# Generators of the orthogonal group of a quadratic form in odd dimension in characteristic 2

In characteristic not $$2$$, the Theorem of Cartan-Dieudonné states:

1. [Grove, Theorem 6.6]: Let $$q$$ be a nondegenerate symmetric quadratic form of dimension $$n$$ in characteristic not $$2$$. Then every element in the orthogonal group $$O(q)$$ can be written as a product of at most $$n$$ reflections.

In a nutshell: I am looking for a similar statement but over a field in characteristic $$2$$ and in particular in odd dimension. At the end I will make the statement that I would like to have more precise.

As I am new to quadratic forms, I got confused by different notions of non-degeneracy, so I want to fix the terminology; I take the one from [Grove]. To a quadratic form $$q$$ on $$V$$ one can associate a bilinear form $$b$$ on $$V$$ via $$b(v,w)=q(v+w)-q(v)-q(w)$$. In characteristic $$2$$, several quadratic forms can give the same bilinear form (e.g. $$y^2+xz$$ and $$xz$$). Let $$\mathrm{rad}_b(V)=\{w\in V\mid b(v,w)=0 ~\forall v\in V\}$$. If $$\mathrm{rad}_b(V)\neq\{0\}$$, then $$q$$ is called

• degenerate if the characteristic is not $$2$$,
• defective if the characteristic is $$2$$.

In characteristic $$2$$ and dimension odd, all quadratic forms are defective! In characteristic $$2$$, we say that $$q$$ is regular if $$q(v)\neq0$$ for all non-zero $$v\in\mathrm{rad}(V)$$. (In [Connors], regular is called nondegenerate).

Example: The two quadratic forms $$q_1=y^2+xz$$ and $$q_2=xz$$ on $$V=k^3$$, where $$k$$ is a field of characteristic $$2$$, give the same bilinear form and they both have $$\mathrm{rad}(V)=\langle(0,1,0)\rangle$$. So they are defective. While $$q_1$$ is regular, $$q_2$$ is not.

From now on, assume that $$k$$ is a field in characteristic $$2$$. Here transvections play the role of reflections. I have found the following statements that go in the right direction:

1. [Grove, Theorem 14.16]: Assume that $$q$$ is regular and non-defective (see the assumption on p. 129 which I believe is still taken in Theorem 14.16). Then $$O(q)$$ is generated by transvections, except for one example over the field $$\mathbb{F}_2$$ where $$V$$ is $$4$$-dimensional.

The above theorem can not be applied in odd dimension. The next statement is exactly what I want, but it is stated only for dimension $$3$$:

1. [Connors, Proposition 3.2]: Let $$q$$ be a defective, anisotropic regular quadratic form in dimension $$3$$. Then every element of $$O(q)$$ is the product of at most $$2$$ transvections.

I am looking for a reference of the following statement (if true): Let $$V$$ be a vector space over any field in characteristic $$2$$, and let $$q$$ be a regular, anisotropic quadratic form on $$V$$ of odd dimension $$n$$ (i.e. $$q$$ defective). Then every element in the orthogonal group $$O(q)$$ can be written as a product of (at most $$n$$) transvections.

[Grove] : Larry C. Grove, "Classical groups and geometric algebra", 2002

[Connors] : Edward A. Connors, ''The Structure of $$O'(V)/DO(V)$$ in the Defective Case'', 1973

I suggest reading the result 6.2.17 in the book written by A.J.Hahn and O.T.O'Meara called "The Classical Groups and K-Theory". The result is a general version of the Cartan-Dieudonné-Scherk theorem (which implies the Cartan-Diuedonné theorem). The result is stated in the language of "Quadratic spaces over form rings". The case of the defective orthogonal groups is a particular case of the general "unitary groups" $$U_n(V)$$. More precisely, in the book you find

6.2.17 Theorem. Suppose $$U_n(V)$$ is not a symplectic group. If $$J=\mathrm{id}_R$$, assume that $$R$$ is not $$F_2$$ and if $$J\neq \mathrm{id}_R$$, assume $$R$$ is not $$F_4$$. Let $$\sigma$$ in $$U_n(V)$$ be non-trivial and let $$S$$ be its residual space.

(i) If $$\sigma$$ is not totally isotropic, $$\sigma$$ is a product of $$\mathrm{dim} S$$ symmetries, but no fewer.

(ii) If $$\sigma$$ is totally isotropic, $$\sigma$$ is a product of $$\mathrm{dim} S+2$$ symmetries, but no fewer.

When $$J=\mathrm{id}_R$$ you can get the defective orthogonal groups and so any $$\sigma$$ will be a product of "symmetries" (i.e. transvections in the case of the defective orthogonal group).

• Thanks for stating the result! I accepted this answer, but I'd still be interested in a reference that uses the language of quadratic form over vector space. – JNS Mar 25 at 9:26

For a reference, you could use Theorem I.5.1 from C. Chevalley, The Algebraic Theory of Spinors, Columbia University Press, New York, 1954. (This also appears in volume 2 of his collected works.) The result is the following (rephrased):

The orthogonal group $$O(V,q)$$ is generated by reflections unless the underlying field has only 2 elements, $$\dim V = 4$$ and $$q$$ is hyperbolic.

In fact, if you only care about anisotropic quadratic forms, then the proof of Cartan-Dieudonné, in any characteristic, is much easier than the general case, and very elegant. (I learned this proof from Richard Weiss but I can't remember the orginal source, unfortunately).

Let $$f$$ be the bilinear form corresponding to $$q$$. For any $$c \in V$$ with $$c \neq 0$$, let $$\pi_c$$ be the corresponding reflection $$\pi_c(v) := v - \frac{f(v,c)}{q(c)} c .$$ Now let $$\varphi \in O(V,q)$$ be arbitrary, and let $$F = \operatorname{Fix}_V(\varphi)$$. We will show by induction on the codimension of $$F$$ that $$\varphi$$ is generated by reflections. If $$\operatorname{codim} F = 0$$, then $$F = V$$ so $$\varphi = 1$$. Now assume that $$\operatorname{codim} F > 0$$ and let $$v$$ be an element of $$V \setminus F$$, so $$w := \varphi(v) - v \neq 0$$. We claim that $$\pi_w$$ fixes $$F$$ and maps $$v$$ to $$\varphi(v)$$.

First, notice that by applying $$q$$ on $$\varphi(v) = v + w$$, we get $$q(w) = - f(v,w)$$, so indeed $$\pi_w(v) = v + w = \varphi(v)$$. Next, if $$u \in F$$, then $$f(u,w) = f(u,v) - f(u,\varphi(v)) = f(u,v) - f(\varphi(u),\varphi(v)) = 0,$$ so $$\pi_w$$ fixes $$u$$. This proves our claim. We conclude that $$\varphi^{-1} \circ \pi_w$$ fixes the space $$\langle F, u \rangle$$, which has codimension one less than $$F$$, so we can apply induction, and we are done.