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.
The intersection of a Grassmannian and an open set
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1It'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 $\Lambda^n(V)$). – 2011-08-31
1 Answers
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 $\Gamma$.
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0But why a subspace $\Lambda$ complementary to $\Gamma$ corresponds to a linear map from $V/\Gamma$ to $\Gamma$? – 2011-08-31