# How to prove the distance isometry is a smooth on a Riemannian manifold?

Suppose $(M,g)$ is a Riemannian manifold, $d$ the induced metric (by geodesic lengths), and $f$ is an isometry on $M$ in the metric space sense. I want to prove $f$ is a diffeomorphism using the exponential map.

But the first thing is I want to prove smoothness. One fact I know is that $f$ sends geodesics to geodesics. So here I go.

For each $p$, consider some $\epsilon>0$ so that we have a geodesic ball $\exp_p(\bar B_\epsilon(0))$ around $p$ (in which $\bar B$ means closed balls in $T_pM$). Within this ball, I want to show $f$ is smooth. For each $q\in \exp_p(\bar B_\epsilon(0))$ there exists a unique $v_q\in T_pM$ so that $\|v_q\|\le\epsilon$ and that $\exp_p(v_q)=q$. Thus $$f(q)=\exp_{F(p)}(d(f)_p v_q)$$ since $d(f)_p$ preserves lengths in tangent vectors. (Here $df$ denotes the differential of $f$, not to be confused with the metric $d$.)

However, I'm having trouble showing why $q\mapsto v_q\in T_pM$ is smooth. Is there any explicit expression of this map? (I know it is continuous of course (and indeed an isometry) since $d(p,q)=\|v_q\|$. But how to show the direction of $v_q$ also varies smoothly with $q$?)

• Your argument has a huge hole in it: You are assuming that $df_p$ exists and this is not at all obvious (since all what you know is that $f$ is Lipschitz). Start by reading here: en.wikipedia.org/wiki/Myers-Steenrod_theorem – Moishe Kohan May 25 '17 at 3:39
• @MoisheCohen yeah I later realised that. However, what I already proved was that $f$ sends geodesics geodesics, so we have a well defined "directional derivative" $D_p^F$ which acts the same as $d(f)_g$ i.e. takes $v_q\in T_pM$ to the initial velocity of the geodesic $F\circ\gamma_{p\to q}$. I could prove the map $D_p^F$ is a linear map. So we can just replace $d(f)_p$ with $D_p^F$ and conclude. – Vim May 25 '17 at 3:44
• Linearity is hard. – Moishe Kohan May 25 '17 at 4:08
• Look here: math.stackexchange.com/questions/1486882/…. This lemma is the key step. You use it to prove that $df$ is an isometric map of each tangent space. Then linear algebra/elementary geometry shows that it $df$ is linear at each point. – Moishe Kohan May 25 '17 at 15:18
• You can do this by elementary geometry. Start with the 2-dimensional case. This is a pleasant Euclidean geometry exercise. Now prove that general isometries $E^n\to E^n$ send planes to planes. – Moishe Kohan May 26 '17 at 15:55

There is a very explicit expression of the map $q\mapsto v_q$: it's the inverse of the exponential map $\exp_p^{-1}$ (well you write $\exp_p (v_q) = q$....). In particular, it is smooth as $\exp_p$ is a local diffeomorphism (I assume you know how to prove that).

Note that $q\mapsto v_q$ is indeed NOT an isometry. Your formula $d(p, q) = \| v_q\|$ holds true, but it works only for $p$. If $q_1, q_2$ are both not $p$, you do not have

$$d(q_1, q_2) = \| v_{q_1} - v_{q_2}\|.$$

Indeed the metric $g_p$ defined on $B_\epsilon(0) \subset T_pM$ is flat, so in general $\exp_p$ cannot be an isometry.

• Thanks! Sorry I was being too flippant, I just meant that it is an "isometry" with fixed $p$ (wouldn't make sense if we didn't fix $p$ anyway, I suppose?). – Vim May 24 '17 at 7:31

OK, here is a proof that each isometric embedding $f: E\to E$ of a finite-dimensional Euclidean space to itself is an affine transformation with orthogonal linear part. As I said in a comment, it is a pleasant elementary geometry exercise. (Lemmata 1 and 2 are essentially from John Stillwell's "Four Pillars of Geometry".)

Lemma 1. Pick $n+1$ generic (i.e. not in a proper affine subspace) points $A_0, A_1, ..., A_n$ in the Euclidean $n$-space $E^n$. Then the tuple of distances $\tau(P):=(|A_0P|, |A_1P|,..., |A_nP|)$ determines the position of a point $P\in E^n$ uniquely.

Proof. Suppose that there are two distinct points $P, Q\in E^n$ such that $\tau(P)=\tau(Q)$. Then all $n+1$ points $A_0, A_1,..., A_n$ belong to the perpendicular bisector of the segment $PQ$. Hence, $A_0, A_1,..., A_n$ all belong to a proper affine subspace. qed

Lemma 2. Each isometric embedding $f: E^n\to E^n$ is an affine isometry with orthogonal linear part.

Proof. Let $A_0...A_n\subset E^n$ be the vertex set of a simplex with unit edges. Since $f(A_0,...,A_n)$ is also the vertex set of a simplex with unit edges, there exists an affine isometry (with orthogonal linear part) $g: E^n\to E^n$ such that $f(A_0)=g(A_0), f(A_1)=g(A_1),... f(A_n)=g(A_n)$. Since both $f$ and $g$ preserve distances, Lemma 1 implies that $f(P)=g(P)$ for all $P\in E^n$. qed

• Thanks this is absolutely stunning! Just a simple followup question: you claimed $f(A_0\cdots A_n)$ is also a simplex with unit edges, but is it obvious from the mere definition of an isometry? I think it deserves some more explanation. – Vim May 27 '17 at 3:01
• @Vim: I meant just the vertex set, that's all what you need. – Moishe Kohan May 27 '17 at 3:07
• Yeah I know you meant just the vertices. To make my question more concrete, given that $\|f(A_i)f(A_j)\|=\|A_iA_j\|,\,i,j=0,\cdots,n$ is it enough to conclude $f(A_k)$ form the vertices of a simplex? In two dimensional, yes, we have SSS congruence criterion, but what about higher dimensions? – Vim May 27 '17 at 3:09
• @Vim: The simplex is the convex hull of these vertices. If you are asking how to construct $g$: Just assume that $A_0=f(A_0)=0$ (after a translation) and take the unique linear map sending each $A_i$ to $f(A_i)$. It will be orthogonal since inner products between basis vectors determine the entire inner product. – Moishe Kohan May 27 '17 at 3:28
• Thanks! That makes sense and is also what I was thinking about just now: as you said, let $f$ preserves zero, then define a linear map $g$ so that $g(e_k)=f(e_k),k=1,\cdots,n$. Next, since $$(g(e_i),g(e_j))=(f(e_i),f(e_j))=(e_i,e_j)$$ (the last equality is due to the two-dimensional cosine theorem) and hence $g$ is a linear map that preserves inner products and must be orthogonal. – Vim May 27 '17 at 3:31