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The test case: flat Riemannian manifolds November 12, 2009

Posted by Akhil Mathew in differential geometry, MaBloWriMo.
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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.

One way is easy to check by computation.  We show that zero R implies flatness.

First of all, note that on {\mathbb{R}^n} with the usual metric (or an open submanifold thereof), there is a family of {n} vector fields {X_1, \dots, X_n} ({=\frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n}}) which at each point form an orthonormal basis. Moreover, they commute. By contrast, if vector fields with this condition exist on a Riemannian manifold, then it is necessarily locally isometric to {\mathbb{R}^n}. For the commutation condition means that we can locally choose coordinates {x_1, \dots, x_n} such that {X_i = \frac{ \partial}{\partial x_i}}. Then by orthonormality the metric must be given by {\sum dx^i \otimes dx^i}. So if {R \equiv 0}, we have to construct such vector fields. Pick {p \in M} and {(X_1)_p, (X_2)_p, \dots, (X_n)_p} forming an orthonormal basis for {T_p(M)}. I claim that these can be extended to vector fields {X_1, \dots, X_n} as above. First of all, we may assume that as a manifold {M} is an open subset of {\mathbb{R}^n}, just not necessarily with the standard metric, and that {p=\mathbf{0}}. Define {X_1} first along the line {x_2=\dots=x_n=0} by parallel translation of {(X_1)_p} along the {x_1}-axis. Then define {X_1(x_1, x_2, 0, 0 ,\dots, 0)} by starting with {X_1(x_1,0,0,\dots,0)} and parallel translating along the line from {(x_1,0, \dots, 0) \rightarrow (x_1, x_2, 0, \dots, 0).} Keep using parallel translation in this manner to make these into vector fields.

I claim now that {\nabla_X X_1 \equiv 0} for all {X}. By linearity over smooth functions, it is sufficient to show the case of {X = \frac{\partial}{\partial x_i}}. We treat the case {n=2}. Then we already have

\displaystyle \frac{D}{\partial x_2} X_1 \equiv 0

since {X_1} was obtained by parallel translation along “vertical” lines and

\displaystyle \frac{D}{\partial x_1} X_1 = 0 \ \mathrm{on \ the \ line} \ x_2 = 0.

Thus

\displaystyle \frac{D}{\partial x_2} \frac{D}{\partial x_1} X_1 = \frac{D}{\partial x_1} \frac{D}{\partial x_2} X_1 \equiv 0

where we have used the Clairaut-like theorem for connections whose curvature tensor vanishes! Now this implies {\frac{D}{\partial x_1} X_1} is parallel in {x_2}, hence identically zero because it vanishes on the horizontal line {x_2=0}. This proof can be extended to higher {n}.

We do the same for {X_i} for {i>1}. Then the {X_i} are orthonormal because parallel translation preserves inner products. I claim that {[X_i,X_j]=0}. To see this, note that

\displaystyle \nabla_{X_i} X_j - \nabla_{X_j} X_i - [X_i,X_j] = 0

by symmetricity. Since the {X_i} are parallel in every direction, we get {[X_i,X_j]=0}. Consequently the test case is proved.

(The term “test case” is borrowed from Michael Spivak.)

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Comments»

1. TK - February 22, 2012

Did you prove the “only if” part?


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