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--- blog:2023-06-20 [2023/06/20 19:54] – pzhou
+++ blog:2023-06-20 [2023/06/25 15:53] (current) – external edit 127.0.0.1
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 ====== 2023-06-20 ======
+  * muse on euler class
+===== euler class =====
 Yesterday I read about BFN's construction. One thing that strikes me is the appearance of 'euler class'.
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   * 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.
+===== what should be true =====
+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.