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The question listed above (in the context of the complex numbers, but it is a reasonable question to ask in any dimension) was asked by a student in my complex analysis class, and I did not have an immediate answer. Closed subgroups of Euclidean space $({\mathbf R}^n, +)$ are of course just Lie subgroups (by Cartan's theorem), and thus split into the direct sum of a linear subspace and a discrete group by standard Lie algebra theory, and the connected closed subgroups are then nothing more than the linear subspaces; but if one just assumes connectedness but not closedness initially, I'm not sure how to get started since the group is now not obviously locally compact and most of the techniques I am aware of are not available.

Perhaps the problem is marginally simpler if one upgrades connectedness to path connectedness, but again I don't see how to proceed (except in one dimension, which is easy).

One could also pose the question in other Lie groups than Euclidean spaces, but I am expecting the Euclidean case to contain most of the essence of the difficulty.

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    $\begingroup$ Well, at least for $n=1$ that's true. :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 2 at 23:05
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    $\begingroup$ From what I remember, there are connected dense subgroups (assuming ZFC). On the other hand, path connected subgroups of Lie groups are (immersed) Lie subgroups. I will try to dig up references. $\endgroup$
    – Moishe Kohan
    Commented Oct 2 at 23:07
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    $\begingroup$ See mathoverflow.net/questions/3157/non-lie-subgroups $\endgroup$
    – Bjorn Poonen
    Commented Oct 2 at 23:18
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    $\begingroup$ Regarding the final paragraph: in some Lie groups it is very easy to produce connected non-closed subgroups, e.g. R embedded inside T^2 by taking a "line with irrational slope". But I don't immediately see how this helps with the original question. (I fear I have misread/misunderstood something.) $\endgroup$
    – Yemon Choi
    Commented Oct 2 at 23:19
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    $\begingroup$ Huh, was not expecting the connected case to be so much more pathological than the path-connected case! Need to adjust my intuition in this area... $\endgroup$
    – Terry Tao
    Commented Oct 3 at 1:24

2 Answers 2

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Here is an article with a short proof of Jones' theorem on existence of dense connected proper subgroups of $\mathbb R^2$:

Maehara, Ryuji, On a connected dense proper subgroup of $\mathbb R^2$ whose complement is connected, Proc. Am. Math. Soc. 97, 556-558 (1986). ZBL0593.54037. (In unrestricted access at AMS site)

On the other hand, a path-connected subgroup of a Lie group is always a Lie subgroup (not necessarily closed):

Yamabe, Hidehiko, On an arcwise connected subgroup of a Lie group, Osaka Math. J. 2, 13-14 (1950). ZBL0039.02101.

On the other hand, every Lie subgroup of $\mathbb R^n$ is closed (since the exponential map is a diffeomorphism); hence, every path-connected subgroup in $\mathbb R^n$ is closed.

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  • $\begingroup$ I can never remember the terminology; does "Lie subgroup" imply closed? If so, then the conclusion can be upgraded to the non-existence of non-closed, path-connected subgroups of $\mathbb R^n$ (the natural generalisation of the question asked), not just the non-existence of dense such subgroups, right? $\endgroup$
    – LSpice
    Commented Oct 2 at 23:53
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    $\begingroup$ @LSpice In a general group, there can be non-closed path-connected subgroups, as Yemon Choi points out. mathoverflow.net/questions/479975/… In $\mathbb R^n$, tehre are no non-closed path-connected subgroups. The definition used in the paper is "a continuous isomorphic image of a connected Lie group". In $\mathbb R^n$, such things must be closed subgroups by a simple argument. $\endgroup$
    – Will Sawin
    Commented Oct 3 at 0:53
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    $\begingroup$ @LSpice: it is a nice exercise to check that every Lie subgroup of $\mathbb R^n$ is closed (just look at the exponential map). $\endgroup$
    – Moishe Kohan
    Commented Oct 3 at 1:19
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    $\begingroup$ Thanks for this nice answer! At least I don't feel too embarrassed that I could not see a simple way to answer my student's question either way, it appears to be a somewhat subtle question (in particular, dependent on choice, and on the precise type of connectedness assumed). $\endgroup$
    – Terry Tao
    Commented Oct 3 at 1:29
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    $\begingroup$ Jones' paper is quite informative too on what is possible with discontinuous additive functions and their graphs. $\endgroup$
    – KP Hart
    Commented Oct 3 at 11:39
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A small amendment of the result of Hidehiko mentioned in the answer of Moishe Kohan: Theorem 2.2 in this paper implies that every continuum-connected subgroup of $\mathbb R^n$ is a (closed) linear subspace of $\mathbb R^n$.

A topological space $X$ is defined to be continuum-connected if any points of $X$ are contained in a compact connected subset of $X$. It is clear that every path-connected topological space is continuum-connected (but not vice-versa).

Therefore, the answer to the problem depends on the type of connectedness: for usual connectedness the answer is negative and for continuum-connectedness it is affirmative.

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