Topologists loops versus algebraists loops

Question

Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}_m$. The affine Grassmannian case also provides positive evidence.

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. Regarding putting the classical topology of X((t))(ℂ), one should not be scared of the ind-scheminess. ℂ((t)) has a natural structure of a topological ring, and hence we topologise X(ℂ((t))) in the usual manner, taking the subspace topology using a closed embedding into affine n-space for some n.

[paragraph redacted]

Answer 1

Here's an example constructed using moonface's idea without leaving the smooth realm: Take an affine curve $X$ whose smooth projective model $\overline{X}$ has genus $g > 0$. Define $S^1_a = \mathrm{Spec}(\mathbf{C}((t)))$, and $D^2_a = \mathrm{Spec}(\mathbf{C}[[t]])$.

Claim: The map $X((t))(\mathbf{C}) \to LX$ is not a homotopy equivalence. In fact, it is not even surjective on $\pi_0$.

Proof: The (split) fibration $\Omega(X) \to LX \to X$ shows that $\pi_0(LX) = \pi_0(\Omega(X)) = \pi_1(X)$. So it suffices to show that the natural map $X(S_a^1) \to \pi_1(X)$ is not surjective. As $\pi_1(X) \twoheadrightarrow \pi_1(\overline{X})$, it even suffices to show that not every element in $\pi_1(\overline{X})$ is realised by a map $f:S^1_a \to X$. Given such an $f$, the composite map $S^1_a \to X \to \overline{X}$ factors as a map $S^1_a \to D^2_a \to \overline{X}$ by the valuative criterion. In particular, the induced map on fundamental groups is trivial as $D^2_a$ is simply connected. As $g > 0$, we are done.

[ It seems that $\mathrm{Spec}(\mathbf{C}((t)))$ has a Hodge structure of Tate type and, consequently, cannot detect loops except those of weight $0$, i.e., those that come from removing divisors. Does anyone know if Hodge theory makes sense for such big objects? ]

Answer 2

Just a metacomment on Bhargav's answer: it's not always true that $\pi_0(LX) = \pi_1(X)$, namely when $X$ is not simply connected (as is certainly the case in this example). In general $\pi_0(LX)$ is the set of conjugacy classes of elements of $\pi_1(X)$ -- think about the change-of-basepoint isomorphism in $\pi_1$.

However, this certainly doesn't break the argument: there are lots of nontrivial conjugacy classes in $\pi_1(X)$.

Answer 3

I don't really understand all the bits of the question, but this is a bit long for a comment. It's sparked by one of David Ben-Zvi's comments above. If we take a compact semisimple Lie group, $G$, with trivial centre then we can consider polynomial loops in $G$ by embedding $G$ in some matrix group, for example via the adjoint representation on its complexified Lie algebra, and considering loops that are polynomial in the corresponding matrix algebra. Then this space is homotopy equivalent to the space of smooth loops in $G$ (it doesn't matter if we take based or free so long as we're consistent). This is proved in, for example, Pressley and Segal's book Loop Groups (the exact statement is Proposition 8.6.6, but of course there's considerable build up to that). And, of course, smooth is homotopy equivalent to continuous.

Is this in any way relevant to the question?

Answer 4

This is not an answer, but these are my thoughts so far and hopefully they will lead someone to a correct answer (hence the community wiki). My vague recollection is that the algebraic loop space only sees stuff "near" the constant loops, which is consistent with moonface's comments. I apologize if I am missunderstanding anything (I'm one of the struggling topologists). Basically I want to look at an example, which I think will elucidate the matter.

Let's take $X = \mathbb{G}_m$, the multiplicative group. Then $ \mathbb{G}_m(\mathbb{C}) = Spec \mathbb{C}[b, b^{-1}]$. As an analytic space I think this is just $\mathbb{C}^\times$, so on the topological side we get an interesting loop space. We have a fibration sequence,

$$\Omega \mathbb{C}^\times \to L\mathbb{C}^\times \to \mathbb{C}^\times $$

and since topologically $\mathbb{C}^\times \simeq S^1$, this shows that $\pi_0(L \mathbb{C}^\times) \cong \mathbb{Z}$. This is something that we should be able to detect if the algebraic version of the loop space is similar to the topological one, just count the number of components.

So what is the algebraic loop space in this case? Well, I guess by definition it is $Spec \; \mathbb{C}((t))[b,b^{-1}]$. Now remind me, how do we turn this into a space? and how many components does it have?

If you try to take the $\mathbb{C}$-points of it, i.e. homomorphisms, $$ \mathbb{C}((t))[b, b^{-1}] \to \mathbb{C}$$ don't you just get $\mathbb{C}^\times$? This seems to suggest that it is an infinitesimal thickening of the constant loops.

Answer 5

While the answer to your question is negative in general, as pointed out before, the answer is positive for certain type of loop groups. This can be proved using topological twin buildings. See Linus Kramer, Loop Groups and Twin Building. (It is not stated very explicitly, but the topology used on the twin building and hence the algebraic loop group is meant to be the ind-topology coming from the Bruhat cell decomposition.)