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

${I_2}$

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.

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.

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…)