= Vanishing odd complete contractions of an antisymmetric matrix
{title2=$\operatorname{tr}(A^{2j+1})=0,\quad A^T=-A$}
For a real <antisymmetric matrix>, <transpose> gives $\operatorname{tr}(A^{2j+1})=\operatorname{tr}((-A)^{2j+1})=-\operatorname{tr}(A^{2j+1})$, so this <trace> vanishes. Every complete pairwise <tensor contraction> of copies of $A$, using the Euclidean <Kronecker delta>, splits into closed index cycles. Each cycle is a trace of a power of $A$, up to a sign from index order. An odd total number of copies forces an odd-length cycle, so the whole contraction vanishes. This assertion concerns pairwise index contraction with the metric; inserting additional tensors would be a different claim.
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