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Today in a talk, it has been mentioned that there exists algebraic groups over the local field $\mathbb{R}$ such that the finite central extension can not be defined algbraically over $\mathbb{R}$ or its algebraic closure $\mathbb{C}$. I guess already covers of $SL(2)$, which is even defined over $\mathbb{Z}$, and the metaplectic group are such an example!?

I am curious, what is the (intuitive) reason for this lack. And, how to proof it rigorously?

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    $\begingroup$ Discrete group central extensions are classified by the second group cohomology. Find out the cohomology classifying central algebraic group extensions. There will probably be a comparison homomotphism from this cohomology to group cohomology which won't be surjective in general, even in degree $2$. An element which is not in the image of the comparison homomorphism will give you a non-algebraic group extension. $\endgroup$
    – Fernando Muro
    Commented Jul 7, 2011 at 19:48
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    $\begingroup$ It seems to me that central extensions of an algebraic group should be "non-algebraic until proven algebraic" rather than the other way around. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 7, 2011 at 20:41
  • $\begingroup$ Following your suggestion, I edited the title. $\endgroup$
    – Marc Palm
    Commented Jul 20, 2011 at 8:00
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    $\begingroup$ Finite dimensional representations of the Lie algebra of $SL(2,\mathbb R)$ are easily classified and they are all representations of $SL(2,\mathbb R)$. If the double cover were algebraic it would have a faithful finite dimensional representation. $\endgroup$
    – Torsten Ekedahl
    Commented Jul 20, 2011 at 12:40

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The double cover of $SL(2,\mathbb R)$ is not algebraic.

This can be blamed on the fact that the map $$\pi_1\big(SL(2,\mathbb R)\big)\cong \mathbb Z\quad\longrightarrow\quad \pi_1\big(SL(2,\mathbb C)\big)=0$$ is not injective.

If the double cover of $\pi_1(SL(2,\mathbb R))$ were algebraic, it would have a complexification, which would be a double cover of $SL(2,\mathbb C)$. But $SL(2,\mathbb C)$ doesn't have any double covers since its fundamental group is trivial.

Using that method, you can actually detect which covers are algebraic:
Let $G$ be a real algebraic Lie group, and let $A$ be a finite abelian group. A central extension of $G$ by $A$ determines a homomorphism $\pi_1(G)\to A$. The cover is algebraic iff that homomorphism extends to a homomorphism $\pi_1(G_{\mathbb C})\to A$.

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    $\begingroup$ Is it clear that the "complexification" functor preserves covering maps? (so that the complexification of a cover is $\endgroup$
    – Pierre
    Commented Jul 20, 2011 at 16:25
  • $\begingroup$ oop, is a cover of the complexification). $\endgroup$
    – Pierre
    Commented Jul 20, 2011 at 16:25
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    $\begingroup$ I depends how you define "complexification". At the Lie algebra level, it's pretty easy to complexify. At the Lie group level, it's not so clear what you mean in general... and indeed, some real Lie groups cannot be complxified. If the group is algebraic, then of course, you can complexify without problem, and covers (aka Lie algebra isomorphisms) complexify to covers. $\endgroup$
    – André Henriques
    Commented Jul 20, 2011 at 17:35
  • $\begingroup$ oh OK, complexifying preserves Lie algebra isomorphisms so it preserves covers, nice -- that double covers are turned into double covers is a little more interesting (though your argument is fine), indeed the change in $\pi_1$ is interesting. If by "complexifying" you mean going from compact to complex reductive (say from $U_n$ to $GL_n( \mathbb{C}$) then the fundamental group (indeed the homotopy type) is unchanged; but if by "complexifying" you include the passage from $SL_2(\mathbb{R})$ to $SL_2(\mathbb{C})$, then things happen, as you point out. I had never thought of this. $\endgroup$
    – Pierre
    Commented Jul 20, 2011 at 18:39

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