Taking trace of vector valued differential forms

Can anyone kindly give some reference on taking trace of vector valued differential forms?

Like if $A$ and$B$ are two vector valued forms then I want to understand how/why this equation is true?

$dTr(A\wedge B) = Tr(dA\wedge B) – Tr(A\wedge dB)$

One particular case in which I am interested in will be when $A$ is a Lie Algebra valued one-form on some 3-manifold. Then I would like to know what is the precise meaning/definition of $Tr(A)$ or $Tr(A\wedge dA)$ or $Tr(A\wedge A \wedge A)$?

In how general a situation is a trace of a vector valued differential form defined?

It would be great if someone can give a local coordinate expression for such traces.

Any references to learn this would be of great help.

Solutions Collecting From Web of "Taking trace of vector valued differential forms"

I do not know what you mean by vector-valued form, exactly. But your equation follows from two facts:

  • $d$ satisfies the (graded) Leinbiz equation $$d(A\wedge B) = dA\wedge B + (-1)^{\text{degree of $A$}}A\wedge dB,$$
  • and $\mathrm{Tr}$ is linear and commutes with $d$, so that $$d\,\mathrm{Tr}(A)=\mathrm{Tr}\,dA.$$

You can define a trace on any vector space $V$ where there is a representation of $V$ on some other space $W$ just by picking a basis on $W$, defining the trace on $V$ as matrix trace (every element of $V$ becomes a matrix with respect to the basis of $W$) and proving that under a change of bases, the trace stays the same.

Now what to do in the case of $V$-valued differential forms? First, let us assume that there is a representation of $V$ on some vector space. Without the assumption there is no meaning of a trace, I believe. And at least for finite dimensional Lie algebras, there always is a representation, the adjoint representation.
So we have a linear map $\operatorname{tr}: V \to \mathbb{R}$.

Recall that a $V$-valued differential form on $M$ is a smooth map $\omega : TM \to V$ such that $\omega$ restricted to any tangent space $T_p M$ is an element of the $V$-valued exterior algebra $\Lambda^n (T_p M, V)$ of $T_p M$.
That is, the restriction $\omega_p$ is a completely antisymmetric map $\omega_p : T_p M \times T_p M \times \cdots \times T_p M \to V$.

By $\operatorname{tr}(\omega)$, we just mean the composition $\operatorname{tr} \circ \omega$. We just feed whatever the differential form gives us into the trace operation. It is a real valued differential form.

Now, if you also have a multiplication defined on $V$, as will be the case if there is a representation (just ordinary matrix multiplication), you can also define the wedge product $\wedge$ analogously to the real-valued case, just inserting the $V$-multiplication instead of the ordinary scalar multiplication.
As Mariano already explained, it satisfies the Leibniz equation.
Mariano also explained that tr is linear and therefore we can pull the $-$ through the trace.

To your special cases: Be careful with Lie algebra valued differential forms! There are at least two possible $\wedge$ products, depending on whether you define it upon multiplication in the adjoint representation or the Lie bracket! The difference is normally only is a factor, but still one should be clear about what $\wedge$ one uses. So please clarify this.