I am simply going to jump into the proof of the Myers theorem and refer to yesterday’s post for background. In particular, to prove it we only need to prove the following lemma:


Lemma 1 If {M} is a complete Riemannian manifold of dimension {n} and Ricci curvature bounded below by {C>0}, then a geodesic of length {L} does not minimize energy if {L> \pi \sqrt{ \frac{n-1}{C}}}.


If we choose a geodesic {\gamma: [0,L] \rightarrow M} parametrized by unit length, we then will be done if we find an endpoint-preserving variation {\gamma_u} of {\gamma} with {\frac{d^2}{d^2 u} E(\gamma_u)_{u=0} > 0}. In other words if we can find a vector field {V(t)} along {\gamma} with {V(0)=V(L)=0} and

\displaystyle \boxed{I_2(V) := \int_0^L R\left( \dot{\gamma}, V, \dot{\gamma}, V\right) - g\left( V, \frac{D^2V}{Dt^2} \right) > 0 }

 we will obtain a contradiction; cf. the initial discussion in yesterday’s post and the second variation formula.

Construction of V

Choose parallel vector fields {E_2, \dots, E_n} along {\gamma} which, together with {\gamma'}, form an orthonormal basis for {T_{\gamma(t)}(M)} at each {t}. To do this choose {E_i(0)} for {i>1}, and then parallel translate. Then since {R} is skew-symmetric in the last two variables and {R(\gamma',\gamma',\gamma',\gamma')=0}

\displaystyle \rho(\gamma',\gamma') = \sum R( E_i,\gamma', E_i, \gamma' ) \geq C. 

In particular, we could use the sum of the {E_i} to get the vector field {V} satisfying the boxed inequality—except {E_i(0) \neq 0}. So we define

\displaystyle F_i(t) := \sin\left( \frac{\pi t}{L} \right)E_i(t) . 

By the product rule for covariant differentiation and the paralellism of {E_i},

\displaystyle \frac{D^2}{Dt^2} F_i(t) = -\frac{\pi^2}{L^2} \sin\left( \frac{\pi t}{L} \right)E_i(t). 

So using orthonormality and {R(F_i, \gamma',F_i,\gamma') \geq C \sin^2\left( \frac{\pi t}{L} \right)}, we can get that

\displaystyle \sum I_2(F_i) = \int_0^L C \sin^2\left( \frac{\pi t}{L} \right) - \frac{(n-1)\pi^2}{L^2} \sin^2\left( \frac{\pi t}{L} \right) dt \leq 0

 if {\gamma} minimizes energy, so {L^2 \leq \frac{(n-1)\pi^2}{C}}.

Where next?

This ends my MaBloWriMo series on differential geometry; there’s one more day in November, of course, but I need to first learn more new material before I can go further.  Besides, it is probably healthy both for this blog and for myself to cover some other topics.

The posting over the next few weeks will probably be less structured than these entries.  I’m not yet quite sure what I want to discuss, but likely the topic will be one or two out of Riemann surfaces, Koszul complexes and depth, spectral sequences, and singular integrals.  Feel free to suggest something.

I edited this post to fix some sign issues.  (11/29)

I now want to discuss a result of Myers, which I can summarize as follows:

If {M} is a complete Riemannian manifold with positive, bounded-below curvature, then {M} is compact.

This is a very loose summary—Myers’ theorem actually gives a lower bound for the diameter of {M}. Moreover, I haven’t explained what “bounded below curvature” actually means. To say that the sectional curvature is bounded below is sufficient, but we can do better.

I will now outline how the proof works.

The first thing to notice is that any two points {p,q \in M} can be joined by a length-minimizing geodesic {\gamma}, by the Hopf-Rinow theorems. In particular, if we can show that every sufficiently long geodesic (of length {>L}, say) doesn’t minimize length, then {M} is necessarily of diameter at most {L}.

All the same, the length function as a map {\mathrm{Curves} \rightarrow \mathbb{R}} is not so easy to work with; the energy integral is much more convenient. Moreover, we know that if {\gamma} minimizes length and is a geodesic, it also minimizes the energy integral.

If {\gamma} is a geodesic that minimizes the energy integral (among curves with fixed endpoints), then in particular we can consider a variation {\gamma_u} of {\gamma}, and consider the function {E(u):=E(\gamma_u)}; this necessarily has a minimum at {u=0}. It follows that {E''(0) \leq 0}. If we apply the second variation formula, we find something involving the curvature tensor that looks a lot like sectional curvature.

Before turning to the details, I will now define the refinement of sectional curvature we can use:

Ricci curvature

Given a Riemannian manifold {M} with curvature tensor {R} and {p \in M}, we can define a linear map {T_p(M) \rightarrow T_p(M)},

\displaystyle X \rightarrow -R(X,Y)Z  

that depends on {Y,Z \in T_p(M)}. The trace of this linear map is defined to be the Ricci tensor {\rho(Y,Z)}. This is an invariant definition, so we do not have to do any checking of transformation laws.

A convenient way to express this is the following: If {E_1, \dots, E_n} is an orthonormal basis for {T_p(M)}, then by linear algebra and skew-symmetry

\displaystyle \boxed{\rho(Y,Z) = \sum R(E_i, Y, E_i, Z) .}

The Ricci curvature has many uses. Considered as a {(1,1)} tensor (by the functorial isomorphism {\hom(V \otimes V, \mathbb{R}) \simeq \hom(V, V)} for any real vector space {V} and the isomorphism {T(M) \rightarrow T^*(M)} induced by the Riemannian metric), its trace yields the scalar curvature, which is just a real-valued function on a Riemannian manifold. It is also used in defining the Ricci flow, which led to the recent solution of the Poincaré conjecture.  I may talk about these more advanced topics (much) later if I end up learning about them–I am finding this an interesting field, and may wish to pursue geometry further.

Statement of Myers’ theorem

Theorem 1 Let {M} be a complete Riemannian manifold whose Ricci tensor {\rho} satisfies \displaystyle \rho(X,X) \geq C|X|^2   

for all {X \in T_p(M), p \in M}. Then

 \displaystyle \mathrm{diam}(M) \leq \pi \sqrt{ \frac{n-1}{C}}.  (more…)