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!
 A: 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.
