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Local invariants

  1. Sep 3, 2009 #1


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  3. Oct 3, 2009 #2


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    Last edited by a moderator: Apr 24, 2017
  4. Oct 3, 2009 #3


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    Ok, and generally speaking, a local invariant for Riemannian geometry is a mathematical object O(p) associated to a point p on a Riemannian manifold (M,g) such that if there exists f:(M,g)-->(M',g') a local isometry at p, then O(f(p))= O(p).

    A local invariant of topological manifolds is dimension.

    A local invariant for symplectic manifolds would be an object O(p) associated to a point p on a symplectic manifold (M,ω) such that if there exists f:(M,ω)-->(M',ω') a local symplectomorphism at p, then O(f(p))= O(p). But by Darboux's theorem, there is no local invariant that is "properly symplectic", in the sense that every symplectic invariant would also be a smooth invariant, and thus would not be helpful in classifying symplectic manifolds. for this, we can only rely on global invariant.

    Thank you for this great link mma!

    Question: Are there properly smooth invariants (local or global)? i.e. do we know of a way to associate an object to a smooth manifold such that this object is the same for two diffeomorphic manifolds, but not necessarily so if the smooth manifolds are simply homeomorphic?
  5. Oct 5, 2009 #4
    the gauss curvature for an embedded surface in R3 can be calculated from the unit normal.
    this is an extrinsic definition. gauss proved that it could be calculated from the internal metric as well. this makes it intrinsic and a invariant of a local isometry. i am not sure if this the works for hypersurfaces in higher dimensions. that is an interesting question perhaps.
  6. Oct 5, 2009 #5
    in the differentiable category being differentiable at a point is a local property preserved under local diffeomorphism but not homeomorphism.
  7. Oct 9, 2009 #6
    There are no local differentiable invariants: locally, every manifold without boundary is diffeomorphic to Euclidean space. If you add extra structure, say a metric, one might ask if every manifold without boundary is locally isometric to Euclidean space. It turns out that this is not the case.

    Global invariants (like Euler number, compactness, homology, etc.) are only interesting (/defined) when you discuss the entire manifold (i.e. it wouldn't make sense to discuss the Euler number at a point). The question you ask (topological vs. smooth) becomes difficult immediately, because it turns out that many invariants defined in terms of a fixed smooth structure or metric don't depend at all on the choices you make.

    Defining meaningful smooth invariants is very difficult (the Donaldson and Seiberg-Witten invariants spring to mind, but there are others).
  8. Oct 9, 2009 #7
    that was very helpful. are there local invariants of conformal structures?
  9. Oct 14, 2009 #8
    I'm not too familiar with the definition, but I'll take a shot. I'm guessing that conformal structure is related to the conformal equivalence of metrics. In this case, there would be local conformal invariants (i.e. the round metric on the sphere is not conformally equivalent to a flat metric in a neighborhood of any point). Pushing the analogy a bit, if you take a metric with strictly positive sectional curvature on a compact manifold, any metric conformally equivalent to it would also have strictly positive sectional curvature. I.e. the sign of sectional curvature would also seem to be a local conformal invariant.
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