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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeAug 25th 2011

    started Euler class

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeMar 19th 2019
    • (edited Mar 19th 2019)

    added statement of the Whitney sum formula for Euler classes:


    The Euler class of the Whitney sum of two oriented real vector bundles to the cup product of the separate Euler classes:

    χ(EF)=χ(E)χ(F). \chi( E \oplus F ) \;=\; \chi(E) \smile \chi(F) \,.

    diff, v5, current

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeApr 18th 2019
    • (edited Apr 18th 2019)

    added references on Euler forms:


    diff, v10, current

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeApr 27th 2019
    • (edited Apr 28th 2019)

    finally added this kind of remark, to the Properties-section:


    For EE a vector bundle of even rank rank(E)=2krank(E) = 2 k, the cup product of the Euler class with itself equals the kkth Pontryagin class

    χ(E)χ(E)=p k(E). \chi(E) \smile \chi(E) \;=\; p_k(E) \,.

    (e.g. Walschap 04, p. 187)

    When the Euler class is represented by the Euler form of a connection \nabla on EE, which then is fiber-wise proportional to the Pfaffian of the curvature form F F_\nabla of \nabla, the above relation corresponds to the fact that the product of a Pfaffian with itself is the determinant: (Pf(F )) 2=det(F )\big( Pf(F_\nabla) \big)^2 = det(F_\nabla).

    diff, v12, current

    • CommentRowNumber5.
    • CommentAuthorDavidRoberts
    • CommentTimeApr 27th 2019

    Why F F_\nabla and F AF_A?

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeApr 28th 2019

    Thanks for catching this! Fixed now.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeApr 28th 2019
    • (edited Apr 28th 2019)

    I have added statement of the following proposition, for which I am citing (Walschap 04, Chapter 6.6, Thm. 6.1, p. 201-202)


    Let XX be a smooth manifold and EπXE \overset{\pi}{\longrightarrow} X an oriented real vector bundle of even rank, rank(E)=2k+2rank(E) = 2k + 2.

    For any choice of connection \nabla on EE (SO(dim(X))SO(dim(X))-connection), let χ( E)Ω 2k(X)\chi(\nabla_E) \in \Omega^{2k}(X) denote the corresponding Euler form.

    Then the pullback of the Euler form χ( E)\chi(\nabla_E) to the unit sphere bundle S(E)S(π)XS(E) \overset{S(\pi)}{\longrightarrow} X is exact

    (S(π)) *χ( E)=dΩ \big( S(\pi) \big)^\ast \chi(\nabla_E) \;=\; d \Omega

    such that the trivializing form has (minus) unit integral over any of the (2k+1)-sphere-fibers S x 2k+1ι xS(E)S^{2k+1}_x \overset{\iota_x}{\hookrightarrow} S(E):

    S 2k+1ι x *Ω=1. \int_{S^{2k+1}} \iota_x^\ast \Omega \;=\; -1 \,.

    diff, v15, current

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeApr 28th 2019

    finally added this classical reference (also at Pfaffian):

    • Shiing-Shen Chern, A Simple Intrinsic Proof of the Gauss-Bonnet Formula for Closed Riemannian Manifolds, Annals of Mathematics Second Series, Vol. 45, No. 4 (1944), pp. 747-752 (jstor:1969302)

    diff, v16, current

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