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I'm fairly new to algebraic geometry, so this is may just be fairly simple question about when two varieties are "the same" variety.

In this question, I noted that the Grassmannian can be expressed as an algebraic variety in (at least) two ways:

  1. Using the Plücker embedding, where the subspace spanned by any orthonormal $v_1, v_2, ..., v_n$ is given by $v_1 \wedge v_2 \wedge ... \wedge v_n$
  2. Using the projective embedding, where a subspace is identified with the unique orthogonal projection matrix to that subspace

Or as matrices, if the matrix $M = [v_1 | v_2 | ... | v_n]$, where the $v_i$ are column vectors, then

  1. The Plücker embedding can be identified with the $n$'th compound matrix $C_n(M)$
  2. The projective embedding can be identified with the projection matrix $MM^\dagger$, where $M^\dagger$ is the pseudoinverse

These would appear to be two totally different algebraic varieties, defined by different sets of polynomial equations, embedded in two different Euclidean spaces, with two different metrics. Yet somehow, people often talk of the "Grassmann variety" as though there were a unique variety associated with the Grassmannian. So this leads to my question:

  1. Is there some sense in which these two varieties are "the same" variety?
  2. In general, what intrinsic properties can one use to determine if two varieties are "the same variety," independent of their particular embedding into a Euclidean space?
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    $\begingroup$ As everywhere in mathematics, there is a notion of isomorphism. A simpler example of your phenomenon is that embedding of $\Bbb P^1$ as a smooth conic in $\Bbb P^2$ or as a twisted cubic in $\Bbb P^3$. $\endgroup$ Jun 1, 2019 at 0:16
  • $\begingroup$ In what sense are the two varieties isomorphic though? They aren't defined by the same equations nor have the same metric. I guess they'd both be homeomorphic as topological spaces? Is there nothing stronger than that? $\endgroup$ Jun 1, 2019 at 1:11
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    $\begingroup$ en.wikipedia.org/wiki/Morphism_of_algebraic_varieties $\endgroup$ Jun 1, 2019 at 3:18

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Yes, the simple answer was given by Eric Wofsey in the comments. They are isomorphic as algebraic varieties, so that there is a regular polynomial map going from one to another. So that is the sense in which they are "the same" variety.

This is interesting as it shows that just because these different incarnations of the Grassmannian are isomorphic as algebraic varieties, they aren't isomorphic regarding every possible structure that people often want on a Grassmannian. In particular, in these two cases, they aren't isometrically isomorphic, nor do they span subspaces of the same dimension in their respective embeddings. So that is, while the variety structure is preserved via regular map isomorphisms, everything about the metric changes, so that these are "different" metric spaces on "the same" variety.

A thorough review of the different types of metric that can be placed on the Grassmannian can be found in this paper: http://sci-hub.tw/https://doi.org/10.1137/040607605

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