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If $G$ is a compact Hausdorff topological group then every neighborhood of the identity contains a neighborhood $U$ which is invariant under conjugation. That is, $gUg^{-1}=U$ for all $g\in G$.

Proof: The map $f:G \times G \rightarrow G$ defined by $f(g, k) = gkg^{-1}$ is continuous. If $V$ is an open neighborhood of the identity, then $G-V$ is compact and $U=G-f(G \times (G-V))$ is open, contained in $V$, invariant under conjugation, and contains the identity.

Now, consider the (non-compact) subgroup $T$ of $\mathrm{SL}(2,\mathbb{R})$ consisting of upper triangular matrices, i.e. $2\times 2$ matrices of the form $$\begin{pmatrix}a& b\\0& 1/a\end{pmatrix},\;\;\text{$a\in\mathbb{R}-\{0\}$ and $b\in\mathbb{R}$.}$$

Is the conclusion of the first paragraph false for $\mathbf{T}$?

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As far as I can tell, $T$ isn't compact. Am I being crazy? – Alex Youcis Aug 2 '13 at 2:44
Yes, $T$ is clearly non-compact. That is precisely why I am asking. – John Aug 2 '13 at 2:50
up vote 5 down vote accepted

If $U$ is any neighbourhood of $1$, then it contains an element $n := \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$ for some $b \neq 0$. If we conguate $n$ by the matrix $t := \begin{pmatrix} a & 0 \\ 0 & a^{-1} \end{pmatrix}$ we get $\begin{pmatrix} 1 & a^2 b \\ 0 & 1 \end{pmatrix},$ and $a^2 b$ can be made arbitrarily large by choosing $a$ large enough. Thus if $U$ has compact closure, it cannot be conjugation invariant.

Since $T$ is certainly locally compact, we conclude that it does not satisfy the property of the first paragraph of the OP.

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And that is how it is done. Nice! – John Aug 2 '13 at 2:56

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