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Let $(M,g)$ be an $m$-dimensional Riemannian symmetric space and $d(\cdot,\cdot)$ be the geodesic distance function. Fix any $\alpha\in M$ and $v\in T_\alpha M$ with $\|v\|=1$.

Q1: Define $$ g(t)=\int_M f(t,x)e^{-f(t,x)}dx, $$ where $f(t,x)=d^2(Exp_\alpha(tv),x).$ Question: Is $g$ second-order differentiable at $t=0$?

Answer: Here $g$ is a constant function so the answer is yes. This was given by Willie Wong.

Q2: There was a typo in the above question. To fix it, we define $$ h(t)=\int_M f(t,x)e^{-f(0,x)}dx, $$ where $f(t,x)=d^2(Exp_\alpha(tv),x).$ Question: Is $h$ second-order differentiable at $t=0$?

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Consider the function $G:M\to \mathbb{R}$ given by $$ G(p) = \int_M d^2(p,x) \exp( - d^2(p,x)) ~dx $$ your $g(t)$ is $g(t) = G(\exp_\alpha(tv))$.

As $M$ is a symmetric space, given $p,q$ there exists an isometry that maps $p$ to $q$. So we see obviously that $G(p) = G(q)$.

Hence your function $g$ is the constant function.

Hence the answer to your question is "yes, it is twice differentiable at $t = 0$".

(If you intended to ask about a general complete Riemannian manifold $M$ instead of a symmetric space, it would require more work and you should ask a new question.)

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  • $\begingroup$ Thanks for your answer! I just found my question had a typo. In its current form, the answer is in fact trivial. I will revise my question now. Sorry about the confusion! $\endgroup$
    – Hengchao Chen
    Commented Feb 3 at 15:16

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