# bilinear symmetric form

Let $$b$$ be a symmetric bilinear form on a $$k$$-dimensional vector space $$V_k(\mathbb{F}_q)$$, with $$char(\mathbb{F}_q)=2$$ and $$k$$ odd. Theorem 3.15 of the book S. Ball, Finite Geometry and Combinatorial applications states that $$V_k(\mathbb{F}_q)=E \oplus F,$$ where the restriction to $$E$$ of $$b$$ is an alternating form and $$F$$ is a non-isotropic one-dimensional subspace.

In Corollary 3.10, it says that a non-degenerate alternanting form $$b$$ on $$V_k(\mathbb{F}_q)$$ ($$k=2r$$ even) is, up to a choice of a basis, $$b(u,v)=\sum_{i=1}^r(u_{2i-1}v_i-u_{2i}v_{2i-1}).$$

Now, in Corollary 3.16, it states that a non-degenerate symmetric bilinear form $$b$$ on $$V_k(\mathbb{F}_q)$$ ($$k=2r+1$$) is, up to a choice of a basis, $$b(u,v)=\sum_{i=1}^r(u_{2i-1}v_i+u_{2i}v_{2i-1})+u_{2r+1}v_{2r+1}.$$ Why this last statement is true?

First, subtraction is the same as addition, since we work over a field of characteristic $$2$$.

Applying the theorem, we get $$V=E\oplus\langle f\rangle$$ with $$b(f,f)\ne 0$$ and $$b|_E$$ being alternating.

Then, by the previous corollary, we can choose an appropriate basis on $$E$$, and let $$w\notin E$$ be a nonzero element of $$E^\perp=\{x\mid\forall e\in E:\,b(x,e)=0\}$$, which should exist since $$b$$ is nondegenerate. (See below.)

Then we can write $$w=e+\lambda f$$ with some $$e\in E$$ and $$\lambda\ne 0$$, and so, using symmetricity of $$b$$ and characteristic $$2$$, $$b(w,w)\ =\ b(e,e)+2\lambda\, b(w,f)+\lambda^2\,b(f,f)\ =\ \lambda^2\,b(f,f)\ne 0$$ Since in $$\Bbb F_{2^m}$$ every element is square (because $$x=x^{2^m}$$), $$b(w,w)=\alpha^2$$ and we can choose the last basis vector as $$\alpha^{-1}w$$ to get the desired coordinate form.

Update: Finally, to prove that $$b|_E$$ is nondegenerate (which also implies the existence of such $$w$$), assume that $$E\subseteq e^\perp$$ for some nonzero $$e$$, and choose a complementary subspace $$E'\subseteq E$$ with $$\langle e\rangle\oplus E'=E$$
Then $$\dim E'=2r-1$$ is odd, so the alternating bilinear form $$b|_{E'}$$ must be degenerate, yielding another vector $$e'\in E'$$ perpendicular to whole $$E$$.

Because of nondegeneracy, $$b(e,f)\ne 0$$, and thus $$b(e',f)=\lambda\,b(e,f)$$ with some $$\lambda\in F$$, so $$e'-\lambda e$$ would be $$b$$-perpendicular to the whole space.

• Why you get the desidered coordinate form? $$b(u,v)=b(\sum_{i=1}^{r}u_ie_i+u_{2r+1}w,\sum_{i=1}^{r}v_ie_i+v_{2r+1}w)$$ $$=\sum_{i=1}^r(u_{2i-1}v_{2i}+v_{2i-1}u_{2i})+\sum_r b(u_ie_i,v_{2r+1}w)+\sum_r b(u_{2r+1}w,v_ie_i)+u_{2r+1}v_{2r+1}b(w,w)$$. Why $b(w,e_i)=0$? Sep 15, 2021 at 12:02
• Because we chose $w$ from $E^\perp$. Sep 15, 2021 at 13:21
• thank you. Why $\lambda$ is different from $0$? Sep 15, 2021 at 14:39
• But, we don't know if $E^{\perp} \subseteq E$ or not. Morover, in order to apply the result about the alternating form on $E$, first we must prove that $b$ is also non-degenerate on $E$. Sep 15, 2021 at 14:58
• Ok, I think I got it. Updated my answer. Sep 16, 2021 at 20:28