I know this:

There are two non-degenerate quadratic forms on $GF(2)^2r$. The hyperbolic form may be taken to be $Q^+(x)=x_0 x_1 + \cdots +x_{2r-2}x_{2r-1}$ ,

and the elliptic form to be

$Q^-(x)=x^2_0 +x_0x_1 +x^2_1 +x_2x_3 + \cdots +x_{2r-2}x_{2r-1}$.

and my question is:

I want to know more about this, do you know any book or article about this?


1 Answer 1


The short answer: Robert Wilson's nice book "The finite simple groups" has a huge amount of information about quadratic spaces (and other linear algebra structures) over finite fields of all characteristics. (This has nothing to do with "Einstein on the Beach" by a different Robert Wilson.)

The long answer: over any finite field $k$ of any characteristic there are exactly two isomorphism classes of non-degenerate quadratic forms of even rank $n \ge 2$, generalizing what you have listed for even rank $2r$ in characteristic 2, and to see this conceptually it seems best to use the formalism of algebraic groups.

The "split" form $Q_n^+$ is defined to be $$x_1 x_2 + \dots + x_{n-1}x_n,$$ and to define another one we first define $Q^{-}_2$ to be the norm form for the quadratic extension $k'$ of $k$ (given by homogenizing an irreducible quadratic polynomial over $k$). Note that the equation $Q^{-}_2(x,y) = 1$ has $k$-rational solution set of size $(q^2-1)/(q-1) = q+1$ for $q := |k|$, whereas $Q^{+}_2=1$ has solution set of size $q-1$, so indeed $Q^{-}_2 \not\simeq Q^{+}_2$. We define $Q^{-}_n(x_1,\dots,x_n) = Q^{-}_2(x_1,x_2) + Q^{+}_{n-2}(x_3,\dots,x_n)$ for even $n \ge 4$. Witt cancellation is valid in even rank without restriction on the characteristic, so it shows that $Q^{-}_n \not\simeq Q^{+}_n$ for all even $n \ge 2$.

For even $n \ge 2$, to show that there are no more isomorphism classes than these we first recall that the set of isomorphism classes of rank-$n$ non-degenerate quadratic spaces $(V,Q)$ over a field $K$ are in bijection with the pointed Galois cohomology set ${\rm{H}}^1(K,{\rm{O}}_n)$. Thus, the exact sequence (for any even $n$) $$1 \rightarrow {\rm{SO}}_n \rightarrow {\rm{O}}_n \rightarrow \mathbf{Z}/2\mathbf{Z} \rightarrow 1$$ of $K$-groups defines a map of pointed sets ${\rm{H}}^1(K,{\rm{O}}_n) \rightarrow {\rm{H}}^1(K,\mathbf{Z}/2\mathbf{Z})$. Since $\mathbf{Z}/2\mathbf{Z}$ is commutative, by the "twisting method" we see that the fiber through the class of $(V,Q)$ is identified with the image of ${\rm{H}}^1(K,{\rm{SO}}(Q))$ in ${\rm{H}}^1(K,{\rm{O}}(Q))$. For finite $K$ (and still assuming $n$ is even), the pointed set ${\rm{H}}^1(K,{\rm{SO}}(Q))$ vanishes, due to Lang's theorem (since ${\rm{SO}}(Q)$ is smooth and connected). Hence, for finite $k$ and even $n$ the map of sets
$${\rm{H}}^1(k,{\rm{O}}_n) \rightarrow {\rm{H}}^1(k,\mathbf{Z}/2\mathbf{Z})$$ is injective. The target has size 2, so we're done for even $n$.

The case of odd $n$ is easier to understand since over any field $K$ we have ${\rm{O}}_n = {\rm{SO}}_n \times \mu_2$ as $K$-group schemes (where projection to $\mu_2$ is the determinant) with ${\rm{SO}}_n$ smooth and connected. Lang's theorem implies that for finite $k$ the natural map $${\rm{H}}^1(k,{\rm{O}}_n) \rightarrow {\rm{H}}^1(k,\mu_2) = k^{\times}/(k^{\times})^2$$ is bijective. Thus, again the size is 2 when the characteristic is odd, and in addition to the "split" form $Q^+_n = x_0^2 + Q^+_{n-1}$ another non-degenerate one of rank $n$ is $Q^{-}_n = x_0^2 + Q^{-}_{n-1}$. (We have $Q^{-}_n \not\simeq Q^{+}_n$ due to Witt's cancellation theorem once again, with the caveat that in odd rank Witt cancellation is only valid away from characteristic 2.)

The cohomology set is trivial when $k$ has characteristic 2 (so $Q^+_n := x_0^2 + Q^+_{n-1}$ is the only example up to isomorphism of odd rank $n$ over finite fields of characteristic 2; if we make the definition $Q^{-}_n := x_0^2 + Q^{-}_{n-1}$ then it happens to be isomorphic to $Q^{+}_n$ over $k$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.