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Let $(M^n,g)$ be a smooth Riemannian manifold. The distance between two points is the infimum of the lengths of the curves which join the points. Consider the square of the distance function

$d^2\colon M\times M\to \mathbb{R}$

Why is the square of the distance function infinitely smooth near the diagonal?

In other words why does, for all $p \in M$, there exists a neighborhood $U$ of $p$ such that for all $x,y\in U$, $d^2(x,y)$ is infinitely smooth on $U\times U$.

I saw this fact here and it is supposed to be easy but I can't work this out or find a proof of it... If one of the response on the other post actually anwsers the question I would love to have some precisions.

Any help would be appreciated! Thanks.

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    $\begingroup$ It does not seem to me that in the linked post it is proved what the OP asked..... In my opinion this is not duplicate. $\endgroup$
    – Raziel
    Commented Apr 17, 2017 at 12:10
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    $\begingroup$ To second @Raziel : the OP asked the question precisely because he/she does not fully understand the post to which this one was closed as a duplicate. The closure (as duplicate) seems to me rather circular. $\endgroup$
    – Willie Wong
    Commented Apr 17, 2017 at 13:04
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    $\begingroup$ This should follow (if I am not mistaken) from the smoothness of the exponential map $\exp: TM \to M$. The exponential map is smooth from smooth dependence of initial data for ordinary differential equations. In a neighborhood of the zero section you have that $d(x, \exp_x (v))^2 = \langle v, v \rangle_x$ and what you want should follow from the implicit function theorem. $\endgroup$
    – Willie Wong
    Commented Apr 17, 2017 at 15:11
  • $\begingroup$ I vaguely remember that the Hessian of the squared distance function at a point $(p,p) \in M \times M$ is related to the Riemannian metric tensor at $p$. (More exactly, the Hessian has the kernel of dimension $n$, which is the tangent space to the diagonal. On the normal bundle of the diagonal the Hessian should be the metric tensor.) Since the metric is infinitely smooth, the squared distance function is smooth. $\endgroup$
    – Ivan Izmestiev
    Commented Apr 17, 2017 at 20:07
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    $\begingroup$ @AntonPetrunin: done. $\endgroup$
    – Willie Wong
    Commented Dec 8, 2022 at 21:22

1 Answer 1

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(copy of my comment above to get this off the unanswered list)

Given that $(M,g)$ is smooth Riemannian, then the exponential map $\exp:TM\to M$ is smooth due to the smooth dependence on initial data for ordinary differential equations (with smooth coefficients).

Looking at $\exp_x:T_xM\to M$, it sends $\exp_x(0) = x$ and $d\exp_x$ has full rank. So by the implicit function theorem, there exists an open neighborhood $\Omega$ of the zero section of $TM$ and an open neighborhood $U$ of the diagonal in $M\times M$, such that $\exp: \Omega \to U$ is a diffeomorphism. The inverse mapping sends $(x,y)\in U \to (x,v)\in\Omega$ with $x\in M$ and $v\in T_xM$; this mapping is smooth.

Now observe that $d(x,y)^2 = \langle v,v\rangle_{x}$ (the RHS is the Riemannian inner product on $T_xM$), which manifestly is a composition of smooth functions.

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