If I have a singular matrix $X$ with components $X_{\mu\nu}$:
$t^{\nu}X_{\mu\nu}=0$
By considering now $X_{\mu\nu}$'s as components of a 2-form can I say that:
$X\wedge X=0$ ?
If yes, how?
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1 Answer
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Your condition on $X$ is that it has a kernel, and that by itself does not mean that $X \wedge X$ doesn't have to vanish. For instance in five dimensions, you could have $$X = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\\\ -1 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 \\\\ 0 & 0 & -1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}.$$ Then letting $t = (0,0,0,0,1)$, your condition is satisfied, but $X \wedge X$ is not zero.
However, it is true in $2n$ dimensions that a non-zero $t$ exists if and only if $X^{\wedge n} = 0$. That's because $X^{\wedge n}$ is proportional to the Pfaffian of $X$, which is a certain square root of the determinant of $X$. (In odd dimensions, the determinant of an antisymmetric matrix is zero by calculation, while the Pfaffian is set to zero by definition. So $t$ always exists in this case.)
answered Dec 23, 2009 at 16:57
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2$\begingroup$ Typo? You meant $X^{\wedge n}$, right? $\endgroup$– t3sujiCommented Dec 23, 2009 at 17:33
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$\begingroup$ Yes, thanks t3suji! This is the second time that you have improved my answer. $\endgroup$ Commented Dec 23, 2009 at 18:41
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$\begingroup$ thank you very much, I felt that what I was calculating
$X\wedge X$looks like a Pfaffian but I did not see the relationship. Now I can continue my study of the Plebanski action :-) $\endgroup$– PedroCommented Dec 23, 2009 at 21:50