How to solve Aschbacher's exercise on symmetric forms

Question

Aschbacher's Finite Group Theory is an excellent textbook, and its exercise 7.9.10 on page 104 is used to justify several assumptions in the chapter on classical groups.

Let $F$ be a field and $f$ a sesquilinear form on $V$ with respect to the automorphism $\theta$ of $F$, such that for all $x,y \in V$, $f(x,y)=0$ if and only if $f(y,x)=0$. Prove that either:

1. $f(x,x)=0$ for all $x \in V$, $\theta=1$, and $f$ is skew symmetric, or

2. there exists $x \in V$ with $f(x,x) \neq 0$ and one of the following holds:

   (a) $\theta=1$ and $f$ is symmetric

   (b) $|\theta|=2$ and $f$ is similar to a hermitian symmetric form

   (c) $|\theta|>2$ and $\operatorname{Rad}(V)$ is of codimension $1$ in $V$.

A function $f:V\times V \to F$ is called sesquilinear with respect to the field automorphism $\theta$ iff $f(ax+y,z) = a\cdot f(x,z) + f(y,z)$ and $f(x,ay+z) = \theta(a) f(x,y) + f(x,z)$ for all $a \in F$ and $x,y,z \in V$.

While trying to write up my own justification, I noticed this exercise is false. Take $V$ to be two-dimensional, $f(x,y)=0$ for all $x,y \in V$, and $\theta \neq 1$. The fix cannot be to assume $f$ is nondegenerate, since (2c) is explicitly degenerate. I hope however, the fix is simply a 3rd possibility, that $f(x,y)=0$ for all $x,y \in V$, but $|\theta|>1$ (I call such a form null).

I have shown that a skew symmetric form is either null or $\theta=1$. However, I am having trouble with case (2) [ so we assume $f$ is not skew-symmetric ].

If $\theta=1$, then I get that each individual $f(x,y)=\pm f(y,x)$, but I'm having trouble proving that only one sign occurs.

If $\theta\neq 1$, then I don't have much useful. In particular, I haven't shown $\theta^2=1$ or that $f(x,x)$ lies in the fixed field of $\theta$ (or at least that they all lie in the same coset).

Answer 1

For 2 (a), you have $f(x,x) = a \ne 0$ and you need to prove that there do not exist $y,z$ with $f(y,z) = b = -f(z,y)$.

Looking at the matrix of the form restricted to the space spanned by $x,y,z$, there seem to be two types that you need to rule out

$\left(\begin{array}{rr}a&b\\-b&c\end{array}\right)$ and $\left(\begin{array}{rrr}a&c&d\\c&0&-b\\d&b&0\end{array}\right)$, with $a,b$ nonzero in both cases.

In the first case, putting $u=(1,0)$, $v=(1,1)$ gives $f(u,v)=a+b$, $f(v,u)=a-b$, and we cannot have $f(u,v) = \pm f(v,u)$.

In the second case, considering $u=(1,1,1)$, $v=(0,1,0)$ forces $c=0$ and similarly $d=0$. and then $u=(1,1,0)$, $v=(1,0,1)$ gives a contradication.

I haven't had time to think about (b) and (c) yet!