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I'm trying to understand the geometric meaning of (symmetric) bilinear forms.

I'm reading parts of "Symmetric Bilinear Forms", in particular, the appendix mentions what I'm interested in: on page 100 they write

"Let $M = M^{2n}$ be a closed manifold of dimension $2n$, and let $F_2$ be the field with two elements. If $x,y$ are homology classes in $H_n(M, F_2)$, the intersection number $$ x \cdot y = y \cdot x \in F_2$$ is defined. The Poincaré duality theorem, see e.g. [Spanier], implies that $H_n(M,F_2)$ is an inner product space over $F_2$ using the intersection number as inner product."

  1. What exactly is the intersection number? For example: I don't know how to think about $n>1$ but I think if $n=1$ we think of the elements in $H_1$ as equivalence classes of paths, namely, two paths are equivalent if they differ by a boundary which means that $p_1 - p_2 = \partial U$ where $U \subset M$ is a submanifold (is this the correct term?) of $M$. If $M$ is for example the torus, we see that any two cycles around it form the boundary of a cylinder, hence are equivalent. Similarly, any two cycles around the centre hole form the boundary of an annulus, hence are equivalent. Hence the first homology group is generated by two elements and hence we get $H_1(T) = \mathbb Z \oplus \mathbb Z$. Now assume we pick two arbitrary representatives $x,y$ of each equivalence class. Now I don't know the actual definition of intersection number but assuming it means number of points of intersection, is it correct that $x \cdot y = 1$ in the example of the torus? And what about a sphere? Then it should be $x \cdot y = 0$ because we don't have any holes. Can you please give me a rigorous definition of intersection number?

  2. Where does the Poincaré duality come in here? As far as I know it tells us that $H^k (M^n) = H_{n-k}(M^n)$. I dont's see where we need this to compute intersection numbers.

Thank you for your help.

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2 Answers

up vote 8 down vote accepted

The role of Poincare duality is that it provides one way to define the intersection pairing:

As countinghaus notes, for an $n$-dimensional connected closed manifold, the intersection pairing (with mod $2$ coefficients) is a pairing $H_{n-i} \times H_{n-j} \to H_{n - i - j}.$ One way to define it is geometrically, as in countinghaus's answer, by choosing well-behaved representatives for cycles, moving them within their respective homology classes so as to be transverse, and then intersecting them to obtain a new cycle. (Because we are working mod $2$, orientations don't matter.)

To prove that this process is well-defined, and to derive its basic properties (e.g. that it is bilinear) takes some effort, and so many authors prefer to put that effort elsewhere, e.g. into stating and proving Poincare duality. Once you have Poincare duality, you can do the following:

Cup product gives a map $H^i \times H^j \to H^{i+j}$. Poincare duality (with mod $2$ coefficients) identifies $H^i$ with $H_{n-i}$. Thus we can rewrite this as $H_i \times H_{n-i} \to H_{n - i - j}.$ It turns out that this is the intersection pairing described geometrically above. (To see why, look at this answer --- it treats the case of $\mathbb Q$-coefficients rather than $\mathbb F_2$-coefficients, but the idea is the same.)

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Dear Matt. Thank you for this clear answer! –  Matt N. Jun 15 '12 at 20:56
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  1. Your definition is correct with a small caveat. In general, if $M$ is a smooth manifold, we get a pairing $H_{n-k}(M) \times H_{n-i}(M) \to H_{n-k-i}(M)$ in the following way. Given two oriented cycles $\gamma$ and $\gamma'$, we can move them by a homotopy until they intersect transversely. We then take the oriented cycle $\gamma \cap \gamma'$ in the intersection to be their intersection product. In general orientation introduces a sign into the pairing which complicates things when you intersect odd-dimensional cycles. If $k + i = n$, our intersection is just a bunch of points, which we simply count up with the appropriate signs (since $H_0 = \mathbb{Z}$).

  2. The authors cite Poincare duality to show that this is an "inner product" (positive definite bilinear pairing) and not just an arbitrary pairing. I like to think of this as the statement of Poincare duality: you give me a cycle, I can find an ($\mathbb{R}$-linear combination of) cycle(s) of appropriate codimension to get intersection number 1 with it.

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Dear countinghaus, Note that the OP considers homology with mod $2$ coefficients, not $\mathbb Z$ coefficients, so the issue with signs and orientations go away. (Another small remark: you may want to mention that $M$ should be closed and orientable in order to define the intersection pairing with $\mathbb Z$-coefficients.) Regards, –  Matt E Jun 15 '12 at 20:20
    
Dear @countinghaus Could you point me to a source (not necessarily online) where this pairing is explained in more detail? –  Matt N. Jun 15 '12 at 20:26
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