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arxiv: 1601.08115 · v1 · pith:MPPG65UDnew · submitted 2016-01-29 · 🧮 math.AG · math.CO

A geometric approach to alternating k-linear forms

classification 🧮 math.AG math.CO
keywords uparrowmathcalthenlinearalternatingemptysetleastmathbb
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Given an $n$-dimensional vector space $V$ over a field $\mathbb K$, let $2\leq k < n$. There is a natural correspondence between the alternating $k$-linear forms $\varphi$ of $V$ and the linear functionals $f$ of $\bigwedge^kV$. Let $\varepsilon_k:{\mathcal G}_k(V)\rightarrow {\mathrm{PG}}(\bigwedge^kV)$ be the Plucker embedding of the $k$-Grassmannian ${\mathcal G}_k(V)$ of $V$. Then $\varepsilon_k^{-1}(\ker(f)\cap\varepsilon_k(\mathcal{G}_k(V)))$ is a hyperplane of the point-line geometry ${\mathcal G}_k(V)$. All hyperplanes of ${\mathcal G}_k(V)$ can be obtained in this way. For a hyperplane $H$ of ${\mathcal G}_k(V)$, let $R^\uparrow(H)$ be the subspace of ${\mathcal G}_{k-1}(V)$ formed by the $(k-1)$-subspaces $A\subset V$ such that $H$ contains all $k$-subspaces that contain $A$. In other words, if $\varphi$ is the (unique modulo a scalar) alternating $k$-linear form defining $H$, then the elements of $R^\uparrow(H)$ are the $(k-1)$-subspaces $A = \langle a_1,\ldots, a_{k-1}\rangle$ of $V$ such that $\varphi(a_1,\ldots, a_{k-1},x) = 0$ for all $x\in V$. When $n-k$ is even it might be that $R^\uparrow(H) = \emptyset$. When $n-k$ is odd, then $R^\uparrow(H) \neq \emptyset$, since every $(k-2)$-subspace of $V$ is contained in at least one member of $R^\uparrow(H)$. If every $(k-2)$-subspace of $V$ is contained in precisely one member of $R^\uparrow(H)$ we say that $R^\uparrow(H)$ is spread-like. In this paper we obtain some results on $R^\uparrow(H)$ which answer some open questions from the literature and suggest the conjecture that, if $n-k$ is even and at least $4$, then $R^\uparrow(H) \neq \emptyset$ but for one exception with ${\mathbb K}\leq{\mathbb R}$ and $(n,k) = (7,3)$, while if $n-k$ is odd and at least $5$ then $R^\uparrow(H)$ is never spread-like.

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