Let $G$ be a compact Lie group with Haar measure $dg$ and (finite-dimensional real) Lie algebra $\frak g$. Endow $\frak g$ with an $\hbox{Ad}$-invariant norm $\|\cdot\|_{\frak g}$ so that $\frak g$ becomes a normed space (one can use the norm induced by the negative of the Killing form).

Define the function $$ f:{\frak g}\to{\frak g},~~x\mapsto\int_Ggxg^{-1}dg, $$ where the integral is the Bochner integral associated with the norm $\|\cdot\|_{\frak g}$.

$f(x)$ exists for each $x\in\frak g$ because the norm of the integrand is bounded: $$ \|gxg^{-1}\|_{\frak g}=\|\hbox{Ad}_gx\|_{\frak g}=\|x\|_{\frak g}. $$ $f$ is $\hbox{Ad}$-invariant because $f(gxg^{-1})=f(x)$ for all $x\in\frak g$ and $g\in G$. Finally, some direct calculations using explicit parametrizations of $G$ and $\frak g$ show that $f$ is identically zero for certain groups $G$ and nonzero for others.

My questions are thus the following:

1) Can we find a general class of compact Lie groups $G$ such that $f$ is identically zero, or nonzero?

2) Is $f$ identically zero for all semisimple compact Lie groups G?

(maybe some Weyl integration formula or some argument based on Casimir functions can be used, but I have not been able to do it...)

Thank you for the help!

  • 1
    $\begingroup$ If $G$ is compact, then $\mathfrak g$ is the direct sum of simple and of abelian Lie algebras. Your formula simply projects onto the abelian part. If you need details, please ask at math.stackexchange. $\endgroup$ Apr 13, 2016 at 16:59

1 Answer 1


(this is an edit made after user65432 caught me assuming implicitly that $G$ is connected.)

In case $G$ is connected, $f=0$ iff $\mathfrak{g}$ has no center. In the general case, $G/G^0$ acts on the center $\mathfrak{z}<\mathfrak{g}$ and $f=0$ iff there are no invariant vectors for this action ($G^0$ denotes here the connected component of $G$).

An example for the latter case is $G=\{-1,1\}\ltimes S^1$, where $f=0$ though $\mathfrak{g}$ has a one dimensional center.

The proof is standard:

For every finite dimensional representation $\rho\colon G\to \text{GL}(V)$, $f(v)=\int \rho(g)vdg$ gives a linear operator on $V$ which is the projection on the subspace of invariants $V^G$ (clearly the image of $f$ consists of invariant vectors and $f$ is the identity on invariant vectors).

Specializing to $\rho=\text{Ad}$ and noting that $\mathfrak{z}=\mathfrak{g}^{G^0}$ gives the answer.

  • $\begingroup$ Thanks. There is just a detail I am not sure to understand: To have that ${\frak g}^G=\hbox{center of }{\frak g}$ don't we need to assume that $G$ is connected? Don't we use the exponential ${\frak g}\to G$ in the proof, which is surjective only in the connected case? $\endgroup$
    – user65432
    Apr 13, 2016 at 18:11
  • $\begingroup$ My bad. I was assuming $G$ to be connected. Thanks for catching my mistake. It is corrected now. $\endgroup$
    – Uri Bader
    Apr 13, 2016 at 19:16
  • $\begingroup$ Great. Thank you very much for the explanations. $\endgroup$
    – user65432
    Apr 13, 2016 at 21:18

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