2
$\begingroup$

$\DeclareMathOperator\SO{SO}$Recall that the closed-subgroup theorem (Wikipedia link) says that a closed subgroup of a Lie group is a Lie group.

I am pretty sure that this theorem should have a "local" generalisation. I'll formulate it for $\SO(n)$.

Generalisation? Fix small $\varepsilon>0$. Let $X\subset \SO(n)$ be a closed subset contained in the closed ball $\overline B_{e}(\varepsilon)$ centred at the identity $e$ of $\SO(n)$. Suppose $X$ has the following properties.

  1. $X=X^{-1}$.

  2. $X\cdot X\cdot X\cdot X\cap \overline B_{e}(\varepsilon)=X$.

Then there exists a Lie group an $G$, a morphism $\varphi: G\to \SO(n)$ and a finite subset $S\subset \SO(n)$ such that $$X=\overline B_{e}(\varepsilon)\cap S\cdot \varphi(\overline B_{G,e}(10 \varepsilon)),$$

where $\overline B_{G,e}(10 \varepsilon)$ is a small ball in $G$. In particular, $X$ is a submanifold of the ball (with finite number of connected components).

Are you aware of some statement of this kind? (the power $4$ is random, maybe it should be replaced by some other power)

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ What if (a trite example) $X = \{(x, \sqrt{2}x) : x \in [-\epsilon, \epsilon]\} \subset \mathbf{T}^2$ (and the torus is embedded in $\mathrm{SO}(n)$)? $\endgroup$
    – Sean Eberhard
    Commented May 6, 2021 at 15:42
  • $\begingroup$ Dear Sean, thanks! This is indeed a counter-example, but it am not afraid of it. I'll modify the question so that there is still a meaning, which is good enough for me $\endgroup$
    – aglearner
    Commented May 6, 2021 at 16:24
  • 1
    $\begingroup$ I'm not sure why the OP is modifying the question. In the original question, the OP did not ask that the Lie subgroup $G\subset \mathrm{SO}(n)$ be compact, i.e., closed, just that it exist, and the 'skew-line' that Sean writes down is obviously part of a (non-closed) Lie subgroup. Did the OP mean to require that $G$ be a closed Lie subgroup? $\endgroup$
    – Robert Bryant
    Commented May 6, 2021 at 20:31
  • 2
    $\begingroup$ @RobertBryant The problem is that the non-compact skew line still doesn't work, because its intersection with a nbd of 1 is dense there $\endgroup$
    – Sean Eberhard
    Commented May 6, 2021 at 21:10
  • 1
    $\begingroup$ @SeanEberhard: Ah, you are right. Thanks for the explanation! $\endgroup$
    – Robert Bryant
    Commented May 7, 2021 at 0:55

1 Answer 1

Reset to default
4
$\begingroup$

See Tao's book on Hilbert's fifth problem -- https://terrytao.wordpress.com/books/hilberts-fifth-problem-and-related-topics/ -- Theorem 3.1.7, where it is referred to as local Cartan's theorem.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Thanks a lot Sean! It looks exactly as what I was looking for. $\endgroup$
    – aglearner
    Commented May 6, 2021 at 22:17
  • $\begingroup$ The theorem @SeanEberhard mentions is also available in Tao's online version of his notes (perhaps the same as the book?) as Theorem 12 of Notes 2 - Building Lie structure from representations and metrics. It states "If $H$ is a locally compact local subgroup of a local Lie group $G$, then there is an open neighbourhood $H'$ of the identity in $H$ that is a smooth submanifold of $G$, and is thus also a local Lie group." $\endgroup$
    – LSpice
    Commented Aug 3, 2021 at 8:22

You must log in to answer this question.

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