Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbf{T}$?

share|improve this question
    
As far as I can tell, $T$ isn't compact. Am I being crazy? –  Alex Youcis Aug 2 '13 at 2:44
2  
Yes, $T$ is clearly non-compact. That is precisely why I am asking. –  John Aug 2 '13 at 2:50
add comment

1 Answer

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.

share|improve this answer
    
And that is how it is done. Nice! –  John Aug 2 '13 at 2:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.