Riemannian Geometry

2 Theorems

Let \(M\) be a connected Riemannian manifold.

Definition 2.1. For a continuous function from an interval to a metric space, the length is the supremum of the sum of the distances for partitions of the interval.

Proposition 2.2. A length minimizing path \(x \to y\) in \(M\) is a geodesic up to reparameterization.

Proof. Locally, geodesics are length minimizing, essentially by Gauss’s Lemma. Since lengths are additive under subdivision, we see by subdividing that the path has to coincide locally with minimizing geodesics. □

The essence of Hopf-Rinow is in the following proposition, which can be seen as a refinement of it:

Proposition 2.3. If \(\exp _p\) is defined in a ball of radius \(r\) for a point \(p\), then any point of \(M\) of distance \(r\) from \(p\) is connected to \(p\) by a minimal geodesic.

Proof. First the proposition will be proved for \(B = \exp _p(B_r)\), which is compact. For a point \(q\) in the interior, choose a sequence of arclength parameterized piecewise geodesics in \(B\) from \(p\) to \(q\) converging to one of minimal length. By Arzela Ascoli, these converge to a path of minimal length, which is a minimal geodesic. By passing to limits, this is true even not in the interior.

Now let \(q\) be any point of distance \(r\) from \(p\) in \(M\), and consider a sequence of paths in \(M\) from \(p\) to \(q\), whose lengths converge to the distance. WLOG, they can be assumed to be minimal geodesics until the first point they leave \(M\). Their initial segment is then given by exponentiating something in the ball of radius \(r\), so these converge to some point \(q'\) by compactness. But then the distance from \(q\) to \(\exp (q')\) is \(0\) by additivity of lengths, so \(q \in \exp _p(B_r)\). □

Theorem 2.4 (Hopf-Rinow). TFAE for a (connected) Riemannian manifold \(M\):
- 1. \(M\) is geodesically complete
- 2. \(\exp _p\) is completely defined for some \(p\)
- 3. Closed and bounded sets in \(M\) are compact
- 4. \(M\) is complete as a metric space.
Moreover these imply \((5)\): Any two points are connected by a minimal geodesic.

Proof. \((1) \implies (2)\) is trivial. \((2) \implies (5)\): Follows from the proposition. We thus see that \(M\) is exhausted by the compact sets \(\exp _p(B_r)\), showing \((2) \implies (3)\). \((3) \implies (4)\) is trivial. \((4) \implies (1)\): The obstruction to extending a geodesic is it being defined on an interval \([0,b)\) and it not convergint at \(b\). □

From now one, \(M\) will usually be complete.

Definition 2.5. A map between Riemannian manifolds is \(\textbf {expanding}\) if the derivative doesn’t decrease the length of tangent vectors.

Lemma 2.6. If \(f:M \to N\) is an expanding map, \(M\) is complete, then \(f\) is a covering map.

Proof. It suffices to show the smooth path lifting property. But the obstruction to lifting a path is the lift going off to infinity, which would give it infinite length since \(M\) is complete. but an expanding map cannot decrease the length of a curve so this isn’t possible. □

Proposition 2.7. A point \(p\) in \(M\) with no conjugate points (also called a pole) has the exponential map a covering map.

Proof. The exponential map is étale so we can give the tangent space at \(p\) the pullback metric. This is complete by Hopf Rinow since the lines going through the origin are still complete geodesics. Then the exponential map with this metric is expanding so we are done by the previous lemma. □

Theorem 2.8 (Cartan-Hadamard). If \(M\) has nonpositive sectional curvature, the exponential map at any point \(p\) is a covering map.

Proof. Using the Jacobi equation we will first see that there are no conjugate points to \(p\). The point is that if \(J\) is a Jacobi field, then

\[\langle J,J\rangle ''(t) = 2\langle J,J'\rangle ' = 2\langle J,J''\rangle +2\langle J',J'\rangle \]

The second term is nonnegative, and the first term is twice \(-\langle J, R_{\gamma ',J}\gamma '\rangle \) which is nonnegative by hypothesis. Then since for small \(t\) there are no conjugate points, this shows that it is true globally, so the exponential map is étale. By the previous proposition we are done. □

Theorem 2.9 (Bonnet-Myers). If \(M\) has \(\Ric \geq \frac {n-1} {r^2}g\), then the diameter of the universal cover is \(\leq \pi r\), and in particular it is compact.

Proof. One can pass to the universal cover without affecting the hypotheses. One needs to show that a unit speed parameterized geodesic \(\gamma \) of length \(l\geq \pi r\) cannot be minimal. We can wiggle the curve and see it reduces length by the second variation formula. Choose an orthonormal parallel frame of the normal bundle of \(\gamma \): \(E_1,\dots E_{n-1}\), and let \(F_i = \sin (\frac {\pi } lt) E_i\). \(\sum _i H(F_i,F_i) = \sum _i \int \langle R_{\gamma ',F_i} \gamma ' - F_i'',F_i\rangle = \int -\Ric (\gamma ',\gamma ') + \frac {(n-1)\pi ^2}{l^2}\sin ^2(\frac {\pi }{l}t) < 0\) by assumption on \(\Ric ,l\). Thus the trace of \(H\) is negative on the subspace spanned by \(F_i\) so there is a negative eigenvalue, i.e a variation reducing length. □

Lemma 2.10. An orthogonal transformation of \(\RR ^n\) with determinant \((-1)^{n-1}\) has a fixed point.

Proof. If \(n\) odd, then the characteristic polynomial being odd degree has a zero, so there is a real eigenvalue. Complex conjugate pairs contribute positive terms to the determinant, so there is an eigenvalue of \(1\). If \(n\) even, then again since conjugate pairs contribute positive terms, there must be an odd number of \(-1\) eigenvalues. Since there is an even number of real eigenvalues, there is an eigenvalue of \(1\). □

Theorem 2.11 (Weinstein-Synge). If \(M^n\) is compact, oriented, and has positive sectional curvature and has a conformal self map \(f\) inducing \((-1)^n\) on \(H^n(M;\ZZ )\), then there is a fixed point.

Proof. Let \(p\) minimize \(d(p,f(p)) = l\), which we assume is positive. Choose a minimal geodesic \(\gamma \) from \(p\) to \(f(p)\). Let \(A\) be the orthogonal map given by applying \(df\) to \(T_p(M)\) and parallel transporting back to \(p\) along \(\gamma \). Since \(\gamma \) is minimal, \(d(\gamma (t),f(\gamma (t)) \geq l)\) which together with the triangle inequality with \(\gamma (t), f(p), f(\gamma (t))\) shows that \(\gamma ,f(\gamma )\) glue together to form a geodesic. This implies that \(A\) is the identity on \(\gamma '\), so by the lemma, \(A\) has a fixed line orthogonal to \(\gamma ', v\). Let \(E_v\) be the parallel field corresponding to \(v\). Now applying the second variation formula to \(E_v\), we get \(H(E_v,E_v) = \int \langle R_{\gamma ',E_v}\gamma ',E_v\rangle \leq 0\) because positive sectional curvature. But then for points \(q\) slightly in the direction of \(E_v\), \(d(q,f(q)) <l\), a contradiction. □

Question 2.12. Is this true if \(f\) is a diffeomorphism?

Corollary 2.13 (Synge). Let \(M^n\) have possitive sectional curvature. If its dimension is even, it is simply connected. If it is odd, it is orientable.

Proof. Apply the previous theorem to an appropriate cover, noting that \(M^n\) must be compact. □

Theorem 2.14. Let \(\Delta \) be the Bochner Laplacian. Then for a compact \(M\), subharmonic functions \(\Delta f\geq 0\) are constant.

Proof. By passing to a cover, we can assume \(M\) is orientable. Then for any \(g\), via the Cartan homotopy formula we have \(\int _M \Delta g \omega _{vol} = \int _M \diverg \nabla g \omega _{vol} = \int _M L_{\nabla g}\omega _{vol} = \int _M d\iota _{\nabla g} \omega _{vol} = 0\) by Stoke’s theorem. Plugging in \(g = f^2\) we find \(0 = \int _M \Delta f^2 \omega _{vol} = \int _M (2f \Delta f+\langle \nabla f, \nabla f\rangle ) \omega _{vol}\) so since \(f\) is subharmonic, we see \(f\) is constant. □

Proposition 2.15. Having \(Ric \geq (n-1)\kappa g\) is equivalent to the Bochner inequality \(\frac 1 2 \Delta |\nabla u |^2 \geq \frac {(\Delta u)^2} n + \nabla u (\Delta u) + (n-1)\kappa \Vert \nabla u\Vert ^2\) for all \(u\).

Proof. Note that the Cauchy Schwarz inequality for \(I,\Hess (u)\) gives \(|\Hess (u)|^2 \geq \frac {(\Delta u)^2} n\). This together with the Bochner formula and the bound gives the inequality. We can always find a function \(u\) so that at a given point \(p\), \(\Hess u(p)\) is diagonal, and \(\nabla u(p)\) is arbitrary. Then the inequality with the Bochner formula gives the Ricci curvature bound. □