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YCor
YCor
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typo
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Geoff Robinson
Geoff Robinson
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Is it known for which $n, k\in\Bbb N$ there exists a matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^n)$ that

  • acts transitively on $\mathrm{Gr}(k,n)$, i.e., on the $k$-dimensional subspaces of $\Bbb R^n$, but
  • acts not transitively on $\mathrm{Gr}(k+1,n)$, i.e., on the $(k+1)$-dimensional subspaces of $\Bbb R^n$?

For example, $\mathrm U(n)$ (acting on $\Bbb C^n\cong\Bbb R^{2n}$) acts tansitivelytransitively on $\mathrm{Gr}(1,\Bbb R^{2n})$ but not on $\mathrm{Gr}(2,\Bbb R^{2n})$.

Is it known for which $n, k\in\Bbb N$ there exists a matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^n)$ that

  • acts transitively on $\mathrm{Gr}(k,n)$, i.e., on the $k$-dimensional subspaces of $\Bbb R^n$, but
  • acts not transitively on $\mathrm{Gr}(k+1,n)$, i.e., on the $(k+1)$-dimensional subspaces of $\Bbb R^n$?

For example, $\mathrm U(n)$ (acting on $\Bbb C^n\cong\Bbb R^{2n}$) acts tansitively on $\mathrm{Gr}(1,\Bbb R^{2n})$ but not on $\mathrm{Gr}(2,\Bbb R^{2n})$.

Is it known for which $n, k\in\Bbb N$ there exists a matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^n)$ that

  • acts transitively on $\mathrm{Gr}(k,n)$, i.e., on the $k$-dimensional subspaces of $\Bbb R^n$, but
  • acts not transitively on $\mathrm{Gr}(k+1,n)$, i.e., on the $(k+1)$-dimensional subspaces of $\Bbb R^n$?

For example, $\mathrm U(n)$ (acting on $\Bbb C^n\cong\Bbb R^{2n}$) acts transitively on $\mathrm{Gr}(1,\Bbb R^{2n})$ but not on $\mathrm{Gr}(2,\Bbb R^{2n})$.

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M. Winter
M. Winter
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Groups that act transitively on $\mathrm{Gr}(k,\Bbb R^n)$ but not transitively on $\mathrm{Gr}(k+1,\Bbb R^n)$

Is it known for which $n, k\in\Bbb N$ there exists a matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^n)$ that

  • acts transitively on $\mathrm{Gr}(k,n)$, i.e., on the $k$-dimensional subspaces of $\Bbb R^n$, but
  • acts not transitively on $\mathrm{Gr}(k+1,n)$, i.e., on the $(k+1)$-dimensional subspaces of $\Bbb R^n$?

For example, $\mathrm U(n)$ (acting on $\Bbb C^n\cong\Bbb R^{2n}$) acts tansitively on $\mathrm{Gr}(1,\Bbb R^{2n})$ but not on $\mathrm{Gr}(2,\Bbb R^{2n})$.