Bilinear Forms associated With a Quadratic Form over Z/2

[Bacle]

Hi, All:

Given a quadratic form Q(x,y) over a field of characteristic different from 2, we can

find the bilinear form B(x,y) associated with Q by using the formula:

(0.5)[Q(x+y)-Q(x)-Q(y)]=B(x,y).

I know there is a whole theory about what happens when we work over fields of
characteristic 2, with the Arf -Invariant , Artin's and other's books on Geometric
Algebra and everything, which I am looking into.

Still, I wonder if someone knows the quick-and-dirty on how to transform an
actual, specific quadratic q form over Z/2 into its associated bilinear form.

Thanks.

[Hurkyl]

The reason for all the theory is that there isn't a symmetric bilinear form satisfying Q(x) = B(x,x), except for special Q's. Observe that B(x,x) is a linear function of x....

[Bacle]

Thanks, Hurkyl:

Would you give me some idea on the conditions under which the
associated B(x,y) exists?

[Bacle]

I'm thinking specifically of the case in which the bilinear form is (x,y)_2 ; the intersection form in H_1(Sg,Z/2) ; all defined on a symplectic basis for Sg ---Sg is the orientable, genus -g surface, and a symplectic basis {x1,y1,x2,y2,...,x2g,y2g} for Sg is one in which (xi,yi)_2=1 and (xi,yj)=0 if i=/j .

We then say that q(x) is a quadratic form associated with the given bilinear form, if :

q(x+y)-q(x)-q(y)=(x,y)_2

And then we seem to classify these forms by their arf invariant; there seem to be 8 forms with Arf invariant 1 and 8 with Arf invariant 0 (the Arf invariant when working over Z/2 is an element of Z/2); given a choice of symplectic basis as above, the Arf invariant
is defined as : (q(x1)q(y1)+q(x2)q(y2) ).

Still, I don't know what the issue is with the forms with Arf invariant 1 . I know that the Arf invariant classifies the quadratic forms mod2, i.e., two forms defined over F_2 are equivalent iff they have the same Arf invariant ; just like we
classify quadratic forms over fields of characteristic different from 2 by their resolvent, i.e., all quadratic forms over fields of characteristic different from 2 can be diagonalized ( I think by symmetry) , and the sum of the square of their diagonals is an invariant , i.e., if forms Q,Q' are equivalent, then they will have the same resolvent.