Lie algebra (invariant)-action on bilinear form Let $(\pi, V)$ be a rational representation of $G$ a Lie algebra, and let $B$ be a bilinear form defined on $V$.
Then there is an action of $G$ on $B$ given by $$ g \cdot B(u,v) = B(g^{-1}u, g^{-1}v).$$
We can choose to write $B(u,v) = v^t \Gamma u$, so that succinctly $ g \cdot \Gamma = g^{-1} \Gamma g^{-1}$.
I want to show that the action of $X \in \mathfrak{g}$ on $B$ is given by $$ X \cdot B(u,v) = - B(d\pi(X)u,v) - B(u, d\pi(X)v).$$
This form looks related to the fact that the differential should be a derivation, but I am having trouble making this clear (this might be related to my other question here).
Otherwise, as in another of my questions, I am familiar with the definition of the differential via both the exponential map and vector fields, however I am unsure how to apply it here to obtain the desired action on the bilinear form.
Furthermore, is it true that if $B$ is $G$-invariant then it is also $\mathfrak{g}$-invariant? And/or if this is true/not true, is it clear from the solution to my first question?
 A: To be clear, the action of $G$ is on the vector space of all bilinear forms
$$ \operatorname{Bil}(V) = \{B \colon V \times V \to \mathbb{C} \mid B \text{ bilinear}\}.$$
For a hand-wavy derivation of the action of $\mathfrak{g}$, suppose that $G$ is embedded inside $\operatorname{GL}_n$, so that we can consider a small perturbation $\pi(I + tX)$ for some $X \in \operatorname{Mat}_n$: by the definition of the differential we should have
$$ \pi(I + tX)(v) = v + t d \pi_I(X)(v) + O(t^2),$$
the term with the $t$ in front gives us the action of $X \in \mathfrak{g}$ on $V$, namely $X \cdot v = d \pi_I(X) v$.
Now let $\rho(g)(B)(u, v) = B(\pi(g^{-1})u, \pi(g^{-1})v)$, so considering a perturbation again we get
$$\begin{aligned} (\rho(I + tX)B)(u, v) 
&= B(\pi(I + tX){^-1} u, \pi(I + tX)^{-1} v) \\
&= B(\pi(I - tX + O(t^2))u, \pi(I - tX + O(t^2))v) \\
&= B(u - t d\pi_I Xu + O(t^2), v - t d\pi_I Xv + O(t^2)) \\
&= B(u, v) - t B(d\pi_I X u, v) - t B(u, d \pi_I X v) + O(t^2).\end{aligned}$$
Here we have approximated $(I + tX)^{-1} = I - tX + t^2 X^2 - t^3 X^3 + \cdots$ by $I - tX + O(t^2)$, and the last inequality uses the bilinearity of $B$. So we find that the action of $X \in \mathfrak{g}$ is $$(X \cdot B)(u, v) = - B(d\pi_I(X) u, v) - B(u, d \pi_I(X), v).$$
Any "real" proof of this result is essentially upgrading this simplified result to be more rigorous (in fact for affine algebraic groups, this is basically the real proof, once one knows how to interpret $t^2 = 0$ appropriately). But if you are working over differentiable manifolds you can make this rigorous using your definition of the differential.
Another way to see this is to first understand how the Lie algebra acts on the dual $V^*$, and on tensor products; then one uses the canonical isomorphism $V^* \otimes V^* \to \operatorname{Bil}(V)$ taking $\varphi \otimes \psi$ to the bilinear form $(u, v) \mapsto \varphi(u) \psi(v)$. Of course then we have that
$$ X \cdot (\varphi \otimes \psi) = -d\pi_I(X)^t \varphi \otimes \psi - \varphi \otimes d\pi_I(X)^t \psi,$$
which is the same bilinear form.
