• If $A$ is a differential one form then $A\wedge A .. (more\text{ }than\text{ }2\text{ }times) = 0$ Then how does the $A\wedge A \wedge A$ make sense in the Chern-Simon's form, $Tr(A\wedge dA + \frac{2}{3} A \wedge A \wedge A)$ ?

    I guess this anti-commutative nature of wedge product does not work for Lie algebra valued one-forms since tensoring two vectors is not commutative.

  • If the vector space in which the vector valued differential form is taking values is $V$ with a chosen basis ${v_i}$ then books use the notation of $A = \Sigma _{i} A_i v_i$ where the sum is over the dimension of the vector space and $A_i$ are ordinary forms of the same rank.

I would like to know whether this $A_i v_i$ is just a notation for $A_i \otimes v_i$ ?

  • Similarly say $B$ is a vector valued differential form taking values in $W$ with a chosen basis ${w_i}$ and in the same notation, $B = \Sigma _j B_j w_j$. Then the notation used is that, $A \wedge B = \Sigma _{i,j} A_i \wedge B_j v_i \otimes w_j$

    I wonder if in the above $A_i \wedge B_j v_i \otimes w_j$ is just a notation for $A_i \wedge B_j \otimes v_i \otimes w_j$ ?

  • $A$ and $B$ are vector bundle valued differential forms (like say the connection-1-form $\omega$ or the curvature-2-form $\omega$) then how is $Tr(A)$ defined and why is $d(Tr A) = Tr( d A)$ and $Tr(A \wedge B) = Tr(B \wedge A)$ ?

  • Is $A \wedge A \wedge A \wedge A = 0$ ? for $A$ being a vector bundle valued $1$-form or is only $Tr(A \wedge A \wedge A \wedge A) = 0$ ?

  • If A and B are two vector bundle valued $k$ and $l$ form respectively then one defines $[A,B]$ as , $[A,B] (X_1,..,X_{k+l}) = \frac{1}{(k+l)!} \Sigma _{\sigma \in S_n} (sgn \sigma) [A (X_{\sigma(1)},X_{\sigma(2)},..,X_{\sigma(k)}) , B (X_{\sigma(k+1)},X_{\sigma(k+2)},..,X_{\sigma(k+l)})]$

    This means that if say $k=1$ then $[A,A] (X,Y) = [A(X),A(Y)]$ and $[A,A] = 2A \wedge A$. The Cartan structure equation states that, $d\Omega = \Omega \wedge \omega - \omega \wedge \Omega$.

    But some people write this as, $d\Omega = [\Omega,\omega]$.

    This is not clear to me. Because if the above were to be taken as a definition of the $[,]$ then clearly $[A,A]=0$ contradicting what was earlier derived.

  • 4
    $\begingroup$ The equation $A\wedge A = 0$ is not always true. For example, consider a 4 dimensional vector space $V$ with basis $e_i$. Then the form $A = e_1\wedge e_2$ + $e_3\wedge e_4$ does not satisfy $A\wedge A = 0$. If you'd like a differential form version, the de Rham cohomology ring of $\mathbb{C}P^2$ is $\mathbb{R}[\omega]/\omega^3 = 0$, where $\omega$ is (represented by) a certain 2 form (the Poincare dual to the canonical $\mathbb{C}P^1$ in $\mathbb{C}P^2$, if you wish). So, $\omega$ is a 2-form that satisfies $\omega\wedge \omega \neq 0$. $\endgroup$ – Jason DeVito Oct 3 '10 at 19:42
  • $\begingroup$ @Jason The $A$ you are considering is an ordinary $2$ form and hence obviously $A \wedge A$ is not 0. But for ordinary $1$-form A, $A\wedge A = 0$. Anyway I have further put in some edits into the question. Would like to know your answers to them. Thanks for the reply. $\endgroup$ – Anirbit Oct 3 '10 at 19:58
  • $\begingroup$ Sorry, I missed the word "one" in "one form". I'll think about what your asking, but I'm not too terribly familiar with it. In particular, I have no idea what A is in the Chern Simons form. $\endgroup$ – Jason DeVito Oct 3 '10 at 20:04
  • $\begingroup$ @Jason $A$ is a Lie algebra valued $1$-form in the Chern-Simon's form. It is a connection on a principle G-bundle. It would be great if you can help me with this notion of "Trace" and its various consequences as I have explained in the question. Waiting for your responses. $\endgroup$ – Anirbit Oct 3 '10 at 20:21
  • 4
    $\begingroup$ Small point but you've asked 10 questions without accepting the answers to any of them. I suspect you'd get more participation in your threads if you read the responses and accepted some answers. $\endgroup$ – Ryan Budney Oct 3 '10 at 20:57

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?


I'll respond to your comments in order.

  1. 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 $$ A\wedge A = \begin{pmatrix}a&b\\c&d\end{pmatrix} \wedge \begin{pmatrix}a&b\\c&d\end{pmatrix} $$ $$ = \begin{pmatrix} a\wedge a + b\wedge c & a\wedge b + b\wedge d \\ c\wedge a + d\wedge c & c\wedge b+d\wedge d\end{pmatrix} $$ $$ = \begin{pmatrix} b\wedge c & b\wedge (d-a) \\c\wedge (a-d) & c\wedge b \end{pmatrix}. $$
  2. 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 $C^\infty(M,\mathbb{R})$.
  3. 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).
  4. 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. $Tr(ABC)=Tr(CAB)=Tr(BCA)$).
  5. 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.
  • $\begingroup$ @Aaron Thanks for your reply. If I am working on a vector bundle where $X$ is some chosen vector field and ${s^i}$ are a basis of sections then the connection one-forms $A ^i _j$ can be defined as maps taking vector fields $X$ to numbers $A^ i _j (X)$ satisfying $\nabla _X s^i = A^ i _j (X) s^j$. Then one says that the connection form $A$ is the ``matrix of ordinary 1-forms" whose entries are $A^i _j$ . But I don't know how to interpret $A$ itself as a map. The wedge product as I understand it is well-defined between ordinary forms. How do you defined that between such matrix of 1-forms? $\endgroup$ – Anirbit Oct 4 '10 at 8:31
  • $\begingroup$ @Aaron For the Chern-Simon's theory $A$ is a connection on a principle $G$-bundle. There $A$ is a composition of two maps (pointwise), first projecting the tangent space of the total space to the vertical subspace and then mapping it to the Lie Algebra of the structure group. Given such a map how do you defined a wedge product between them? Can such maps also be thought of as a matrix of ordinary forms? I don't understand your $\omega \otimes s$ decomposition. $\endgroup$ – Anirbit Oct 4 '10 at 8:36
  • $\begingroup$ @Aaron Using the fact that $Hom(V,W) = V^* \otimes W$ one can write a connection on a principle $G$-bundle as $\Sigma _a \omega ^a \otimes T^a$ (sum over dimension of the Lie Algebra) where $omega _a$ are ordinary-forms on the total space (?) and T^a are a basis of the Lie Algebra. How does this relate to your way of decomposing the connection? $\endgroup$ – Anirbit Oct 4 '10 at 8:38
  • $\begingroup$ @Aaron. Let me explain where the wedge of 4-A's came from. Given a connection $A$ on the G-bundle one defines its curvature as $F = dA + A\wedge A$ and then one wants to show that $Tr(F\wedge F) = d(Chern-Simon's\text{ }Form)$ I seem to get the LHS and the RHS differing by $Tr(A \wedge A \wedge A \wedge A)$ and hence I am guessing that it is 0. $\endgroup$ – Anirbit Oct 4 '10 at 8:42
  • $\begingroup$ @Aaron May be you can make the notion of $Tr$ a little more explicit. Like given a rank k vector valued differential form say $\Omega$ what is $Tr(\Omega)$. $Tr(\Omega)$ has to be an ordinary form mapping a k-fold product of the domain vector space to a number and we know that $\Omega$ maps a k-fold product of the domain vector space to an element of the target vector space. How does trace manage to generate a number out of this? $\endgroup$ – Anirbit Oct 4 '10 at 8:45

Let me note that http://en.wikipedia.org/wiki/Vector-valued_form seems to be highly relevant. They essentially define the bracket. Note that in the case of a 1-form A valued in $\mathfrak{g}$ you get $[A,A]=[A\wedge A (v \otimes w)]= A(v)\otimes A(w)- A(w)\otimes A(v)$, which has value in $\mathfrak{g} \otimes \mathfrak{g}$. To combine them you use the bracket in $\mathfrak{g}$, so you get $2 [A(x), A(y)]$.

Your definition for $[A, B]$ has a factor of $\frac{1}{(k+l)!}$, which in the case of your vector bundle being over a point does not reproduce the bracket on $\mathfrak{g}$. Maybe it should be $\frac{1}{(k!l!)}$, as in Wikipedia?

Edit: Mmm, this was nonsense. Over a point you'd be doing k=l=0, so it would not matter. Anyhow, Wikipedia definition seems to give constants more consistent with the rest. But I can't do arithmetic, so no guarantees.

On the other hand, as suggested by Aaron, you can view A as a matrix of 1-forms, and wedge-multiply the matrices. Then $A\wedge A (x, y) = A(x)A(y)-A(y)A(x)= [A(x), A(y)]$ and indeed, $2 A\wedge A= [A,A]$.

I am a bit confused by this wedge notation you give in the question. $A_i$ and $B_j$ differential forms - components of A and B. We then tensor things up, using wedge instead of product, getting a vector in forms valued in the tensor product. We would then need a map $V \otimes V=\mathfrak{g} \otimes \mathfrak{g}$ to $V=\mathfrak{g}$ to get a form with values in $V$ again. Presumably this is the Lie bracket on V, and the result is the 'wedge multiplication of matrices' as above. The 'issue with skew symmetrization' is that this wedge operation is not the bracket operation on forms, but it's less skewsymmetrized version.

The Cartan equation should work out fine. We have 2-form $\Omega$ and 1-form $\omega$. Their bracket on the triple $a,b,c$ is a sum of 6 terms, with a factor of 1/2 in front. These 6 terms pair up into 3 pairs of 2 equal terms each(as $\Omega$ is antisymmetric). So you get 3 terms with coefficient 1, which we then bracket. On the other hand, wedge multiplying the matrices for $\Omega$ and $\omega$ in different order results in same 3 bracket terms (wedging 2 form and 1 form produces 3 distinct terms). One just needs to write down the formulas. (I'm sure there is a general formula for bracket in terms of antisymmetrization of the wedge).

Maybe another point is that $[A,A]=0$ for 'ordinary' $\mathbb{R}$ valued forms is a consequence of the Lie bracket on $\mathbb{R}$ being trivial.

Now, trace is evidently defined as before - viewing A as a matrix of forms, trace is the summ of the diagonal forms. It is then clear why it commutes with the differential. However, $Tr(A\wedge B) = (-1)^{deg A \times deg B} Tr(B\wedge A)$. This follows from the usual formulas for trace and commuting the forms $A_{i,j}$ past $B_{j,k}$.

Finally, the trace of $A\wedge A\wedge A\wedge A$. It is $\Sigma_i A^4_{ii}= \Sigma_i \Sigma_j A^2_{ji} \wedge A^2_{ij}$. $A^2_{kl}=\Sigma_m A_{km} \wedge A_{ml}$ is in general non-zero (just take A a 3 by 3 matrix with 0 in $A_{12}$ entry and $A_{2,3}\wedge A_{3,1} \wedge A_{1,3} \wedge A_{3,2}$ non 0). BUT, you are on a 3-manifold (or you'd be doing a differnt C-S form), so there is no 4-form at all!

  • $\begingroup$ @Max The issue you point out about wedge product of vector valued forms not producing an element in $g$ without another map that takes a k-fold tensor product of $g$ to $g$ in some canonical way is central to the whole confusion about defining "Trace" of vector valued forms. If one thinks in terms of matrices as Aaron suggested then it makes sense computationally without telling what this map is. You know of a good reference which explain the notion of trace (hopefully in some explicitly basis invariant way!) $\endgroup$ – Anirbit Oct 22 '10 at 5:39
  • $\begingroup$ @Max But then again it is not clear to me as to how one can think of any rank k vector valued form as a matrix of ordinary k-forms. The connection and the curvature form seem to be very special in the sense that they admit such a representation. $\endgroup$ – Anirbit Oct 22 '10 at 5:57
  • $\begingroup$ I'm not sure I understand what you are confused about, but here is an attempt: The forms you are considering are not just valued in a vector space. They are valued in a Lie algebra, which in most cases one thinks of as a matrix algebra - a subalgebra of GL(V) for some vector space. Each component of such matrix is a scalar valued form. The trace makes perfect sense, as for any matrix. And, indeed, for arbitrary vector bundle valued form there is no trace (there is a completely unrelated story in the presence of a metric on the base or some such, but that's not relevant here). $\endgroup$ – Max Oct 23 '10 at 17:39
  • $\begingroup$ Also, you don't need k fold tensor product of anything, just a map $\mathfrk{g} \otimes \mathfrk{g} \mapsto \mathfrk{g}$. Which in your context is pretty much always the Lie bracket (aka the commutator). $\endgroup$ – Max Oct 23 '10 at 17:40
  • $\begingroup$ @Max For defining a wedge product between k Lie Algebra valued 1-forms, won't you need a map from Lie(G)⊗Lie(G)⊗..k−times..⊗Lie(G)→Lie(G) It is not clear to me as to what this natural map is. Except for the connection and the curvature form, its not clear that any arbitrary lie-algebra valued k-form can be thought of as a matrix of ordinary forms so that the computations Aaron suggested can be done $\endgroup$ – Anirbit Oct 25 '10 at 6:48
  1. This statement does make sense. Simply take some linear combinations of the co-basis vectors (i.e., co-vectors that constitute a basis for the cotangent bundle) and calculate the exterior product. Note: It suffices to treat the case tangent bundle, however, you can also use a more general vector bundle.

  2. Certainly, the two notations you outlines aren't something different. This is due to the definition of the tensor product.

  3. You will have to regard the forms which are vector bundle valued, i.e., which are $\in\Gamma(Hom(V,V)) = \Gamma(V^{\star}⊗V)$, where V denotes your vector bundle. By choosing a local representation of such a section and using the definition of a fiber of a vector bundles, one btains that locally, one can write this as a matrix form. Since this however is possible everywhere, simpley choose a partition of unity. But no one actually calculates that explicitly.

  4. $\wedge_{i=0}^{4} A \neq 0$ in general. But the trace has to vanish. Simply write down this as an (antisymmetric) matrix, for which it is clear that the trace vanishes.

  5. What you have written is a completely unknown formulation of cartan's structure equation to me. I know the equation like this: $\Omega = d\omega + \dfrac{1}{2}[\omega,\omega]$


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