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Let $A$ be a $n \times n$ matrix of 1-forms (for example, a connection form). Note that $A \wedge A$ is not $0$, but by using the anti-symmetry of the wedge product applied to the entries of $A$ we can check that $\text{Tr}(A \wedge A)=0$.

Let $\sigma_i(A)$ denote the $i$-th symmetric polynomial of the eigenvalues of $A$, so $\text{det}(I+tA)=1+t\sigma_1(A)+t^2\sigma_2(A)+...+t^n\sigma_n(A)$. My claim above is equivalent to saying that $\sigma_1(A \wedge A)=0$ for any matrix $A$ of 1-forms. My question is, is it true that $\sigma_i(A \wedge A)=0$ for all $i$?

Thanks!

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I figure out an answer to my own question: Note that $\sigma_i$ is an invariant polynomial, in the sense that for any invertible matrix $B$ we have $\sigma_i(B^{-1}AB)$; this follows since the determinant is invariant under similarity.

Let $A$ be a diagonal matrix of 1-forms, then $A \wedge A$ is a diagonal as well. Hence, $\sigma_i(A \wedge A)$ is just the $i-th$ elementary symmetric polynomial on the diagonal entries of $A \wedge A$. Now using the definition of the elementary symmetric polynomials and the anti-symmetry of the wedge product for 1-forms we get that $\sigma_i(A \wedge A)=0$.

Let $B$ is a invertible matrix of 1-forms and consider $B^{-1} \wedge A \wedge B$. By the invariance of $\sigma_i$ under similarity we get

$\sigma_i(A \wedge A)= \sigma_i(B^{-1} \wedge (A \wedge A) \wedge B)= \sigma_i((B^{-1} \wedge A \wedge B) \wedge (B^{-1} \wedge A \wedge B))$.

Since any upper triangular with distinct diagonal entries is similar to a diagonal matrix, the claim is true for these upper triangular with distinct diagonal entries. By continuity, it is true for upper triangular matrix. Finally, since any square matrix is similar to an upper triangular matrix, the claim is true for any matrix.

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