How do the subgroups of $(\mathbb C,+)$ which are connected in $\mathbb C$ look like ? Can they be characterized in some way ? (like for example , subgroups of $(\mathbb C\setminus \{0\} , .)$ , that are compact in $\mathbb C$ , are always contained in $S^1$ ) . If this class is too big , then what if we assume the subgroups to be path connected ? ( for one thing , it must be uncountable )

  • $\begingroup$ can we suppose connecting paths with a stronger regularity, like $C^1$ (looking at $\mathbb{C}$ like $\mathbb{R}^2$)? $\endgroup$ – Andrea Marino Mar 3 '17 at 18:59
  • $\begingroup$ @AndreaMarino : ah off-course if you have some answer assuming some extra regularity you are very welcome to post it $\endgroup$ – user228168 Mar 3 '17 at 19:02

There are some utterly weird connected subgroups of $(R^2,+):

F. B. Jones, Connected and disconnected plane sets and the functional equation $f(x)+f(y)=f(x+y)$. Bull. Amer. Math. Soc. 48, (1942) 115–120.

Ryuji Maehara, On a connected dense proper subgroup of $R^2$ whose complement is dense. Proc. AMS 97 (1986) 556–558.

Given such examples, I am very skeptical that one can reasonably classify connected subgroups.

On the other hand, if you assume that your subgroup is path-connected, then Yamabe's theorem shows that each path-connected subgroup of a Lie group $G$ is analytic, i.e. is obtained by exponentiating a subalgebra in the Lie algebra of $G$:

H. Yamabe, On an arcwise connected subgroup of a Lie group. Osaka Math. J. 2, (1950) 13–14.

and a more detailed (and readable) proof:

M. Goto, On an arcwise connected subgroup of a Lie group. Proc. Amer. Math. Soc. 20 1969 157–162.

In your case, the Lie algebra is $R^2$, regarded as a vector space (the Lie bracket is identically zero) and its Lie subalgebras are vector subspaces of $R^2$. The Lie-exponentiation in this case is the identity map. It follows that path-connected subgroups of $(R^2, +)$ are linear subspaces of $R^2$.

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  • $\begingroup$ um so ... that path connected subgroup must be a linear subspace ... can we prove this in some more elementary way ? $\endgroup$ – user228168 Mar 3 '17 at 18:48
  • $\begingroup$ @SaunDev: You are welcome to read the proof and adapt it to your case. It will be simpler due to commutativity of the group. $\endgroup$ – Moishe Kohan Mar 3 '17 at 18:59

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