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Let $V$ be a vector space of dimension $>n$, and define the subset $$ K:=\{ ([\omega],v)\mid v\wedge\omega=0 \}\subset\mathbb{P}(\Lambda^nV)\times V\, . $$ Denote also by $\pi:K\longrightarrow \mathbb{P}(\Lambda^nV)$ the restriction of the canonical projection onto the first factor of $\mathbb{P}(\Lambda^nV)\times V$. Then, $\pi$ is a linear bundle over $\mathbb{P}(\Lambda^nV)$ with variable fibre dimension.

QUESTION 1: is it true that $$\{[\omega]\in \mathbb{P}(\Lambda^nV)\mid \dim\pi^{-1}([\omega])=n\}\quad\quad\quad (^*)$$ is precisely the Plucker image of the Grassmanniann $\mathrm{G}(n,V)$?

In the affirmative case, I'd like to know whether such a "bundle" $\pi$ ever appeared, perhaps in other guises, and ask the next

QUESTION 2: what if I replace $n$ with an arbitrary $i\neq n$ in the above $(^*)$? What do I obtain?

I'm expecting some subset of $\mathbb{P}(\Lambda^nV)$ intrinsically related to $\mathrm{G}(n,V)$, like its tangent/secant variety.

(I'm aware of another MO question, concerning the Plucker embedding and the tautological bundle, and another one, concerning its ring of functions, though neither can help here.)

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    $\begingroup$ The answer to your first question is yes (you can see this e.g. by taking taking a basis $v_i$ of $V$ with $v_1,v_2,\cdots v_n$ a basis of $\pi^{-1}([\omega])$ and writing $\omega$ in terms of the induced basis of $\wedge^nV.$) $\endgroup$ – dhy Jul 30 '15 at 6:04
  • $\begingroup$ Ok, let me try. If $[\omega]\in \mathrm{G}(n,V)$, then $\omega=v_1\wedge\cdots\wedge v_n$, and $\pi^{-1}([\omega])=\langle v_1,\ldots,v_n\rangle$, which is the easy implication. Conversely, knowing that $\dim \pi^{-1}([\omega])=n$, you choose a basis of $V$ such that $\{v_1,\ldots,v_n\}$ generates $\pi^{-1}([\omega])$: now you suggest to write down $\omega$ as a linear combination of $v_1\wedge\cdots\wedge v_n$ and other terms, and show that the coefficients in front of the latter must necessarily be zero. Yes, it should work, thanks! Do you have any clue about question #2? $\endgroup$ – Giovanni Moreno Jul 30 '15 at 17:13

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