# Exponential map of a Lie group around $0$

I found the following on my notes of differential geometry:

Proposition: Let $G$ be a Lie group with Lie algebra $\mathfrak{g}=T_1G \cong \mathbb{R}^n$. The exponential map $\text{exp} \colon \mathfrak{g} \to G$ is smooth and $D_0\text{exp} \colon \mathfrak{g} \to \mathfrak{g}$ is the identity.

Proof: Identifying $T_0\mathfrak{g} \cong \mathfrak{g}$ we get $D_0\text{exp}(x) = \frac{d}{dt} \mid _{t=0}\text{exp}(tx)=x.$

My first question is: is it so obvious that exp is smooth that it is not even worth mentioning the proof, or is it just that one should check it with the charts and it's boring?

My second question is: $t \to \text{exp}(tx)$ is a function $\mathbb{R} \to G$, so what does the derivative $\frac{d}{dt}$ exactly mean? Why is the derivative at $t=0$ equal to $x$?

• For your second question, since $t\mapsto \text{exp}(tx)$ is a smooth curve on $G$, the derivative is what assigns a tangent vector to that curve. The derivative at 0 then assignes a tangent vector to that curve at $e$, so an element of $\mathfrak{g}$. Since the exponential map essentially generates a unique one-parameter subgroup of $G$, whose tangent vector at the identity is $x$, it is not surprising that differentiating this curve at the identity will give you $x$. – Bence Racskó Jul 30 '16 at 22:18