I'm having problems with this simple question.

Consider $G$ a connected Lie Group and $\mathfrak g = \text{Lie}(G)$. Let $k:\mathfrak g \times \mathfrak g \rightarrow \mathbb R$ be a symmetric bilinear form that is $\text{ad}$-invariant (i.e. $k([W, X], Y) = -k(X, [W, Y])$ for all $X, Y, W \in \mathfrak g$). I want to show that $k$ is $\text{Ad}$-invariant, i.e. $k(\text{Ad}_g(X), \text{Ad}_g(Y)) = k(X, Y)$ for all $g \in G$. I believe that somehow it is necessary to relate $\text{ad}$ and $\text{Ad}$, (possibly by the formula $\exp(\text{ad}_X)= \text{Ad}_{\exp(X)}$) since $G$ is connected is generated by $\exp( \mathfrak g)$. However, I was unsuccessful in trying to do this. Could someone give me a tip?

## Best Answer

Since $G$ is connected, it is generated by a neighborhood of $e$. We can also find a neighborhood $V\subseteq \mathfrak{g}$ of $0$ and a neighborhood $U\subseteq G$ such that $\exp:V\rightarrow U$ is a diffeomorphism.

So for $g\in U$, $g= \exp(W)$ for some $W\in \mathfrak{g}$, which means $\text{Ad}_g = \exp(\text{ad}_W)$. If $k$ is ad-invariant, then that means $k(\text{ad}_WX, Y) = k(X,-\text{ad}_WY)$. Since $\text{ad}_W$ and $-\text{ad}_W$ certainly commute, we have $$\exp(-\text{ad}_W)\exp(\text{ad}_W) = \text{exp}(0_\mathfrak{g}) = \text{id}_G$$

Now for any $g\in U$, $g^{-1} = \exp(-W)$, and so $\text{Ad}_{g^{-1}} = \exp(\text{ad}_{-W}) = \exp(-\text{ad}_W)$, and hence:

\begin{align*}k(\text{Ad}_gX, Y) &= k(\exp(\text{ad}_W)X, Y)\\ &= \sum_{j=0}^\infty\frac{1}{j!}k(\text{ad}_W^nX, Y)\\ &= \sum_{j=0}^\infty\frac{1}{j!}k(X,(-\text{ad}_W)^nY)\\ &=k(X,\exp(-\text{ad}_WY))\\ &=k(X,\text{Ad}_{g^{-1}}Y) \end{align*}

By the way, if you look up toe coadjoint and coAdjoint representations, you'll see that what we just showed was that $\text{ad}_W^* = -\text{ad}_W$ and $\text{Ad}_g^* = \text{Ad}_{g^{-1}}$.

Ad-invariance follows immediately then, since $\text{Ad}:G\rightarrow GL(\mathfrak{g})$ is a homomorphism, meaning $\text{Ad}_{g^{-1}} = \text{Ad}_g^{-1}$, and so $$k(\text{Ad}_gX, \text{Ad}_gY) = k(X, \text{Ad}_{g^{-1}}\text{Ad}_gY) = k(X,Y)$$.

We're actually done now because any $g\in G$ can be written as a product of elements in $U$ and so you can check this holds $\forall g\in G$.