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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?

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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?

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  • $\begingroup$ 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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