2
$\begingroup$

$\DeclareMathOperator\Gr{Gr}$The Grassmnnian variety $\Gr(k,n)$ is the set of $k$-dimensional subspaces of $\mathbb{C}^n$. The coordinate ring $\mathbb{C}[\Gr(k,n)]$ is generated by Plucker coordinates subject to Plucker relations.

Denote by $\Gr(k, \pm \infty)$ the set of $k$-dimensional subspaces of $\mathbb{C}^{\infty}$. The coordinate ring $\mathbb{C}[\Gr(k,\pm \infty)]$ is generated by Plucker coordinates $P_J$, $J \subset \mathbb{Z}$, subject to Plucker relations. Has $\mathbb{C}[\Gr(k,\pm \infty)]$ been defined in the literature? Thank you very much.

$\endgroup$
5
  • 3
    $\begingroup$ Your variety is the classifying space for the unitary group $\operatorname{\rm U}(k)$, and is much studied in this context. $\endgroup$ Commented Nov 14, 2023 at 12:53
  • 2
    $\begingroup$ Why include the $\pm$ symbol? $\endgroup$ Commented Nov 14, 2023 at 12:54
  • $\begingroup$ @SamHopkins, the column indices of a matrix in $Gr(k, \pm \infty)$ can be any negative and positive numbers. I use it to distinguish $Gr(k, +\infty)$ in which the matrices have column indices positive. $\endgroup$ Commented Nov 14, 2023 at 16:58
  • 1
    $\begingroup$ That seems like an unusual convention. Normally I think of $\mathbb{C}^\infty$ as having a canonical ordered basis $e_1,e_2,\ldots$. Especially when writing $\mathbb{C}^\infty = \cup_{n=0}^{\infty} \mathbb{C}^n$ as a direct limit. $\endgroup$ Commented Nov 14, 2023 at 17:03
  • $\begingroup$ There are lots of subtly different versions of this construction, especially if you are coming from an algebraic geometry perspective. Some good references to start might be arxiv.org/abs/1806.11233 and kurims.kyoto-u.ac.jp/~kenkyubu/kashiwara/the-flag.pdf $\endgroup$
    – Oliver
    Commented Nov 16, 2023 at 18:36

1 Answer 1

4
$\begingroup$

You should have a look at the Appendix to this paper of mine:

https://arxiv.org/abs/1212.3528

https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/jdt064

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .