So this question is a continuation of the following one

[1] On the determination of a quadratic form from its isotropy group

For some motivations and relevant backgrounds related to this question see K. Conrad's answer in [1].

So let $K$ be a field of characteristic $2$ and let $V$ be an $n$-dimensional vector space over $K$. Recall that a function $Q:V\rightarrow K$ is a quadratic form on $V$ if

(1) $Q(av)=a^2v$ for all $a\in K$ and $v\in V$,

(2) $(v,w)\mapsto B(v,w):=(Q(v+w)-Q(v)-Q(w))$ is a $K$-bilinear form.

Define $O(Q):=\{A\in GL(V):Q(Av)=Q(v)\;\;\mbox{for all $v\in V$}\}$ to be the isotropy group of $Q$ (or isometry group of $Q$).

We say that $Q$ is non-degenerate if $Q(v)=0$ and $B(v,V)=0$ imply that $v=0$.

Q: If $Q$ and $Q'$ are non-degenerate quadratic forms on $V$ and $O(Q)=O(Q')$ does this imply that there exists $\lambda\in K^{\times}$ such that $Q=\lambda Q'$ (in such a case we say that $Q$ and $Q'$ are associated)?

Note that the answer is positive if $char(K)\neq 2$. So you may try to complete K. Conrad's argument (which breaks down when $char(K)=2$) or you may try to provide an example of two non-degenerate non-associated quadratic forms $Q$ and $Q'$ such that $O(Q)=O(Q')$.


My recent paper with A. Ruozzi includes a proof of this statement that covers the characteristic 2 case:

Classifying forms of simple groups via their invariant polynomials


The theory in general characteristic is covered by these nice notes of Casselman's

  • $\begingroup$ Dear Igor, this is indeed a nice document but I could not dig the answer to the question $\endgroup$ – Hugo Chapdelaine May 26 '12 at 2:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.