# Is the set of images of an open subset of full-rank matrices an open subset of the Grassmannian?

$$\DeclareMathOperator\Gr{Gr}$$ 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.

• At least one thing seems wrong in what you said: the Grassmannian is a projective variety, not an affine variety. Feb 16, 2020 at 1:46
• @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)
– Ben
Feb 16, 2020 at 3:36
• 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." Feb 16, 2020 at 3:38
• 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?
– Ben
Feb 16, 2020 at 19:01

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.