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.