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Let $V\in G(k,n)$. A family of $k$-dimensional subspaces (of $\Bbb{C}^n$) $\rho:V\to B$ parameterized by a variety $B$ induces a map of sets $B\to G(k,n)$ (taking a point $b\in B$ to the class of the fiber $\rho^{-1}(b)$). This map of sets is infact an algebraic morphism.

  1. Will every $k$-subspace $V$ have a morphism to any arbitrary variety $B$?
  2. What is "class of the fiber $\rho^{-1}(b)$? Does it mean that every $b\in B$ is mapped to all its pre-images with respect to $\rho$, across all elements of $G(k,n)$?
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    I feel something has been lost in translation here. A family of $k$-dimensional subspaces over $B$ should not itself be a map from a single $k$-dimensional subspace unless $B$ is a point. The idea of a family is to have a map $\rho: X \rightarrow B$ such that you have one of something (here, $k$-dim subspaces) over each $b \in B$, i.e. as $\rho^{-1}(b)$. – aes Dec 10 '14 at 04:58
  • I'm not sure why we would map from a single subspace either. – Matt Samuel Dec 10 '14 at 14:29

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