Peng Zhou

stream of notes

User Tools

Site Tools


blog:2023-06-27

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Next revision
Previous revision
blog:2023-06-27 [2023/06/27 16:54] – created pzhoublog:2023-06-27 [2023/06/27 18:32] (current) pzhou
Line 4: Line 4:
 ===== BFN approach of matter ===== ===== BFN approach of matter =====
  
 +==== diversion on topology ====
 Let's tell the story just one more time. We have GL1(C)GL_1(\C) acting on C\C in the standard way. We have two stacks, one is C(K)/GL1(K)\C(K) / GL_1(K), one is C(O)/GL1(O)\C(O) / GL_1(O). It is a set with some group action.  Let's tell the story just one more time. We have GL1(C)GL_1(\C) acting on C\C in the standard way. We have two stacks, one is C(K)/GL1(K)\C(K) / GL_1(K), one is C(O)/GL1(O)\C(O) / GL_1(O). It is a set with some group action. 
  
Line 14: Line 15:
 one thing that I noticed is that, if we have a quotient map by group action XX/GX \to X/G, then usually the natural topology on X/GX/G (say it is discrete) and the one on XX are different. For example, consider BB acting on G/BG/B. So, the topology on GL1(K)GL_1(K) and that on $GL_1(K)/GL_1(O)$ is not compatible. That is a bit confusing.  one thing that I noticed is that, if we have a quotient map by group action XX/GX \to X/G, then usually the natural topology on X/GX/G (say it is discrete) and the one on XX are different. For example, consider BB acting on G/BG/B. So, the topology on GL1(K)GL_1(K) and that on $GL_1(K)/GL_1(O)$ is not compatible. That is a bit confusing. 
  
-What's the topology on GL1(K)GL_1(K) and KK? We have inclusion GL1(K)=K\{0}GL_1(K) = K \RM \{0\}. KK is an infinite dimensional vector space+What's the topology on GL1(K)GL_1(K) and KK? We have inclusion GL1(K)=K\{0}GL_1(K) = K \RM \{0\}We can ask, what is the topology of KK. K=C((t))K = \C( (t) ). The abstract metric completion of a metrized space equal to the induced closure after isometrically embedded into a complete metric space. We define the tt-adic valuation on KK, which is not the usual one induced from polynomial.  
 + 
 +The subset c0+tC[[t]]c_0 + t \C[ [t] ] is itself both open and closed,  since it is is the open and closed ball of radius r0r_0 centered at c0c_0, where 1>r0>e11 > r_0 > e^{-1}.  
 + 
 +So, the space KK is disconnected. Or it is totally disconnected. wiki says, all non-archimedian division rings are toally disconnected. fine.  
 + 
 +So, why does affine Grassmannian have interesting topology? What topology are we talking about even? The analytic topology over C\C? Or some other ones? The naive picture that LGLG is a L+GL^+G torsor over GrGGr_G seems to be correct.  
 + 
 +If GG is GL1GL_1, then there is no contradiction. But if G=GL2G=GL_2, then GrGGr_G is not a totally disconnected space, take a connected component, take two points, connect them by a (real segment), lift it up (using some local trivialization of LGGrGLG \to Gr_G),  then it means that GL2(K)GL_2(K) is not totally disconnected. Even GL2(O)GL_2(O) may not be totally disconnected. The case for GL1(O)GL_1(O) is a coincidence.  
 + 
 +What is the topology on GL2(O)GL_2(O)? We have the matrix M2(O)M_2(O). It is a nice enough vector space.  
 + 
 +OK, the t-adic metric on F((t))F( (t) ) is viewing FF as a totally disconnected space, like Fq\F_q. Even treating C\C as a disconnected space. And that is too fine.  
 + 
 +==== OK then, how about the weird convolution euler index ==== 
 +slow down. let's work over C\C, as real two dimensional space.  
 + 
 +Consider a rank rr vector space over C\C, with C\C^*-diagonal action. First, consider the non-equivariant BM homology of the space. by definition, this is the relative homology of the one point compactification, relative to the unique point at infinity. So here, it is $H_*(S^{2r}, pt)$, which is generated by the fundamental class.  
 + 
 +Now, we turn on C\C^* or S1S^1 action. There are many invariant subspaces. We also have S1S^1 action on S2rS^{2r}, coming from the suspension of S1S^1 acting on S2r1S^{2r-1}.  
 + 
 +What can we say then? If we want to compute the equivariant cohomology (which is equal to equivariant homology up to shift) of C\C acted by S1S^1, then we pass to the Borel model, we form EG×GCEG \times_G \C, G=S1G=S^1, and we get a line bundle over BG=CPBG=\C \P^\infty. The top dimensional 'homology', or 0-degree (degree = codim) cohomology is defined. That is [CG][\C_G], the fundamental (homology) class of the universal associated bundle.  Then, we can ask, what is the homology of [0G][0_G], the universal zero-section's fundamental class? By Thom isomorphism, we know **it is equivalent to something in the base, times the full fiber of [CG][0G][\C_G] \to [0_G]**, by perturbation and stretching. This is equal to computing the self-intersection of the zero-section.  
 +[0G]=[CG]u [0_G] = [\C_G] \cdot u  
 +where uu is cohomological degree 22. [0G]=[CGr]ur [0_G] = [\C^r_G] \cdot u^r. This make sense at least degreewise.  
 + 
 +So, the actual non-compact space fundamental class is 
 +[CGr]=[0G]ur [\C^r_G]= \frac{[0_G]}{u^r}  
 + 
 +OK, this summarizes my understanding of equivariant localization between a vector space and the zero section.  
 + 
 +For the abelian case, there should be several approaches, all leading to the correct consistent answers.  
 +BFN and VV both defined some convolution, maybe even BDG.  
 + 
 +==== BFN convolution space ==== 
 +Recall what is matrix multiplication: given matrices of sizes n1×n2n_1 \times n_2, n2×n3n_2 \times n_3, called M12M_{12} and M23M_{23}, we can form their product, which is given by 
 +(M13)ij=k(M12)ik(M12)kj (M_{13})_{ij} = \sum_k (M_{12})_{ik} (M_{12})_{kj}  
 + 
 +This can be considered as composing correspondences. We need a way to do composition (multiplication), and a way to do summation (over all possible kk, intermediate state).  
 + 
 +If we package this in terms of BM homology, we  
 + 
 + 
 + 
 + 
 + 
 + 
 + 
  
-  
  
  
blog/2023-06-27.1687884854.txt.gz · Last modified: 2023/06/27 16:54 by pzhou