Oh right, this is supposed to be a math blog, isn’t it?

So, I was trying to understand the proof in do Carmo (which is a great book, incidentally) of Synge-Weinstein, and I tried to write an expository paper on it.  Unfortunately I can’t get the formatting perfect in WordPress since some of the equations go out of the margins, but most if it should look ok.   I thus  made a PDF of this too.  I dressed it up in formal article formatting just for the heck of it.

We give an exposition of the proof of a few results in global Riemannian geometry due to Synge and Weinstein using variations of the energy integral.

1. Introduction

One of the big refrains of modern Riemannian geometry is that curvature determines topology. Recall, for instance, the basic Cartan-Hadamard theorem that a complete, simply connected Riemannian manifold of nonpositive curvature is diffeomorphic to ${\mathbb{R}^n}$ under the exponential map. We proved this basically by showing that ${\exp_p}$ is nonsingular under the hypothesis of nonnegative curvature (using Jacobi fields) and that it was thus a covering map (the latter part was relatively easy). More difficult, and relevant to the present topic, was the Bonnet-Myers theorem, which asserted the compactness of a complete Riemannian manifold with bounded-below, positive Ricci curvature. The proof there showed that a long enough geodesic could not minimize energy (by using the second variation formula—recall that the second variation formula is intimately connected with curvature), and therefore could not minimize length. Since the distance between two points in a complete Riemanninan manifold is the length of the shortest geodesic between them (Hopf-Rinow!), this implied a bound on the diameter.

Today, however, we’re going to assume at the outset that the manifold in question is already compact. One of the theorems will be that a compact, even-dimensional orientable manifold of positive curvature is simply connected. In particular, there is no metric of everywhere positive sectional curvature on the torus ${\mathbb{T}^2}$. (more…)

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’m going to keep the same notation as before.  In particular, we’re studying how the energy integral behaves with respect to variations of curves.  Now I want to prove the second variation formula when ${c}$ is a geodesic.Now to compute ${\frac{d^2}{d^2 u} E(u)|_{u=0}}$, for further usage. We already showed $\displaystyle E'(u) = \int_I g\left( \frac{D}{dt} \frac{\partial}{\partial u} H(t,u), \frac{\partial}{\partial t} H(t,u) \right)$ Differentiating again yields the messy formula for ${E''(u)}$: $\displaystyle \int_I g\left( \frac{D}{du} \frac{D}{dt} \frac{\partial}{\partial u} H, \frac{\partial}{\partial t} H \right) + \int_I g\left( \frac{D}{dt} \frac{\partial}{\partial u} H, \frac{D}{du}\frac{\partial}{\partial t} H(t,u) \right).$

Call these ${I_1(u), I_2(u) }$. ${I_2}$

Now ${I_2(0)}$ is the easiest, since by symmetry of the Levi-Civita connection we get $\displaystyle I_2(0) = \int g\left( \frac{D}{dt} \frac{\partial}{\partial u} H(t,u), \frac{D}{dt}\frac{\partial}{\partial u} H(t,u) \right) = \int g\left( \frac{D}{dt} V, \frac{D}{dt} V \right).$ For vector fields along ${c}$ ${E,F}$ with ${E(a)=F(a)=E(b)=F(b)=0}$, we have $\displaystyle \int g\left( \frac{D}{dt} E, \frac{D}{dt} F \right) = - \int g\left( \frac{D^2}{dt^2} E , F \right).$ This is essentially a forum of integration by parts. Indeed, the difference between the two terms is $\displaystyle \frac{d}{dt} g\left( \frac{D}{dt} E, F \right).$

So if we plug this in we get $\displaystyle \boxed{ I_2(0) = -\int g\left( \frac{D^2}{dt^2} V , V \right).}$ ${I_1}$

Next, we can write $\displaystyle I_1(0) = \int_I g\left( \frac{D}{du} \frac{D}{dt} \frac{\partial}{\partial u} H(t,u) |_{u=0}, \dot{c}(t) \right)$ Now ${R}$ measures the difference from commutation of ${\frac{D}{dt}, \frac{D}{du}}$. In particular this equals $\displaystyle \int_I g\left( \frac{D}{dt} \frac{D}{du} \frac{\partial}{\partial u} H(t,u) |_{u=0}, \dot{c}(t) \right) + \int_I g\left( R(V(t), \dot{c}(t)) V(t), \dot{c}(t)) \right).$

By antisymmetry of the curvature tensor (twice!) the second term becomes $\displaystyle \int_I g\left( R( \dot{c}(t), V(t)) \dot{c}(t), V(t), \right).$

Now we look at the first term, which we can write as $\displaystyle \int_I \frac{d}{dt} g\left( \frac{D}{du} \frac{\partial}{\partial u} H(t,u), \dot{c}(t)\right)$

since ${\ddot{c} \equiv 0}$. But this is clearly zero because ${H}$ is constant on the vertical lines ${t=a,t=b}$. If we put everything together we obtain the following “second variation formula:”

Theorem 1 If ${c}$ is a geodesic, then $\displaystyle \boxed{\frac{d^2}{du^2}|_{u=0} E(u) = \int_I g\left( R( \dot{c}(t), V(t)) \dot{c}(t) - \frac{D^2 V}{Dt^2}, V(t) \right).}$

Evidently that was some tedious work, and the question arises: Why does all this matter? The next goal is to use this to show when a geodesic cannot minimize the energy integral—which means, in particular, that it doesn’t minimize length. Then we will obtain global comparison-theoretic results.