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$\DeclareMathOperator\Gr{Gr}$ Let $\Gr(k,n)$ be the set of $k$-dimensional subspaces in affine space $\mathbb{A}^n$ over an algebraically closed field. If $U\subseteq (\mathbb{A}^n)^{\times k}$ is an open subset of $n \times k$ matrices of rank $k$, is $\{\text{Im}(u): u \in U\} \subseteq \Gr(k,n)$ open?

Here is my convoluted attempt at a solution. Via the Plücker embedding, one can view $\Gr(k,n)$ as the set of totally decomposable elements of $\bigwedge^k_n$, which can be shown to form a projective variety.

$\Gr(k,n)$ can also be viewed as the set of $n \times k$ matrices of rank $k$, quotiented by the (equivalence relation induced by) the right action of $\text{GL}(k)$. Since $\text{GL}(k)$ is a reductive algebraic group, I believe this quotient is again an affine variety with induced topology given by the quotient topology. Is this correct? If so, then I don't think it's hard to prove that these two varieties corresponding to $\Gr(k,n)$ are isomorphic, which would seem to imply my desired result.

Related question.

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  • $\begingroup$ At least one thing seems wrong in what you said: the Grassmannian is a projective variety, not an affine variety. $\endgroup$ Feb 16, 2020 at 1:46
  • $\begingroup$ @SamHopkins Yep, its projective. I meant that I am taking Gr(k,n) to be the projective variety of k-planes in affine n-space, as opposed to projective n-space (although the distinction doesn’t really matter) $\endgroup$
    – Ben
    Feb 16, 2020 at 3:36
  • $\begingroup$ I was referring to this bit: "Since GL(k) is a reductive algebraic group, I believe this quotient is again an affine variety with induced topology given by the quotient topology." $\endgroup$ Feb 16, 2020 at 3:38
  • $\begingroup$ Oh, thanks for catching that. As a set, Gr(k,n) can be viewed as the set of rank k $n\times k$ matrices, quotiented by the right action of GL(k). Can we view this set as a projective variety? Does the induced quotient topology agree with the topology of the Plücker embedding? $\endgroup$
    – Ben
    Feb 16, 2020 at 19:01

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The map $U \to \mathbb{G}(k,n)$ that you are interested in is flat, and a flat finitely presented map is well known to be open.

Flatness can be checked in several ways: for example, the map is generically flat, by the generic flatness theorem. But it is also $\mathrm{GL}_n$-equivariant, and the action of $\mathrm{GL}_n$ on $U \to \mathbb{G}(k,n)$ is transitive. Alternatively, it follows from the fact that the fibers are equidimensional, $\mathbb{G}(k,n)$ is regular and $U$ is also regular, hence Cohen--Macaulay.

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