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Suppose $G$ is a compact group endowed with Haar measure $\mu$. If $g \in G$, then denote by $g^G$ the conjugacy class of $g$ in $G$.

Is there anything that can be said in general about $\mu(g^G)$?

If not, are there further restrictions we can put on $G$ in order to ensure that we can say something?

Thank you.

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up vote 6 down vote accepted

Some minor observations: If $G$ is a connected Lie group, then $g^{G}$ is a manifold of lower dimension than $G$, hence of measure $0$.

If $G$ is $O(2)$, the group of $2 \times 2$ orthogonal matrices, then topologically $O(2) \cong S^1 \sqcup S^1$, with one circle being the rotations and the other being the reflections. Any two reflections are conjugate to each other, so this conjugacy class has measure $1/2$.

Although it technically doesn't meet the description of your question, you might be looking for the Weyl integral formula. Let $K$ be a compact connected Lie group, with maximal torus $T$ and Weyl group $W$, so the conjugacy classes of $K$ are given by $T/W$. Let $U$ be a small open neighborhood in $T/W$. Weyl integration tells you the volume of the subset of $K$ whose conjugacy classes lie in $U$. Here is an example in physicists' notation: The set of matrices in $SU(2)$ whose conjugacy class is $\left( \begin{smallmatrix} e^{i \alpha} & 0 \\ 0 & e^{-i\alpha} \end{smallmatrix} \right)$ with $\alpha$ between $\theta$ and $\theta + d \theta$ has volume $\sin^2 \theta \ d \theta$.

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Your examples are helpful, thank you. They add to my suspicion that little can be said in general. When $G$ is non-compact, things get worse. If I recall correctly, one can have a dense conjugacy class even when $G$ is locally compact! – Bob Heffernan Oct 15 '11 at 7:32

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