Let $Q\subset \mathbb P^n$ ($n\geq 3$) be a smooth quadric. Let us denote $$Z=\{ [P]\in G(3,n+1), \exists l\subset Q\mathrm{\ line\ s.t.\ } 2l\subset Q\cap P\}$$ the locus of osculating planes. Is there a "nice" way to see $Z$ in $G(3,n+1)$, as, for example the zero locus of a vector bundleor something like that?
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$\begingroup$ You need to change the conditional of your set to $Q\cap P \supseteq 2l$, since the set you wrote will contain in its closure the set of $[P]$ such that $P$ is contained in $Q$ (once $n\geq 5$). $\endgroup$– Jason StarrCommented Aug 8, 2016 at 16:36
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$\begingroup$ The "corrected" set $Z$ is a symmetric determinantal variety as in Harris-Tu, at least if the characteristic is not $2$. If $S\subset \mathcal{O}^{\oplus (n+1)}$ denotes the universal (locally split) subbundle of rank $3$ on $G(3,n+1)$, then the quadratic form induces a symmetric homomorphism $f:S\to S^\vee$. The set $Z$ is the rank $\leq 1$ locus of $f$. $\endgroup$– Jason StarrCommented Aug 8, 2016 at 16:40
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This depends on your definition of "nice". On $G(3,n+1)$, we have a tautological subbundle $E\hookrightarrow \mathcal{O}_G^{n+1}$. The quadratic form $q$ defining $Q$ induces a quadratic form on $E$, and $Z$ is the locus where this form has rank $\leq 1$ (well, the closure of $Z$ -- this includes the planes which are contained in $Q$). This gives a codimension 3 locus in $G(3,n+1)$. It is not the zero set of a vector bundle, but it still has nice properties, see this paper of Harris and Tu.
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$\begingroup$ Thank you very much. Are there similar results for the locus where a trilinear symmetric form is of rank $\leq r$ (for a given $r$)? $\endgroup$– pi_1Commented Aug 12, 2016 at 13:41