I've been studying differential geometry using Do Carmo's book. There's the notion of exponential map, but I don't understand why it is called "exponential" map. How does it has something to do with our common notion of exponentiation?

I read from the book The road to reality (by R. Penrose) that it is related to taking exponentiation when making (finite) Lie group elements from Lie algebra elements. It seems like using Taylor's theorem on a manifold so we have, for example, there was the following equation explaining why it is the case.

$f(t) = f(0) + f'(0)t + \frac{1}{2!}f''(0) t^2+\cdots = (1+t\frac{d}{dx}+\frac{1}{2!}t^2\frac{d^2}{dx^2}+\cdots)f(x)|_{x=0} = e^{t\frac{d}{dx}}f(x)|_{x=0}$.

The differential operator can be thought of as a vector field on a manifold, and it is how Lie algebra elements (which are vectors, on a group manifold (Lie group), in a tangent space at the identity element). If I understood correctly, the truth is that this is exactly the exponential map that sends a vector on a tangent space into the manifold in such a way that it becomes the end point of a geodesic (determined by the vector) having the same length.

Why is the above Taylor expansion valid on a manifold? Why is the exponential map the same as taking exponentiation?


2 Answers 2


The idea of an exponential is the continuous compounding of small actions. Suppose you start with an object $p$, perform an action on it $v$, and then add the result back to the original object. What happens if you instead take half as much action but do it twice? What about if you take one tenth the action but do it ten times? The exponential function tries to capture this idea: $$\exp (\text{action}) = \lim_{n \rightarrow \infty} \left(\text{identity} + \frac{\text{action}}{n}\right)^n.$$

On a differentiable manifold there is no addition, but we can consider this action as pushing a point a short distance in the direction of the tangent vector, $$``\left(\text{identity} + \frac{\text{v}}{n}\right)"p := \text{push }p\text{ by} \frac{1}{n} \text{ units of distance in the }v \text{ direction}.$$

Doing this over and over, we have $``\left(\text{identity} + \frac{\text{v}}{n}\right)^n"p$ means push $p$ by $\frac{1}{n}$ units of distance in the $v$ direction, then push it again in the same direction you already pushed it, and keep doing so until you have pushed it $n$ times.

So long as $\frac{1}{n}$ is small enough that pushing points and vectors in a tangent direction makes sense, what we end out doing is pushing the point $p$ a total of $1$ unit of distance along the geodesic generated by $v$.

  • $\begingroup$ Wow. Thank you for the clear explanation! $\endgroup$
    – Henry
    Oct 31, 2014 at 15:01
  • 1
    $\begingroup$ @MarineGalantin A geodesic is the curve whose tangent vector at a given point is the parallel transport of the tangent vector at the initial point. See, for example dspace.mit.edu/bitstream/handle/1721.1/36859/8-962Spring2002/NR/… This process of pushing a point then pushing it again is like some sort of Riemman integral approximation to this process $\endgroup$
    – Nick Alger
    Jun 10, 2018 at 3:39
  • 1
    $\begingroup$ Thank you, Nick! One question, is the last sentence connected to the algebraic exp where e~2.718? It seems here it's a different operation, using 1/n * n. $\endgroup$ Apr 12, 2019 at 18:30
  • 1
    $\begingroup$ @Rethliopuks Not sure I understand the question. It is true that $e = \lim_{n \rightarrow \infty} \left(1+1/n\right)^{n}$, if that is what you are saying $\endgroup$
    – Nick Alger
    Apr 13, 2019 at 19:28
  • 1
    $\begingroup$ Absolutely perfect. If I may add also: in the reverse, logs are what transform multiplications into addition and this is precisely what people needed while they were stuck in ships in the 1600's and 1700's! $\endgroup$ Jan 25 at 18:31

Consider the following:

Let $a\in\mathbb{R}$, and define a smooth vector field $X$ on $\mathbb{R}_+$, by $X(p)=ap$. Note that $X$ is left-invariant, when thinking of $\mathbb{R}_+$ as a Lie group. Now let $\gamma$ be a trajectory of $X$, with $\gamma(0)=1$. By solving a simple ODE, one verifies that $\gamma(1)=e^a$, thus the term "exponential map" for Lie gropus coincides with what we know as exponential map on $\mathbb{R}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.