Let {M} be a Riemannian manifold with metric {g}, Levi-Civita connection {\nabla}, and curvature tensor {X,Y,Z \rightarrow R(X,Y)Z}. Define\displaystyle R(X,Y,Z,W) := g\left( R(X,Y)Z, W\right). Apparently people sometimes refer to this as the curvature tensor, though it is probably not too confusing.

Some algebra

Recall also the following three identities, proved here:

  • (Skew-symmetry) {R(X,Y,Z,W) = -R(Y,X,Z,W)}
  • (Skew-symmetry) {R(X,Y,Z,W) = -R(X,Y,W,Z)}
  • (Bianchi identity) {R(X,Y,Z,W) + R(Z,X , Y, W) + R(Y, Z, X, W) = 0}

I claim now that there is a type of symmetry:

\displaystyle \boxed{ R(X,Y,Z,W) = R(Z,W,X,Y).} This is in fact a general algebraic lemma.

Lemma 1 Let {V} be a real vector space and let {R:V \times V \times V \times V \rightarrow \mathbb{R}} be a quadrilinear map satisfying the three bulleted identities. Then it satisfies the boxed one.

The proof is some slightly messy algebra, which I’ll only sketch. There is a geometrical way of thinking about this that Milnor presents in Morse Theory.

Let the Bianchi identity as written above be denoted by {Rel_{X,Y,Z,W}}. If we add {Rel_{X,Y,Z,W}, Rel_{W,Y,Z,X}, Rel_{X,W,Z,Y}, Rel_{X,Y,W,Z}} and use skew-symmetry several times, we obtain

\displaystyle R(W,Y,Z,X) + R(Z,W,Y,X) + R(X,W,Z,Y) = 0. 

Now using {Rel_{W,Y,Z,X}} gives

\displaystyle R(X,W,Z,Y) = R(Y,Z,W,X)  

and suitably interchanging all the variables gives the result.

Sectional curvature

Notation as above, for a 2-dimensional subspace {S \subset T_p(M)}, define the sectional curvature {K(S)} as

\displaystyle K(S) := R( E, F, F, E)  

if {(E,F)} form an orthonormal basis for {S}.I now claim this is well-defined.

Proposition 2 If {X,Y \in T_p(M)} span {S}, then\displaystyle K(S) = \frac{ R(X,Y,Y,X) }{(X,X)(Y,Y) - (X,Y)^2}. In particular, {K(S)} depends only on {S} and is well-defined.


Indeed, write {X = x_1 E + x_2 F, Y = y_1 E + y_2 F}. Then this is a computation depending on the previous identities already proved. (Messy algebra and blogging do not mix well.)  

Sectional curvature determines {R}

The sectional curvature actually encodes all the information contained in {R}. Indeed, if we had two Riemannian metrics on the same manifold with curvature tensors {R,R'} with the same sectional curvature in all two-dimensional planes, then

\displaystyle R'(X,Y,Y,X) = R(X,Y,Y,X)  for all {X,Y \in T_p(M)}.

Consider the difference {R' - R}. Then it satisfies all the four identities in the first section of this post, along with {B(X,Y,X,Y) = 0} (by skew-symmetry again). Also by the fourth identity, {B(X,Y,X,Y')} is symmetric in {Y,Y'}, so the skew-symmetry just proved implies

\displaystyle B(X,Y,X,Y') \equiv 0. In particular, we get another skew-symmetric identity:

\displaystyle R(X,Y,Z,W) = -R(Z,Y,X,W)=R(Y,Z,X,W) . 

Applying this again gives

\displaystyle R(X,Y,Z,W) = R(Z,X,Y,W) , and the Bianchi identity clearly gives that {B \equiv 0}.

I should say something about the geometric interpretation about all this.  If S \subset T_p(M), then \exp_p(S \cap U) for U a small neighborhood of 0 \in T_p(M) is a surface in M.  Then the Gauss curvature of that surface is K(S).

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) }.


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).}


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.

Apologies for some initial bugs in the formulas–I have now corrected them.

Today I will prove the Cartan-Hadamard theorem.

Nonvanishing of Jacobi fields

The key lemma is that (nontrivial) Jacobi fields do not vanish.

Lemma 1 Let {M} be a Riemannian manifold of negative curvature, {\gamma: [0,M]} a geodesic on {M}, and {J} a Jacobi field along {\gamma} with {J(0)=0}. If {\frac{D}{dt}V(t)|_{t=0} \neq 0}, then {J(t) \neq 0} for all {t > 0}.

Indeed, we consider { \frac{d^2}{dt^2} \left \langle J(t), J(t)\right \rangle}, which equals

\displaystyle 2\frac{d}{dt} \left \langle \frac{D}{dt} J(t), J(t) \right \rangle = 2\left \langle \frac{ D^2}{dt^2} J(t), J(t)\right \rangle + 2 \left| \frac{D}{dt} V(t) \right|^2.

I claim that this second derivative is negative, which will follow if we show that

\displaystyle \left \langle \frac{ D^2}{dt^2} J(t), J(t)\right \rangle \geq 0.

 But here we can use the Jacobi equation and the antisymmetry of the curvature tensor to turn {\left \langle \frac{ D^2}{dt^2} J(t), J(t)\right \rangle } into

\displaystyle \left \langle R(\dot{\gamma}(t), J(t)) \dot{\gamma(t)}, J(t) \right \rangle = -\left \langle R( J(t),\dot{\gamma}(t) ) \dot{\gamma(t)}, J(t) \right \rangle \geq 0.

 (The last inequality is from the assumption of negative curvature.)

This proves the claim.

Now there are arbitrarily small {t} with { \left \langle J(t), J(t) \right \rangle \neq 0} because {\frac{D}{dt} J(t)|_{t=0} \neq 0}, so in particular there must be arbitrarily small {t} with { \frac{d}{dt} \left \langle J(t), J(t) \right \rangle > 0}. In particular, this derivative is always positive. This proves the claim.

I followed Wilkins in the proof of this lemma.

Proof of the Cartan-Hadamard theorem

By yesterday’s post, it’s only necessary to show that {\exp_p} is a regular map. Now if {X,Y \in T_p(M)}

\displaystyle d(\exp_p)_X(Y) = J(1)

 where {J} is the Jacobi field along the geodesic {\gamma(t) = \exp_p(tX)} with {J(0)=0, \frac{D}{dt} J(t)|_{t=0} = Y}. This is nonzero by what has just been proved, which establishes the claim and the Cartan-Hadamard theorem.

Recall that two Riemannian manifolds {M,N} are isometric if there exists a diffeomorphism {f: M \rightarrow N} that preserves the metric on the tangent spaces. The curvature tensor  (associated to the Levi-Civita connection) measures the deviation from flatness, where a manifold is flat if it is locally isometric to a neighborhood of {\mathbb{R}^n}.

Theorem 1 (The Test Case) The Riemannian manifold {M} is locally isometric to {\mathbb{R}^n} if and only if the curvature tensor vanishes. (more…)

It turns out that the curvature tensor associated to the connection from a Riemannian pseudo-metric {g} has to satisfy certain conditions.  (As usual, we denote by \nabla the Levi-Civita connection associated to g, and we assume the ground manifold is smooth.)

First of all, we have skew-symmetry

\displaystyle R(X,Y)Z = -R(Y,X)Z.

This is immediate from the definition.

Next, we have another variant of skew-symmetry:

Proposition 1 \displaystyle g( R(X,Y) Z, W) = -g( R(X,Y) W, Z)  (more…)

Today I will discuss the Riemann curvature tensor. This is the other main invariant of a connection, along with the torsion. It turns out that on Riemannian manifolds with their canonical connections, this has a nice geometric interpretation that shows that it generalizes the curvature of a surface in space, which was defined and studied by Gauss. When {R \equiv 0}, a Riemannian manifold is flat, i.e. locally isometric to Euclidean space.

Rather amusingly, the notion of a tensor hadn’t been formulated when Riemann discovered the curvature tensor. 

Given a connection {\nabla} on the manifold {M}, define the curvature tensor {R} by

\displaystyle R(X,Y)Z := \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z.

There is some checking to be done to show that {R(X,Y)Z} is linear over the ring of smooth functions on {M}, but this is a straightforward computation, and since it has already been done in detail here, I will omit the proof.

The main result I want to show today is the following:

Proposition 1

Let {M} be a manifold with a connection {\nabla} whose curvature tensor vanishes. Then if {s: U \rightarrow M} is a surface with {U \subset \mathbb{R}^2} open and {V} a vector field along {s}, then\displaystyle \frac{D}{\partial x} \frac{D}{\partial y} V = \frac{D}{\partial y} \frac{D}{\partial x} V. (more…)