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 $GL_1(\C)$ acting on $\C$ in the standard way. We have two stacks, one is $\C(K) / GL_1(K)$, one is $\C(O) / GL_1(O)$. It is a set with some group action.  Let's tell the story just one more time. We have $GL_1(\C)$ acting on $\C$ in the standard way. We have two stacks, one is $\C(K) / GL_1(K)$, one is $\C(O) / GL_1(O)$. 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 $X \to X/G$, then usually the natural topology on $X/G$ (say it is discrete) and the one on $X$ are different. For example, consider $B$ acting on $G/B$. So, the topology on $GL_1(K)$ 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 $X \to X/G$, then usually the natural topology on $X/G$ (say it is discrete) and the one on $X$ are different. For example, consider $B$ acting on $G/B$. So, the topology on $GL_1(K)$ and that on $GL_1(K)/GL_1(O)$ is not compatible. That is a bit confusing. 
  
-What's the topology on $GL_1(K)$ and $K$? We have inclusion $GL_1(K) = K \RM \{0\}$. $K$ is an infinite dimensional vector space+What's the topology on $GL_1(K)$ and $K$? We have inclusion $GL_1(K) = K \RM \{0\}$. We can ask, what is the topology of $K$. $K = \C( (t) )$. The abstract metric completion of a metrized space equal to the induced closure after isometrically embedded into a complete metric space. We define the $t$-adic valuation on $K$, which is not the usual one induced from polynomial.  
 + 
 +The subset $c_0 + t \C[ [t] ]$ is itself both open and closed,  since it is is the open and closed ball of radius $r_0$ centered at $c_0$, where $1 > r_0 > e^{-1}$.  
 + 
 +So, the space $K$ 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 $\C$? Or some other ones? The naive picture that $LG$ is a $L^+G$ torsor over $Gr_G$ seems to be correct.  
 + 
 +If $G$ is $GL_1$, then there is no contradiction. But if $G=GL_2$, then $Gr_G$ 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 $LG \to Gr_G$),  then it means that $GL_2(K)$ is not totally disconnected. Even $GL_2(O)$ may not be totally disconnected. The case for $GL_1(O)$ is a coincidence.  
 + 
 +What is the topology on $GL_2(O)$? We have the matrix $M_2(O)$. It is a nice enough vector space.  
 + 
 +OK, the t-adic metric on $F( (t) )$ is viewing $F$ as a totally disconnected space, like $\F_q$. Even treating $\C$ as a disconnected space. And that is too fine.  
 + 
 +==== OK then, how about the weird convolution euler index ==== 
 +slow down. let's work over $\C$, as real two dimensional space.  
 + 
 +Consider a rank $r$ vector space over $\C$, with $\C^*$-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 $\C^*$ or $S^1$ action. There are many invariant subspaces. We also have $S^1$ action on $S^{2r}$, coming from the suspension of $S^1$ acting on $S^{2r-1}$.  
 + 
 +What can we say then? If we want to compute the equivariant cohomology (which is equal to equivariant homology up to shift) of $\C$ acted by $S^1$, then we pass to the Borel model, we form $EG \times_G \C$, $G=S^1$, and we get a line bundle over $BG=\C \P^\infty$. The top dimensional 'homology', or 0-degree (degree = codim) cohomology is defined. That is $[\C_G]$, the fundamental (homology) class of the universal associated bundle.  Then, we can ask, what is the homology of $[0_G]$, 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 $[\C_G] \to [0_G]$**, by perturbation and stretching. This is equal to computing the self-intersection of the zero-section.  
 +$$ [0_G] = [\C_G] \cdot u $$ 
 +where $u$ is cohomological degree $2$. $ [0_G] = [\C^r_G] \cdot u^r$. This make sense at least degreewise.  
 + 
 +So, the actual non-compact space fundamental class is 
 +$$ [\C^r_G]= \frac{[0_G]}{u^r} $$ 
 + 
 +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 $n_1 \times n_2$, $n_2 \times n_3$, called $M_{12}$ and $M_{23}$, we can form their product, which is given by 
 +$$ (M_{13})_{ij} = \sum_k (M_{12})_{ik} (M_{12})_{kj} $$ 
 + 
 +This can be considered as composing correspondences. We need a way to do composition (multiplication), and a way to do summation (over all possible $k$, 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