Hi there! I have a very simple question, which requires an expert in multilinear algebra.

$V$ is an $n$-dimensional vector space, and $\omega\in V^\ast\wedge V^\ast$ is a skew-symmetric form on it. Then, mimicking symplectic geometry, I call isotropic a subspace $L\subseteq V$ such that $\omega|_L\equiv0$.

QUESTION. Let $L$ be an hyperplane in $V$, and $\alpha\in V^\ast$ a dual covector for $L$, i.e., such that $L=\ker\alpha$: which conditions on $\alpha$ correspond to the fact that $L$ is isotropic?

In other words, I'm trying to characterize the covectors $\alpha\in V^\ast$ such that $\omega(x,y)=0$ for all $x,y\in\ker\alpha$. My instinct says that the condition I'm looking for is

$\alpha\wedge\omega=0,\quad\quad (*)$

but my poor skills in multilinear algebra are not sufficient to prove this - that's why I'm seeking for advice.

My question ends here, though I have some curiosity on this matter, which might be answered by an expert. Namely, in symplectic geometry there is the notion of Lagrangian Grassmannian... so here there should be that of Isotropic Grassmannian (I cannot type curly braces below):

$I_r(V,\omega)=(L\in G_r(V)\mid \omega|_L\equiv 0 )$

Where I can find some information (if any) about properties/applications of this $I_r(V,\omega)$ (smoothness, tangent spaces, canonical structures on it, etc.)? In terms of Isotropic Grassmannians, the main question reads: how to characterize the image of $I_{n-1}(V,\omega)$ by means of the isomorphism $G_{n-1}(V)\to \mathbb{P}V^\ast$? This is what puzzled me: $I_{n-1}(V,\omega)$ seems to be a quadric variety, but condition $(*)$ above is linear!

  • 1
    $\begingroup$ If there is an isotropic hyperplane and $\omega$ is nonzero, then it has rank 2 hence decomposes as $\alpha_1\wedge\beta_1$, and the Grassmannian is simply a $P^1$, namely the hyperplanes containing $\ker\omega=\ker\alpha_1\cap\ker\beta_1$. $\endgroup$
    – BS.
    Aug 2, 2012 at 15:33

1 Answer 1


Your guess is right. Knowing that insertion of vectors acts as a derivation you get

$i_X i_Y (\alpha\wedge\omega)=i_Y\alpha\cdot i_X\omega+i_X\alpha\cdot i_Y\omega +\alpha\cdot i_Xi_Y\omega$

If $X,Y\in \ker \alpha$ then the right hand side reduces to $\alpha \cdot i_X i_Y\omega$ and if you assume $\alpha \wedge \omega=0$ this implies that $i_X i_Y\omega=0$ which shows one direction. For the converse direction you might choose a basis of $V$ complementing a basis of the kernel of $\alpha$ and insert any three basis vectors into $\alpha\wedge \omega$ to see its zero by the above computation.

I haven't thought about the remaining question.


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