# When this Ad-invariant function on a Lie algebra is zero?

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!

• 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. Apr 13, 2016 at 16:59

(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.

• 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? Apr 13, 2016 at 18:11
• My bad. I was assuming $G$ to be connected. Thanks for catching my mistake. It is corrected now. Apr 13, 2016 at 19:16
• Great. Thank you very much for the explanations. Apr 13, 2016 at 21:18