Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G(k, V)$ be the Grassmannian of all subspaces of dimension $k$ of a vector space $V$ with dimension $n$. Let $\Gamma \subset V$ be a subspace of dimension $n-k$. If $w_1, \ldots, w_k$ is a basis of $\Gamma$, then $\omega = w_1 \wedge \cdots \wedge w_{n-k}$ is an element of $\Lambda^{n-k}V = \Lambda^{k}V^{*}$. Think $\omega$ as a homogeneous linear form on $\mathbb{P}(\Lambda^{k}V)$. Let $U$ be the affine open subset where $\omega \neq 0$. Why the intersection of $G(k, V)$ with $U$ is the set of $k$-dimensional subspaces $\Lambda \subset V$ complementary to $\Gamma$? Why $G(k, V) \cap U \simeq Hom(V/\Gamma, \Gamma)$? This question is from page 65 of the book Algebraic Geometry. Thank you very much.

share|cite|improve this question
Have you tried the cases $n=3, k=1$ or $k=2$ by hand? As a further hint: What exactly is the identification between $\wedge^{n-k}V$ and $(\wedge^k V)^*$ again? – Jyrki Lahtonen Aug 23 '11 at 20:13
It's worth noting that the identification of $\Lambda^{n-k}(V)$ and $(\Lambda^k V)^{\ast}$ isn't completely canonical: it depends on a choice of volume form (a basis for $\Lambda^n(V)$). – Qiaochu Yuan Aug 31 '11 at 16:33
up vote 0 down vote accepted

Let $v_1, \ldots, v_n$ be a basis of $V$. Then $w_i=\sum_{j}a_{ji}v_j$ for some $a_{ji}$. $\omega=w_1\wedge \cdots \wedge w_{n-k} = \sum_{i_1\cdots i_k}c_{i_1\cdots i_{n-k}}v_{i_1}\wedge \cdots \wedge v_{i_{n-k}}$. $v_{i_1}\wedge \cdots \wedge v_{i_{n-k}} \mapsto f$ where $f(v_{i_1}\wedge \cdots \wedge v_{i_{n-k}})=1$ and $f(v_{j_1}\wedge \cdots \wedge v_{j_{n-k}})=0$ if $\{j_1, \ldots, j_{n-k}\} \neq \{i_1, \ldots, i_{n-k}\}$. Under this identification, $\omega \in \Lambda^{k}V^*$. Since $U=\{w_1\wedge \cdots \wedge w_k \in \Lambda^{k}V \mid \omega(w_1\wedge \cdots \wedge w_k) \neq 0 \}$, if $w \in G(k, V) \cap U$, then $\omega(w) \neq 0$. Since only $k$-dim subspaces $w$ complementary to $\Gamma$ satisfy the condition $\omega(w) \neq 0$, the intersection of $G(k, V)$ with $U$ is the set of $k$-dim subspaces complementary to $\Gamma$.

share|cite|improve this answer
But why a subspace $\Lambda$ complementary to $\Gamma$ corresponds to a linear map from $V/\Gamma$ to $\Gamma$? – LJR Aug 31 '11 at 16:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.