Lie group structure on exotic $\mathbb{R}^4$

Question

Are there Lie group structures on exotic $\mathbb{R}^4$s?

By the theorem that every continuous group homomorphism of two Lie groups is smooth, we can conclude that if $G$ and $H$ are two Lie groups that are isomorphic as groups, then if they are homeomorphic, they are also diffeomorphic.

From this we can conclude that the topological group $(\mathbb{R}^4,+)$ only admits one smooth structure turning it into a Lie group (namely the usual one). Now there are uncountably many other smooth structures on $\mathbb{R}^4$ (but still homeomorphic), namely the so called exotic $\mathbb{R}^4$s. By our above arguments we know that exotic $\mathbb{R}^4$s do not admit a Lie group structure when $+$ is the operation.

But does there exist a different group operation on some exotic $\mathbb{R}^4$ turning it into a Lie group?

Answer 1

No, there are no such Lie groups. One way to answer is to observe that if $G$ is a contractible (as a topological space) Lie group then its exponential map is a diffeomorphism, see for instance Yves' answer here and references therein. In dimension 4 one can prove the result you are asking about without quoting any deep theorems and using the basic structure theory of Lie groups and Lie algebras. Namely, if $G$ is a simply-connected Lie group, consider its Levi-Malcev decomposition $S\rtimes H$, where $S$ is solvable and $H$ is semisimple. Then check that both $S$, $H$ have to be simply-connected and if $G$ is contractible, so are $S$ and $H$. There are few cases to consider then, the most interesting one is when $H$ is nontrivial. There are just two nontrivial semisimple Lie algebras of dimension $\le 4$, namely, $o(3)$ and $sl(2,\mathbb R)$ (it is a nice exercise which you can solve without appealing to root systems). The first one will not give you a contractible Lie group. The second one will give you Lie groups locally isomorphic to $SL(2,\mathbb R)$, leaving you with the universal cover of this group. The latter is easily seen to be diffeomorphic to $\mathbb R^3$ (start by proving that $SL(2,\mathbb R)$ is diffeomorphic to $S^1\times \mathbb R^2$). Lastly, taking a semidirect product with $\mathbb R$ (which is actually a direct product in this case) gives you a group diffeomorphic to $\mathbb R^4$.