Let $q$ be a quadratic form on the finite vector space $V$ over $\mathbb{F}_2$.

Question

And let $q$ be such that the bilinear form $B: (x, y) \mapsto q(x+y)+q(x)+q(y)$ is nondegenerate. Let $$A(q)=\frac{\sum_{x\in V}(-1)^{q(x)}}{\sqrt{|V|}}\in \mathbb{R}.$$ I need prove that $A(q)$ is $1$ or $-1$. And now I haven't got any idea.

Answer 1

It is not too hard to see that if $q=q_1\perp q_2$ then $A(q)=A(q_1)\cdot A(q_2)$.

Since the polar form of $q$ is nondegenerated, we have $q=q_1\perp \dots\perp q_r$ with $q_i = [a_i,b_i]$ (so $q_i(x,y)=a_ix^2+xy+b_iy^2$), so we only need to show that $A([a,b])=\pm 1$. There are three cases to consider, since $a$ and $b$ can be $0$ or $1$ in $\mathbb{F}_2$, and you can check them easily yourself. You can check that $A([a,b])$ is $1$ if $a$ or $b$ is $0$, and is $-1$ when $a=b=1$.

As someone pointed out in the comments, a direct calculation shows that this is more or less the Arf invariant of $q$, in the sense that $A(q)=(-1)^{\Delta(q)}$ where $\Delta(q)\in \mathbb{F}_2$ is the Arf invariant of $q$ (so maybe it can be considered a multiplicative version). I had never seen that characterization though, and I guess it would be awkward to generalize to other fields of characteristic 2.