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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?

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Yes. See for example. One of the key things to understand is the you shouldn't think of $\exp: \mathbf{R}\to\mathbf{R}$; instead $\exp: \mathbf{R} \to \mathbf{R}_+$ is the "right" way of thinking about the exponential function. Then you realise that $\exp$ is a group homeomorphism from $(\mathbf{R},+) \to(\mathbf{R}_+,\times)$. Now $(\mathbf{R}_+,\times)$ is a 1D Lie group and hence a Riemannian manifold, and its tangent space is isomorphic to $\mathbf{R}^1$. Then you see that the $\exp$ on numbers coincide with the $\exp$ map in Riemannian geometry – Willie Wong Nov 8 '11 at 9:42
But, in you setting, consider $exp_0(1) = \gamma(1)$, where $\gamma$ is a curve with constant speed 1 starting at 0. Hence $\gamma(1) = 1$, and not $\gamma(1) = e$ as I was expecting... So in what sense $exp$ on numbers coincide with the $exp$ of riemannian geometry? Can you be more precise? – Abramo Nov 8 '11 at 9:53
I think the problem is that your metric on $(\mathbb{R}_+,\times)$ should be (left-)invariant under the group action on itself. – Aaron Mazel-Gee Nov 8 '11 at 10:19
Like @Aaron wrote, the Riemannian metric on $(\mathbf{R}_+,\times)$ which you need to use has the line element $\mathrm{d}s^2 = y^{-2} \mathrm{d}y^2$, $y\in\mathbf{R}_+$. In the wikipedia link I gave above, it is worked out in more detail in the Section on Relationships. – Willie Wong Nov 8 '11 at 11:15

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