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authorJaron Kent-Dobias <jaron@kent-dobias.com>2020-12-07 14:15:51 +0100
committerJaron Kent-Dobias <jaron@kent-dobias.com>2020-12-07 14:15:51 +0100
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Added lazy paragraph defending our choice of constraint by smearing the alternative.
Diffstat (limited to 'bezout.tex')
-rw-r--r--bezout.tex8
1 files changed, 7 insertions, 1 deletions
diff --git a/bezout.tex b/bezout.tex
index 02158a0..897e79d 100644
--- a/bezout.tex
+++ b/bezout.tex
@@ -31,7 +31,7 @@
\begin{equation} \label{eq:bare.hamiltonian}
H_0 = \frac1{p!}\sum_{i_1\cdots i_p}^NJ_{i_1\cdots i_p}z_{i_1}\cdots z_{i_p},
\end{equation}
-where the $z$ are constrained by $z\cdot z=N$ and $J$ is a symmetric tensor
+where $z\in\mathbb C^N$ is constrained by $z^2=N$ and $J$ is a symmetric tensor
whose elements are complex normal with $\langle|J|^2\rangle=p!/2N^{p-1}$ and
$\langle J^2\rangle=\kappa\langle|J|^2\rangle$ for complex parameter
$|\kappa|<1$. The constraint is enforced using the method of Lagrange
@@ -41,6 +41,12 @@ multipliers: introducing the $\epsilon\in\mathbb C$, this gives
\end{equation}
At any critical point $\epsilon=H/N$, the average energy.
+When compared with $z^*z=N$, the constraint $z^2=N$ may seem an unnatural
+extension of the real $p$-spin spherical model. However, a model with this
+nonholomorphic spherical constraint has a disturbing lack of critical points
+nearly everywhere, since $0=\partial H/\partial z^*=-p\epsilon z$ 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
only if it is also a critical point of its imaginary part. The number of
critical points of $H$ is therefore the number of critical points of