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In a concrete manifold of $\mathbb{R}^3$ (namely, the $2$-sphere of radius $a$), I obtained he following exponential map $$ \exp_pX=p\cos(\frac{\parallel X \parallel}{a}) + \frac{X}{\parallel X \parallel}a\sin(\frac{\parallel X \parallel}{a}) $$ And now I want to prove that its differential at zero is the identity.

My attempt:

I have

$$d_0\exp_p:T_0(T_pM)=T_pM\rightarrow T_{exp_p0=p}M$$

This is a linear map. If I prove that the associated matrix is the identity, I'm done. Now, a basis for $T_pM$ is $\{ \partial_\phi, \partial_\theta \}$. $T_pM$ is natural homeomorphic to $\mathbb{R}^2$ with a global chart $(x^1,x^2)\overset{\Psi}{\rightarrow} x^1\partial_\phi + x^2\partial_\theta$. Now, the entries of the Jacobian matrix are $\frac{\partial(\exp_p \circ \Psi)_i}{\partial x^j}$. I have $$ \frac{\partial(\exp_p \circ \Psi)}{\partial x^1}= \frac{\partial}{\partial x^1} \Big(p\cos(\frac{\parallel (x^1,x^2) \parallel}{a}) + \frac{(x^1,x^2)}{\parallel (x^1,x^2) \parallel}a\sin(\frac{\parallel (x^1,x^2) \parallel}{a})\Big) $$

I can certainly derive the second term. But the first one does not make sense: how do I deal with $p$? I cannot write it in the coordinates of the chart... Also, the derivative of the second term gives something very far from $(1,0)$ (which we would expect since this is the first row of the Jacobian matrix).

Any hints?

(Context: Manifolds in $\mathbb{R}^n$ (i.e not in the context of general smooth manifolds)).

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  • $\begingroup$ The problem with your approach — as far as it goes — is that you're nowhere using your explicit chart to work on $S^2$ (in the image). That is, you need to be differentiating $\Psi^{-1}\circ\exp_p\circ\Psi$. The geometry of spherical coordinates is most naturally adapted to looking near the north pole, but of course it is not a valid chart at the north pole itself. $\endgroup$
    – Ted Shifrin
    Commented May 19, 2017 at 0:14
  • $\begingroup$ @TedShifrin Yes, I should differentiate $\Psi^-1\circ\exp_p\circ\Psi$. I think that I should have written $$ \frac{\partial(\Psi^-1\circ\exp_p \circ \Psi)}{\partial x^1}= \frac{\partial}{\partial x^1} \Big(p\cos(\frac{\parallel (x^1,x^2) \parallel}{a}) + \frac{(x^1,x^2)}{\parallel (x^1,x^2) \parallel}a\sin(\frac{\parallel (x^1,x^2) \parallel}{a})\Big) $$ But I still have the same problem. $\endgroup$
    – soap
    Commented May 19, 2017 at 14:59

1 Answer 1

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The way I usually think about the exponential map is in terms of radial lines through the origin. That is instead of trying to calculate the Jacobian in coordinate charts what one can do is consider a fixed $X \in T_{p}M$ and consider the curve $\gamma(t) = tX \in T_{p}M$. Then to show that the differential of $\exp_{p}: T_{p}M \to S^{2}$ is the identity it suffices to show that$\frac{d}{dt}\bigg\vert_{t = 0} \exp_{p}(\gamma(t)) = X$. To see this think about what the chain rule tells you.

To elaborate, the chain rule tells you that $\frac{d}{dt}\bigg\vert_{t = 0} \exp_{p}(\gamma(t)) = d_{0}\exp_{p}(\gamma'(0)) = d_{0}\exp_{p}(X)$. From there use the formula you've already provided to see what the curve $\exp_{p}(\gamma(t))$ is and calculate its derivative.

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  • $\begingroup$ What definition of differential are you using? It seems I am having trouble with the chain rule $D(f\circ g)(p)(v)=\Big(Df(g(p))\circ Dg(p)\Big)(v)$: $$ \frac{d}{dt}\bigg\vert_{t=0} \exp_{p}(\gamma(t)):= D(\exp_p\circ\gamma)(0)(1)=\Big( D\exp_p(\gamma(0)=0)\circ D\gamma(0) \Big)(1) $$ And this does not seem to be what you got. That is, your first inequality confuses me. $\endgroup$
    – soap
    Commented May 19, 2017 at 15:28
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    $\begingroup$ I'm using the same definition as you, if you unravel notation, what you got is the same as what I got. The key point is that, in your notation $D\gamma(0)(1) = \gamma'(0) = X$. Also, $D\exp_{p}(0)$ in your notation is what I wrote as $d_{0}\exp_{p}$. $\endgroup$
    – user127562
    Commented May 19, 2017 at 15:39
  • $\begingroup$ I saw the differential of a map $f:M\rightarrow N$ defined by $d_pf(v)(h)=v(h\circ f)$, where $v\in T_pM$. Sorry, but I do not see how this is the same as the derivative $Df(p)$... $\endgroup$
    – soap
    Commented May 19, 2017 at 16:16
  • $\begingroup$ Do you see the connection? $\endgroup$
    – soap
    Commented May 24, 2017 at 15:19
  • $\begingroup$ Just learned this myself, one can view the action of a tangent vector $v$ as this time derivative of the composition with the curve starting at $v$. So here $\frac{d}{dt}|_0 exp_p(\gamma(t))=D(exp_p\circ \gamma)(\frac{d}{dt}|_0)\overset{\text{chain rule}}{=} D(exp_p)\circ D(\gamma)(\frac{d}{dt}|_0)=D(exp_p)(\gamma'(0)).$ $\endgroup$
    – manifoldcurious
    Commented Jun 2, 2021 at 9:19

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