Hi, All:

I am trying to see if there is a nice relation between two different definitions of quadratic form q; a topological definition $q_T$, and an algebraic definition $q_A$, and, if there is, how to go from one version to the other; either from version #1 below to version#2, or #2 to #1, or, even better, both ways.The two definitiions. Version #1 (defined below) is the topological version, and #2 (also defined below) is the algebraic version. My main interest here, actually, is to see if we can express the topological format as a homogeneous polynomial of degree 2 in {x,y} This may be embarrasingly-simple for someone with a stronger Abstract-Algebra background than mine:

1)In algebra, a quadratic form $q_A$ is a homogeneous polynomial P(x,y) of degree 2 in {x,y}; e.g., P(x,y)=$x^2+y^2$

2)In Topology/Homology, a quadratic form $q_T$ is defined as a map $q_T:Z\rightarrow Z_2$, where Z is a finitely-dimensional $Z_2$- vector space, associated with an intersection form ; more specifically, in my case of interest, Z is $H_1(Sg,\mathbb Z_2)$,where Sg is the genus-g surface, with the finite generating set (symplectic basis) {$x_1,y_1;x_2,y_2;....;x_2g,y_2g$} , with the intersection pairing $(x,y)_2(mod2)$, defined by: $(x_i,y_j)_2=1$ , if i=j, and 0 otherwise. Then $q_T$ is defined to be a quadratic form , if it satisfies the identity:

```
$q_T(x+y)=q_T(x)+q_T(y)+(x,y)_2$ , for all x,y in Z (##)
```

In addition, we do know the values of $q_T(x_i$) and $q_T(y_i)$ , i.e., we know the value of the quadratic form in the basis elements.

So basically, I would like to be able to express my Topological $q_T$ as a homogeneous, quadratic polynomial. The fact that (##) above looks a lot like the bilinear form associated with a(n) algebraic quadratic form makes me think that this may be possible, but the fact that we are working over $Z_2$ makes me worry that, if possible, it may be difficult.