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  • 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.

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    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 $\omega\wedge \omega \neq 0$.2010-10-03
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    @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.2010-10-03
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    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.2010-10-03
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    @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.2010-10-03
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    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.2010-10-03
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    I think it just takes a little time for the website to update that information.2010-10-04
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    Don't have time to write a full answer now, but: you shouldn't think $A$ as vector valued. You *should* think $A$ as taking value in an Algebra. In an algebra, the multiplication between two objects is defined (not the tensor product); whereas in a vector space it is not. (If you don't have a good concept of an algebra, think *matrices*. The set of all NxN matrices form an algebra. If an algebra is non commutative, then $A_iA_j - A_jA_i$ need not vanish.2010-10-19

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