Re: local classification of riemannian manifolds


On Wed, 19 Oct 2005 13:24:37 -0400, Thomas Mautsch wrote:

> but I was taken aback by David L. Johnson's answer
> that "the curvature determines the metric, due to Riemann".
>
> I looked up Sektion 7 in Spivacs "Differential Geometry", Vol. 2:
> It might be that David meant Corollary 13
> ("The Curvature Determines The Metric"), which says that
> if you are given two Riemannian manifolds and a *linear isometry*
> from one of the tangent spaces of the first manifold
> to one of the tangent spaces of the second manifold
> that in addition preserves all sectional curvatures at these points,
> then the metrics are locally isometric around these two points.
>
> So this theorem only says that knowledge of the metric
> and all sectional curvatures at one point
> determines the germ of the metric around this point. -
> This is in most cases more knowledge
> than knowing only the curvature tensor
> and all of its covariant derivatives.

Yeah, that's what I meant. Page 7-26.

The theorem that Spivak proves states that, if you have a linear isometry
between the tangent spaces at one point, L:T(M,x)-->T(N,y), and if the
sectional curvatures at points Exp(V) and Exp(L(V)) agree for V
sufficiently close to 0, then there is a local isometry between them.
The sectional curvatures have to agree on all 2-planes parallel translated
from x (and y) to Exp(V) and Exp(L(V)) along the geodesic determined by V.

Is this more than knowing the curvature tensor and all its covariant
derivatives? I am not certain, but the parallel translation can
certainly be expressed in terms of covariant derivatives. But if you
only mean covariant derivatives at one point, then certainly you are
right; without the metric being analytic you need something like a
germ.

The conditions about the exponential normalize for change in gauge
(the action of the diffeomorphism group), so it is a stronger statement
than simply a germ-equivalence (defined how?) between curvature tensors.

--

David L. Johnson

__o | You will say Christ saith this and the apostles say this; but
_`\(,_ | what canst thou say? -- George Fox.
(_)/ (_) |



.

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