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Let $G$ be a Lie group and $\mathfrak g$ the respective Lie algebra. It is a well-known statement that the differential $d\exp_0:\mathfrak g\to\mathfrak g$ of the exponential map $\exp:\mathfrak g\to G$ is the identity. However this statement seems rather problematic to me for the following reasons.

Strictly speaking, $d\exp_0$ is a map from $T_0\mathfrak g$ to $T_eG$.

Problem 1: As I see it, for $T_0\mathfrak g$ to exist, $\mathfrak g$ would have to be a manifold. Is this generally true and if so, why?

Let's assume the tangential space of $\mathfrak g$ in the zero vector-field exists or is well-defined or whatever. Of course we can isomorphically identify $\mathfrak g$ with $T_eG$ via the map $\Phi_G:X\mapsto X_e$. However to make sense of $\widehat{d\exp_0}:\mathfrak g\to\mathfrak g$, we'd need a corresponding map from $\mathfrak g$ to $T_0\mathfrak g$ as

$$ \require{AMScd}\begin{CD} \mathfrak g @>{\widehat{d\exp_0}}>> \mathfrak g\\ @VV(?)V @AA\Phi_G^{-1}A \\ T_0\mathfrak g @>{d\exp_0}>> T_eG \end{CD} $$

However, the isomorphism $\Phi$ would "only" identify $T_0\mathfrak g$ with $\mathfrak g(\mathfrak g)$, the latter being the set of left-invariant vector fields on $\mathfrak g$. What would left-invariant even mean as we don't have a multiplication on $\mathfrak g$ to begin with?

Problem 2: Is there another isomorphism between $\mathfrak g$ and $T_0\mathfrak g$ I'm not aware of? How else should one make sense of the statement "$d\exp_0=\operatorname{id}_{\mathfrak g}$"?

I've also checked some literature; for example Warner in "Foundations of Differentiable Manifolds and Lie Groups" Theorem 3.31 states, that "$d\exp:T_0\mathfrak g\to T_eG$ is the identity map (with the usual identifications)" which in the respective proof is "immediate for $tX$ is a curve in $\mathfrak g$ whose tangent vector at $t=0$ is $X$". But on the 20 pages between introducing $\mathfrak g$ and said theorem, I did not find any "usual identification" of $\mathfrak g$ and $T_0\mathfrak g$...

Thanks in advance for any answer or comment!

Frederik vom Ende
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$\mathfrak{g}$ is a finite-dimensional vector space. Any finite-dimensional vector space must be isomorphic to $\mathbb{R}^n$ for some $n$, and this gives you a manifold structure, which does not depend on the choice of an isomorphism since any bijective linear map $\mathbb{R}^n\to \mathbb{R}^n$ is a diffeomorphism. Then it's easy to see that with this manifold structure the tangent space at any point $x$ can be identified with $\mathfrak{g}$, exactly as the tangent space at any point in $\mathbb{R}^n$ is $\mathbb{R}^n$.

Arnaud D.
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  • This really helped me, thanks a lot! Do you know if one can state how $\Psi:\mathfrak g\to T_0\mathfrak g$ explicitely operates, e.g. if $\Psi(X)$ for any $X\in\mathfrak g$ can be written in an easy way? – Frederik vom Ende Mar 01 '18 at 13:32
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    Well, if you define $T_0\mathfrak{g}$ as the set of equivalence classes of curves which have the same tangent vector at $0$ in some chart, then you can simply identify a vector with the equivalence class of the corresponding straight line : for any $X\in \mathfrak{g}$ you just take the equivalence class of $t\mapsto tX$. – Arnaud D. Mar 01 '18 at 13:37