# Connected, one-dimensional Lie groups

Show that a connected, one-dimensional Lie group $G$ is isomorphic to $\mathbb{R}$ or $S^1$.

So far my approach has been to show a non-trivial, one-parameter subgroup of $G$ is surjective, but I have not really made much progress.

I have only just begun studying Lie groups, so my knowledge of theoretical results is basically limited to the definition of the exponential map $\exp_G: T_e G\to G$ and a few results regarding this.

Since $G$ is one-dimensional, its exponential map is of the form $$\exp:\mathbb{R}\to G.$$ Moreover, since $G$ is connected and abelian, the exponential map is surjective. Thus, either $\ker\exp=\{0\}$ and $G\cong\Bbb R$, or $\ker\exp=r\mathbb{Z}$ for some $r>0$ and $\exp$ factors through an isomorphism $S^1\cong \Bbb R/r\Bbb Z\to G$.

• Why do you assume that $G$ is abelian? And even if it is how does it follow that the exponential map is surjective? – freakish Feb 13 '17 at 13:24
• @freakish A Lie group with abelian Lie algebra is abelian (see here). Since $G$ is connected, it is generated by a neighborhood of the identity. Thus, if $g\in G$ then $g=\exp(x_1)\cdots\exp(x_n)$ for $x_i\in\Bbb R$. Now, because $\mathbb{R}$ is abelian, $[x_i,x_j]=0$ for all $i,j$ and hence $g=\exp(x_1+\cdots+x_n)\in\exp(\mathbb{R})$. Thus, $\exp$ is surjective. – Spenser Feb 13 '17 at 13:29
• Thanks, this is prefect! I have one question though: How do you conclude $\ker\exp = \{0\}$ or $\ker\exp = r\mathbb{Z}$? – user114158 Feb 13 '17 at 14:30
• @MadsFriis Let $A=\ker\exp$ and suppose $A\neq\{0\}$. By the comment above, $\exp$ is a group homomorphism. Since $\exp$ is continuous, $A$ is a closed subgroup of $\mathbb{R}$. Let $r=\inf\{a\in A:a>0\}$. Since $\exp$ is injective on some neighborhood of $0$, we have $r>0$. Moreover, $r\in A$ since $A$ is closed. Thus, $r\Bbb Z\subseteq A$ since $A$ is a group. We now show that $A\subseteq r\Bbb Z$. Let $a\in A$ and suppose that $a\notin r\Bbb Z$. Then, there exists $k\in\Bbb Z$ such that $0<a-kr<r$. But $a-kr\in A$ since $r\in A$ so it contradicts the definition of $r$. Thus, $A=r\Bbb Z$. – Spenser Feb 13 '17 at 17:23
• @Spenser yea, this is pretty much what I arrived at after an hours work - thanks! – user114158 Feb 14 '17 at 13:23

One way would be to go to Lie algebras. If $G$ is one-dimensional then its Lie algebra is also one-dimensional. As such it is abelian. So all one-dimensional Lie groups have isomorphic Lie algebras. Therefore by Lie's third theorem they all have isomorphic universal covering groups.

Now it can be easily checked that $\mathbb{R}$ has itself as a universal covering group. Therefore $\mathbb{R}$ is a universal covering group for every Lie group of dimension one. In particular every Lie group of dimension one is a homomorphic image of $\mathbb{R}$. Now you can check that every (proper) closed subgroup of $\mathbb{R}$ is of the form $r\mathbb{Z}$ for some $r\in\mathbb{R}$. Thus the only quotients of $\mathbb{R}$ that could be candidates for a Lie group are $\mathbb{R}$ itself and $S^1$. This completes the proof.

• Thanks! I have not really studied Lie algebras and such yet, so I would prefer something "elementary" only concerning Lie groups. – user114158 Feb 13 '17 at 13:08