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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.

ARichardson
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  • I don't know where you found this "solution," but I definitely did not write it. I would not have claimed that a Lie group isomorphism is also a Lie algebra isomorphism. – Jack Lee Aug 06 '21 at 21:41
  • Wow, did not expect a response from the man himself. I have seen a response somewhere on here referencing this https://wj32.org/wp/wp-content/uploads/2012/12/Introduction-to-Smooth-Manifolds.pdf as being self solutions. Since there seems to be no name attached to the document, I did not question the assertion. – ARichardson Aug 07 '21 at 23:29
  • I don't know who wrote those solutions, but it wasn't me. They should not be considered authoritative. – Jack Lee Aug 08 '21 at 15:00
  • Found it, https://math.stackexchange.com/questions/1089185/topological-covering-local-diffeomorphism-gives-smooth-covering – ARichardson Feb 19 '22 at 23:01
  • See the comment I posted after the one you linked to. I don't know where Tomasso Seneci got the idea that those solutions were posted by me; they weren't. – Jack Lee Feb 20 '22 at 00:04

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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 $GL(n,\Bbb R)$.

José Carlos Santos
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  • Ok, this is super helpful. Just two clarifying questions $g$ is a function, not an element of the group? It seemed like you were using greek for functions and non-greek for elements and I don't want to miss what you are saying here. Second, how do we extend it? I know $GL(n,\mathbb{R}) is supposed to be an open subset of $\mathfrak{gl}(n,\mathbb{R})$, so would this be a time to use bump functions or partitions of unity? And why does the definition allow for this? I guess that is 3, but thank you – ARichardson Aug 06 '21 at 04:08
  • Sorry, that is supposed to read $GL(n,\mathbb{R})$ is supposed to be an open subset of $\mathfrak{gl}(n,\mathbb{R})$ , but I missed the edit time window. Also not sure how they were supposed to be isomorphic, which is why I asked here incase there was something simple was overlooking – ARichardson Aug 06 '21 at 04:16
  • The elements of $GL(V)$ are the invertible linear maps from $V$ into itself; the elements of $GL(n\Bbb R)$ are the invertible linear maps from $\Bbb R^n$ into itself (or the invertible $n\times n$ matrices; it amounts to the same thing). And the way I defined $\varphi$ also makes sense when $g$ is any linear from $V$ into itself. And the set of all such linear maps is precisely $\mathfrak{gl}(V)$. – José Carlos Santos Aug 06 '21 at 07:52
  • Isn't the set of all such maps $GL(V)$, and the algebra on them $\mathfrak{gl}(V)$? – ARichardson Aug 08 '21 at 20:02
  • If you're talking about the group of all invertible linear maps from $V$ to $V$ then, yes, that's $GL(V)$. And its Lie algebra is $\mathfrak{gl}(V)$. – José Carlos Santos Aug 08 '21 at 20:24