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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.

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I put pages 100-105 of Milnor and Husemoller at zakuski.utsa.edu/~jagy/milnor_homology.pdf –  Will Jagy Jul 29 '11 at 0:12
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3 Answers

up vote 4 down vote accepted

There are some quick answers, but what you actually want is chapter V, section 1 in Symmetric Bilinear Forms by Milnor and Husemoller, pages 100-105, the section named "Homology Theory of Manifolds." They even include an appendix 1 called Quadratic Forms, pages 110-113.

I put pages 100-105 at this LINK

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Thanks for the ref., and for being pretty specific. Still, I don't have access to that book, and I won't for a while; any chance you can give me a 'quick-and-dirty' until I get the book? –  Larry Jul 27 '11 at 18:33
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A general notion of quadratic map between abelian groups is defined in MacLane's 1950 ICM Proceedings article. Namely, a set-theoretic map $q: A \to B$ is called quadratic if the following conditions are satisfied:

  1. $q(a+b+c) - q(b+c) - q(c+a) - q(a+b) + q(a) + q(b) + q(c) = 0$ for all $a,b,c \in A$.
  2. $q(a) = q(-a)$ for all $a \in A$.

The algebraic version you wrote comes from the special case where $B$ is a commutative ring, and $A$ is a free $B$-module of rank 2, whose dual module is given the basis $\{ x, y \}$. The polynomial then takes elements of $A$ and applies the functions $x$ and $y$ to produce elements of $B$ that are multiplied and added. This quadratic form is known as binary, because it uses two variables. In addition to the case of binary quadratic forms, there are quadratic forms that take arbitrarily many variables.

The topological version arises from the fact that if $A$ and $B$ are vector spaces over $\mathbb{F}_2$, the criterion for a map to be quadratic simplifies to the condition that the map $A \times A \to B$ defined by $(x,y) \mapsto q(x+y) - q(x) - q(y)$ be bilinear. There is a brief description of this alternative condition (with a spurious factor of $\frac12$) in Wikipedia.

I don't think there is a natural quadratic form on $H_1(S,\mathbb{Z})$. The intersection form in degree 1 is naturally symplectic, and it only becomes quadratic after reduction mod 2. You certainly have the option of making a quadratic polynomial in $2g$ variables over $\mathbb{Z}/2\mathbb{Z}$, though. With an appropriate choice of basis, I think it is $a_1 b_1 + \cdots + a_g b_g$.

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Thanks, all your answers were helpful, but Will's ref. combining algebra, topology and geometry was particularly helpful. –  Larry Aug 10 '11 at 21:27
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Look at Rourke and Sullivan's paper "On the Kervaire Obstruction". Every quadratic form leads to a bilinear form but when $1+1=0$ you can't go the other way around, and the classification of mod-2 quadratic forms mirrors properties of immersions of surfaces.

Given a surface $S_g$ with one boundary component, and $V=H_1(S_g,\mathbb Z/2\mathbb Z)$, a quadratic form which by definition satisfies $$q_T(x+y)=q_T(x)+q_T(y)+(x,y)_2$$ is completely determined up to isomorphism by its Arf invariant, $\varepsilon(q_T)$. Since you know the evaluations at the symplectic basis this is just $$\varepsilon(q_T)=\sum q_T(x_i)q_T(y_i)$$

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I think you may have meant that "every bilinear form leads to a quadratic form but when $1+1 = 0$ you cannot go the other way around." –  Akhil Mathew Aug 28 '12 at 14:44
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