From 35c4e960648856414d3425eddb69881e9028d6f9 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Mon, 28 Oct 2024 16:35:14 +0100
Subject: Clarified notation surrounding gradient and Hessian, and standardized
 arXiv bib entries.

---
 marginal.tex | 13 ++++++++++---
 1 file changed, 10 insertions(+), 3 deletions(-)

(limited to 'marginal.tex')

diff --git a/marginal.tex b/marginal.tex
index d9e1d47..b9fabc9 100644
--- a/marginal.tex
+++ b/marginal.tex
@@ -469,8 +469,9 @@ extremizing the Lagrangian
   L(\mathbf x,\pmb\omega)=H(\mathbf x)+\sum_{i=1}^r\omega_ig_i(\mathbf x)
 \end{equation}
 with respect to $\mathbf x$ and the Lagrange multipliers
-$\pmb\omega=\{\omega_1,\ldots,\omega_r\}$. The corresponding gradient and
-Hessian of the energy associated with this constrained extremal problem are
+$\pmb\omega=\{\omega_1,\ldots,\omega_r\}$. To write the gradient and Hessian of the energy, which are necessary to count stationary points, care must be taken to ensure they are constrained to the tangent space of the configuration manifold. For our purposes, the Lagrangian formalism offers a solution: the gradient $\nabla H:\mathbb R^N\times\mathbb R^r\to\mathbb R^N$ and
+Hessian $\operatorname{Hess} H:\mathbb R^N\times\mathbb R^r\to\mathbb R^{N\times N}$ of the energy $H$ can be written as the simple vector derivatives of
+the Lagrangian $L$, with
 \begin{align}
   &\nabla H(\mathbf x,\pmb\omega)
   =\partial L(\mathbf x,\pmb\omega)
@@ -483,7 +484,13 @@ Hessian of the energy associated with this constrained extremal problem are
   \end{aligned}
 \end{align}
 where $\partial=\frac\partial{\partial\mathbf x}$ will always represent the
-derivative with respect to the vector argument $\mathbf x$.
+derivative with respect to the vector argument $\mathbf x$. Note that unlike
+the energy, which is a function of the configuration $\mathbf x$ alone, the
+gradient and Hessian depend also on the Lagrange multipliers $\pmb\omega$. In situations
+with an extensive number of constraints, it is important to take seriously
+contributions of the form $\frac{\partial^2L}{\partial\mathbf
+x\partial\pmb\omega}$ to the Hessian \cite{Kent-Dobias_2024_On}. However, the cases we study here have
+$N^0$ constraints and these contributions appear as finite-$N$ corrections.
 
 The number of
 stationary points in a landscape for a particular function $H$ is found by
-- 
cgit v1.2.3-70-g09d2

