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.

 

[blue_raver22]

I am not an expert in this particular field, but I can provide some general information on this topic. Bilinear forms and quadratic forms are important tools in mathematics, particularly in linear algebra and geometry. They are used to study quadratic equations and their solutions, and can provide insights into the geometric properties of these equations.

In this case, the content is discussing the relationship between a quadratic form Q(x,y) and a bilinear form B(x,y) over a field of characteristic different from 2. The formula (0.5)[Q(x+y)-Q(x)-Q(y)]=B(x,y) is a way to find the bilinear form associated with a given quadratic form. This means that the two forms are related and can provide similar information about the underlying quadratic equation.

However, when working over a field of characteristic 2 (such as Z/2), there are some differences in the properties of these forms. For example, the Arf-Invariant and Artin's books on Geometric Algebra discuss how these forms behave differently in this context. This is an important area of study and there is a lot of theory and research dedicated to understanding these differences.

To transform a specific quadratic form over Z/2 into its associated bilinear form, you can use the formula mentioned above. This will give you the bilinear form B(x,y) that is associated with the quadratic form Q(x,y). This can be a useful tool in studying the properties of the quadratic equation and its solutions.

In summary, bilinear forms and quadratic forms are important tools in mathematics and have many applications. While there is a whole theory dedicated to studying these forms over fields of characteristic 2, the formula provided can help in quickly finding the associated bilinear form for a given quadratic form over Z/2.