# Non-Lie Subgroups

A result of Borel and Lichnerowicz states that the holonomy group of a connection on a principal $G$-bundle is a Lie subgroup of $G$ (Cartan had earlier asserted this, but apparently without proof).
This restriction, that it be a Lie subgroup, allows for a lot of poorly-behaved subgroups, for example a line with irrational slope on a torus. This subgroup comes from a perfectly fine immersion of the Lie group $\mathbb{R}$, but it's not closed in the induced topology of the torus.

As an example of something that's not a Lie subgroup, let $G= \mathbb{R}$, consider an uncountable set of $\mathbb{Q}$-independent points, none of which are rational, and consider the subgroup they generate. If this were a Lie subgroup it would be the image of an uncountable discrete space (there can't be anything $1$-dimensional, since we left out the rationals), which wouldn't be second countable, hence not a manifold and not a Lie group.

This seems like a pretty contrived example, and I suspect there is more content to 'being a Lie subgroup' than having countably many components. However, I can't seem to pin down something that would illustrate this. Can anyone give me an example of a connected subgroup of a Lie group that is not a Lie subgroup?

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@Sammy: I think it is an immersion, unless I've got my definitions wrong. The derivative is clearly injective on tangent spaces. – Kevin H. Lin Oct 28 '09 at 23:44
You're right, Kevin. I deleted my comment. – Sammy Black Oct 29 '09 at 3:09
Your question inspired this one: mathoverflow.net/questions/3289/… – David Speyer Oct 29 '09 at 17:14

## 3 Answers

As it turns out, any connected subgroup of a Lie group must be a Lie subgroup. See Sigurdur Helgason's book for more details.

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I don't have the book, but it appears that the theorem is for path-connected subgroups: ams.org/mathscinet-getitem?mr=38356. See mathoverflow.net/questions/3289/… for a counterexample that is connected but not path-connected. – Eric Wofsey Oct 30 '09 at 2:46
I see. Thanks for making the distinction! – Aaron Mazel-Gee Oct 30 '09 at 3:04
On a quick skim, all I see is Theorem 2.3 which says that a connected closed subgroup is a Lie subgroup and prop. 2.11, a locally compact subgroup is closed. To my mind, this theorem gives a negative answer to the question Harold may have meant to ask, namely: "Are there are topologically reasonable subgroups which are not Lie subgroups?". – David Speyer Oct 30 '09 at 3:24

Apparently, there is no consensus in the literature on how to define a Lie subgroup of a Lie group. One should be careful not to mix different definitions. The result of Yamabe (1950) states that every arcwise connected subgroup of a real Lie group is Lie subgroup (in the sense of Yamabe). Bourbaki uses the notion of an integral subgroup (Lie Groups and Lie Algebras, Chapter III, §8, Exercise 4). Onishchik/Vinberg use the notion of a virtual Lie subgroup (Foundations of Lie theory in Lie groups and Lie algebras I, p. 39, Theorem 2.4). Note that Onishchik/Vinberg also use the (stronger) notion of a Lie subgroup and present a counterexample not fulfilling their definition on page 14. For more on the technical details one can refer to Bourbaki (Chapter III, §6 and §8) or Onishchik/Vinberg.

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As Aaron and Eric have pointed out, Yamabe's theorem is the statement that an arcwise connected subgroup of a Lie group must be a Lie subgroup. There is also a classical theorem by Cartan and Chevalley, also proved in Hochschild's book, that a connected locally compact topological group admitting a continuous homomorphism into a Lie group which is injective in a neighborhood of the identity must be a Lie group. In particular, if $H$ is a subgroup of a Lie group $G$ and $H$ is connected and locally compact with respect to a topology containing the relative topology, then $H$ is a Lie group (note that Prop. 2.11 in the book by Helgason only addresses local compactness in the relative topology).

I found a very interesting discussion by Shahla Ahdout and Sheldon Rothman in the Australian Mathematical Society Web Site - the Gazette in which they exhibit an example of connected, locally connected subgroup of the additive group $\mathbf R^2$ which contains no arcs and is dense in $\mathbf R^2$, essentially quoting F. Burton Jones, Connected and disconnected plane sets and the functional equation f(x) +f(y) =f(x+y), Bull. Amer. Math. Soc. 49 (1942), 115-120. Such a subgroup is not a Lie group, indicating how essential is the role of arcwise connectivity.

Edit: Regarding Robert's post, it is certainly important to point out the different existing definitions of Lie subgroups. To make myself clear, I am using the definition of Lie subgroup as in the book by Helgason, ch.2 (or Warner, ch. 3), namely, a Lie subgroup of a Lie group is an abstract subgroup which is an (immersed) submanifold and may have a topology finer than the relative topology. I think this definition is very common and has the advantage of yielding, for a given Lie group, a bijective correspondence between Lie subalgebras of its Lie algebra and its Lie subgroups.

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The definition in Helgason: a Lie subgroup of $G$ is a submanifold $H$ which is a subgroup of $G$, and is a topological group. A submanifold means that $H$ is a manifold endowed with an immersion $H\mapsto G$. In this def, any subgroup of $G$ could be considered as a 0-dimensional Lie subgroup, and we need beware that a Lie subgroup is not only determined by its image in $G$, and might be non-connected even if its image in $G$ is connected. A (quite artificial) way to avoid these issues is to assume $H$ has at most countably many components, as Helgason does explicitly in several propositions. – YCor Nov 28 '11 at 17:50
Yes, I also implicitly assumed manifolds are Hausdorff and second countable. I don't think the second countability is so artificial, as otherwise, e.g. $\mathbf R^2$ would have a very strange $1$-dimensional manifold structure by decomposing into a noncountably many copies of $\mathbf R$. It is also known that for a Lie group second countability follows from having countably many components. – Claudio Gorodski Nov 28 '11 at 19:01