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.


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Topologists loops versus algebraists loops
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? ]


