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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.
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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 $A_i \otimes v_i$ ?
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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 $A_i \wedge B_j \otimes v_i \otimes w_j$ ?
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$A$ 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 $Tr(A \wedge B) = Tr(B \wedge A)$ ?
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Is $A \wedge A \wedge A \wedge A = 0$ ? for $A$ being a vector bundle valued $1$-form or is only $Tr(A \wedge A \wedge A \wedge A) = 0$ ?
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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.
Best Answer
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):
EDIT
I'll respond to your comments in order.