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Hello every one!

Here is a question which I would like to ask you:

Let $G=limit G_{i}$ is an ind-group (Shafarevich called it infinitely-dimensional algebraic group). It is well-known that $G$ is smooth. Is it true that for any (or at least for one) $x \in G$ there exist a natural $N$ big enough, such that $G_{i}$ is smooth at $x$ for all $i > N$ (or just for infinitely many $i : i > N$) ?

Thank you for your answers and considerations! Andriy.

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  • $\begingroup$ I don't understand how can this be well know. What if all the G_i are equal to the same non-smooth group? $\endgroup$
    – Moshe
    May 4, 2012 at 17:13
  • $\begingroup$ By definition all G_i are finitely dimensional (affine) varieties. In case all G_i are equal, then G=G_i for some (any) i and G is an algebraic group. It is known (and in fact easy and could be found in any book about algebraic groups) that algebraic group is smooth! I already know that answer for my question is negative! $\endgroup$
    – Andriy
    May 6, 2012 at 23:33
  • $\begingroup$ So the $G_i$ that appear in your question are not algebraic groups? $\endgroup$
    – Moshe
    May 7, 2012 at 14:04
  • $\begingroup$ Yes, they are not groups. They are just affine varieties. Sorry, I did not write it in my question (but it is just definition of ind-group (where it is not required that G_i should be groups)). $\endgroup$
    – Andriy
    May 7, 2012 at 15:57
  • $\begingroup$ I don't know exactly what definition you are working with. Also, I'm not sure what you mean by ``$G_i$ is smooth at $x$'' when $x$ is a point of $G$. If you just mean that $G$ is a group object in the ind-category of algebraic varieties with all morphisms between them, it seems that you could take $G_i=X$, a fixed algebraic variety with one singular point $p$, and with maps constantly mapping $G_i$ to $p$ of the next copy. Then $G$ is a point, which can be give the structure of the trivial group, and the inverse image contains $p$. Anyway, you say you already know the answer... $\endgroup$
    – Moshe
    May 7, 2012 at 16:28

1 Answer 1

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What Shafarevich proved was that, over a field of characteristic zero, an ind-group $G$ is smooth everywhere. However, his definition of ``smooth'' is in fact a version of formally smooth, generalized to infinite dimensions. This does not imply that $G$ is a union of smooth varieties, as asked; according to p. 14 of [Fishel, Grojnowski and Teleman, arXiv:math/0411355v1] the formal loop group $G((z))$ is a counterexample if $G$ is a simple algebraic group over $\mathbb C$.

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  • $\begingroup$ Thanks for the answer! That is very interesting. Let me ask you another question: is it true that there exist $x \in G$ and a natural $N$, big enough, such that $G_i$ is smooth at $x$ for infinitely many $i>N$? $\endgroup$
    – Andriy
    Jun 12, 2012 at 6:18

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