# 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.

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I assume by a vector valued form you mean a section of some $\Lambda^iT^*_M\otimes \mathcal{E}$ where $\mathcal{E}$ is a trivial vector bundle on $M$ with fiber a vector space $E$, say. So for your question to make sense, it seems to me $E$ needs to be equipped with a trace map $\mathrm{Tr}:E\to \mathbf{R}$. This will be the case, for example, when $E$ is the Lie algebra of a Lie subgroup $G\subset \mathrm{GL}_n(\mathbf{R})$. In such a case (and in any reasonable case I can imagine) the trace map will be linear, and so the argument in Mariano's answer works. It's easy to write in coordinates – Sam Lichtenstein Aug 25 '10 at 3:14
(cont.) because the trace is all happening on $\mathcal{E}$. For example, if $e_{ij}$ is the standard basis for $\mathfrak{gl}_n$ and if $x_k$ are coordinates on your manifold, a $\mathfrak{gl}_n$-valued $n$-form locally looks like $\sum f_{ijk_1\ldots k_n} e_{ij}dx_{k_1}\wedge\cdots\wedge dx_{k_n}$, and the trace is then locally the (usual) $n$-form $\sum f_{iik_1\ldots k_n} dx_{k_1}\wedge\cdots dx_{k_n}$. – Sam Lichtenstein Aug 25 '10 at 3:18
@Sam Thanks for this answer. Can you tell me what is the domain and range space of the Tr map of a vector valued k-form? If $\omega$ is such a form then at the point $p \in M$ $\omega$ is mapping $T_pM \times T_pM \times..k-times..T_pM \rightarrow E$. Now what is the $Tr(\omega)$ map? Your coordinate expression seems to indicate that $Tr(\omega)$ is an ordinary k-form? – Anirbit Aug 25 '10 at 6:26
@Sam I guess the rank of the form $k$, the rank of the bundle and the dimension of the manifold are independent quantities. – Anirbit Aug 25 '10 at 6:34
In those terms, $Tr(\omega)$ should (I think) just be the composition of $\omega_p:T_pM\times\cdots\times T_pM\to E$ with the map $Tr:E\to \mathbf{R}$. So yeah, it's an ordinary form. – Sam Lichtenstein Aug 25 '10 at 17:04

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.

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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 $$d\,\mathrm{Tr}(A)=\mathrm{Tr}\,dA.$$
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Thanks a lot for your help. I wanted to understand how for a vector valued differential form $Tr$ is defined? What is say the expression of $Tr(A)$ in coordinates? – Anirbit Aug 24 '10 at 12:44
Well, it depends on what $A$ is! If you edit your question to explain precisely what $A$ and $B$ are, someone might be able to explain. – Mariano Suárez-Alvarez Aug 24 '10 at 13:06
I was thinking of this in the context of A being a Lie-algebra valued 1-form on the Lie Group. For how general a case is $Tr(A)$ definable? – Anirbit Aug 24 '10 at 18:17
@Anirbit, please edit the question and fill in all the details to make sense of it. Make it easy for people to answer your question! :) – Mariano Suárez-Alvarez Aug 24 '10 at 19:42
I have further elaborated in the question what precisely I am stuck with. – Anirbit Aug 24 '10 at 19:58