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Let $V$ and $W$ be two vector spaces of dimension $n$ each. Then is there a decomposition of the Grassmannian $G(r, V\oplus W)$ in terms of $G(r,V)$ and $G(r,W)$ ? I am expecting something similar to the wedge decomposition. My guess is that it is the direct sum of cartesian products of $G(k,V)$ and $G(l,W)$ such that $k+l=r$. Is this true ?

Arctic Char
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icmes
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  • For an answer see here. – Dietrich Burde Apr 04 '17 at 14:30
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    @DietrichBurde That answer is about the entire exterior algebra, while the OP would be interested only in the projectivization of the space of decomposable wedges in a given degree, no? – Stephen Apr 04 '17 at 14:32
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    Yes, I am looking for a similar result for Grassmannian. – icmes Apr 04 '17 at 14:38
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    In some sense one shouldn't expect such a decomposition. If there were one, it would induce a corresponding decomposition of the tangent space at any point $E \in G(r, V \oplus W)$, but we may identify canonically $T_E G(r, V \oplus W)$ with $E^* \otimes ((V \oplus W) / E)$. At a generic point, $E$ is transverse to both $V$ and $W$, and for such $E$ there is no way to rewrite $(V \oplus W) / E$ in a 'decomposed' way. – Travis Willse Apr 04 '17 at 16:06

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I don't think there is a reasonable way to construct $Gr(r,V\oplus W)$ from the $Gr(k,V)$'s and $Gr(l, W)$'s.
For example $Gr(1,\mathbb C\oplus \mathbb C)=\mathbb P^1(\mathbb C)$, a complex projective variety of dimension $1$.
But the $Gr(k,\mathbb C)$'s are sets with just $1$ element if $k=0$ or $k=1$ and $Gr(k,\mathbb C)=\emptyset$ for $k\geq 2$.
How on earth would you build $\mathbb P^1(\mathbb C)$ from pieces that are singleton sets or empty?

Georges Elencwajg
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