I was reading about representations of compact Lie groups and came upon a certain construction of an $Ad$-invariant inner product on the Lie algebra of a compact Lie group. I'm having trouble showing that it is actually Ad-invariant. Here's my thinking so far:

Let $G$ be a compact Lie group with Lie algebra $\mathfrak{g}$, and let $|dg|$ be a left-invariant (hence bi-invariant) smooth density on $G$. The claim is that $$B(x,y) := \frac{1}{\text{vol}(G)} \int_G B' (\text{Ad}_g (x), \text{Ad}_g(y)) \ |dg|$$ is an Ad-invariant inner product on $\mathfrak{g}$ (where $B': \mathfrak{g}\times \mathfrak{g}\to \mathbb{R}$ is an arbitrary inner product).

Fixing $x,y\in \mathfrak{g}$ and letting $ \ \ f:G\to \mathbb{R}$, $g\mapsto B'(\text{Ad}_g(x),\text{Ad}_g(y))$, we compute:

\begin{align*} B(\text{Ad}_h (x), \text{Ad}_h(y)) &= \frac{1}{\text{vol}(G)} \int_G B'(\text{Ad}_g\text{Ad}_h (x), \text{Ad}_g\text{Ad}_h (y)) \ |dg| \\ &= \frac{1}{\text{vol}(G)} \int_G B'(\text{Ad}_{gh} (x), \text{Ad}_{gh} (y)) \ |dg| \\ &= \frac{1}{\text{vol}(G)} \int_G f(gh) \ |dg| \\ \end{align*} So, I need to show that $\int_G f(gh) \ |dg| = \int_G f(g) \ |dg|$ for all $h\in G$ $(\star)$.

Intuitively this is clear, since $|dg|$ is bi-invariant, and right multiplication by $h$ inside the argument of $f$ just "shuffles" the values of $f$ around to different points on $G$. However I can't seem to find an algebraic justification for why this should be true.

Here is a thought I had: Let $\chi:G\to \mathbb{R}^*$, $g\mapsto |\det(\text{Ad}_g)|$ be the so-called "modular function" of $G$. I know that any left-invariant smooth density $\mu$ satisfies $$(R_h)^* \mu= \frac{1}{\chi(h^{-1})}\ \mu$$ Since $G$ is compact we have $\chi\equiv 1$, hence $(R_h)^* \mu= \mu$. Thus, to prove $(\star)$, it suffices to show that the density $f |dg|$ is left-invariant. However, I wasn't able to show this (and I'm not entirely convinced whether it's even true).

Any suggestions would be greatly appreciated!


Some of the details can be removed/abstracted-away: let $G$ be a compact topological group (need not be Lie...), and $V$ a finite-dimensional (real?) representation space for $G$. Then $V$ admits a $G$-invariant inner product. Specifically, given any inner product $\langle,\rangle$ on $V$, the averaged inner product $$ \langle v,w\rangle' \;=\; \int_G \langle gv,\,gw\rangle\;dg $$ with Haar measure $dg$ giving $G$ total mass $1$, is $G$-invariant.

The proof you've sketched, with various explicit details suppressed, succeeds in this context, simply by "changing variables" in the integral. In fact, it succeeds for possibly infinite-dimensional Hilbert spaces $V$, using a bit of elementary functional analysis.

The change-of-variables trick you were hesitant about is exactly the invariance of the functional $f\to \int_G f(g)\,dg$, namely, that for all $h\in G$ we have $\int_G f(gh)\,dg=\int_G f(g)\,dg$. And this invariance is an equivalent of invariance of the Haar measure.

  • $\begingroup$ Thanks, I think I got it now! $\endgroup$ – itinerantleopard Aug 4 '18 at 0:02

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