Exercise from Lee's smooth manifolds As the title states, this is exercise 8.43 in Lee's "Intro to Smooth Manifolds".
If $V$ is any finite-dimensional real vector space, the composition of the canonical isomorphisms yields a Lie algebra isomorphism between
$$lie\big(GL(V)\big)\overset{\chi_{Id}}{\to}T_{Id}GL(V)\to \mathfrak{gl}(V)$$
Now, I am very new to differential geometry, so things that may be obvious are not sinking in as such yet. But, I found a solution Lee had written, starting
"Choose a basis $B$ for $V$ and let $\phi: GL(V) \to GL(n,\mathbb{R})$ be the associated Lie
group isomorphism. It is also a Lie algebra isomorphism $\mathfrak{gl}(V)\to \mathfrak{gl}(n,\mathbb{R})$ since it is linear. ..."
Here is my confusion/question. Is $GL(n,\mathbb{R})$ isomorphic to $\mathfrak{gl}(n,\mathbb{R})$? Or, more generally, is any Lie group isomorphic to its Lie Algebra? Does this question even make sense? From the same book, I know that $lie(GL(n,\mathbb{R})$ is isomorphic to $\mathfrak{gl}(n,\mathbb{R})$, but this is involving the left-invariant smooth vector fields on the group, not necessarily the group elements themselves. I spent a while trying to work through it similarly to the solution he gave for the $GL(n,\mathbb{R})$ to $\mathfrak{gl}(n,\mathbb{R})$ isomorphism before running into this solution, but cannot make sense of that second implication unless $GL(V)$ is isomorphic to $\mathfrak{gl}(V)$, and respectively for the general linear group.
 A: In what sense would $GL(n,\Bbb R)$ and $\mathfrak{gl}(n,\Bbb R)$ would be isomorphic? Are you asking whether the Lie group $GL(n,\Bbb R)$ is isomorphic with the Lie group $(\mathfrak{gl}(n,\Bbb R),+)$? The answer is negative, since the second group is abelian, whereas the first one is not.
Bet, yes, a Lie group and its Lie algebra can be isomorphic in the sense of the previous paragraph. The Lie group $(\Bbb R^n,+)$ is isomorphic to its Lie algebra.
Anyway, if $B=\{v_1,\ldots,v_n\}$ is a basis of $V$ and if you define $\psi\colon V\longrightarrow\Bbb R^n$ by$$\psi(a_1v_1+a_2v_2+\cdots+a_nv_n)=(a_1,a_2,\ldots,a_n),$$then $\psi$ is a vector space isomorphism, which induces an isomorphism $\phi\colon GL(V)\longrightarrow GL(n,\Bbb R)$:$$\phi(g)=\psi\circ g\circ\psi^{-1}.\tag1$$And $\phi$ is a linear map. Actually, the definition $(1)$ allows you to extend $\phi$ to a map $\phi\colon\mathfrak{gl}(V)\longrightarrow\mathfrak{gl}(n,\Bbb R)$ which is a Lie algebra isomorphism.
So, what Lee is claiming is that there is a Lie algebra isomorphism $\phi\colon\mathfrak{gl}(V)\longrightarrow\mathfrak{gl}(n,\Bbb R)$ and that the restriction if $\phi$ to $GL(V)$ is a Lie group isomorphism from $GL(V)$ onto $GL(n,\Bbb R)$.
