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The Grassmannian $\text{Gr}(k,n)$ can be described as the quotient $$\text{Gr}(k,n) \cong \text{GL}(k,k) \ \backslash \ \text{Mat}^*_{\mathbb{R}}(k,n) $$

where $\text{Mat}^*_{\mathbb{R}}(k,n)$ is the set of real $k \times n$ matrices of rank $k$. Every point of $\text{Gr}(k,n)$ can be represented by a $k \times n$-matrix $M$. The Plücker coordinates on $\text{Gr}(k,n)$ are all the $k \times k$-minors of $M$.

I don't really understand this construction. I have been reading about how to construst $\text{Gr}(k,n)$ as a projective variety via embedding into $\mathbb{P}\mathbb{R}$ of dimension $n \choose k$ and expressing the $n \choose k$-minors by vanishing of homogeneous polynomials, but I am not sure how to relate that description to the one above.

I also came across a construction of $\text{Gr}(k,n)$ as quotient of $\text{GL}(k,k)$ by some stabilizer, but this seems to be related to its structure as a manifold, not as a variety, as far as I understood. My goal here is to understand another space, which is a subvariety in $\text{Gr}(k,n)$, but I don't really follow this definition of $\text{Gr}(k,n)$ to begin with.

Any help/explanation would be really great!

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That should be $\text{Mat}^*_{\mathbb{R}}(k,n)/Gl(k,k)$. You might find it helpful to instead think of this as the quotient $\text{Mat}^*_{\mathbb{R}}(k,n)/\sim$, where $\sim$ is the relation by which $A \sim B$ if $A,B$ have the same row-space.

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  • $\begingroup$ Hmmm, weird that there's a typo, because it's mentioned as $\text{Gr}(k,n) \cong \text{GL}(k,k) \ / \ \text{Mat}^*_{\mathbb{R}}(k,n)$ twice in the paper I am reading..Thanks for your answer either way! $\endgroup$ – mike Jun 30 at 20:14
  • $\begingroup$ @mike Is it instead written as $GL(k,k) \setminus \text{Mat}_{\Bbb R}^*(k,n)$? $\endgroup$ – Ben Grossmann Jun 30 at 20:17
  • $\begingroup$ Oh yes, i am def blind. feeling really silly now :D $\endgroup$ – mike Jun 30 at 20:22
  • $\begingroup$ @mike The fact that it's from the left is to emphasize that the action is multiplication from the left by elements of $GL(k,k)$ $\endgroup$ – Ben Grossmann Jun 30 at 20:23
  • $\begingroup$ Hmm Yes, so it's really just re-stating the same construction like with Plücker embedding? I mean, we have a map from $(k,n)$-matrices of rank $k$ to projective space of dimension $n \choose k$, where we consider the projective space in order to remain invariant under action of $\text{GL}(k,k)$, correct? $\endgroup$ – mike Jun 30 at 20:28

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