# How close are the exponential maps on $\mathbb{S}^2$ at two nearby points?

Consider the two dimensional sphere $$\mathbb{S}^2$$ and let $$p, q \in \mathbb{S}^2$$. Let $$\text{exp}_{p}$$ and $$\text{exp}_{q}$$ be the exponential maps on $$\mathbb{S}^2$$ at points $$p$$ and $$q$$ respectively. I am interested in the map $$\psi := \text{exp}_{p}^{-1} \circ \text {exp}_{q}$$ defined on the unit disc $$\mathbb{D} \subset \mathbb{R}^2$$. I expect that if $$p$$ and $$q$$ are nearby points, then the map $$\psi$$ is close to the identity map. My question is: Is there a way to quantify this closeness? More precisely, is it possible to write the Taylor series expansion of $$\psi = (\psi_1, \psi_2)$$ around the origin as \begin{align*} \psi_1(x,y) &= \psi_1(0) + \frac{\partial \psi_1}{\partial x}(0)~x + \frac{\partial \psi_1}{\partial y}(0)~ y + \ldots \\ \psi_2(x,y) &= \psi_2(0) + \frac{\partial \psi_2}{\partial x}(0)~x + \frac{\partial \psi_2}{\partial y}(0)~ y + \ldots \end{align*} and show that the only significant coefficients in the above expansion are $$\frac{\partial \psi_1}{\partial x}(0) \approx 1$$, $$\frac{\partial \psi_2}{\partial y}(0) \approx 1$$ and all other partial derivatives $$\frac{\partial^{m+n} \psi_i}{\partial^{m} x \partial^{n} y}(0) \approx 0$$, for all other choices of $$m,n$$ and any $$i = 1,2$$.

Edit: After reading the comments/ answers of user35593 and Gro-Tsen , I realized that the following comments are essential regarding the above question:

1. First is regrding the comment of user35593 (please read his/her comment below): The question makes sense only when we specify a certain way of identifying the tangent spaces at the points $$p$$ and $$q$$. Consider the rotation $$R :\mathbb{R}^3 \rightarrow \mathbb{R}^3$$ which maps $$p$$ to $$q$$, then it also maps $$T_{p}(\mathbb{S}^2)$$ to $$T_{q}(\mathbb{S}^2)$$ and it gives a way of identifying the two tangent spaces. The above question I asked is for the tangent spaces identified in this manner.

2. Second is regarding the answer of Gro-Tsen : One possible interpretation of $$\frac{\partial^{m+n} \psi_i}{\partial^{m} x \partial^{n} y}(0) \approx 0$$'' which I am interested in is: Given $$\epsilon >0$$, is it possible to find $$\delta >0$$ such that whenever $$d_{\mathbb{S}^2}(p,q)\leq \delta$$ we have \begin{align*} \left| \frac{\partial^{m+n} \psi_i}{\partial^{m} x \partial^{n} y}(0) \right| \leq \epsilon \end{align*} whenever $$m+n \leq 100$$ (except for the cases of $$\frac{\partial \psi_1}{\partial x} (0)$$ and $$\frac{\partial \psi_2}{\partial y} (0)$$ which I expect to be $$\approx 1$$).

Thanks!

• Have you tried writing down explicit formulae? Letting $\mathbb{S}^2 = \{x^2+y^2+z^2=1\} \subseteq \mathbb{R}^3$, we can assume $q=(0,0,1)$ and $p=(\sin δ,0,\cos δ)$, then $\exp_p(u,v) = (\sin r\,\cos θ, \sin r\,\sin θ, \cos r)$ where $(r,θ)$ are the polar coordinates of $(u,v)$, and $ψ$ takes $(u,v)$ to $(u',v')$ where $(\sin r'\,\sin θ', \cos r', \sin r'\,\cos θ') = (-\sin δ\,\cos r + \cos δ\,\sin r\,\cos θ, \sin r\,\sin θ, \cos δ\,\cos r + \sin δ\,\sin r\,\cos θ)$. At least the map $(r,θ)\mapsto (r',θ')$ looks reasonably explicit. – Gro-Tsen Apr 18 at 9:58
• Your map maps the tangent space of q to the tangent space of p. You need to specify how you identify the tangent spaces with $\mathbb{R}^2$, i.e. what bases you choose in order to arrive at a map $\mathbb{R}^2\rightarrow \mathbb{R}^2$. You could for example consider the geodesic from p to q. and identify the tangent vectors to the geodesic at p and q with each other. – user35593 Apr 18 at 10:48
• What should be true is that for fixed $p$, and $q$ nearby, the maps $F_q = \exp_q^{-1} \exp_p$, considered as a map on (a ball in) $\Bbb R^2$ by using a fixed trivialization of $TS^2$ near $p$. These should vary smoothly in $q$, with $F_p$ given by the identity. Since $D^n F_p = 0$ for $n>1$, and the maps $F_p$ are smooth in $p$, the Taylor approximation should give the existence of $C_n$ so that $|D^n F_q| \leq C_n |p-q|$ for $q$ near to $p$. This seems like the sort of smallness estimate you want. Gro-Tsen's answer implies that $C_n \to \infty$, though. – Mike Miller Apr 18 at 11:39
• You can argue that your map and therefore also the partial derivatives at 0 depend smoothly on $p, q$. For $p=q$ you get the identity. Then the coefficients are exactly 1 resp. 0. For $p\noteq$ they are of order $d(p,q)$ away from the identity case. – user35593 Apr 18 at 14:49
• It probably is easiest to consider, given a curve $c$ such that $c(0) = p$, the map $$\psi_t = \exp_p^{-1}\circ\exp_{c(t)}: T_{c(t)}\mathbb{S} \rightarrow T_p\mathbb{S}.$$ In particular, its derivative with respect to $t$ at $t = 0$ is a essentially a Jacobi field and satisfies the Jacobi equation, which here is just $$J'' + J = 0.$$ – Deane Yang Apr 18 at 19:21

I believe the following proves that the partial derivatives of $$\psi$$ at the origin cannot be bounded by a constant (so they certainly cannot be $$\approx 0$$ for any reasonable meaning of this symbol).

Assume on the contrary that (for some fixed $$p,q$$ distinct and not antipodal), the partial derivatives of $$\psi := \exp_p^{-1} \circ \exp_q$$ (defined in some neighborhood of the origin) are all bounded by a constant. Then, by summing the Taylor series expansion of $$\psi$$ at $$0$$ we see that $$\psi$$ extends to a real-analytic function $$\psi\colon\mathbb{R}^2 \to \mathbb{R}^2$$, which by analytic extension must still satisfy $$\exp_p \circ \psi = \exp_q$$. Let me argue why this is impossible.

We can assume w.l.o.g. that the coordinates $$(u,v)$$ on the tangent plane to $$\mathbb{S}^2$$ at $$q$$ were chosen so that $$\exp_q$$ maps the axis $$(u,0)$$ to the great circle connecting $$q$$ and $$p$$, and more precisely, if $$0<\delta<\pi$$ is the distance between $$q$$ and $$p$$ on $$\mathbb{S}^2$$, that $$\exp_q$$ takes $$(\delta,0)$$ to $$p$$. Furthermore, we can similarly assume on the coordinates $$(u',v')$$ of the tangent plane at $$p$$ that $$\exp_p$$ maps the axis $$(u',0)$$ to the same great circle and takes $$(-\delta,0)$$ to $$q$$. Then $$\exp_q(u,0)$$ is the point obtained by traveling a distance $$u$$ on $$\mathbb{S}^2$$ starting from $$q$$ in the direction of $$p$$, and $$\exp_p(u,0)$$ is the point obtained by traveling a distance $$u$$ on $$\mathbb{S}^2$$ starting from $$p$$ in the direction opposite to $$p$$, thus $$\psi(u,0) = (u-\delta,0)$$ for $$u$$ in the neighborhood of $$0$$, hence everywhere by analytic extension.

On the other hand, if $$(u,v)$$ lies on the circle $$C$$ with radius $$\pi$$ around the origin then $$\exp_q$$ takes $$(u,v)$$ to the antipode $$\tilde q$$ of $$q$$. But the inverse image of $$\tilde q$$ by $$\exp_p$$ is discrete (since $$\exp_p$$ is a diffeomorphism outside of circles of radius $$k\pi$$ around the origin which are mapped to either $$p$$ or its antipode $$\tilde p$$, and we are assuming $$p,q,\tilde p,\tilde q$$ distinct); and $$\psi$$ must map $$C$$ (which is connected) inside this inverse image: so $$\psi$$ must be constant on $$C$$. But this contradicts the fact that $$\psi(\pi,0) = (\pi-\delta,0)$$ and $$\psi(-\pi,0) = (-\pi-\delta,0)$$ (as per previous paragraph) are not equal.

• Nice answer! Though I only realized after the comment of user35593 that the way we identify the tangent spaces is very important and the question makes sense only if we identify it appropriately - like your observation above shows. Hence I have edited the question. The expectation is $\psi \approx \text{Id}$ only when $T_{p}(\mathbb{S}^2)$ and $T_{q}(\mathbb{S}^2)$ are identified like I have mentioned in the edited version. This is not the way you have identified the tangent spaces though. – April Apr 18 at 12:08
• It is the way I identified the tangent spaces. However, now that you've edited the question to clarify what you mean by $\approx 0$, and the order of the quanrifiers, this doesn't answer it. But it's not related to the identification of the tangent spaces. – Gro-Tsen Apr 18 at 14:12
• The derivatives being bounded would not necesssrily imply the Taylor series to coverge unless you mean bounded by a common constant in which case we would only have convergence on a square. – user35593 Apr 18 at 14:52

Not a solution but too long for a comment.

Note that the map depends only on the distance $$d(p,q)$$ of $$p,q$$, i.e. we have $$\phi(t)\colon \mathbb{R}^2\rightarrow \mathbb{R}^2$$ where $$t=d(p,q)$$. Note further that $$\phi(t_1+t_2)=\phi(t_1)\circ \phi(t_2)$$, i.e. $$\phi(t)$$ is a one-parameter group. Let $$D=d\phi(t)/dt|_{t=0}$$. Note that then $$d\phi(t)/dt=D(\phi(t))$$. Hence $$\phi(T)(u)$$ can be found by solving the ODE $$f'(t)=D(f(t))$$ on the interval $$[0,T]$$ with initial condition $$f(0)=u$$. Let us now find $$D$$ for our case of the sphere.

The exponential and its inverse on the sphere are $$exp(v)=\begin{pmatrix} sin(|v|)\frac{v}{|v|}\\ cos(|v|) \end{pmatrix}$$ and $$log \left(\begin{pmatrix} x\\ y\\z \end{pmatrix}\right)=\frac{arccos(z)}{\sqrt{1-z^2}}\begin{pmatrix} x\\ y \end{pmatrix}$$. Applying and infinitesimal rotation by $$\alpha$$ after the exponential map yields $$\begin{pmatrix} sin(|v|)\frac{v_1}{|v|}-\alpha \cdot cos(|v|)\\sin(|v|)\frac{v_2}{|v|} \\ cos(|v|)+\alpha \cdot sin(|v|)\frac{v_1}{|v|} \end{pmatrix}$$ Now applying the logarithm using $$d\frac{arccos(z)}{\sqrt{1-z^2}}/dz=\frac{arccos(z)z-\sqrt{1-z^2}}{\sqrt{1-z^2}^3}$$ yields $$v+\alpha \cdot sin(|v|)\frac{v_1}{|v|} \frac{(|v|cos(|v|)-sin(|v|))}{sin(|v|)^3} sin(|v|)\frac{v}{|v|} -\begin{pmatrix} \alpha \frac{cos(|v|)}{sin(|v|)}|v|\\0\end{pmatrix}\\ =v+\alpha \cdot \frac{(|v|cos(|v|)-sin(|v|))v_1}{sin(|v|)|v|^2} v -\begin{pmatrix} \alpha \frac{cos(|v|)}{sin(|v|)}|v|\\0\end{pmatrix}$$ Hence $$D(v)=\frac{(|v|cos(|v|)-sin(|v|))v_1}{sin(|v|)|v|^2} v-\begin{pmatrix} \frac{cos(|v|)}{sin(|v|)}|v|\\0\end{pmatrix}.$$