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started Euler class
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(E⊕F)=χ(E)⌣χ(F).added references on Euler forms:
Varghese Mathai, Daniel Quillen, below (7.3) of Superconnections, Thom classes, and equivariant differential forms, Topology Volume 25, Issue 1, 1986 (10.1016/0040-9383(86)90007-8)
Siye Wu, Section 2.2 of Mathai-Quillen Formalism (arXiv:hep-th/0505003)
Liviu Nicolaescu, Section 8.3.2 of Lectures on the Geometry of Manifolds, 2018 (pdf, MO comment)
finally added this kind of remark, to the Properties-section:
For E a vector bundle of even rank rank(E)=2k, the cup product of the Euler class with itself equals the kth Pontryagin class
χ(E)⌣χ(E)=pk(E).(e.g. Walschap 04, p. 187)
When the Euler class is represented by the Euler form of a connection ∇ on E, which then is fiber-wise proportional to the Pfaffian of the curvature form F∇ of ∇, the above relation corresponds to the fact that the product of a Pfaffian with itself is the determinant: (Pf(F∇))2=det(F∇).
Why F∇ and FA?
Thanks for catching this! Fixed now.
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 X be a smooth manifold and Eπ⟶X an oriented real vector bundle of even rank, rank(E)=2k+2.
For any choice of connection ∇ on E (SO(dim(X))-connection), let χ(∇E)∈Ω2k(X) denote the corresponding Euler form.
Then the pullback of the Euler form χ(∇E) to the unit sphere bundle S(E)S(π)⟶X is exact
(S(π))*χ(∇E)=dΩsuch that the trivializing form has (minus) unit integral over any of the (2k+1)-sphere-fibers S2k+1xιx↪S(E):
∫S2k+1ι*xΩ=−1.finally added this classical reference (also at Pfaffian):
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