If $G$ is a Lie group with algebra $\frak g$ which is not simply connected and has fundamental group $\pi_1(G)$, then there exists a larger Lie group $G_0$, the cover of $G$, which has the same algebra and is simply connected. Moreover, $G\sim G_0/\pi_1(G)$.
I want to apply this theory to $SO(2,\mathbb{R})$. What is the cover of $SO(2,\mathbb{R})$?
1) I know of the relation $S^1\sim \mathbb{R}/\mathbb{Z}$. Is this the answer? Notice that here the group $\mathbb{R}$ is defined with + as its operation.
2) On the other hand, the Lie algebra of $SO(2,\mathbb{R})$ is the real line. Exponential of the real line, $e^{x}$, produces the group $\mathbb{R}_+$, positive reals with $\times$ as operation. Is this the answer?
I am confused.
Yes, the answer is $(\mathbb{R},+)$. The covering map from $\mathbb R$ onto $SO(2,\mathbb{R})$ is simply $t\mapsto\left(\begin{smallmatrix}\cos t&-\sin t\\\sin t&\cos t\end{smallmatrix}\right)$.
Concerning the exponential map, don't attach too much meaning to the this name. If you see $\mathbb R$ as the Lie algebra of the Lie group $(\mathbb{R},+)$, then the exponential map is just the identity (yes, you read it right). And if you see $\mathbb R$ as the Lie algebra of the Lie group $SO(2,\mathbb{R})$, then the exponential map is the map that I defined in my previous paragraph.