The exponential map I'm following a course about riemannian geometry, and I was fascinated with the exponential map.
I was wondering what the reason of this name is... is there any relationship with the real and complex exponential?
 A: Yes, there is, although the name origins from the fact that for the Lie group $GL(n,\mathbb{R})$ the exponential map is the usual matrix exponential. To make this precise:
Consider $GL(n,\mathbb{R})$ as an open set of $\mathbb{R}^{n^2}$. Thus it is a smooth manifold and the usual matrix multiplication is a smooth map.
If we consider the tangent space at the Idendity we find $T_{Id}GL(n,\mathbb{R})=M(n,\mathbb{R})$. Now take an element of this tangent space $x\in M(n,\mathbb{R})$ and define the curve $c:t\mapsto Id+tx$ for some $t\in (-\epsilon,\epsilon)$, which is indeed a curve on $GL(n,\mathbb{R})$ for small enough $\epsilon$. On the other hand we find that the left invariant vector field associated to this $x\in T_{Id}GL(n,\mathbb{R})$ is given by $X(g)=D_eL_g(x)=gx$ (denoting by $L_g$ the usual left multiplication). The next task is to solve the ordinary differential equation
$X(c(t))=c'(t)$ where $c(0)=Id$
The unique solution of this ODE is given by $c(t)=Exp(tx)=\sum\limits_{k=0}^\infty \frac{(tX)^k}{k!}$ 
When remembering that the definition of the Lie Group exponential given by $\exp_G(x)=c(1)$ for the flow associated to a left invariant vector field 
we realize that we exactly resemble the usual exponential map $\exp_{GL(n,\mathbb{R})}(x)=Exp(x)$. Further when regarding Lie groups with biinvariant metrics one also finds that the Lie group exponential is nothing else than the value at the idendity of the more general exponential map defined on Riemannian manifolds.
