Here are some notes on hyperbolic manifolds. The aim is to prove that if $M_1$ and $M_2$ are simply connected, complete Riemannian manifolds having constant sectional curvature of $-1$, then $M_1$ and $M_2$ are isometric.

For simplicity, let's consider a Riemann 2-manifold $M$ with constant Gaussian curvature $-1.$ Pick a point $p\in M$ and let $E(s,t)$ be the exponential map, "parameterising" $M$ with two variables $s,t$, in such a way that the curve $\gamma$ below is a geodesic. The author also quotes that "$dE$ is an isometry at $0.$" ($E(0,0) = p.$) Define the vector field $$ J (t) = \frac{dE}{ds}(s=0,t). $$ along the geodesic $\gamma(t)= E(0,t),$ and let $T$ be the unit velocity field of this curve.

The author claims (on page 5) that $(*)$ $DJ/dt$ is orthogonal to $T$ at $t=0,$ "because $dE|_0$ is an isometry." Here $DJ/dt$ is just the covariant derivative (connection) $D_T J.$

The claim is made in the proof of Gauss's Lemma.

I try to interpret what this means, but I am not sure - I only know the standard definition of isometry here, which makes sense locally in an open set or globally, but certainly not "at a point." Saying that $dE|_0$ is an isometry is also a bit unusual - we usually say that a map is an isometry, rather than the derivative of the map at a point is an isometry.

How to interpret the above claim $(*)$ made by the author properly? How can I prove the claim? Is the claim really obvious?

By the way, the "Gauss's lemma" mentioned in the notes is also a bit unusual - it is not about geodesics being perpendicular to a sphere.

  • 5
    $\begingroup$ Doesn't it just mean that $dE$ is a map of tangent spaces, and, in that capacity, is an isometry? $\endgroup$
    – LSpice
    Jul 26 at 3:09
  • $\begingroup$ Yes this makes sense. But how does this lead to the conclusion that the two vectors mentioned above are perpendicular? $\endgroup$
    – Ma Joad
    Jul 26 at 3:31
  • $\begingroup$ The map $dE$ is a linear isomorphism and an isometry of the inner product on the tangent spaces. $\endgroup$
    – Ben McKay
    Jul 26 at 11:28
  • $\begingroup$ @BenMcKay I see. But how does that lead to the claimed result (*)? $\endgroup$
    – Ma Joad
    Jul 26 at 13:18
  • 1
    $\begingroup$ @DeaneYang Yes, thanks. I have changed the title. $\endgroup$
    – Ma Joad
    Aug 4 at 0:26

(It seems some people find this question too elementary for MO, but I think there is a true difficulty in tracking the different objects in this subject and that answering such questions make MO more useful rather than less.) (Between writing this and finishing my answer, I spent a considerable amount of time, and I think it is far from an obvious question.)

First, there are things to be careful about with these notes. For example, the coordinates $(s,t)$ should be polar coordinates, even though it is not made explicit (otherwise, $t\mapsto E(s_0,t)$ would not usually be a geodesic when $s_0\neq 0$, and $J(0)$ would not be $0$ in general), but it seem at some points that we are assumed to think in Cartesian coordinates.

First, let us discuss the "isometric at a point" part. I like to draw differential geometry with colors. To represent $TM$ in 2D, I represent $M$ in 1D, like this: Picture of M, TM and the exponential map

$E$ is a map from $T_pM$ to $M$, so that $dE$ is a map from $T(T_pM)$ to $TM$. The correct interpretation of "$dE$ is an isometry at zero" seems to be "the evaluation at zero of $dE$ is an isometry", i.e. "$dE_0$ is an isometry" (I do not like the notation $dE_{|0}$, akin to a restriction, while we more usually think $dE$ as a map with two arguments, a point which I put in index and a vector which I put into parentheses).

Now, $dE_0$ is a map from $T_0(T_p M)$ to $T_{E(0)}M = T_pM$. Thing is, $T_pM$ is a affine (even linear) space to begin with, so that each of its tangent spaces (in particular $T_0(T_p M)$) can be canonically identified to $T_pM$ itself, as we often do for calculus in $\mathbb{R}^n$. With this identification, $dE_0$ becomes a (linear) map from $T_pM$ to itself, and it is part of the definition of the exponential that it is an isometry (actually, with this identification $dE_0$ is even the identity).

It is not obvious to me how to interpret the statement (*). We can compare e.g. to Gallot-Hulin-Lafontaine Riemannian Geometry. The Cartan-Hadamard theorem is proven in 3.87, but the proof is slim, most the necessary ingredient having been stated before. The Gauss Lemma is 3.70; it rests on the expression $dE_{(0,t)}(u) = Y(t)$, where $Y$ is the unique Jacobi field along $\gamma$ such that $Y(0)=0$ and $Y'(0)=u/t$. In he current situation, since $J$ is a Jacobi field, this expression simply means that $\frac{D J}{dt}(t=0)=\partial y$ in Cartesian coordinates $(x=\cos(s) t, y=\sin(s) t)$. In particular you get that $\frac{D J}{dt}(t=0)$ is orthogonal to $T(0)=\partial x$.

The mentioned expression is proved in 3.46; basically, it rests on uniqueness of Jacobi fields with specified initial value and derivative (standard second order ODE stuff), and observing that $t\mapsto E(tv)$ is a geodesic, so that varying $v$ makes a geodesic variation, whose derivative must be a Jacobi field.

  • $\begingroup$ Thank you for the answer! $\endgroup$
    – Ma Joad
    Aug 6 at 1:22
  • 1
    $\begingroup$ @Ma Joad. You can also find interesting Manfredo do Carmo's book "Riemannian Geometry". In particular, Chapter 5 about Jacobi fields and Chapter 8, section 4 e.g. Theorem 4.1 . Perhaps also this could help you: math.stackexchange.com/questions/3991778/… $\endgroup$
    – Holonomia
    Aug 19 at 9:54

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