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?



two non-degenerate quadratic forms on $GF(2)^2r$
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 $k$).


