$\text{ad}$-invariant bilinear form and connected Lie group

lie-algebraslie-groups

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?

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 $$\forall g\in G$$.