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Euler characteristic as a total derivative

  1. May 2, 2013 #1
    We all know that the Euler characteristic is a topological invariant. But let's suppose that we don't know this or anything else about algebraic topology for that matter. We are given only the Gauss-Bonnet theorem, which expresses the Euler characteristic in geometrical terms. In his string theory text, Polchinski claims that this expression is a "total derivative" (actually, the integral of one), which I believe is physics-speak for "exact form". He claims that this observation demonstrates that the Euler characteristic does indeed depend only on topology. How does this argument go? It seems that, if what we are given is true, Stoke's theorem should rewrite the Euler characteristic in terms of the value of some form on the boundary. But this seems wrong... no?
  2. jcsd
  3. May 2, 2013 #2


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    I don't think that the argument Polchinski gives there (that the curvature form is locally exact) is a valid argument. He gives the correct argument back on page 16, where he explains that the variation with respect to the metric of the curvature form on a 2-manifold vanishes because ##R_{ab} = (R/2) g_{ab}## by the Bianchi identities. Therefore the integral of the curvature form is independent of the metric and only depends on the topology of the manifold.

    The same argument must also work in higher dimensions. There the correct integral involves powers of the Riemann tensor, Ricci tensor and Ricci scalar. The Riemann tensor is no longer so tightly constrained, by as we vary each term in the expression for the Euler characteristic, we will find that the sum of contributions to the variation vanishes.
  4. May 2, 2013 #3


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    I think this works but check Milnor's Characteristic Classes.

    The Gauss bonnet form on a surface and its generalization to compact manifolds with a Riemannian metric, the Pfiaffian form, is closed and depends only on the Levi-Civita connection. It is a polynomial in the matrix of curvature 2 forms.

    Given two connections compatible with two different Riemannian metrics, the line segment between them,

    t[itex]\nabla_{1}[/itex] + (1-t)[itex]\nabla_{2}[/itex]

    is a 1 parameter family of connections on the manifold. This yields a Pfiaffian form at each point in the interval and all of these forms together are a closed n-form on MxI. It now follows from Stokes theorem the the integral of the Pfiaffian form over the manifold is independent of the connection.

    This argument works for other "Invariant polynomials" in the curvature 2 form and these polynomials do not require the connection to be compatible with a Riemannian metric. Also the vector bundle need not be the tangent bundle of the manifold. These are the Chern classes of complex vector bundles and the Pontryagin classes of real vector bundles (up to torsion). Their periods are all independent of the connection.

    One requires the connection to be compatible with a metric in order for the Pfiaffian form to exist in general. There are flat ( curvature form is identically zero) vector bundles whose connections are not compatible with a metric which never the less do not have a vector field with no zeros.

    I do not see why this argument shows that the Pfiaffian form of the tangent bundle is a topological invariant. For one thing, the form depends on the tangent bundle not just the manifold. Further if the topological manifold has more than one smoothness structure then I am not sure how to relate the two Pfiaffian forms.

    If one defines a topological invariant as something that is preserved by homeomorphisms then showing that a geometric property is a topological property requires translating the geometry into topology. Why would invariance under a Riemannian metric show this?

    Even with the Euler characteristic when defined combinatorially , one must show that it is a topological invariant. I have always seen this done using homology theory.
    Last edited: May 3, 2013
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