Let $V = \mathbb{C}^n$ with basis $e_1,\dots,e_n$, and $U = \langle e_1,\dots,e_k\rangle$. Let $$\Sigma(U)=\{\sigma\in Gr(V,2)\mid \sigma\in U \}$$ be the Schubert cell of $2$-planes contained in $U$. The equations for $\Sigma(U)$ are the Plucker relations, in addition to the relations $$w_{ij}=0\quad 1\leq i\leq k, i< j\leq n$$ where $w_{ij}$ are the induced (by $e_1,\dots,e_n$) coordinates on $\wedge^2 V$. Write $$K = \text{span}(w_{ij})_{1\leq i\leq k, i< j\leq n}\subset (\wedge^2 V)^*\subset \mathbb{C}[w_{ij}]_{1\leq i<j\leq n}.$$ Hence we write the ideal defining $\Sigma(U)$ as $$I(\Sigma(U)) = P+\langle K\rangle$$ where $\langle K\rangle$ denotes the ideal generated by $K$, and $P$ is the ideal of Plucker relations. My question is the following: does there exist a subspace $L\subsetneq K$ such that $$V(P+\langle L\rangle) = \Sigma(U)$$ ($V(P+\langle L\rangle)$ denotes the variety defined by $P+\langle L\rangle$) or equivalently, $$\sqrt{P+L}=P+K.$$ In other words, can we defined $\Sigma(U)$ as the intersection of $Gr(V,2)$ with a subspace larger then $V(K)$?

It is clear that we cannot leave out any of the $w_{ij}$, but I'm having a hard time proving that there is no such $L\subsetneq K$ at all.


1 Answer 1


The dimension of $Gr(V,2)$ is $2 (n-2)$. For each point $x$ of $Gr(V,2)$ not in $\Sigma(U)$, the codimension of the space of subspaces $L$ that vanish on $x$ in the space of $m$-dimensional subspaces $L$ of $K$ is $m$. Hence the codimension of the space of subspaces $L$ that vanish on any point of $Gr(V,2)- \Sigma(U)$ is $m- 2(n-2)$. Thus as soon as $m> 2(n-2)$, there exists a suitable $L$. Because the dimension of $U$ is ${n \choose 2} - {k \choose 2}$, this is viable as soon as $2 \leq k \leq n-3$.


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