Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M$ be an $n$-dimensional compact and oriented manifold. Then one can define the intersection pairing $H_k(M,\mathbb Z) \times H_{n-k}(M,\mathbb Z) \to \mathbb Z$. One possible formulation of the Poncaré duality is the folowing:

Every linear functional $H_{n-k}(M,\mathbb Z) \to \mathbb Z$ is given by intersection with some class $\alpha \in H_k(M,\mathbb Z)$ and if $\beta \in H_k(M,\mathbb Z)$ has interesection number $0$ with every class in $H_{n-k}(M,\mathbb Z) $ then $\beta$ is a torsion element.

This formulation is given on Griffiths & Harris "Principles of Algebraic Geometry" and they use this to define the fundamental class of a closed sumbanifold as follows: if $V \subset M$ is a closed and oriented submanifold of dimendion $k$, intersection with $V$ defines a linear funcional $H_{n-k}(M,\mathbb Z) \to \mathbb Z$ and they say

"the corresponding cohomology class $\eta_V \in H^{n-k}(M)$ is the fundamental class of $V$."

What does they mean by "the corresponding cohomology class"? I can see that $\text{Hom} (H_{n-k}(M,\mathbb Z), \mathbb Z)$ is related to $H^{n-k}(M,\mathbb Z)$ by the universal coefficient theorem, but how is $\eta_V$ obtained explicitly?

share|improve this question
add comment

2 Answers

I thought that definition was already pretty explicit.

You say that one can define a linear functional $H_{n-k}(M,\mathbb{Z})\rightarrow\mathbb{Z}$ by declaring "for $\alpha\in H_{n-k}(M,\mathbb{Z})$, send it to $\alpha\cdot [V]$ where $[V]$ is the homology class of $V$" (where presumably $\alpha\cdot[V]$ has already been defined)

Then the universal coefficient theorem says that $\hom(H_{n-k}(M;\mathbb{Z}),\mathbb{Z})$ is isomorphic to $H^{n-k}(M;\mathbb{Z})$ if Ext$(H_{n-k-1}(M;\mathbb{Z}),\mathbb{Z})=0$. We don't know that it is, but regardless there is a surjection from cohomology to the hom group, so every linear functional comes from a cohomology class. So you can explicitly define $\eta_V\in H^{n-k}(M,\mathbb{Z})$ by

$\eta_V(\alpha)=\alpha\cdot[V]$

(One issue I'm having with this definition is that I'm used to the intersection number being defined using Poicare duality. What is the definition you/they are using for $\alpha\cdot[V]$? Is this actually what you are asking about?)

share|improve this answer
1  
Ok, you are right, i got confused because I thought we needed $H_{n-k}(M,\mathbb Z)$ to be free to get Tor$(H_{n-k}(M;\mathbb{Z}),\mathbb{Z})=0$ but we only need $\mathbb Z$ to be free. The definiton of $\eta_V$ is straightforward then. –  Lucas Kaufmann Mar 3 '12 at 18:23
1  
The definition of the intersection pairing i'm using is the one choosing smooth representatives of each homology class and then counting the number of intersection points taking the orientation into account –  Lucas Kaufmann Mar 3 '12 at 18:26
    
Wait, I got confused again =P. To get $H^{n-k}(X,\mathbb Z) \simeq \text{Hom}(H_{n-k}(M;\mathbb{Z}),\mathbb{Z})$ we need $\text{Ext}(H_{n-k-1}(M;\mathbb{Z}),\mathbb{Z})=0$ right? And we don't have this in general. –  Lucas Kaufmann Mar 3 '12 at 18:37
1  
Oh, whoops.... I think it's still ok, because in the short exact sequence you will still get a surjection from $H^{n-k}$ to hom, so every linear functional "is" a cohomology class –  you Mar 3 '12 at 18:56
    
Do you know a way to describe this cohomology class explicitly? We need to check that this lift is well defined, because we could choose two different classes that map to the same functional. –  Lucas Kaufmann Mar 3 '12 at 20:17
add comment

If $V$ is the closed submanifold of dimension $k$, I would think of $\eta_V$ as the unique closed $n-k$-form with the property that, for all $\gamma \in H_{n-k}(X)$, one has $$ \int_{\gamma} \eta_v = \gamma\cdot V, $$ where $\cdot$ is the intersection product and $\gamma$ is an arbitrary homology class of dimension $n - k$. The integral is well-defined here by Stokes.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.