• muse on euler class

Yesterday I read about BFN's construction. One thing that strikes me is the appearance of 'euler class'.

I have never really understood euler class. the input is some bundle over some space, and the output is some cohomology class on the space itself. The only natural thing that I can think of, is the intersection of the zero section with its own perturbation. However, thinking in terms of perturbation is not very intrinsic. this thing should be defined topologically (like what if your underlying space is algebraic, very rigid, no room for deformation? )

then, what do you mean, when you say euler class of a bundle over a point? I would say, the product of weights of that bundle. a weight, is a $\C^*$-valued function on the torus, and it is also a $\C$-valued function on the Lie algebra of the torus. Infinitesimal weights are elements in $t^*$, so products of them lives in $Sym t^*$. That makes sense.

Also, Euler class make sense. The Euler class of a bundle is in degree of the (real rank) of the bundle.

Finally, this is not just ad-hoc definition, it is a calculation using the Borel model. For example, given a standard representation of $G=S^1$, call it $L$, we get a universal bundle $L$ over $BG$. This bundle restricts to $\P^1$ (already visible there) should be given by $u$.

Let's fix the normalization once and for all, and do the BWB computation once in my life. Consider $\C^*$ acting on $\C$. Then, consider the principal $\C^*$ bundle on $\P^1$, so we should get the total space to be $O(-1)$. oops. Maybe, I should by convention, insert a -1 somewhere. OH well.

Well, now you have this vector bundle, being the direct sum of these line bundles, what do you want to say to them? indeed, Euler class turns direct sum of a bundle to (cup) product of cohomology class, so it is fine.

And, I realized that, there is no difference between homology an cohomology, if your base is just 0-dimensional.

still, I would like to figure out, what is the difference between cohomology and homology. Can we say (BM) homology is just degree shift of cohomology?

• Say, BMH on $\R^n$, is only supported in top degree $n$, $BMH_n$ which we can turn into the $H^0$.
• the usual homology on $\R^n$ is supported on a point (which can move around, but can never escape to infinity), this should corresponds to compactly supported cohomology, a bump form.
• in the above statement, I don't use any duality, I just say by intution, if we want to match the support, then that is that. And the degree matching is using 'codimension'.
• I only know how to do duality between homology and cohomology, using integration, and that does not agree with the above matching, so the above matching is not through integration. Well, you can say it is still through integration.

Well, given a smooth manifold, compact or not, oriented or not, we can consider two kinds of sheaves, constant sheaf, and dualizing sheaf (orientation sheaf with degree shift), and we can consider two kinds to cohomology, sheaf cohomology and compactly suppported sheaf cohomology $p_*, p_!$, where $p: X \to pt$. The two are related, we just have $\C = p^* \C$, and $\omega = p^! \C$.

The usual Poincare duality statement is that, for an orinted compact manifold, so $p_* = p_!$ and $p^! \C = p^* \C [d]$, we have $$ H_k(X) \cong H^{n-k}(X) $$

Wikipedia says, Poincare duality is 'cap product'. Assuming that $X$ is compact orientable, then we top degree homology class $[X]$, then $[X] \cap - $ is the map that takes $H^k(X) \to H_{n-k}(X)$.

I don't know what is cap product in the sheaf world, maybe, in the de Rham model, we can do polyvector field as a local model for homology, and differential form as a local model for cohomology?

So, can we define a de Rham model of polyvector field? But, this might be too far a field. In any case, we should be able to use the pairing to identify $(PV)_k = (\Omega^k)^\vee$, like an honest definition. So, what is the statement?

$$ H^*(X, (\Omega^*)^\vee) \otimes H^*(X, \Omega^*). $$

oh well, it is not smooth polyvector field, is it more severe, called $k$ current. that's the local version of integration chain.

first of all, why we choose to use homology instead of cohomology? is it convention, or is it because the space is singular or non-compact?

when we do convolution product, suppose we have a group, a compact Lie group. let me try to formulate it in both ways.

If we use sheaves on compact group, what do I do? let me try to use homology, just bare homology. So, I have a cycle, and another cycle. I take box product, then I pushforward. but pushing forward is intersect with fiber, no that would be restrict to the fiber. Then, what does pushforward mean? Can I pushforward cohomology class?

No, it is not that complicated. Let's consider the matrix model first. Suppose we have $M_1, M_2, M_3$, and correspondences, $Z_{12}, Z_{23}$. assuming properness all around. Say $Z_{12}$ is of codimension $a = n_1+n_2 - d_{12}$, and $Z_{23}$ of codimension $b = n_2+n_3 - d_{23}$. We pullback these classes, take their cup product, so get degree $a+b$ inside $n_1+n_2+n_3$ space, which is of dimension $d_{12}+d_{23}-n_2$, then we just pushforward.

Now, consider the group situation, I want to say, dimension $d_1$ and $d_2$ add up, then you do the product. so it is literally just pushforward under the product.

In our current situation, we have a group $\Z$, and over it we have the bundle of $N$ sections on the ravioli. pretend that it is finite dimensional. For simplicity, we can rigidify. and just consider basis.