# Sheafification of loop scheme/group

Let $$X$$ be a scheme over $$K = k((t))$$, where $$k$$ is a field. We define the loop scheme $$LX$$ to be the functor from the category of $$k$$-algebras to sets by $$R \mapsto LX(R) := X(Spec (R((t))))$$.

Do we need sheafification in order to make $$LX$$ a sheaf? A particular interest is when $$X$$ is a connected linear reductive group over $$K$$ and fpqc site is chosen.

• I think not. It’s easy to check for affine schemes, including reductive groups, by just using coordinates. – Will Sawin Oct 19 '19 at 16:53

Despite Will Sawin's answer in comment, it was not so obvious for me at first glance, so I give a detailed (at least for $$X$$ affine) proof here.

Lemma. Let $$A$$ be a $$K$$-algebra and let $$A\to B$$ be fpqc. Then the following sequence is exact $$A((t))\to B((t))\rightrightarrows (B\otimes_{A}B)((t)).$$

Proof. Since $$A\to B$$ is fpqc, Grothendieck's fpqc-descent shows that the sequence $$A\to B\rightrightarrows B\otimes_A B$$ is exact. Consider the sequence $$A[[t]]\to B[[t]]\rightrightarrows (B\otimes_{A}B)[[t]].$$ We claim that this sequence is also exact.

The first arrow $$A[[t]]\to B[[t]]$$ is injective since $$A\to B$$ is injective and since $$A[[t]] = \prod_{n\in \mathbf{N}}A$$ and $$B[[t]] = \prod_{n\in \mathbf{N}}B$$ as $$A$$-modules. It remains to show that for $$f(t)\in B[[t]]$$, $$f(t)\otimes 1 = 1\otimes f(t)\in (B\otimes_{A}B)[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]].$$ Suppose that $$f(t) = \sum_{n\in \mathbf{N}} b_n t^n\in B[[t]]$$ satisfies the condition $$f(t)\otimes 1 = 1\otimes f(t)$$. It is enough to show that $$f(t) \mod{t^N} \in A[t]/t^N$$ for each $$N\ge 1$$. Indeed, if we tensorise the first exact sequence by $$\otimes_{A}A[t]/t^N$$, we get $$A[t]/t^N\to B[t]/t^N\rightrightarrows (B\otimes_{A} B)[t]/t^N,$$ which implies that $$b_n\in A$$ for $$n < N$$. Hence $$f(t)\in A[[t]]$$.

Inverting $$t$$, we deduce the exactness of $$A((t))\to B((t))\rightrightarrows (B\otimes_{A}B)((t)).$$ End of Proof

Now if $$X = \mathrm{Spec}\, C$$ is an affine $$K$$-scheme and A,B are as above, then the exactness of the following sequence follows from the lemma $$\mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, (B\otimes_{A} B)((t))).$$ Hence, if $$X$$ is affine, $$LX$$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.

Let me give an answer beynd the case when $$X$$ is affine.

Upshot: Minimally restricting the setup, the answer is "LX is a sheaf, so you don't need to sheafify" for all quasi-projective schemes $$X/K((t))$$, and with respect to a very strong topology.

Suppose $$K$$ is has positive characteristic, and restrict your functor LX to the category $${\rm Perf}_K$$ of perfect $$K$$-algebras. Recall the arc-topology on schemes, introduced by Bhatt--Morrow arXiv:1807.04725. Essentially, a map $$X \rightarrow Y$$ of qcqs schemes is an arc-cover if it's surjective and any specialization relation between points in $$Y$$ lifts to $$X$$. E.g., you could take the spectrum of the product over all possible valuation rings at all points of $$Y$$, thus obtaining a (huge) affine scheme, which maps to $$Y$$, and this would give you an arc-cover. Note that the arc-topology is even stronger than the $$v$$-topology of Bhatt--Scholze arXiv:1507.06490, and "much" stronger than the fpqc-topology. E.g. if $$V$$ is a valuation ring of rank $$2$$ and $$\mathfrak{p}$$ is its unique prime ideal of height $$1$$, then $${\rm Spec} V_{\mathfrak{p}} \,\dot\cup\, {\rm Spec} V/\mathfrak{p} \rightarrow {\rm Spec} V$$ is an arc-cover, but not a $$v$$-cover (and so in particular, not an fpqc-cover), see Corollary 2.9 of arXiv:1807.04725. This said, we have the following result, see Theorem 5.1 of arXiv:2003.04399:

Theorem. Suppose $$X$$ is a quasi-projective $$K((t))$$-scheme. Then LX is a sheaf for the arc-topology on $${\rm Perf}_K$$.

Remark 1. I would guess that this should also extend to the case that $$K$$ has characteristic zero, but the proof in loc.cit. uses perfectoid spaces, and so cannot literally be carried over to the case when $$K$$ has characterstic zero.

Remark 2. This result continues to hold in a mixed characteristic setup, i.e., when $$K((t))$$ is replaced by an appropriate $$p$$-adic field with residue field $$K$$ (e.g. if $$K = \mathbb{F}_p$$, instead of $$K((t))$$ you could take $$\mathbb{Q}_p$$, or some finite totally ramified extension of it).