[Math] Taking trace of vector valued differential forms

differential-geometry

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.

Best Answer

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 $$d\,\mathrm{Tr}(A)=\mathrm{Tr}\,dA.$$

Related Solutions

[Math] Gauge transformations in differential forms

The method is to use the Leibniz rule in the differentiation and and change the sign whenever the exterior derivative moves past an odd form. In addition the following identity must be used $ dgg^{-1} + g dg^{-1} = 0$, and remember that the commutator is is between odd forms thus it is with a plus sign.

Here are the intermediate results:

$\frac{1}{2}[A_g, A_g] = \frac{1}{2}g[A, A] g^{-1} -[dgg^{-1}, g A g^{-1}] + dgg^{-1}\wedge dgg^{-1}$

$dA_g = g dA g^{-1} + [dgg^{-1}, g A g^{-1}] - dgg^{-1}\wedge dgg^{-1}$

Here are the required details:

$ d(gAg^{-1})$

Application of the Leibniz rule (Please observe the minus sign in the last term)

$ d(gAg^{-1}) = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1}$

Using the identities $g g^{-1} = 1$ in the first term and $ dg^{-1} = - g^{-1}dg g^{-1} $ in the last term

$ = dg g^{-1} g \wedge A g^{-1} + g dA g^{-1} +g A g^{-1} \wedge dg g^{-1} $

Collection of the first and last term into a commutator:

$ = g dA g^{-1} +[ dg g^{-1}, g A g^{-1} ]$

$ d(dg g^{-1})$

Application of the Leibniz rule (Please observe the minus sign in the last term)

$ d(dg g^{-1}) = ddg g^{-1} - dg \wedge dg^{-1}$

Using the identities $dd = 0$ and again $ dg^{-1} = - g^{-1}dg g^{-1} $, we obtain:

$ d(dg g^{-1}) = + dg g^{-1}\wedge dg g^{-1}$

[Math] Some questions about differential forms

I've seen some of this stuff before, but I'm definitely not an expert. Here's my take (although I'm likely wrong on at least some of it, hopefully not too much):

You can think of this as a matrix of 1-forms. Then, $A \wedge A$ means "matrix multiplication, where multiplication is wedge product".
Yes, I think $A_i v_i$ really means $A_i \otimes v_i$.
I think also $A_i \wedge B_j v_i \otimes w_j$ means $A_i \wedge B_j \otimes v_i \otimes w_j$, although there might be some issue of skew-symmetrization.
A vector-bundle-valued differential k-form locally looks like $\omega \otimes s$, where $\omega$ is a usual k-form and $s$ is a section. Then $d$ should act by the Leibniz rule (once you've fixed a connection). I'd imagine that the trace is just trace in the usual sense, and then it's obvious that this commutes with $d$ and that $Tr$ admits cyclic permutations of the factors in its argument.
I think that whether $A\wedge A\wedge A\wedge A$ must be zero depends on the dimension of the vector bundle where it takes its values. If for example it's the constant 1-dimensional bundle, then this is no different from usual 1-forms.
Your last question is a little unclear, sorry. Could you clarify exactly what you're referring to?

EDIT

I'll respond to your comments in order.

If $A$ takes a vector field and spits out a matrix, we can consider each of the entries of the matrix (in a fixed basis, of course) as an ordinary 1-form. If you want to wedge two matrix-valued forms, you just write out the matrices, act like you're multiplying matrices like you learned a long time ago (i.e. take the dot product of rows with columns), but then when you want to simplify the expression, you consider the product to be the wedge. So, e.g. if $$ A = \begin{pmatrix}a&b\\c&d\end{pmatrix} $$ is a 2x2 matrix-valued 1-form, this means that $a$, $b$, $c$, and $d$ are just ordinary 1-forms. So evaluating on a vector $v$ gives the 2x2 matrix (of numbers) $$ A(v) = \begin{pmatrix} a(v) & b(v) \\c(v) & d(v) \end{pmatrix}. $$ If you want to calculate $A\wedge A$, this is the 2x2 matrix-valued 2-form $$ A\wedge A = \begin{pmatrix}a&b\\c&d\end{pmatrix} \wedge \begin{pmatrix}a&b\\c&d\end{pmatrix} $$ $$ = \begin{pmatrix} a\wedge a + b\wedge c & a\wedge b + b\wedge d \\ c\wedge a + d\wedge c & c\wedge b+d\wedge d\end{pmatrix} $$ $$ = \begin{pmatrix} b\wedge c & b\wedge (d-a) \\c\wedge (a-d) & c\wedge b \end{pmatrix}. $$
This should be more or less answered by (1). The decomposition of the form $\omega \otimes s$ is just saying that a vector-bundle-valued form can be thought of as an ordinary form (that's $\omega$), except that you need to decide which direction in the fiber it takes you (that's $s$). This specializes to the case where the bundle is the trivial $\mathbb{R}$-bundle. Then $s$ is actually just a function, and we're recovering the fact that $n$-forms are a module over $C^\infty(M,\mathbb{R})$.
Like you said in your first comment, a connection takes in tangent vectors to the total space. From here, this is exactly what I said in (2).
Well, any form is nilpotent -- if you wedge it with itself enough times, you eventually get 0 for dimensional reasons. (This assumes that the (co)tangent space is finite-dimensional, of course. I have no idea what happens in the case that it isn't.) The only obvious reason (to me) that $A\wedge A\wedge A\wedge A$ would be traceless is if it's 0. But then again, I don't know anything about Chern-Simons theory. Or maybe it's a trick along the lines of the proof that the commutator of two square matrices is always traceless (? I'm pretty sure that's true, and follows from the fact that $Tr(AB)=Tr(BA)$, which you should note does not imply that $Tr(ABC)=Tr(ACB)$, only that you can cyclically permute the matrices so e.g. $Tr(ABC)=Tr(CAB)=Tr(BCA)$).
All I know is that back in linear algebra, you learn that the trace of a matrix is invariant under basis change. In other words, we can have a well-defined notion of the trace of a linear map. Of course, this requires that the dimensions of the domain and codomain agree, which doesn't seem like it'll be true in the example you gave (of a vector-valued differential form). However, if they do agree, then (in a basis) you can consider a form to be matrix-valued, and then the trace is in all likelihood just the sum of the diagonal entries, which are themselves ordinary differential forms.

Best Answer

Related Solutions

[Math] Gauge transformations in differential forms

[Math] Some questions about differential forms

Related Question