Definition: Let $\mathfrak{g}$ be a Lie algebra and $(\pi ,V)$ be a representation of $\mathfrak{g}$ on $V$. A bilinear form $B$ on $V$ is said to be invariant under $\pi$ if for any $x\in \mathfrak{g}$ and any $y,z\in V$, one has the following:$$B(\pi (x)y,z)+B(y,\pi (x)z)=0.$$
Now let $\mathfrak{g}$ and $\pi$ be as above and let $V=\mathfrak{g}$. Consider the bilinear form $B_\pi$ on $\mathfrak{g}$ as follows:$$B_\pi (x,y)=\operatorname{Trace}(\pi (x)\pi (y)),\quad x,y\in \mathfrak{g}.$$
Claim: $B_\pi$ is invariant under $\pi$.
Need:$$\operatorname{Trace}(\pi (\pi (x)y)\pi (z))+\operatorname{Trace}(\pi (y)\pi (\pi (x)z))=0,\qquad \forall x,y,z\in \mathfrak{g}.$$
But I can't see why it has to be true. Any help in this regard would be greatly appreciated.
Thanks for your time.
We have $B_{\rho}(x,y)={\rm tr} (\rho(x)\rho(y)) = {\rm tr} (\rho(y)\rho(x))=B_{\rho}(y,x)$. The form is ad-invariant for the Lie bracket, because we have \begin{align*} B_{\rho}([x,y],z) & = {\rm tr} (\rho([x,y])\rho(z)) \\ & = {\rm tr}(\rho(x)\rho(y)\rho(z))-{\rm tr} (\rho(y)\rho(x)\rho(z)) \\ & = {\rm tr}(\rho(y)\rho(z)\rho(x))-{\rm tr} (\rho(y)\rho(x)\rho(z)) \\ & = {\rm tr} (\rho(y) \rho ([z,x])) \\ & = - B_{\rho}(y,[x,z]). \end{align*} Since $V$ is invariant, we have $\rho(x)y\in V$, so that $\rho$ is not defined on $\rho(x)y\in V$, if not $V=\mathfrak{g}$.