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I read (in Structure and Geometry of Lie Groups by Hilgert and Neeb, I think) that it is possible for a Lie group to admit a connected yet not-path-connected subgroup. Specifically it said $(\mathbb{R/Z})^2$ admits such a subgroup.

I did not manage to construct such a subgroup. Can someone help?

Obviously such a subgroup can't be closed, or even a Lie subgroup, because connected Lie groups are path-connected (as all manifolds).

The only group I know which is connected but not path-connected is the solenoid, but it is compact, so if it were a subgroup of $\mathbb{T}^2$ it would have been closed, hence a Lie subgroup and therefore path-connected, which is a contradiction.

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  • $\begingroup$ Do you at least know what that subgroup should look like? Have you seen examples of connected, but not path connected spaces? $\endgroup$ – xyzzyz Aug 17 '16 at 19:56
  • $\begingroup$ I have no idea what it shoud look like. I know it can't be a Lie subgroup, and hence can't be closed (by Cartan's theorem). The only group I know which is connected but not path-connected is the solenoid. But it is compact, so if it were a subgroup of $\mathbb{T}^2$ it would have been closed, hence a Lie subgroup and therefore path-connected, which is a contradiction. $\endgroup$ – Cronus Aug 17 '16 at 20:02
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Apparently, similar questions have been asked in MathOverflow. I will quote them here in case anyone else is interested in this.

New Answer

Here Gabriel C. Drummond-Cole gave a very nice answer: the subgroup $\{(x+\mathbb{Z}^2,\pi x+\mathbb{Z}^2)\}\cup\{(x+\frac{1}{2}+\mathbb{Z}^2,\pi x+\mathbb{Z}^2)\}$ (for $x\in\mathbb{R}$). A direct computation shows this is a subgroup (using the fact $1/2+1/2=0$ in the torus). It has two path connected components: clearly $\{(x+\mathbb{Z}^2,\pi x+\mathbb{Z}^2)\}$ and $\{(x+\frac{1}{2}+\mathbb{Z}^2,\pi x+\mathbb{Z}^2)\}$ are path-connected (as the images of $\{(x,\pi x\}$ and $\{(x+\frac{1}{2},\pi x\}$), and there is no path connecting $(0+\mathbb{Z}^2,0+\mathbb{Z}^2)$ + $(\frac{1}{2}+\mathbb{Z}^2,0+\mathbb{Z}^2)$ since such a path will lift to a path from $(0,0)$ to $(\frac{1}{2},0)$ which lies in the preimage of this subgroup in $\mathbb{R}^2$, i.e. in $\{(x,\pi x)\}\cup\{(x+\frac{1}{2},\pi x\} + \mathbb{Z}^2\subseteq\mathbb{R}^2$. Obviously in this set $(0,0)$ and $(\frac{1}{2},0)$ are in different path-connected components, so this is a contradiction. So our subgroup is not path-connected. However, it is connected: since it has precisely two path-component, we just need to show neither of them is clopen. It's enough to show neither is closed because there are just finitely many (two) of them. But each of them is known to be dense in $(\mathbb{R}/\mathbb{Z})^2$, so you can approach a point of one from points in the other, so they are not closed.

This is a very nice and simple answer.

Old Answer

This question asks in general about subgroups which are not Lie subgroups, but for nontrivial reasons (i.e. not just because they must have uncountably many connected components). Essentially only Claudio Gorodski answered:

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.

The paper by Jones solves the problem. He proves (Theorem 5, page 118) there is a function $f:\mathbb{R}\to\mathbb{R}$ which is additive (i.e. $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$) and which satisfies the following:

  1. $f$ is not continuous,
  2. The graph of $f$ is connected.
  3. The graph of $f$ is not path-connected (follows from Property $3$ on page 119, since any path lying inside the graph is a bounded connected subset. This also follows from Property $2$ on the same page, since any nontrivial path is a nondegenerate continuum (i.e. a connected compact subset containing more than one point)).

To be precise, he constructs an additive function $f:\mathbb{R}\to\mathbb{R}$ which satisfies $1$ and $2$, and asserts property $3$ follows. For this he referred to two articles in German, unfortunately.

However, if we believe there is such a function $f$, then its graph is a subgroup of $\mathbb{R}^2$ (by additivity of $f$) which is connected and not path-connected, and hence this solves my question.

I am not particularly satisfied with this answer: I was hoping for a more pleasant (and more complete) argument. Maybe there isn't such.

In any case, there was a second question in MathOverflow, inspired by the first. It asks whether there are connected non-path-connected subgroups of $\mathbb{R}^2$ by explicitly asking for a construction of $f$ as above.

Martin M. W. gave the following answer, which maybe is simpler than Jones's proof (although it also omits the explanation why $3$ is true):

I think the answer is, yes, the graph can be connected.

By definition, if the graph G is not connected, then we can find disjoint nonempty open sets A and B, such that G is contained in A union B. In particular, this implies no point in G can be contained in the boundary of A. So if we can construct an additive function f whose graph intersects the boundary of any potential separating open set A, we'll have shown the graph is connected.

Before constructing this function, note a technical point. Not all open sets are candidates for separating G. If G = A union B for nonempty open sets A,B, then the projections proj(A) and proj(B) onto the x-axis are both open, and must intersect. In turn this implies the projection of the boundary of A contains an interval. Call open sets with this property "candidate sets".

To make a function f whose graph intersects the boundary of all candidate sets, consider a basis H for R as a vector space over Q. This set has cardinality of the reals. Now note that the set of all open sets in R^2 also has cardinality of the reals. (http://en.wikipedia.org/wiki/Cardinality_of_the_continuum)

Put these two sets (basis H, all open sets) in 1-1 correspondence, so for each h in H, we have an open set O(h). If O(h) is not a "candidate set," let f(h)=0. Otherwise, using the fact that O(h) is a candidate set, we can always find a nonzero rational q, and a real y such that (qh,y) is in the boundary of O(h). Define f(qh)=y. Doing this for all elements of H will determine a unique additive function f on the reals.

The graph of f, by construction, is connected since it intersects the boundary of every candidate separating open set in R^2. (And it's not continuous--if it were, it would miss the boundaries of a lot of open sets!)

There was a second answer, by Andrew Clifford, which linked to two papers: one by Thomas and one by Maehara.

Maehara again constructs an additive function $f:\mathbb{R}\to\mathbb{R}$ satisfying $1$ and $2$ but doesn't prove $3$.

Thomas supposedly proves exactly what we (I?) want: a Lie group of dimension greater than $1$ always admits a connected subgroup which is not a Lie subgroup (and hence isn't path connected, since connected Lie subgroups are path-connected). However, Andrew Clifford (who gave the link to this paper) also wrote "Math Reviews says that there is an error in it, but I think the theorem still holds for Abelian groups." Read at your own risk.


Briefly, the most satisfying answer is the paper by Thomas, if indeed it is true for Abelian groups. If not, I am still hoping to find a cleaner argument. After all, this was supposed to be an exercise.

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