# How do the subgroups of $(\mathbb C,+)$ which are connected (path connected) in $\mathbb C$ look like?

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 )

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

## 1 Answer

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\$.

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