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title: 2024-04-03
created: 2023-03-26T11:59
updated: 2024-05-03T23:22
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# 2024-04-03

> Talk by Akhil Matthew, IAS


- **Theorem** (Lagrange): $n\in \ZZ: \forall n, n = x_1^2 + x_2^2 + x_3^2 + x_4^2$ for some $x_i\in \ZZ$
- **Theorem** (Legendre): $n\in \ZZ: n\not\equiv 7 \mod 8 \implies n = x_1^2 + x_2^2 + x_3^2$ for some $x_i\in \ZZ$
- **Theorem** (Fermat) $n = \prod p_k^{n_k} \in \ZZ: n = x_1^2 + x_2^2 \iff p_k \equiv 3 \mod 4 \implies n_k$ is even.
- **Theorem** (Artin): $f\in \RR(x_1, \cdots, x_n): f = \sum f_i^2 \iff f\geq 0$ where defined.
	- **Theorem** (Pfister): One needs at most $2^n$ functions $f_i$. This is not known to be optimal.
- Quadratic form: $q: F^n \to F$ where $q(\vector x) = \sum x_i a_{ij} x_j$.
- Bilinear form: $B: V\times V \to F$ where $B(\wait, v), B(v,\wait)$ are $F\dash$linear.
- Symmetric: $B(v, w) = B(w, v)$
- Skew-symmetric: $B(v, w) = -1\cdot B(w, v)$
	- Graded-symmetric: $B(v, w) = (-1)^{\abs v \cdot \abs w} B(w, v)$?
- Nondegenerate: $\forall v\in V,\exists w\in V$ such that $B(v, w)\neq 0$.
- Gram matrix: $G_B = (B(e_i, e_j))$ in a basis $\ts{e_i}$.
	- Thus quadratic forms are equivalent to symmetric matrices in $\GL_n(F)$.
- Given an inner product space $(V, \ip\wait\wait)$, there is an associated quadratic space: $q:V\to F$ defined by $q(V) = \ip v v$.
- Polarization: $2vw = q(v+w)-q(v)-q(w)$.
- **Question**: given $v$, what elements of $F$ are of the form $v^2$?
- **Theorem** (Hasse-Minkowski): ? #todo 
- $W^\perp = \ts{v\in V\st vw=0}$.
	- Always have $\dim W + \dim W^\perp = \dim V$
	- NB $W\intersect W^\perp \neq \emptyset$ in general.
	- If $\ro B W$ is nondegenerate, then $V  = W \oplus W^\perp$ and $\ro B {W^\perp}$ is also nondegenerate.
- $\Orth(V, q) = ?$
- Set $\gens{a}$ to be the inner product space $\gens{e_1}_F$ where $e_1^2 = n$; then $q(x) = xax$.
	- Any IP space $V$ one has $V \cong \bigoplus \gens{a_i}$ with $a_i\in F\units$.
	- The $a_i$ are non-unique: $\gens{a_i} \cong \gens {a_i u_i^2}$ for any $u_i\in F\units$. #todo/why 
- Diagonalisation: for any $V\slice \RR$, one has $V\cong \gens{1}\sumpower p \gens{-1}\sumpower q$.
	- Note that $B(x, y) = x^2 + y^2$
- Define $\mathrm{index}(V/\RR) \da p-q$ and $\signature(V/\RR) = (p, q)$.
- **Theorem** (Sylvester): over $\RR$ the invariants are the dimension and index.
- If $F$ has all square roots, $V\cong \gens{1}\sumpower{n}$.
	- Thus over $F =\bar F$, the only invariant of $(V, q)$ is the dimension
- Define the orthogonal group $\Orth(V, q) \da \ts{f: V\to V \st q\circ f = f}$.
	- Let $\cat C = \mods{F}\slice F$, then $\Orth(V, q) = \Aut_{\cat C}(V)$.
- Define $R_v(x) \da x - {2xv\over v^2}v$ for $v^2\neq 0$.
	- NB $\ro{R_v}{\gens{v}_F} = -1$ and $\ro{R_v}{\gens{v}_F^\perp} = 1$. Check $R_v(v)$.
- Define $\mathrm{Ref}(V) = \gens{R_v\st v\in V} \leq \Orth(V, q)$
- **Theorem** (Cartan–Dieudonné):
	- $\Orth(V, q) = \mathrm{Ref}(V)$.
- $U$ diagonalises to $\gens{1}\oplus\gens{-1}$.