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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

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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).

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  • $\begingroup$ 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. $\endgroup$ – JNS Mar 25 at 9:26
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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.

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