Can any of the exotic differentiable structures on $\mathbb R^4$ make $GL(\mathbb R^2)$ into an 'exotic' Lie group structure?

I am just beginning to learn about Lie groups and am made somewhat uncomfortable by the textbook's handwavy decision to talk about Lie groups $GL(V)$ where $V$ is some $n$-dimensional real vector space. It is not clear to me that when when $V\cong\mathbb R^2$ and $End(V)$ is made isomorphic to $\mathbb R^4$ after a choice of basis for $V$, that that $\mathbb R^4$ cannot be exotic (as I know nothing about exotic things except their existence). I believe that maybe the requirement that matrix multiplication be differentiable might settle the issue, but I have no idea to go about showing this.

Otherwise, if I choose to believe that all $n$-dimensional vector spaces for $n\neq 4$ have a unique differentiable structure (they have unique topologies for sure), it becomes obvious that talking about $GL(V)$ for arbitrary real vector spaces is perfectly sensical.

So in short, can $GL(\mathbb R^2)$ ever be a Lie group not diffeomorphic to standard $\mathbb R^4$?

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The $\mathbb{R}^4$ cannot be exotic because the isomorphism to a standard $\mathbb{R}^4$ is linear, and in particular smooth. $\text{GL}_2(\mathbb{R})$ is not even homotopy equivalent to $\mathbb{R}^4$. – Qiaochu Yuan Jul 20 '14 at 16:42

Another way of seeing the answer is "no" is the following theorem:

Let $G$ and $H$ be two compact Lie groups. Suppose $f:G\rightarrow H$ is a continuous homomorphism. Then $f$ is smooth.

Letting your $G = GL(2)$ and $H = GL(2) (exotic)$, the identity map $i:GL(2)\rightarrow GL(2)$ is clearly continuous and a homomorphism, hence it is smooth. The exact same argument shows in the inverse is smooth, so $i$ is a diffeomorphism.

To prove the theorem to begin with, I'd begin with Cartan's Theorem which states that any (topologically) closed subgroup of a Lie group is automatically a smooth submanifold. The proof of this fact can be found in, for example, John Lee's book Introduction to Smooth Manifolds on pg. 526.

Now, given a continuous homomorphism $f:G\rightarrow H$ consider the graph of $f$ $K=\{(g,f(g))\}\subseteq G\times H$. Because $f$ is a homomorphism, $K$ is a subgroup. Because $f$ is continuous, $K$ is a closed subset. By Cartan's Theorem, $K$ is an embedded Lie subgroup.

Next, I claim that if $\pi_1:G\times H\rightarrow G$ is the projection, then $\pi_1|_K:K\rightarrow G$ is actually a diffeomorphism. It is 1-1 because $f$ is a function and onto because the domain of $f$ is all of $G$. It is smooth since it is the restriction of a smooth function to a smooth submanifold (thanks to Cartan's theorem). The fact that the inverse is smooth takes more work. One way to see it is that every smooth homomorphism has constant rank and and by Sard's theorem, the map has full rank somewhere. (If you'd like, I can expand on why the inverse if smooth.)

Finally, notice then that if $\pi_2:G\times H\rightarrow H$ is the other projection map, then $f = \pi_2 \circ \pi_1^{-1}$ is a composition of smooth maps, so is smooth.

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