We know that every finite dimensional Lie group has no small subgroups. Is there any infinite dimensional Lie group which has small subgroups? Please explain your answer by refrence. Thanks so much.
closed as offtopic by YCor, Wolfgang, Myshkin, JanChristoph SchlagePuchta, Alexey Ustinov May 21 '16 at 14:49
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2$\begingroup$ Presumably, by the existence of "small subgroups", you mean that for every neighborhood of the identity $e\in G$, there should be a subgroup contained in that neighborhood. $\endgroup$ – André Henriques Feb 12 '13 at 12:40

1$\begingroup$ hossein, please see mathoverflow.net/howtask I think you are more likely to get help with your questions if you can demonstrate that you already know some details and have made some effort yourself $\endgroup$ – Yemon Choi Feb 12 '13 at 17:47
The locally convex space $\mathbb R^{\mathbb N}$ (product of a countable number of copies of $\mathbb R$) equipped with the product topology can be made into an Abelian Lie group with respect to addition and the standard manifold structure. A local basis at the origin consists of open sets of the form $\Pi U_i$ where $U_i$ is an open neighborhood of $0$ in $\mathbb R$ and $U_i\neq\mathbb R$ for fintely many indices $i$ only. Such a neighborhood of $0$ contains nontrivial vector subspaces, so that $G$ contains small subgroups. This example is discussed in "Towards a Lie theory of locally convex groups", KarlHermann Neeb, Japan. J. Math. 1, 291–468 (2006).

$\begingroup$ I think your link to n "Towards a Lie theory of locally convex groups", KarlHermann Neeb, Japan. J. Math. 1, 291–468 (2006). is defunct. $\endgroup$ – Duchamp Gérard H. E. May 18 '16 at 10:36