November 25, 2009

Sectional curvature

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: , , |
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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).

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November 24, 2009

The second variation formula

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: , , , , |
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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.

 

November 22, 2009

Proof of the Cartan-Hadamard theorem

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: , , , |
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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.

 

November 12, 2009

The test case: flat Riemannian manifolds

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: , , , |
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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…)

 

November 11, 2009

Identities for the curvature tensor

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: , , , |
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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…)

 

November 9, 2009

The Riemann curvature tensor

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: , |
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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…)