Emphasize the rank of G, which will constrain the shape of the LU decomposition

Gram-matrix-parameterization.md

@@ -2,28 +2,13 @@
 
 ### Construction elements as vectors
 
-In [inversive coordinates](https://code.studioinfinity.org/glen/dyna3/src/branch/main/notes/inversive.md), points and generalized spheres are represented, respectively, by timelike and spacelike vectors in $\mathbb{R}^{1,4}$. If we normalize these vectors to pseudo-length $\pm 1$, and choose a lightlike vector on the 1d subspace representing the point at infinity, a lot of the constraints we care about can be expressed by fixing the Lorentz products between vectors.
+Take a 5d vector space $V$ with a bilinear form $(\_\!\_, \_\!\_)$ of signature $-++++$, which we'll call the *Lorentz form*. In [inversive coordinates](https://code.studioinfinity.org/glen/dyna3/src/branch/main/notes/inversive.md), points and generalized spheres are represented, respectively, by timelike and spacelike vectors in $V$. If we normalize these vectors to pseudo-length $\pm 1$, and choose a vector on the lightlike 1d subspace representing the point at infinity, a lot of the constraints we care about can be expressed by fixing the Lorentz products between vectors.
 
 ### Constraints as Gram matrix entries
 
-#### Introducing the Gram matrix
-
-The vectors $a_1, \ldots, a_n \in \mathbb{R}^{1,4}$ representing the elements of our construction can be encoded in a linear map $A \colon \mathbb{R}^n \to \mathbb{R}^{1,4}$, whose matrix is
+The vectors $a_1, \ldots, a_n \in V$ representing the elements of our construction can be encoded in a linear map $A \colon \mathbb{R}^n \to V$, whose matrix is
 \[ \left[\begin{array}{cccc} \rule{0.5pt}{16pt} & \rule{0.5pt}{16pt} & & \rule{0.5pt}{16pt} \\ a_1 & a_2 & \cdots & a_n \\ \rule{0.5pt}{16pt} & \rule{0.5pt}{16pt} & & \rule{0.5pt}{16pt} \end{array}\right] \]
-We can then express constraints by fixing elements of the Gram matrix $G = A^\top A$, where $\top$ is the adjoint with respect to the inner product $\langle\_\!\_, \_\!\_\rangle$ on $\mathbb{R}^n$ and the Lorentz form $(\_\!\_, \_\!\_)$ on $\mathbb{R}^{1,4}$.
-
-#### The geometry of the Gram matrix
+We can then express constraints by fixing elements of the Gram matrix $G = A^\top A$, where $\top$ is the adjoint with respect to the inner product $\langle\_\!\_, \_\!\_\rangle$ on $\mathbb{R}^n$ and the Lorentz form. Since the inner product and the Lorentz form are both non-degenerate, the rank of $G$ matches the dimension of the image of $A$.
 
 The symmetric bilinear form $\langle\_\!\_, G\_\!\_\rangle$ is the pullback of the Lorentz form along $A$:
-\[\begin{align*} \langle\_\!\_, G\_\!\_\rangle & = \langle\_\!\_, A^\top A\_\!\_\rangle \\ & = (A\_\!\_, A\_\!\_). \end{align*}\]
-To confirm that $G$ is the Gram matrix of $a_1, \ldots, a_n$, observe that
-\[\begin{align*} G_{jk} & = \langle e_j, G e_k \rangle  \\  & = (A e_j, A e_k) \\ & = (a_j, a_k), \end{align*}\]
-where $e_1, \ldots, e_n$ is the standard basis for $\mathbb{R}^n$.
-
-#### The rank of the Gram matrix
-
-Since inner products and Lorentz forms are both non-degenerate, the kernel of $\langle\_\!\_, G\_\!\_\rangle$ is $\ker A$. The form $\langle\_\!\_, G\_\!\_\rangle$ on $(\ker A)^\top$ is isometric, through $A$, to the Lorentz form on $\operatorname{im} A$. It follows that if $A$ is onto, then $\langle\_\!\_, G\_\!\_\rangle$ has signature
-
-$-$ | $+$ | $\cdot$
----|---|---
-$1$ | $4$ | $n-5$
+\[\begin{align*} \langle\_\!\_, G\_\!\_\rangle & = \langle\_\!\_, A^\top A\_\!\_\rangle \\ & = (A\_\!\_, A\_\!\_). \end{align*}\]