# Explanation of perpendicularity of a Jacobian vector field

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.

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

$$E$$ 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.

• Thank you for the answer! Aug 6 at 1:22
• @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/… Aug 19 at 9:54