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The last question is too general, this is a modification.

Let $M$ be an $n$ dimensional Riemannian manifold. Fix $m$ points $x_1,...,x_m$. Suppose $y$ is not in the cut locus of $x_i$ for $1 \leqslant i \leqslant m$. Then the function $g(x)=\sum_{i=1}^m d^2(x_i,x)$ is $C^2$ near $y$. Thus there is a small neighborhood $B_r(y)$ of $y$ and a constant $C$ such that $|\nabla^2 g(x)|\leqslant C$ for any $x \in B_r(y)$.

My question: Is the map $x \mapsto \exp_ x(\nabla g(x))$ Lipschitz in $B_r(y)$?

My understanding (I add an condition $sec(M) \geqslant 0$ to simplify my explanation, and I am only interested in manifolds with lower curvature bound): To prove $\exp_x(\nabla g(x))$ is Lipschitz, we just need to show that for any point $x\in B_r(y)$, any unit geodesic $\gamma(t)$ with $\gamma(0)=x$, we have $$ \frac{d}{dt}|_{t=0} dist(\exp_{\gamma(t)}(\nabla g(\gamma(t))), exp_x(\nabla g(x)) \leqslant C_1, $$ and $C_1$ is the same for all the points in the ball. If this is done, then $\exp_z(\nabla g(x))$ is $C_1$-Lipschitz.

Since $|\nabla^2 g| \leqslant C$, we have $|\nabla_{\dot{\gamma}(0)}\nabla g|\leqslant C$, that is, $$ \frac{d}{dt}[{P_t}^{-1}(\nabla g(\gamma(t))]\leqslant C $$ Where $P_t: T_xM \mapsto T_{\gamma(t)}M$ is the parallel transportation. Then \begin{equation*} \begin{array}{ll} dist(\exp_x (\nabla g(x)),\exp_{\gamma(t)} (\nabla g(\gamma(t)))&\leqslant dist\left(\exp_x (\nabla g(x)),\exp_x ({P_t}^{-1}(\nabla g(\gamma(t))\right)\\ &+dist\left(\exp_x ({P_t}^{-1}(\nabla g(\gamma(t)), \exp_{\gamma(t)} (\nabla g(\gamma(t)\right) \end{array} \end{equation*} If the curvature $secM \geqslant 0$, then the exponential map is a contracting map (i.e. 1-Lipschitz), the first term on the RHS is less than $$ |\nabla g(x), {P_t}^{-1}(\nabla g(\gamma(t))|\leqslant Ct+o(t). $$

But for the second term on the RHS, I don't know how to estimate. This is what I want to ask.

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Since $g(x)$ is a $C^2$ map, the map $M \to TM$ given by $x \mapsto \nabla_x g$ is $C^1$. The map $\exp : TM \to M$ is a smooth map.

The function you are asking about is the composition, and so is $C^1$, and hence locally Lipschitz.

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  • $\begingroup$ :It seems to me that $\exp:TM\mapsto M\times M$ is a diffeomorphism only from a small neighborhood of $(y,0_y)$, while $\nabla g(x)$ may not be short. $\endgroup$
    – mafan
    Sep 4, 2015 at 15:32
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    $\begingroup$ But though it is only a local diffeomorphism, it is still smooth everywhere, right? Because the geodesic equation is a smooth ODE, and the solution varies smoothly with the initial conditions. What you are referring to is the possibility that the differential of the map is not of maximal rank, which does not mean non-smoothness. $\endgroup$
    – John Harvey
    Sep 4, 2015 at 18:31
  • $\begingroup$ I understand in this way: Consider $\exp:TM\mapsto M \times M$, calculate the differential at $(y,\nabla g(y)$, the diagonal of the matrix is identity and $(dexp_y)(\nabla g(y))$, so if the last term is not zero, i.e. $exp_y(\nabla g(y)$ is not the conjugate point of $y$. Then a neighhood of $(y,\nabla g(y))$ and $(y,exp_y(\nabla g(y))$ would be diffeomorphism. I know in a local chart geodesic varies smoothly with the initial condition. While for "long geodesic", the existence is proved by a covering argument. I don't know whether we have smooth dependence. $\endgroup$
    – mafan
    Sep 5, 2015 at 5:31
  • $\begingroup$ The smooth dependence is global. Suppose you had a smooth curve in $TM$, but its image was not smooth. You could isolate the failure of smoothness to see that things were breaking down in one particular chart. Every standard reference will tell you the exponential map is globally $C^{\infty}$. See, for example, section 5.4 of Petersen. The value of the differential is not relevant, just that it exists and varies continuously (since you only need $C^1$). The diffeomorphism issue is irrelevant. $x^2$ is not a diffeo near 0, but it's a smooth function nevertheless. $\endgroup$
    – John Harvey
    Sep 5, 2015 at 16:00
  • $\begingroup$ Do you need any further clarification on this? $\endgroup$
    – John Harvey
    Sep 7, 2015 at 19:33

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