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It is known that almost every pair of elements in a connected compact Lie group (topologically) generates the group.

Obviously this isn't true for non-connected groups but

Given a compact Lie group $G$, is it true that almost every pair of elements of $G$ generates a subgroup containing the connected component $G^\circ$ of 1?

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    $\begingroup$ No need to vote for close, without leaving the OP time to clarify the question. It's clear that "the connected component" meant "the connected component of 1". $\endgroup$
    – YCor
    Mar 18, 2017 at 1:54
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    $\begingroup$ @IgorRivin: Is it trivial? Note that the question is (currently) about almost every pair of elements of $G$, not of $G^\circ$. Anyway, if the closed subgroup generated by $x,y$ contains any connected component of $G$, then it contains $G^\circ$. $\endgroup$
    – Nate Eldredge
    Mar 18, 2017 at 3:11
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    $\begingroup$ The infinite dihedral group $O_2({\bf R})$ is a famous counterexample to "close enough to Haar measure": All elements of the non-identity component are involutions! But it still does not refute the OP's conjecture. $\endgroup$
    – Noam D. Elkies
    Mar 18, 2017 at 5:21
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    $\begingroup$ @IgorRivin this is why when you think a question is trivial, writing a comment and waiting for answers can be useful before voting to close. It often occurs to me too. (Btw you should retract your closing vote.) $\endgroup$
    – YCor
    Mar 19, 2017 at 2:40
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    $\begingroup$ @IgorRivin The only possibly unclear thing was whether some subgroup "contains the connected component of $G$". Since a subgroup of any topological group contains a connected component iff it contains the connected component of 1, this leaves little ambiguity. Anyway you have still left your closing vote. $\endgroup$
    – YCor
    Mar 19, 2017 at 14:23

2 Answers 2

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The closed subgroups of $G$ not containing the identity component lie in countably many conjugacy classes of subgroups. So it is sufficient to show that for each closed subgroup $H$ not containing the identity component, the probability that Haar-random $g_1$ and $g_2$ both lie in some conjugate of $H$ vanishes.

Such pairs are parameterized by the manifold of triples $x \in G/H$, $g_1 \in x H x^{-1}, g_2 \in x H x^{-1}$, which is a manifold of dimension $(\dim G - \dim H) + 2 \dim H = \dim G + \dim H$.

The image of this manifold in $G \times G$ under the projection $(x,g_1,g_2)\mapsto (g_1,g_2)$ must have measure $0$, as it is the image of a smaller-dimensional manifold (as $H$ does not contain the identity we have $\dim H < \dim G$) under a smooth map (by Sard's theorem).

For $H$ a subgroup, the same argument just barely fails to show that the probability that a single $g$ is contained in a conjugate of $H$ vanishes. Indeed in this case both manifolds have dimension $\dim(G)$. This is convenient as that statement is false, because we could take $H$ to be a maximal torus, or alternately the subgroup generated by a reflection in an infinite dihedral group, as in Noam's example.

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  • $\begingroup$ What theorem do you use for image of positive dimension under smooth maps to have measure zero? For instance say for closed manifolds $A,B$ under the smooth projection $A\times B\to A$, the image of $A\times\{b\}$ is all of $A$ even if $A$ is a positive codimension submanifold. $\endgroup$
    – YCor
    Mar 21, 2017 at 23:50
  • $\begingroup$ @Ycor I think he meant that the dimension of $G/H \times H \times H$ is lower than the dimension of $G \times G$, so the image is null by Sard's theorem. $\endgroup$
    – Jack the Ripper
    Mar 22, 2017 at 0:19
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    $\begingroup$ @Jack Let $M$ be the set of triples $(x,g_1,g_2)$ as I describe. Then the projection map to $G \times G$ from $M$ clearly exists. I claim $M$ is a manifold of dimension $ 2 \dim H+ (\dim G-\dim H)$. The key point is that locally on $G/H$, we can choose a section $s: G/H \to G$ that defines a coset representation, and then $(xH, h_1,h_2) \to (xH, s(x) h_1 s(x)^{-1}, s(x) h_2 s(x)^{-1})$ is perfectly well-defined. $\endgroup$
    – Will Sawin
    Mar 22, 2017 at 0:35
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    $\begingroup$ The last paragraph was more clear before the revision. $\endgroup$
    – R. van Dobben de Bruyn
    Mar 22, 2017 at 3:37
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    $\begingroup$ @WillSawin I see, so you use this local section to show the set you defined is a submanifold, is that right? (Oh and by the way - you accidentally wrote "as $H$ does no contain the identity..." instead of "identity component"). $\endgroup$
    – Jack the Ripper
    Mar 22, 2017 at 6:08
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In fact, a much stronger result (due to Ito and Kawada), see Theorem 2.3 in Emmanuel Breuillard's notes. To wit, if the support of a measure is not contained in a proper closed subgroup then a random walk is eventually equidistributed. If it is contained in a proper closed subgroup, there are two cases: the first is that the subgroup is of codimension zero (in which case it contains the identity component, and we are done), or it is of positive codimension (which is obviously non-generic).

See also Stromberg 1960:

MR0114874 (22 #5692) Reviewed 
Stromberg, Karl
Probabilities on a compact group. 
Trans. Amer. Math. Soc. 94 1960 295–309.
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    $\begingroup$ @user44191 see the edit. $\endgroup$
    – Igor Rivin
    Mar 19, 2017 at 3:57
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    $\begingroup$ I'm still not sure I understand your proof here. What is the measure that you're having generate the random walk? $\endgroup$
    – user44191
    Mar 19, 2017 at 4:49
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    $\begingroup$ @user44191 and Igor no you not only misread Breuillard's assumption but modify it in your comment. No Breuillard's notes are not confusing, but you seem to be confused anyway and this comment in your post is confusing. It's not "generated by neighborhoods of 1" (which is unclear and could be interpreted as "generated by any neighborhood of 1" but "generated by a compact neighborhood of 1", which is also known as "compactly generated", meaning that some compact neighborhood of 1 generates $G$, and is trivially fulfilled if $G$ is any compact group. $\endgroup$
    – YCor
    Mar 19, 2017 at 14:28
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    $\begingroup$ I don't think I understood your answer. What you say is obvious seems to be exactly my question (why isn't the support of the measure - i.e. the two elements in question - not contained in a subgroup of positive codimension?) and what you deduce from Ito-Kawada seems irrelevant to me (albeit very neat). $\endgroup$
    – Jack the Ripper
    Mar 19, 2017 at 15:21
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    $\begingroup$ @Igor yes, thanks a lot although I turned out to know this. I was just illustrating to explain why a theorem of this sort (Ito-Kawada) is not enough to answer the question. $\endgroup$
    – YCor
    Mar 20, 2017 at 17:50

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