January 1, 2013

Curvature III

Posted by Akhil Mathew under differential geometry | Tags: , , |
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Let {(M, g)} be a Riemannian manifold. As before, one associates to it the curvature tensor

\displaystyle R: TM \otimes TM \otimes TM \rightarrow TM, \quad X, Y, Z \mapsto R(X, Y) Z.

In the previous post, we saw a quantitative expression of how the curvature is a measure of the deviation from the flatness of {M}. Given {M}, one can try to choose local coordinates around a point {p \in M} which make the metric look like the euclidean metric to order 2 at {p}, i.e. local coordinates such that the coefficients near {p} are given by

\displaystyle g_{ij} = \delta_{ij} + O(|x|^2).

However, we saw that the quadratic terms involve precisely the values of the curvature tensor at {p}. Even in the best coordinates, one can’t generally make the coefficients of a metric look euclidean to order 3: the obstruction is precisely the curvature at {p}. Today, I’d like to describe the interpretation of curvature in terms of geodesics. Once again, the material is standard and can be found in introductory textbooks on Riemannian geometry.

1. Curvature and geodesic deviation

There’s another way to think of curvature, which also leads to this: curvature measures how nearby geodesics spread. To think about this, suppose we have a one-parameter family {\gamma_s} of geodesics in {M}, where {\gamma = \gamma_0} is the starting point of the variation. One then has a vector field

\displaystyle V = \left( \frac{d}{ds} \gamma_s\right)_{s = 0}

along the curve {\gamma}, which measures the infinitesimal “spreading” of the one-parameter family {\gamma_s}. Now, a computation shows that {V} satisfies the equation

\displaystyle \frac{D^2}{dt^2} V(t) + R( V, \dot{\gamma}(t)) \dot{\gamma(t)} = 0,

in other words that {V} is a Jacobi field. Here {\frac{D}{dt}} is covariant differentiation along the curve {\gamma}. (more…)

 

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 20, 2009

Variations of curves, Jacobi fields, and the differential of the exponential map

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: , , |
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It is of interest to consider functions on the space of curves {I \rightarrow M}, where {I} is an interval and {M} is a smooth manifold. To study maxima and minima, it is of interest to consider variations of curves, holding the endpoints fixed. Let {c:[a,b] \rightarrow M} be smooth. A variation of {c} is a smooth map\displaystyle H: [a,b] \times (-\epsilon, \epsilon) \rightarrow M

with {H(t,0) = c(t)}, and {H(a,u)=c(a), H(b,u)=c(b)} for all {t,u}. For a variation {H} of {c}, define the variation vector field (which is an analog of a “tangent vector”)

\displaystyle V(t) = \frac{\partial}{\partial u} H ; this is a vector field along {c}. Similarly we can define the “velocity vector field” {\dot{c}} along {c}. If {M} is provided with a connection, we can define the “acceleration vector field” {A(t) = \frac{D}{dt} \dot{c}}, where {\frac{D}{dt}} denotes covariant differentiation.

Given a vector field {V} along {c}, we can construct a variation of {c} with {V} as the variation vector field: take {(t,u) \rightarrow \exp_{c(t)}(u V(t))}.

Variations of geodesics and Jacobi fields Let {M} now be a manifold with a symmetric connection {\nabla}.Let {H} be a variation of a geodesic {c} such that for any {u \in (-\epsilon,\epsilon)}, {t \rightarrow H(t,u)} is a geodesic as well. Then the variation vector field satisfies a certain differential equation. Now\displaystyle \frac{D^2}{dt^2 } V(t) = \frac{D}{dt} \frac{D}{dt} \frac{\partial}{\partial u } H |_{u=0} = \frac{D}{dt} \frac{D}{du} \frac{\partial}{\partial t} H |_{u=0}.

 We have used the symmetry of {\nabla}. Now we can write this as

\displaystyle \frac{D}{du} \frac{D}{d t} \frac{\partial}{\partial t} H |_{u=0} + R\left( \frac{\partial H}{\partial t}, \frac{\partial H}{\partial u}\right) \frac{\partial H}{\partial t} |_{u=0}. By geodesy, the first part vanishes, and the second is

\displaystyle R( \dot{c}(t), V(t)) \dot{c}(t) We have shown that {V} satisfies the Jacobi equation

\displaystyle \boxed{ \frac{D^2}{dt^2 } V(t) = R( \dot{c}(t), V(t)) \dot{c}(t).} Any vector field along {c} satisfying this is called a Jacobi field.

The differential of the exponential map Let {p \in M}, and consider the exponential map {\exp_p: U \rightarrow M} where {U} is a neighborhood of the origin in {T_p(M)}. Let {X,Y \in T_p(M)}. Now the map\displaystyle H: (t,u) \rightarrow \exp_p( tX + ut Y)  takes horizontal lines to geodesics in {M} when {t,u} are small enough. This can be viewed as a vector field along the geodesic {t \rightarrow \exp(tX)}. The variation vector field {J} is thus a Jacobi field, and also at {t=1} is {(\exp_p)_{*X}(Y) \in T_{\exp_p(X)}(M).} Note that {J} satisfies {J(0)=0} and

\displaystyle \frac{D}{dt} J(t)|_{t=0} = \frac{D}{du} \frac{\partial}{\partial t} H(t,u) |_{t,u=0} = \frac{D}{du} (X + uY)|_{u=0} = Y.

Proposition 1 Suppose {X,Y} are sufficiently small. Let {J} be the Jacobi field along the geodesic {\gamma(t) := \exp_p(tX)} with {J(0) = 0, \frac{D}{dt} J(t)|_{t=0} = Y} (i.e. using the ODE theorems). Then\displaystyle J(1)= (\exp_p)_{*X}(Y) \in T_{\exp_p(X)}(M).

 

I will next explain how to use this fact to prove the Cartan-Hadamard theorem on manifolds of negative curvature.

Source: These notes, from an introductory course on differential geometry.