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Does there exists an example of an extension of topological groups $1 \to N \to E \to G \to 1$ admitting a section $s:G \to E$ which is continuous (or continuous in a neighbourhood of identity) and satisfy the propery that $s(x^{-1}) =s(x)^{-1}$. One can see that the extension $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{S}^1 \to 1$ does not admit such a section.

For discrete groups there are many extensions admitting such a section. For example, consider the extension $0 \to \mathbb{Z} \stackrel{i}{\to} \mathbb{Z} \times \mathbb{Z}_2 \stackrel{p}{\to} \mathbb{Z}_4 \to 0$ with $i(x)= (2x, [x])$ and $p(x, [y])=[x + 2y]$. Clearly, the section $s$ defined by $s([0])=(0,[0])$, $s([1])=(-1,[1])$, $s([2])=(0,[1])$ and $s([3])=(1,[1])$ satisfy the property that $s(x^{-1})= s(x)^{-1}$. I am looking for an example in the setting of compact connected topological groups.

edited Jan 1 2011 at 9:44

David Roberts
11.9k●2●18●56

asked Dec 24 2010 at 8:14

gurs
63●4

Yes: a semi-direct product. Or do you want some sort of non-triviality condition? – David Roberts Dec 24 2010 at 8:36

Thanks. But I want a non-trivial extension. – gurs Dec 24 2010 at 9:04

Let me write my question again. Does there exist an extension of topological groups $1 \to N \to E \to G \to 1$ admitting a section $s:G \to E$ which is continuous (or only continous in a neighbourhood of identity) and satisfy the property that $s(x^{-1})=s(x)^{-1}$ for all $x \in G$ but is not a homomorphism. – gurs Dec 24 2010 at 9:13

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Take the extension $1 \to \mathbb{R} \to \mathbb{R}^2 \to \mathbb{R} \to 0$ and the section $x \mapsto (x,x^3)$. – Johannes Ebert Dec 24 2010 at 10:06

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gurs says (moved from an answer): Can someone give an example in the setting of compact topological groups or in the setting of finite dimensional Banach spaces. – Anton Geraschenko♦ Dec 31 2010 at 18:16

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Theo Buehler · Accepted Answer · 2010-12-24 12:07:44Z UTC

A very easy example (inspired by an answer by Laurent Moret-Bailly that he removed immediately after posting): Take $0 \to 3\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/3\mathbb{Z} \to 0$ and define a section $s: \mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}$ by $s(0) = 0$, $s(1) = 1$ and $s(2) = -1$.

Here's an example from functional analysis that I happen to like, but it certainly is rather involved:

Take a non-complemented subspace $F$ of a Banach space $E$. The quotient map $p:E \to E/F$ admits a continuous and homogeneous right inverse $\sigma$ by a theorem of Bartle-Graves and Michael, see E. Michael "Continuous selections, I", Ann. Math. Vol. 63, No. 2 (Mar., 1956), pp. 361-382, Proposition 7.2. Homogeneity yields in particular $\sigma(-x) = -\sigma(x)$, but $\sigma$ cannot be linear because $F$ is not complemented in $E$.

You can take for instance $F = c_{0}$, the space of sequences converging to zero, inside the Banach space $E = \ell^{\infty}$ of bounded sequences.

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