Back to action. I still want to pin down the two patches gluing philosophy, why it works.

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.

on the level of set. What topology do I have on $\C( (t) ) )$?

• on $\C[t]$, which as a vector space is the direct sum of $\oplus \C t^n$, the topology is generated by $U \cdot t^n$ for various open subset $U \In\C$ and various $n$.
• on $\C[ [t] ]$, it is defined as the inverse limit of $\C[t] / (t^n)$. This is defined using the product topology on $\prod_n \C[t] / (t^n)$ (then imposing some consistency condition). A subset is open if it can be generated by cylinder set. (which I think is the same as requiring, the projection map is open map).
• Then, we have $\C( (t) )$, this is the colimit of $t^{-k} \C[ [t] ]$, as $k \to \infty$.

This seems to be irrelevant, but I do want to know how to think about these infinite dimensional vector spaces, or open subset of vector spaces.

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 $GL_1(K) = K \RM \{0\}$. 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.

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.

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