Better describe the geometry of the Gram matrix

Vectornaut 2024-05-20 20:11:52 +00:00
parent b12d117dfd
commit d82c84464d

@ -6,11 +6,24 @@ In [inversive coordinates](https://code.studioinfinity.org/glen/dyna3/src/branch
### Constraints as Gram matrix entries ### 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 $a_1, \ldots, a_n \in \mathbb{R}^{1,4}$ 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
\[ \left[\begin{array}{cccc} \rule{0.5pt}{16pt} & \rule{0.5pt}{16pt} & & \rule{0.5pt}{16pt} \\ a_1 & a_2 & \cdots & a_n \\ \rule{0.5pt}{16pt} & \rule{0.5pt}{16pt} & & \rule{0.5pt}{16pt} \end{array}\right] \] \[ \left[\begin{array}{cccc} \rule{0.5pt}{16pt} & \rule{0.5pt}{16pt} & & \rule{0.5pt}{16pt} \\ a_1 & a_2 & \cdots & a_n \\ \rule{0.5pt}{16pt} & \rule{0.5pt}{16pt} & & \rule{0.5pt}{16pt} \end{array}\right] \]
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 product $(\_\!\_, \_\!\_)$ on $\mathbb{R}^{1,4}$. The bilinear form $\langle\_\!\_, G\_\!\_\rangle$ is the pullback of the Lorentz form along $A$: 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
The symmetric bilinear form $\langle\_\!\_, G\_\!\_\rangle$ is the pullback of the Lorentz form along $A$:
\[\begin{align*} \langle\_\!\_, G\_\!\_\rangle & = \langle\_\!\_, A^\top A\_\!\_\rangle \\ & = (A\_\!\_, A\_\!\_). \end{align*}\] \[\begin{align*} \langle\_\!\_, G\_\!\_\rangle & = \langle\_\!\_, A^\top A\_\!\_\rangle \\ & = (A\_\!\_, A\_\!\_). \end{align*}\]
To confirm that $G$ is the Gram matrix of $a_1, \ldots, a_n$, observe that To confirm that $G$ is the Gram matrix of $a_1, \ldots, a_n$, observe that
\[\begin{align*} G_{jk} & = \langle e_j, G e_k \rangle \\ & = (A e_j, A e_k) \\ & = (a_j, a_k), \end{align*}\] \[\begin{align*} G_{jk} & = \langle e_j, G e_k \rangle \\ & = (A e_j, A e_k) \\ & = (a_j, a_k), \end{align*}\]
where $e_1, \ldots, e_n$ is the standard basis for $\mathbb{R}^n$. 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
$-$ | $+$ | $\cdot$
---|---|---
$1$ | $4$ | $n-5$