This is a question for mathematicians. I am not asking how I can think of tr. I am literally only asking if my precise mathematical defintion is correct. Thats all.

Okay thank you very much for your comments. But still, this is not really my question. My question has nothing to do with Killing forms at all. I just wanted to know if the term "$\mathrm{tr}(B\wedge F[A])$" above is defined as $\mathrm{tr}(\alpha\wedge\beta)\vert_{U}:=\sum_{a,b=1}^{\mathrm{dim}(G)}(\alpha^{a}\wedge\beta^{b})\langle e_{a},e_{b}\rangle_{\mathrm{Ad}(P)}$, where $\langle\cdot,\cdot\rangle_{\mathrm{Ad}(P)}$ is the induced bundle metric from the given inner product $\langle\cdot,\cdot\rangle_{\mathfrak{g}}$. Thats all I want to know...

Sorry, maybe I should ask more precisely: I am asking if my precise definition of $\mathrm{tr}(\cdot\wedge\cdot):\Omega^{d-2}(\mathcal{M},\mathrm{Ad}(P))\times\Omega^{2}(\mathcal{M},\mathrm{Ad}(P))\to\Omega^{d}(\mathcal{M})$ above using the induced bundle metric, local frames, ect. is correct in this very general setting, where $G$ does not need to be a simple Lie group and hence $\langle\cdot,\cdot\rangle$ does not need to be a Killing form...

Yes, that is clear that $\langle\cdot,\cdot\rangle$ is proportional to the Killing form, but is the rest also correct?

Thank you very much for your answer and for providing some references!

Okay, great! Thank you again!

Okay, thank you very much for your answer. So, to understand correctly, $I_{j_{1}j_{2}}^{j_{3}}$ in your notation is the (unique up to multiple) intertwiner of the form $I_{j_{1}j_{2}}^{j_{3}}:V_{j_{3}}\to V_{j_{2}}\otimes V_{j_{2}}$ and hence, using your formula, the matrix coefficients of this intertwiner are exactly given by $(-1)^{-j_{1}+j_{2}+m_{3}}\sqrt{2j_{3}+1}$ times the $3j$-symbol?

Sorry, maybe I should add the definition of the Clebsch-Gordon coefficients...$J$ are the eigenvalues of the "total spin" appearing in the definition of the Clebsch-Gordon coefficients...