7
$\begingroup$

Let $ K $ be a field. Let $ K^n $ be an $ n $ dimensional vector space over $ K $. Let $ KP^{n-1} $ be the projective space of lines in $ K^n $. Let $ GL(n,K) $ be the group of invertible $ n \times n $ matrices over $ K $. What is the smallest subgroup of $ GL(n,K) $ that acts transitively on $ KP^{n-1} $?

Context: When $ K=\mathbb{C} $ then I think the smallest subgroup of $ GL(n,\mathbb{C}) $ acting transitively on $ \mathbb{C}P^{n-1} $ is $ SU(n) $. When $ K=\mathbb{R} $ then I think the smallest subgroup of $ GL(n,\mathbb{R}) $ acting transitively on $ \mathbb{R}P^{n-1} $ is $ SO(n) $. I was wondering if this is true and also what the corresponding group is for other choices of $ K $.

$\endgroup$
5
  • $\begingroup$ In the second case you probably mean "$K=\mathbb R$"? $\endgroup$ Feb 20 at 23:17
  • $\begingroup$ @red_trumpet yes fixed! $\endgroup$ Feb 20 at 23:18
  • 2
    $\begingroup$ If $n$ is even, then $\operatorname{SU}\left(\frac n2\right) < \operatorname{SO}(n)$ acts transitively on $\Bbb R P^{n - 1}$, and if $n \equiv 0 \pmod 4$, then $\operatorname{Sp}\left(\frac n4\right) < \operatorname{SO}(n)$ acts transitively on $\Bbb R P^{n - 1}$. For $n = 7$, $\operatorname{G}_2 < \operatorname{SO}(7)$ (dimension $14$) acts transitively on $\Bbb R P^6$. See Borel's classification of Lie groups acting transitively on spheres. $\endgroup$ Feb 20 at 23:37
  • $\begingroup$ Oh that's very interesting so $ S^7 $ has a transitive action by $ SO(8) $ of dim $ 28 $ and $ SU(4) $ of dim $ 15 $ and $ Sp(2) $ of dim $ 10 $ and is even an H space in a natural way by identifying it with the unit octonions which have dimension $ 7 $ (although they are not a group because they lack associativity) $\endgroup$ Feb 20 at 23:54
  • $\begingroup$ Yes, and $S^7$ also admits a transitive action by $\operatorname{Spin}(7)$ (dimension $21$); that phenomenon is special to that dimension for reasons connected to triality. en.wikipedia.org/wiki/Triality $\endgroup$ Feb 21 at 1:47

2 Answers 2

6
$\begingroup$

Over finite fields you get much more. For example (images of) the Singer cycle (and overgroups). As a somewhat random example, there are (up to conjugacy) 7 minimal transitive subgroups of $PGL(4,3)$, one of which is the image of a Singer cycle of order $40$ and the others up to order $360$.

$\endgroup$
4
$\begingroup$

For $K = \Bbb R$ it follows from Montgomery & Samelson, "Transformation Groups of Spheres" that one can do better than $\operatorname{SO}(n)$ when $n \equiv 0 \pmod 2$ (and $n > 2$) or $n = 7$.

  • If $n \equiv 0 \pmod 2$ and $n > 2$, $\operatorname{SU}\left(\frac n2\right)$, which has dimension $\frac{1}{4} n^2 - 1$, acts transitively on $\Bbb R P^{n - 1}$.

  • If $n \equiv 0 \pmod 4$, $\operatorname{Sp}\left(\frac n4\right)$, which has dimension $\frac18 n (n + 2)$, acts transitively on $\Bbb R P^{n - 1}$.

    • In the special case $n = 16$, $\operatorname{Spin}(9)$, which has the same dimension ($36$) as $\operatorname{Sp}(4)$, also acts transitively on $\Bbb R P^{15}$.
    • In the special case $n = 8$, $\operatorname{Spin}(7)$, which has dimension $21$, acts transitively on $\Bbb R P^7$; it is smaller than $\operatorname{SO}(8)$ (dimension $28$) but it's still larger than $\operatorname{Sp}(2)$ (dimension $10$).
  • If $n = 7$, $\operatorname{G}_2$, which has dimension $14$, acts transitively on $\Bbb R P^6$.

Montgomery, Deane, and Hans Samelson. "Transformation Groups of Spheres." Annals of Mathematics, vol. 44, no. 3, 1943, pp. 454–70. JSTOR, https://doi.org/10.2307/1968975.

$\endgroup$
2
  • $\begingroup$ Maybe also worth mentioning are the nonminimal groups $\operatorname{Sp}\left(\frac n4\right) \cdot \operatorname{Sp}(1)$, of dimension $\frac18 n (n + 2) + 3$, which respectively act transitively on $\Bbb R P^{n - 1}$, $n \equiv 0 \pmod 4$. $\endgroup$ Feb 21 at 2:03
  • $\begingroup$ Interestingly, the sequence $a_n := \operatorname{min}\{\dim G \mid \textrm{$G$ acts transitively on $\Bbb S^{n - 1}$}\}$, $0,1,3,3,10,8,14,10,36,24,\ldots$, does not appear in the OEIS. $\endgroup$ Feb 21 at 16:29

You must log in to answer this question.

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