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Let G be a semisimple algebraic group.

Following work of Matsumoto [1], Brylinski and Deligne [2] constructed a central extension of the functor
G : RingsGroups by the second algebraic K-theory functor.

Plugging in ℂ((t)) into those functors, we get the well known central extension $\widetilde{G\big(\mathbb C((t))}\big)$ of the loop group G(ℂ((t))) by the multiplicative group ℂ*=K2(ℂ((t))).
It is interesting to note that the above group comes from an algebraic group defined over the subfield ℂ of ℂ((t)). Namely, $\widetilde{G\big(\mathbb C((t))}\big)$ = $\widetilde{LG}(\mathbb C)$.

Doing all this with ℚp instead of ℂ((t)), we get a central extension $\widetilde{G(\mathbb Q_p)}$ of G(ℚp) by the group K2(ℚp) = Fp*. Now, here's an idea: maybe that central extension is defined over... the subfield F1 of ℚp?...

My questions:
• Has this been considered before?
• If yes, among all the exitsing notion of "defined over F1", which one(s) make this possible?
• If no: is my heuristic argument is convincing?


References:
[1] Matsumoto, "Sur les sous-groupes arithmétiques des groupes semi-simples déployés".
[2] Brylinski, Deligne, "Central extensions of reductive groups by $K_2$".

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I like this question, but... Brylinski and Deligne consider $K_2$ as a sheaf on the big Zariski site over a field. The first question should be whether, in some sense (Borger?), $K_2$ is a sheaf on the big Zariski site over the field with one element. Is this known? –  Marty May 13 '10 at 18:22
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1 Answer 1

up vote 9 down vote accepted

First a small thing. I am pretty sure we don't have $K_2(\mathbb C((t)))=\mathbb C^*$, we have a surjective residue homomorphism $K_2(\mathbb C((t)))\rightarrow \mathbb C^*$ but, I believe, with a non-trivial kernel. In any case, we can look at the induced central extension and then the rest of what you say is OK. Similarly, we have a surjective map $K_2(\mathbb Q_p)$.

Disrergarding this, there is a much simpler analogy between the two cases which on the one hand, I think, makes the analogy that you want less likely and on the other hand can be proven... To begin with it is not quite true that even $G(\mathbb C((t)))$ is defined over $\mathbb C$ at least not as a group scheme. What happens is that $G(\mathbb C[[t]])$ is a group scheme, it is the inverse limit of the $G(\mathbb C[t]/(t^n))$ and these have a natural structure of algebraic group over $\mathbb C$ (through the Greenberg functor). then $G(\mathbb C[[t]])$ as the inverse limit of algebraic groups is a group scheme (it is not of finite type hence convention forces us to call it a group scheme rather than algebraic group). Now, if we try to pass to $G(\mathbb C((t)))$ we get into trouble. It is an infinite union of schemes (bound the valuations of the entries of the elements of $G(\mathbb C((t)))$ in some faithful linear representation of $G$) but an infinite union of schemes does in general not have a scheme structure. There are ways of extending the scheme notion to cover this case and what we get is what is called an ind-group scheme over $\mathbb C$. Also the loop group type extension of $G(\mathbb C((t)))$ by $\mathbb C^*$ has such an extension (as does every Kac-Moody type group).

The situation for $G(\mathbb Q_p)$ is almost identical; $G(\mathbb Z/p^n)$ are the $\mathbb Z/p$-points of a $\mathbb Z/p$-algebraic group, $G(\mathbb Z_p)$ are the $\mathbb Z/p$-points of a group scheme over $\mathbb Z/p$ and $G(\mathbb Q_p)$ are the $\mathbb Z/p$-points of an ind-group scheme over $\mathbb Z/p$. I think that the same thing is true for the central extension. The upshot is that there is a close analogy to the $\mathbb C$ case but in that analogy $\mathbb C$ is replaced by $\mathbb F_p$ not by $\mathbb F_1$.

Note that in the Connes-Consani version of $\mathbb F_1$ $G$ is defined over $\mathbb F_{1^2}$ so perhaps that is the place to look for a version of the Brylinski-Deligne result.

Addendum: Just to add even more concreteness to George's answer about the explicit form of Greenberg's functor for $\mathbb G_m$. We have that $W_n(B)$ is just $B^n$ with a funny multiplication and addition. They are however given by polynomials (which are independent of $B$). The units in this ring are the tuples of the form $B^\ast\times B^{n-1}$ and multiplication is given by polynomials. This means that the algebraic group associated to this is just $\mathbb G_m\times\mathbb A^{n-1}$ as scheme but with a funny product structure. In particular its $\mathbb F_p$-points are just $(\mathbb Z/p^n)^\ast$.

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Thank you Torsten for your very nice answer. You say that G(Z/p^n) are the Z/p points of a Z/p-algebraic group. Could you maybe illustrate that claim in the simple case G=G_m and n=2. What is the algebraic group in that case? –  André Henriques May 13 '10 at 19:34
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There might be a bit more to the original question. From the lambda/Witt point of view on $\mathbf{F}_1$, "$G(\mathbf{Z}_p)$" (or rather the Greenberg functor applied to $G$) is not just an a group scheme over $\mathbf{F}_p$, it also the base change to $\mathbf{F}_p$ of a group scheme over the "$p$-typical $\mathbf{F}_1$". In this context, this just means that "$G(\mathbf{Z}_p)$", if we view it as a flat group scheme over $\mathbf{Z}_p$ (rather than just looking at its special fiber) has a lift of the Frobenius map. (continued) –  JBorger May 14 '10 at 4:59
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I haven't ever really thought about the passage to the group-ind-scheme $G(\mathbf{Q}_p)$, so I don't know whether it's possible to descend $G(\mathbf{Q}_p)$ similarly, but it seems pretty reasonable. I also don't know anything about the extension, but it seems like a reasonable thing to consider. Also, I have heard that William Haboush has a manuscript on $G(\mathbf{Q}_p)$-like group-ind-schemes, so you might consult him. –  JBorger May 14 '10 at 5:03
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I was too cautious. "$G(\mathbf{Q}_p)$" is definitely a group-ind-scheme defined over the $p$-typical $\mathbf{F}_1$ as long as $G$ is an affine group scheme. (It would take a bit of space to explain this.) You should try to make the central extension then! –  JBorger May 14 '10 at 5:11
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You can see the final section of arxiv:0906.3146, but unfortunately there's not much there. I think we should talk. –  JBorger May 14 '10 at 10:14
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