switch long display formulas to more readable triple-backtick block syntax

Gram-matrix-parameterization.md

@@ -46,11 +46,21 @@ Write the loss function as
 \end{align*}
 ```
 where $\Delta = G - \mathcal{P}(A^\top Q A)$. Differentiate both sides and simplify the result using the transpose-invariance of the trace:
-\[ \begin{align*} df & = \operatorname{tr}(d\Delta^\top \Delta) + \operatorname{tr}(\Delta^\top d\Delta) \\ & = 2\operatorname{tr}(\Delta^\top d\Delta). \end{align*} \]
+```math
+\begin{align*} 
+   df & = \operatorname{tr}(d\Delta^\top \Delta) + \operatorname{tr}(\Delta^\top d\Delta) \\
+       & = 2\operatorname{tr}(\Delta^\top d\Delta).
+\end{align*}
+```
 To compute $d\Delta$, it will be helpful to write the projection operator $\mathcal{P}$ more explicitly. Let $\mathcal{C}$ be the set of indices where the Gram matrix is unconstrained. We can express $C$ as the span of the standard basis matrices $\{E_{ij}\}_{(i, j) \in \mathcal{C}}$. Observing that $E_{ij} X^\top E_{ij} = X_{ij} E_{ij}$ for any matrix $X$, we can do orthogonal projection onto $C$ using standard basis matrices:
 \[ \mathcal{P}(X) = \sum_{(i, j) \in \mathcal{C}} E_{ij} X^\top E_{ij}. \]
 It follows that
-\[ \begin{align*} d\mathcal{P}(X) & = \sum_{(i, j) \in \mathcal{C}} E_{ij}\,dX^\top E_{ij} \\ & = \mathcal{P}(dX). \end{align*} \]
+```math
+\begin{align*}
+   d\mathcal{P}(X) & = \sum_{(i, j) \in \mathcal{C}} E_{ij}\,dX^\top E_{ij} \\
+                               & = \mathcal{P}(dX).
+\end{align*}
+\]
 Since the subspace $C$ is transpose-invariant, we also have
 \[ \mathcal{P}(X^\top) = \mathcal{P}(X)^\top. \]
 We can now see that
@@ -61,9 +71,25 @@ d\Delta & = -\mathcal{P}(dA^\top Q A + A^\top Q\,dA) \\
 \end{align*}
 \]
 Plugging this into our formula for $df$, and recalling that $\Delta$ is symmetric, we get
-\[ \begin{align*} df & = 2\operatorname{tr}(-\Delta^\top \big[\mathcal{P}(A^\top Q\,dA)^\top + \mathcal{P}(A^\top Q\,dA)\big]) \\ & = -2\operatorname{tr}\left(\Delta\,\mathcal{P}(A^\top Q\,dA)^\top + \Delta^\top \mathcal{P}(A^\top Q\,dA)\big]\right) \\ & = -2\left[\operatorname{tr}\big(\Delta\,\mathcal{P}(A^\top Q\,dA)^\top\big) + \operatorname{tr}\big(\Delta^\top \mathcal{P}(A^\top Q\,dA)\big)\right] \\ & = -4 \operatorname{tr}\big(\Delta^\top \mathcal{P}(A^\top Q\,dA)\big), \end{align*} \]
+```math
+\begin{align*}
+   df & = 2\operatorname{tr}(-\Delta^\top \big[\mathcal{P}(A^\top Q\,dA)^\top + \mathcal{P}(A^\top Q\,dA)\big]) \\
+       & = -2\operatorname{tr}\left(\Delta\,\mathcal{P}(A^\top Q\,dA)^\top + \Delta^\top \mathcal{P}(A^\top Q\,dA)\big]\right) \\
+       & = -2\left[\operatorname{tr}\big(\Delta\,\mathcal{P}(A^\top Q\,dA)^\top\big) 
+              + \operatorname{tr}\big(\Delta^\top \mathcal{P}(A^\top Q\,dA)\big)\right] \\ 
+       & = -4 \operatorname{tr}\big(\Delta^\top \mathcal{P}(A^\top Q\,dA)\big),
+\end{align*}
+```
 using the transpose-invariance and cyclic property of the trace in the final step. Writing the projection in terms of standard basis matrices, we learn that
-\[ \begin{align*} df & = -4 \operatorname{tr}\left(\Delta^\top \left[ \sum_{(i, j) \in \mathcal{C}} E_{ij} (A^\top Q\,dA)^\top E_{ij} \right] \right) \\ & = -4 \operatorname{tr}\left(\sum_{(i, j) \in \mathcal{C}} \Delta^\top E_{ij}\,dA^\top Q A E_{ij}\right) \\ & = -4 \operatorname{tr}\left(dA^\top Q A \left[ \sum_{(i, j) \in \mathcal{C}} E_{ij} \Delta^\top E_{ij} \right]\right) \\ & = -4 \operatorname{tr}\big(dA^\top Q A\,\mathcal{P}(\Delta)\big) \\ & = \langle\!\langle dA,\,-4 Q A\,\mathcal{P}(\Delta) \rangle\!\rangle. \end{align*} \]
+```math
+\begin{align*}
+   df & = -4 \operatorname{tr}\left(\Delta^\top \left[ \sum_{(i, j) \in \mathcal{C}} E_{ij} (A^\top Q\,dA)^\top E_{ij} \right] \right) \\
+       & = -4 \operatorname{tr}\left(\sum_{(i, j) \in \mathcal{C}} \Delta^\top E_{ij}\,dA^\top Q A E_{ij}\right) \\
+       & = -4 \operatorname{tr}\left(dA^\top Q A \left[ \sum_{(i, j) \in \mathcal{C}} E_{ij} \Delta^\top E_{ij} \right]\right) \\
+       & = -4 \operatorname{tr}\big(dA^\top Q A\,\mathcal{P}(\Delta)\big) \\
+       & = \langle\!\langle dA,\,-4 Q A\,\mathcal{P}(\Delta) \rangle\!\rangle.
+\end{align*}
+```
 From here, we get a nice matrix expression for the negative gradient of the loss function:
 \[ -\operatorname{grad}(f) = 4 Q A\,\mathcal{P}(\Delta). \]
 This matrix is stored as `neg_grad` in the Rust and Julia implementations of the realization routine.
@@ -73,8 +99,14 @@ This matrix is stored as `neg_grad` in the Rust and Julia implementations of the
 Recalling that
 \[ -d\Delta = \mathcal{P}(dA^\top Q A + A^\top Q\,dA), \]
 we can express the derivative of $\operatorname{grad}(f)$ as
-\[ \begin{align*} d\operatorname{grad}(f) & = -4 Q\,dA\,\mathcal{P}(\Delta) - 4 Q A\,\mathcal{P}(d\Delta) \\ & = 4 Q\big[{-dA}\,\mathcal{P}(\Delta) + A\,\mathcal{P}(-d\Delta)\big]. \end{align*} \]
+```math
-In the Rust and Julia implementations of the realization routine, we express $d\operatorname{grad}(f)$ as a matrix in the standard basis for $\operatorname{End}(\mathbb{R}^n)$. We apply the cotangent vector $d\operatorname{grad}(f)$ to each standard basis matrix $E_{ij}$ by setting the value of the matrix-valued 1-form $dA$ to $E_{ij}$.
+\begin{align*}
+   d\operatorname{grad}(f) & = -4 Q\,dA\,\mathcal{P}(\Delta) - 4 Q A\,\mathcal{P}(d\Delta) \\
+                                            & = 4 Q\big[{-dA}\,\mathcal{P}(\Delta) + A\,\mathcal{P}(-d\Delta)\big].
+\end{align*}
+```
+In the Rust and Julia implementations of the realization routine, we express $d\operatorname{grad}(f)$ as a matrix in the standard basis for
+$\operatorname{End}(\mathbb{R}^n)$. We apply the cotangent vector $d\operatorname{grad}(f)$ to each standard basis matrix $E_{ij}$ by setting the value of the matrix-valued 1-form $dA$ to $E_{ij}$.
 
 #### Finding minima