-2}-|\kappa|}\right)^2 + <\frac{p(p-1)}{2a^{p-2}} +\end{equation} +where $\theta=\frac12\arg\kappa$ \cite{Nguyen_2014_The}. The eigenvalue +spectrum of $\partial\partial H$ therefore is than of an ellipse whose center +is shifted by $p\epsilon$. + +The eigenvalue spectrum of the Hessian of the real part, or equivalently the +eigenvalue spectrum of $(\partial\partial H)^\dagger\partial\partial H$, is the +singular value spectrum of $\partial\partial H$. This is a more difficult +problem and to our knowledge a closed form for arbitrary $\kappa$ is not known. +We have worked out an implicit form for this spectrum using the saddle point of +a replica calculation for the Green function. blah blah blah\dots + +The transition from a one-cut to two-cut singular value spectrum naturally corresponds to the origin leaving the support of the eigenvalue spectrum. Weyl's theorem requires that the product over the norm of all eigenvalues must not be greater than the product over all singular values. Therefore, the absence of zero eigenvalues implies the absence of zero singular values. + \bibliographystyle{apsrev4-2} \bibliography{bezout} |