I have a problem understanding how to compute the differential of the exponential map. Concretely I'm struggling with the following concrete case:

Let $M$ be the unit sphere and $p=(0,0,1)$ the north pole. Then let $\exp_p : T_pM \cong \mathbb{R}^2 \times \{0\} \to M $ be the exponential map at $p$. How do I now compute:

1) $\mathrm{D}\exp_p|_{(0,0,0)}(1,0,0)$

2) $\mathrm{D}\exp_p|_{(\frac{\pi}{2},0,0)}(0,1,0)$

3) $\mathrm{D}\exp_p|_{(\pi,0,0)}(1,0,0)$

4) $\mathrm{D}\exp_p|_{(2\pi,0,0)}(1,0,0)$

where $\mathrm{D}\exp_p|_vw$ is viewed as a directional derivative. I really have no clue how to do this. Can anyone show me the technique how to handle that calculation?


I' ll assume we are talking about the exponential map obtained from the Levi-Civita connection on the sphere with the round metric pulled from $\mathbb R^3$. If so, the exponential here can be understood as mapping lines through the origin of $\mathbb R^2$ to the great circles through the north pole. Its derivative then transports the tangent space at the north pole to the corresponding downward-tilted tangent spaces.

For example, in $(3)$ we map to the tangent space at the south pole (we have traveled a distance of $\pi$). But since this tangent space has been transported along the great circle in the $(x,z)$ plane, the orientation of its $x$-axis is reversed with respect to the north pole. So the result here is $(-1,0,0)$. Similarly, in $(2)$ we travel $\pi/2$ along the same circle and end up in a tangent space parallel to the $(y,z)$ plane. The vector $(0,1,0)$ points to the same direction all the time.

Can you work out the answer for $(1)$ and $(4)$ yourself now?

  • $\begingroup$ Thank you for your illustration. As far as I understand the answers to 1) and 4) should now be: 1) (1,0,0) because we don't travel and so nothing changes 4) (1,0,0) because we travel in a full circle and end up in the tangent space on the northpole again $\endgroup$ – alexlo Jan 26 '13 at 0:38
  • $\begingroup$ Correct. Glad I could help. $\endgroup$ – Marek Jan 26 '13 at 0:56

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