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The exponential function as I know it is defined as: $$\exp:\mathfrak{g}\to G$$ and it gives each element $X$ the value of $\exp_X(1)$ where $\exp_X$ is the unique $\mathbb{R}\to G$ homomorphism that has $d(\exp_X)_0 (0)= X$.

This definition makes sense to me (when I look for example at matrix groups or more simply at $\mathbb{R}$). Despite that, in all simple examples the exponential map is a $G \to G$ map. Of course in these simple cases $\mathfrak{g} \simeq G$ so it still coincides with the definition above. Even so, I see no good reason that the generalization of the exponential map from matrix groups to all group be a $\mathfrak{g}\to G$ map.

Despite that, on matrix groups $\exp$ could be defined as a $G \to M_n(\mathbb{F})$ map, and then you will need to restrict it to a $\exp^{-1}(G) \to G$ map, where for some reason we will have $\exp^{-1}(G)\simeq \mathfrak{g}$. Even so, I see no good reason that the generalization of the exponential map from matrix groups to all group be a $\mathfrak{g}\to G$ map.

What is the reasoning for that?

Alexey
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    Do you know about the exponential map for $G=S^1$? Is its domain $G$? This is really the simplest example. – Moishe Kohan May 02 '22 at 13:47
  • @MoisheKohan I looked it up now and it indeed seems like the image of $exp$ isn't $G$. The explanation I saw was from here. And it implicitly "claimed" that $log G = \mathfrak{g}$. What is the reasoning\intuition behind that? – Alexey May 02 '22 at 14:10
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    You are looking at a wrong place. I suggest you at least read the Wikipedia article, https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory) – Moishe Kohan May 02 '22 at 15:02
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    You're certainly wrong to say that even in the simple cases you have $\mathfrak g \cong G$. If $G=GL(n,\Bbb R)$, what is $\mathfrak g$? – Ted Shifrin May 02 '22 at 16:44

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One way to think of this would be via ODE's, which is in line with the definition of $\exp$ you are using. Indeed, if we have an anonymous manifold $M$, a point $x\in M$, and a (sufficiently regular) vector field $X:M\to TM$ on it, one can consider the initial value problem

$$\dfrac{d\phi}{dt}(t)=X(\phi(t)), \phi(0)=x$$

where $\phi:\mathbb{R}\to M$ is the unknown function. The Existence and Uniqueness Theorem in ODE's guarantees that this IVP has a unique solution $\phi$, which is as regular as the vector field. One can then evaluate $\phi$ at a fixed time $t$ to get a point in $M$. Thus given the base point $x$ and a time parameter $t$, we have a unique map from the vector fields to the manifold.

This applies in particular to a Lie group $G=M$, which is in particular a manifold, with $x=e_G$ the identity element and $t=1$. The only discrepancy is that the Lie algebra $\operatorname{lie}(G)=\mathfrak{g}$ of $G$ is not exactly a space of vector fields on $G$, however one can, in a unique manner, produce a vector field on $G$ out of any element of $\mathfrak{g}$ by simply pushing it forward by the left multiplication operation on $G$; the resulting vector field would be sufficiently regular (it would be real analytic) (see e.g. Left invariant vector field surjects onto the Lie algebra, What is the advantage of defining Lie Algebras by left-invariant vector fields of a Lie Group?, Why is left-invariant vector fields needed to construct a Lie algebra from a Lie group?).

Regarding the simplest example, note that formally speaking we have that, according to our definition of the exponential map of a Lie group, $\exp^{\mathbb{R}_{>0}}:\mathbb{R}\to\mathbb{R}_{>0},X\mapsto e^X$, which is the standard exponential function, and/but $\exp^{\mathbb{R}}:\mathbb{R}\to\mathbb{R}, X\mapsto X$, where $\exp^G$ is the exponential map of the Lie group $G$. Although as you observed these two maps are the same up to a change of coordinates, there is some intrinsic exponential growth of velocities encoded in the first one, whereas there is no growth of velocities in the second one.

Alp Uzman
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