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\begin{document}% \tiny \scriptsize \footnotesize \small \normalsize \large \Large \LARGE \huge \Huge 
\begin{center}{\LARGE Recovering a quadratic form from its orthogonal group}\bigskip 

Pierre-Yves Gaillard\footnote{DOI 10.5281/zenodo.14513720. \href{https://tinyurl.com/3r5vscy8}{My texts at Zenodo}. ORCID \url{https://orcid.org/0000-0002-7960-1698}.} 
\end{center}%\centerline{Recovering a quadratic form from its orthogonal group}

\begin{thm}\label{T1}
Let $K$ be a field of characteristic different from 2, let $V$ be a $K$-vector space of finite dimension $n\ge2$, let $q$ and $r$ be two quadratic forms on $V$, let $\OO(q)$ and $\OO(r)$ be the respective orthogonal groups, and let $Q,R:V\times V\to K$ be the respective associated symmetric bilinear forms. Then we have 
$$
\OO(q)=\OO(r)\iff Q^{-1}(0)=R^{-1}(0)\iff Kq=Kr.
$$ 
\end{thm} 

The only non-trivial ingredient we will use is Witt's Extension Theorem. For completeness sake we give a proof of Witt's Extension Theorem at the end of the text. (I found this proof in Serre's book \textbf{A Course in Arithmetic}.) We will also use the following lemma: 

\begin{lem}\label{L2}
Let $K,V,q,Q$ and $\OO(q)$ be as in Theorem \ref{T1}. Assume that $q$ is degenerate. Then the radical $W=\rad(q)$ of $q$ is the smallest nonzero $\OO(q)$-invariant subspace of $V$. 
\end{lem} 

\begin{proof}
Let $U$ be a nonzero $\OO(q)$-invariant subspace which does not contain $W$. It suffices to derive a contradiction. Let $W'$ and $V'$ be subspaces of $V$ satisfying $U+W=U\oplus W',V=U\oplus W'\oplus V'$. Note that $U\ne0\ne W'$. Let $f'$ be a nonzero linear map from $U$ to $W'$ and define $f,g:V\to V$ by $f(u+w'+v')=f'(u)$ and $g=1+f$. We have $f^2=0$, and thus $1-f=g^{-1}$. Moreover $g$ is easily seen to be in $\OO(r)=\OO(q)$. For $u\in U$ such that $f'(u)\ne0$ we have $gu=u+f'(u)\notin U$, contradiction. 
\end{proof} 

To prove the theorem, we will show $\OO(q)=\OO(r)\implies Q^{-1}(0)=R^{-1}(0)\implies Kq=Kr\implies \OO(q)=\OO(r)$, the last implication being obvious.\medskip

\noi$\bullet\quad\OO(q)=\OO(r)\implies Q^{-1}(0)=R^{-1}(0)$\medskip

\noi We consider three (non-exclusive) cases, denoted A, B and C.\medskip

\noi\textbf{Case A:} $q$ and $r$ are non-degenerate. 

\begin{lem}\label{L3}
Suppose that $v,w\in V$ are linearly independent. Then $Q(v,w)=0\iff$ there is a $g$ in $\OO(q)$ such that $gv=v,gw=-w$. 
\end{lem} 

\begin{proof}
$\implies$: Define $h\in\GL(Kv+Kw)$ by $hv=v,hw=-w$. Then $h$ is in $\OO(q|Kv+Kw)$, where $q|Kv+Kw$ is the restriction of $q$ to $Kv+Kw$. By Witt's Extension Theorem, $h$ extends to $g\in\OO(q)$. 

\noi$\impliedby$: We have $Q(v,w)=Q(gv,gw)=Q(-v,w)=-Q(v,w)$. 
\end{proof} 

\begin{lem}\label{L4}
Suppose that $v\in V$ is nonzero. Then $q(v)\ne0\iff$ there is a $g$ in $\OO(q)$ such that $g^2=1$ and the $(-1)$-eigenspace of $g$ is $Kv$.  
\end{lem} 

\begin{proof}
$\implies$: Define $g\in\GL(V)$ by $gv=-v$ and $gw=w$ if $Q(v,w)=0$. Then $g$ is in $\OO(q)$. 

\noi$\impliedby$: Let $W$ be fixed subspace of $g$. Then we have $V=Kv\oplus W$. Moreover, $w\in W$ implies $Q(v,w)=Q(gv,gw)=Q(-v,w)=-Q(v,w)$, and thus $Q(v,w)=0$. Since this holds for all $w$ in $W$, and $q$ is non-degenerate, we have $q(v)\ne0$. 
\end{proof} 

Lemmas \ref{L3} and \ref{L4} imply the statement when $q$ and $r$ are non-degenerate.\medskip 

\noi\textbf{Case B:} $q$ is non-degenerate. It suffices to show that $r$ is non-degenerate too. Assume by contradiction that $r$ is degenerate. 

\noi Case 1: $r=0$. We have $\OO(q)=\OO(r)=\GL(V)$. Then $\OO(q)$ acts transitively on the nonzero vectors, and on the pairs of linearly independent vectors. This implies $q(v)=q(w)$ for all $v\ne0\ne w$, hence $q(v)\ne0$ for all $v\ne0$, hence there are linearly independent $v$ and $w$ such that $Q(v,w)=0$, hence $Q(v,w)=0$ for all linearly independent $v$ and $w$, hence $0=Q(v,v+w)=Q(v,v)\ne0$ for such $v$ and $w$, contradiction. 

\noi Case 2: $r$ is nonzero (but degenerate). The radical $W$ of $r$ satisfies $0\subsetneqq W\subsetneqq V$. Set $G:=\OO(q)=\OO(r)$. Then $W$ is $G$-invariant and the only $G$-invariant subspaces of $W$ are 0 and $W$. Thus, the intersection of $W$ with the $q$-orthogonal $U$ of $W$ is equal to 0 or $W$. 

\noi Case 2.1: $U\cap W=0$. We have $V=U\oplus W$, and $U$ is $G$-invariant. This contradicts Lemma~\ref{L2}. 

\noi Case 2.2: $U\cap W=W$, i.e. $q|W=0$. Let $w\ne0$ be in $W$. There is a $v\in V$ which is not $q$-orthogonal to $w$ (in particular $v\notin W$). Setting $a:=q(v),b:=Q(v,w)\ne0$, we have $q(2bv-aw)=0$, and Witt's Extension Theorem implies the existence of a $g$ in $G$ such that $gw=2bv-aw\notin W$, contradiction. \medskip

\noi\textbf{Case C:} $q$ and $r$ are degenerate. % previous version https://docs.google.com/document/d/1KSmVNpB0pYjJMraG07WNAvTg75BKEYYqJYlmDNyXeYI/edit?tab=t.0

By Lemma~\ref{L2}, the radical $\rad(q)$ of $q$ is the smallest nonzero $\OO(q)$-invariant subspace of $V$, and similarly for $r$; in particular $W:=\rad(r)=\rad(q)$ and $\rho:=\rk(r)=\rk(q)$. We leave the case $\rho\le1$ to the reader and assume $\rho\ge2$. Let $U$ be a linear complement to $W$ in $V$, let $q'$ and $Q'$ be the respective restrictions of $q$ to $U$ and of $Q$ to $U\times U$, and define $r'$ and $R'$ similarly. We get $\OO(q')=\OO(r')$, hence $Q'^{-1}(0)=R'^{-1}(0)$, and the equality $Q^{-1}(0)=R^{-1}(0)$ follows from Case~A. \medskip

\noi$\bullet\quad Q^{-1}(0)=R^{-1}(0)\implies Kq=Kr$. This implication will follow from: 

\begin{prop}\label{P5}
Let $L$ be a field of arbitrary characteristic, let $V_1,\ldots,V_n$ be $L$-vector spaces, let $f,g:V_1\times\ldots\times V_n\to L$ be $n$-linear maps such that $f^{-1}(0)\subset g^{-1}(0)$. Then there is a $\lambda$ in $L$ such that $g=\lambda f$. 
\end{prop} 

\begin{lem}\label{L6}
Let $V$ be an $L$-vector space and $f$ an endomorphism of $V$. Assume that, for all $v$ in $V$, there is a $\lambda(v)$ in $L$ such that $f(v)=\lambda(v)\,v$. Then there is a $\lambda$ in $L$ such that $f(v)=\lambda\,v$ for all $v$.  
\end{lem} 
The proof is left to the reader. 

\begin{lem}\label{L7}
Let $f,g:V\to W$ two $L$-linear maps. Assume that, for all $v$ in $V$, there is a $\lambda(v)$ in $L$ such that $g(v)=\lambda(v)\,f(v)$. Then there is a $\lambda$ in $L$ such that $g(v)=\lambda\,f(v)$ for all $v$.  
\end{lem} 

\begin{proof}
We have $f^{-1}(0)\subset g^{-1}(0)$ and $g(V)\subset f(V)$. If $f',g':V/f^{-1}(0)\to f(V)$ are the induced maps, then $f'$ is bijective, and we prove the result by applying Lemma~\ref{L6} to $g'\circ f'^{-1}$. 
\end{proof} 

\begin{proof}[Proof of Proposition \ref{P5}] 
We induct on $n\ge1$, the case $n=1$ being trivial. Let $W$ be the space of $(n-1)$-linear forms on $V_1\times\ldots\times V_{n-1}$, define $f':V_n\to W$ by $f'(v_n)(v_1,\ldots,v_{n-1}:=f(v_1,\ldots,v_n)$, and define $g'$ similarly. By the induction hypothesis, there is a $\lambda(v)$ in $L$ such that $g'(v_n)=\lambda(v_n)\,f'(v_n)$ for all $v_n$, and we can and apply Lemma~\ref{L7}. 
\end{proof} 

We end with a statement and a proof of Witt's Extension Theorem. We use the following terminology and notation. Let $V$ be a finite dimensional vector space over a field $K$ whose characteristic is not 2. Assume that $V$ is endowed with a symmetric bilinear form, denoted simply by $(x,y)\mapsto xy$ (we even write $x^2$ for $xx$). We call such $V$ a \textbf{quadratic module}. A map $f$ between quadratic modules is \textbf{metric} if it is linear and satisfies $f(x)f(y)=xy$ for all $x,y$. 

\begin{thm}[Witt's Extension Theorem]\label{TW}
Let $V$ be a non-degenerate quadratic module, $W$ a vector subspace of $V$, and $f:W\to V$ an injective metric map. Then $f$ extends to a metric automorphism of $V$. 
\end{thm} 

\begin{lem}\label{L9}
In the above setting, assume that $W$ is degenerate. Then $f$ extends to a metric injection $f_1:W_1\to V$ with $W_1\supsetneqq W$. 
\end{lem} 

\begin{proof}
Let $w_0$ be a nonzero vector in $\rad(W)$ and $\ell$ a linear form on $W$ such that $\ell(w_0)=1$. There is a $v_0$ in $V$ such that $v_0\,w=\ell(w)$ for all $w$ in $W$. On replacing $v_0$ by $v_0-(y^2/2)\, w_0$, we can assume $(v_0)^2=0$. Set $W_1:=W+Kv_0$. We have in particular $W_1\supsetneqq W$. 

We do the same with $W':=f(W),\ w_0':=f(w_0),\ \ell':=\ell\circ f^{-1}$ instead of $W,w_0,\ell$, obtaining $v_0'$ in lieu of $v_0$, we set $W_1':=W'+Kv_0'$ and $f_1(v_0):=v_0'$, and we check that $f_1$ does the job. 
\end{proof} 

\begin{proof}[Proof of Theorem \ref{TW}]
In view of the lemma, we can assume that $W$ is non-degenerate. We argue by induction on $\dim W$. \smallskip

\noi Case 1: $\dim W=1$. Let $w_0$ be a nonzero vector in $W$, and note $(w_0)^2\ne0$. Set $v_0:=f(w_0)$; in particular $(v_0)^2=(w_0)^2\ne0$, which implies $(w_0+\eps\,v_0)^2\ne0$ for some $\eps\in\{1,-1\}$. Setting $u:=w_0+\eps\,v_0$, we get $V=Ku\oplus(Ku)^\perp$. Define the linear map $s:V\to V$ by $s(u)=-u$ and $s(v)=v$ if $v\perp u$. Note that $w_0-\eps\,v_0\perp u$, and thus $s(w_0-\eps\,v_0)=w_0-\eps\,v_0$. Since $s(w_0+\eps\,v_0)=-w_0-\eps\,v_0$, we get $s(w_0)=-\eps\,v_0=-\eps\,f(w_0)$, and we can set $g:=-\eps\,s$. \smallskip

\noi Case 2: $\dim W\ge2$. Let $W=W_1\oplus W_2$ be an orthogonal direct sum with $W_1\ne0\ne W_2$. Write $f_i:W_i\to V$ for the restriction of $f$ to $W_i$. By the induction hypothesis $f_1$ extends to a metric automorphism $g_1$ of $V$. The map $f':=g_1^{-1}\circ f:W\to V$ satisfies $f'(w_1)=g_1^{-1}(f(w_1))=w_1$ for $w_1\in W_1$, and thus $f'(W_2)\subset W_1^\perp$ (where $W_1^\perp$ is the orthogonal of $W_1$ in $V$). Again by the induction hypothesis, $f'$ extends to a metric automorphism $g_2$ of $W_1^\perp$. Define $g':V\to V$ by $g'(w_1+w_1^\perp)= w_1+g_2(w_1^\perp)$ (obvious notation). Then $g'$ is a metric automorphism of $V$. Finally set $g:=g_1\circ g'$ (again a metric automorphism of $V$). We get, for $w_i\in W_i$, 
$$
g(w_1)=g_1(g'(w_1))=g_1(w_1)=f_1(w_1)=f(w_1),
$$ 
$$
g(w_2)=g_1(g'(w_2))=g_1(g_2(w_2))=g_1(f'(w_2))=g_1(g_1^{-1}(f(w_2)))=f(w_2).
$$ 
This shows that $g$ is indeed a metric automorphism which extends $f$. 
\end{proof} 

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