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Consider a general skew-symmetric $(n+1)\times (n+1)$ matrix $Z$, and let su map $Z$ to the point of $\mathbb{P}^n$ determined by $[pf_0(Z):\dots:pf_n(Z)]$ where the $pf_i(Z)$ are the principal Pfaffians of order $n$ of $Z$.

Is the closure of the image in $\mathbb{P}^n$ of this rational map a known variety?

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  • $\begingroup$ What is the domain of the map? $\endgroup$ – Sasha Mar 18 '18 at 20:56
  • $\begingroup$ The projective space $\mathbb{P}(\bigwedge^{2}V^{n+1})\cong \mathbb{P}^{\binom{n+1}{2}-1}$ parametrizing $(n+1)\times (n+1)$ skew-symmetric matrices. $\endgroup$ – J_Cole Mar 18 '18 at 21:39
  • $\begingroup$ One easy observation is that the map is $PGL(n+1)$-equivariant, hence the image is $PGL(n+1)$-invariant, so there is no other possibility except for $P^n$ itself. $\endgroup$ – Sasha Mar 19 '18 at 7:31
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At least in characteristic zero, the image of this map is all of $\mathbb P^n$. In finite characteristic, I am not sufficiently familiar with Pfaffians to say whether the argument carries through.

Indeed, start by noting that the map is only well-defined for $Z$ of rank $n$, since otherwise all principal Pfaffians of rank $n$ vanish. For $Z$ of maximal rank, consider instead the map that sends $Z$ to $\left[(-1)^0pf_0(Z):\dotsm:(-1)^npf_n(Z)\right]$, which agrees with your map up to an automorphism of $\mathbb P^n$. This map sends $Z$ to its kernel, and every line arises as the kernel of some skew-symmetric matrix.

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  • $\begingroup$ You are completely right. In fact, I was interested in the case $n$ even. Thank you very much. $\endgroup$ – J_Cole Mar 18 '18 at 21:41

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