Counting non-degenerate bicharacters

Question

Let $\mathbb{F}_2^n$ be a finite dimensional vector space over $\mathbb{F}_2$.

A map $b:\mathbb{F}_2^n\times \mathbb{F}_2^n\rightarrow \{\pm 1\}$ is called bilinear or a bicharacter if $$b(a_1+a_2,a_3)=b(a_1,a_3)b(a_2,a_3)\ \textrm{and}\ b(a_1,a_2+a_3)=b(a_1,a_2)b(a_1,a_3)$$ for all $a_1,a_2,a_3\in \mathbb{F}_2^n$. A bilinear $b$ as above is non-degenerate if for every $a_1\in \mathbb{F}_2^n$ there exists $a_2\in \mathbb{F}_2^n$ such that $b(a_1,a_2)\neq 1$.

How many non-degenerate bilinear maps $b:\mathbb{F}_2^n\times \mathbb{F}_2^n\rightarrow \{\pm 1\}$ are there?

It is not too hard to check that there are $2^{(n^2)}$ bilinear maps $\mathbb{F}_2^n\times \mathbb{F}_2^n\rightarrow \{\pm 1\}$, but I haven't found a way of counting the non-degenerate ones!

also $k^\times=k\setminus{0}$. The image of a linear map would contain $0$. — Chris Sanders, Mar 16 '22 at 06:55
Sorry about the confusion! As might have guessed, this question arose in the more general comntext of bilinear maps $A\times A\rightarrow k^{\times}$ for any (finite) abelian group $A$. — JeCl, Mar 16 '22 at 12:28

Jyrki Lahtonen · Accepted Answer · 2022-03-22T05:10:07.570

So with $\{e_1,\ldots,e_n\}$ a basis of $\Bbb{F}_2^n$ we have $$ b(\sum_i x_ie_i,\sum_jy_je_j)=(-1)^{\sum_{i,j}x_ia_{ij}y_j}, $$ where the matrix $A=(a_{ij})\in M_{n\times n}(\Bbb{F}_2)$ is determined by $$b(e_i,e_j)=(-1)^{a_{ij}}.$$

The exponent of $-1$ in general is thus $(x_1,\ldots,x_n)A(y_1,\ldots,y_n)^T$. It follows that $b$ is non-degenerate if and only if the matrix $A$ has full rank. That is, if and only if $A\in GL_n(\Bbb{F}_2)$.

It is well known that $$\#GL_n(\Bbb{F}_2)=(2^n-1)(2^n-2)(2^n-2^2)\cdots(2^n-2^{n-1}).$$

Counting non-degenerate bicharacters

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