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--- blog:2023-06-20 [2023/06/20 20:02] – 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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 $$ H_k(X) \cong H^{n-k}(X) $$
+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.