I have a couple of questions about the explanations on the Grassmannian from Harris' and Griffith's "Principles of algebraic geometry" on page 194/195. We consider for $k \le n$ the $k$-th Grassmannian $G(k, n)$ over an $n$-dimensional complex vector space $V$. Consider the following:

The Cell Decomposition

Recall that the cell decomposition

$$\mathbb{P}^n =\mathbb{C}^n \cup \mathbb{C}^{n-1} \cup ... \cup \mathbb{C}^1 \cup \mathbb{C}^0 $$ of $\mathbb{P}^n = G(1, n+1)$ is obtaines by choosing a flag

$$V =(V_1 \subset ... \subset V_{n-1} \subset V_n \subset \mathbb{C}^{n+1}) $$

of linear subspaces of $\mathbb{C}^{n+1}$ and taking $W_i \cong \mathbb{C}^{n-1} = \{ l \subset \mathbb{C}^{n+1} : l \subset V_i, l \not \subset V_{i-1} \}$; $l$ line. The same technique works to give a cell decomposition of the Grassmannian: if we set $V_i = \{e_1,..., e_i \} \subset \mathbb{C}^n$, then the set of $\Lambda \in G(k,n)$ consider the increasing sequence of subspaces

$$ 0 \subset \Lambda \cap V_1 \Lambda \cap V_2 \cap ... \Lambda \cap V_{n-1} \Lambda \cap V_n= \Lambda .$$

For generic (???) $\Lambda$, $\Lambda \cap V_i$ will be zero for $i \le n-k$, and $(i+k-n)$-dimensional thereafter - indeed, we have seen that the set of such $\Lambda$ is just the open set $U_I \cong \mathbb{C}^{k(n-k)} \subset G(k,n)$. ...

short explanation on $U_I$: we choose $I :=\{i_1,...,i_k\} \subset \{1,...,n\}$ and let $V_I \subset \mathbb{C}^n$ be the $(n-k)$-plane in $\mathbb{C}^n$ spanned by vectors $\{e_j: j \not \in I \}$. then $U_I := \Lambda \in G(k,n): \Lambda \cap V_I = \{0 \} \}$

...Now, for any sequence of subspace of integers $a_1,..., a_k$, set

$$W_{a_1,..., a_k}= \{ \Lambda \in G(k,n): \dim(\Lambda \cap V_{n-k+i -a_i})=i \}.$$

We observe that $\dim(\Lambda + V_{n-k+i -a_i}) =n-a_i$, and conseqently $W_{a_1,..., a_k}$ will be empty unless $a_1,..., a_k $ is anonincreasing sequence of integers $\le n-k$. Since $\dim(\Lambda \cap V_{n-k+i -a_i})=i$ if and only if the rank of the last $k \times (k+a_i -i)$ minor of a matrix representaative of $\Lambda$ is exactly $k-i$ it follows that the closure (???)

$$\overline{W_{a_1,..., a_k}}= \{ \Lambda \in G(k,n): \dim(\Lambda \cap V_{n-k+i -a_i}) \ge i \}$$

is an analytic subvariety of $G(k,n)$.


Q1: what does the author mean here by an "generic" element $\Lambda \in G(k,n)$? my understanding of generic points based on algebro geometrical background is that they are dense in their topological space. does somebody understand which meaning of generic elements the author presents in this text?

Q_2: Why the closure $\overline{W_{a_1,..., a_k}}$ of $W_{a_1,..., a_k}$ equals $ \{ \Lambda \in G(k,n): \dim(\Lambda \cap V_{n-k+i -a_i}) \ge i \}$? I think that we are working with the standard topology of $\mathbb{C}^{k(n-k)}$.


1 Answer 1

  1. A generic element or a general element means one from a open Zariski-dense subset. That's what's going on here - for most $\Lambda$, $\Lambda\cap V_i$ is going to be zero unless their dimensions add up to more than $n+1$. For example, think about a line and a plane in $\Bbb R^3$: the chance they intersect is zero, but if you think about two planes in the same space, then they must intersect in a line.

  2. $\dim(\Lambda\cap V_{n-k+i-a_i})\geq i$ is equivalent to the last $k\times(k+a_i-i)$ minor $M$ having rank $\leq k-i$, which is exactly the condition that the determinants of all $(k-i+1)\times(k-i+1)$ minors of $M$ vanish. This is clearly a closed condition. To show that the closure of $W_{a_1,\cdots}$ is indeed this set, note that the condition that the rank of the matrix being exactly $k-i$ means that there's a minor of size $k-i$ which has nonvanishing determinant. But we can pick a sequence of such matrices so that in the limit, all the minors of size $k-i$ have vanishing determinant: this is just repeated applications of the statement that the set $\{f\neq 0\}$ is dense, which is true in both the zariski and usual topology. This shows that the closure of $W_{a_1,\cdots}$ contains matrices so that the final $k\times(k+a_i-i)$ minor is of rank $k-i-1$. By inductively applying this fact and the knowledge that you only need to take the closure once, we see that the closure of $W_{a_1,\cdots}$ is as claimed.

  • $\begingroup$ the observation that the condition $\dim(\Lambda\cap V_{n-k+i-a_i})\geq i$ is exactly the condition that the determinants of all $(k-i+1)\times(k-i+1)$ minors of $M$ vanish tells me at first only that $K:= \{ \Lambda \in G(k,n): \dim(\Lambda \cap V_{n-k+i -a_i}) \ge i \}$ is closed. trivially it contains $W_{a_1,..., a_k}$ and therefore it also contains $\overline{W_{a_1,..., a_k}} $. why holds also the inclusion $K \subset \overline{W_{a_1,..., a_k}} $; i.e. why $W_{a_1,..., a_k}$ is dense in $K$? $\endgroup$
    – user526728
    Sep 13, 2019 at 13:20
  • $\begingroup$ Updated, my apologies for the omission. $\endgroup$
    – KReiser
    Sep 13, 2019 at 18:45

You must log in to answer this question.