# Invariant Inner Product on Lie Algebra

Let $G$ be a Lie group, $\frak{g}$ its Lie algebra. Suppose $\mathcal{D}$ a representation of $G$ on $V$, $d$ the associated Lie algebra representation. Suppose $V$ is endowed with an inner product. When (if at all) is the following statement true?

The inner product is invariant under $\mathcal{D}$ iff it is invariant under $d$.

I can't find any way of proving it, nor can I think of a counterexample. The question was motivated by the fact that many books prove that the Killing form on a Lie algebra is $ad$-invariant by stating that it's obviously $Ad$-invariant.

Any insight would be much appreciated!

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This should follow readily using $\exp$. – Hagen von Eitzen Feb 15 '13 at 15:53
Ah - I had an incorrect definition for Lie algebra invariance! I thought I needed to verify that $<d(A)X,d(A)Y>=<X,Y>$ for all $X,Y,A$ in the Lie algebra. But in fact that is wrong as this shows! – Edward Hughes Feb 15 '13 at 17:26

Here's how to prove one direction, namely if the inner product $\langle,\rangle$ on $V$ is invariant under $\mathcal{D} : G \to \text{GL}(V)$ then it is invariant under $d : \mathfrak{g} \to \mathfrak{gl}(V)$. Take any $X \in \mathfrak{g}$ and $v,w \in V$. Then we want to show that

$$\langle d(X)v,w\rangle = - \langle v,d(X)w\rangle.$$ Now $e^{tX} \in G$ for all $t \in \Bbb{R}$ and so

$$\langle \mathcal{D}(e^{tX})v,\mathcal{D}(e^{tX})w \rangle = \langle v,w\rangle.$$

Since every bilinear form $B(v,w)$ on a finite dimensional vector space is given as

$$B(v,w) = v\cdot Aw$$

for some $\dim V \times \dim V$ matrix $A$ (where the $\cdot$ denotes scalar dot product) we can say that

$$\mathcal{D}(e^{tX})v \cdot A\left( \mathcal{D}(e^{tX})w \right) = v \cdot Aw$$

for some suitable matrix $A$. The secret now is to differentiate both sides with respect to $t$ and set $t = 0$ to get

$$\mathcal{D}(e^{tX})v \cdot \frac{d}{dt} A\left( \mathcal{D}(e^{tX})w \right)\bigg|_{t=0} + A\left( \mathcal{D}(e^{tX})w \right) \cdot \frac{d}{dt}\mathcal{D}(e^{tX})v \bigg|_{t=0} =\\ \hspace{1in} \mathcal{D}(e^{tX})v \cdot A\left( \frac{d}{dt}\left(\mathcal{D}(e^{tX})w\right) \right)\bigg|_{t=0} + A\left( \mathcal{D}(e^{tX})w \right) \cdot \frac{d}{dt}\mathcal{D}(e^{tX})v \bigg|_{t=0}=0.$$

We now use the identity $\frac{d}{dt}\mathcal{D}(e^{tX})\bigg|_{t=0}= d(X)$ and the relationship between $G$ and $\mathfrak{g}$ to get

$$v \cdot A(d(X)w) + A(w) \cdot d(X)v = v \cdot A(d(X)w) + d(X)v \cdot A(w)$$ and so $$\langle v,d(X)w \rangle + \langle d(X)v,w \rangle = 0$$

as desired.

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