Riemannian Geometry

3 Comparison

Lemma 3.1. If \(f\) solves \(f'' = -kf\), then \(\frac {f'}{f}\) solves \(\rho ' + \rho ^2=-k\).

Proof. Calculation. □

Let \(\sn _k\) denote the solution of \(f''=-kf\) with \(\sn _k(0) = 0, \sn '_k(0) = 1\)

Lemma 3.2 (Riccati Comparison). If \(\rho _1,\rho _2: (0.b) \to \RR \) are smooth functions satisfying \(\limsup _{t\to 0} (\rho _2(t)-\rho _1(t))\geq 0\) and \(\rho '_1 + \rho _1^2 \leq \rho '_{2} + \rho _2^{\prime 2}\), then \(\rho _1(t) \leq \rho _2(t)\). The same holds if all inequalities are strict.

Proof. The second hypothesis implies that \(e^{\int \rho _1 + \rho _2}(\rho _2-\rho _1)\) is increasing. The first hypothesis shows it is nonnegative for arbitrarily small \(t\), so we are done. □

Definition 3.3. A distance function \(\rho \) is one with \(|\nabla \rho | \equiv 1\).

For example, the distance from a submanifold is distance function on some open set. Given a distance function, \(II\) is the second fundamental form on the level sets of \(\rho \), \(m\) is their mean curvature. Then the Bochner formula reads:

\[ |II|^2 + \partial _r m + \Ric (\partial _r,\partial _r)=0\]

where \(\partial r = \nabla \rho \).

Applying the Cauchy Schwartz inequality and \(\Ric \geq (n-1)\kappa g\), this becomes the Riccati inequality

\[ \frac {m^2}{n-1} + \partial _r m + \kappa (n-1)\leq 0\]

If this were an equality, the solution would be \(\frac {(n-1)\sn '_\kappa }{\sn _\kappa }\).

Proposition 3.4 (Mean Curvature Comparison). Suppose \(\Ric \geq (n-1)\kappa g\), \(m(r)\) is the mean curvature of the geodesic sphere of radius \(r\) around a point \(p\), and \(m_\kappa (r)\) is that for the model space. Then \(m(r) \leq m_\kappa (r)\).

Proof. Apply the Ricatti comparison to the inequality above. □

Remark 3.5. I think if you use Gauss’s equation you can show that the above is an equality iff the sectional curvatures of the radial vectors are \(\kappa \).

Remark 3.6. A stronger theorem is true, namely the inequality holds in the weak sense for the Laplcian of any increasing or decreasing radial function (doesn’t have to be distance).

This gives a second proof of Bonnet-Myers.

Corollary 3.7 (Bonnet-Myers 2). 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. Again suppose there is a minimal geodesic \(\gamma \) of length \(l>\pi r\). The distance from one end point is a distance function near \(\gamma \). The mean curvature in the model space goes to \(-\infty \) as the distance goes to \(\pi r\) so by the comparison theorem we get a contradiction. □

Definition 3.8. The cut locus at \(p\), denoted \(\tilde {\cut }_p(M)\) is the region of \(T_p(M)\) consisting of all points \(v\) where \(\exp (v)\) is a minimizing geodesic, but \(\exp (tv)\) isn’t for \(t>1\).

The image of \(\tilde {\cut }_p(M)\) in \(M\) is also called the cut locus and is denoted \(\cut _p(M)\). A point in the cut locus is a cut point.

Lemma 3.9. For a cut point \(v\), either \(\exp (v)\) has its first conjugate point at \(t=1\) or there is another geodesic of the same length going to \(v\).

Proof. Let \(\gamma \) be a limit of a subsequence of the curves \(\exp (t_iv)\), where \(t_i \to 1^+\). If \(\gamma = \exp (v)\) then □

Lemma 3.10. The cut locus has measure \(0\).

Theorem 3.11 (Rauch Comparison). Let \(M,M'\) be manifolds with Jacobi fields \(X,\tilde {X}\) on geodesics \(\gamma ,\tilde {\gamma }\), and suppose that \(X(0) = \tilde {X}(0)= 0\).

Things to add: Cut locus stuff Bishop-Gromov volume comparison Rauch Comparison

Things to potentially include:

• "things are determined by curvature" and symmetric spaces
• Cheng’s theorem on sphere
• Louiville’s theorem on conformal transformations
• Construction of spaces of constant curvature
• Maybe some basic things like Killing’s equation, Jacobi equation First/Second Variation formulae.