Some questions about differential forms

• 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.