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I don't understand a passage in the famous article of milnor and kervaire: let $\xi : E \to S^n$ be a vector bundle (of rank $k$) and let $[\xi] \in \pi_{n-1}(SO_k)$ the map associated to $\xi$. Let $X(\xi)$ be the euler class of $\xi$. Prove that the map: $$ [\xi] \mapsto X(\xi) $$ is a homomorphism. ("this is clearly an additive funtion")

The proof consists on three lines in lemma 7 of the article of milnor and kervaire: "a procedure of killing homotopy groups of differentiable manifolds" but I am really not able to understand it. I think that the problem is to understand exactly which definition of euler class is used!

www.maths.ed.ac.uk/~aar/papers/milnorsurg.pdf

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What don't you understand about the proof? It's pretty clear. Do you understand classifying spaces and naturality of characteristic classes? –  Henry T. Horton Jan 23 '13 at 2:06
    
for example, euler class is, according to wikipedia, defined only for vector bundles, but classifiyng spaces are associated to principal bundles, which are not vector bundles. Naturality is clear to me. Maybe "classifiying space" is the problem –  fritz Jan 23 '13 at 2:13
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Isomorphism classes of principal $\mathrm{SO}(m)$-bundles over $B$ are in one-to-one correspondence with isomorphism classes of oriented rank $m$ vector bundles over $B$ via the associated bundle construction, and the Euler classes for each corresponding pair are the same cohomology class $e \in H^m(B; \Bbb Z)$. –  Henry T. Horton Jan 23 '13 at 2:46
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thank you. And which definition of "euler class" makes the map quoted "clearly additive"? I ask this because there are many definitions of euler class, and the concept is new to me! –  fritz Jan 23 '13 at 18:57

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