Talk:Euler class

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I'm not familiar with the relation between the Vandermonde polynomial and the Euler class. This section needs more explanation, and a reference would help.

-Ian Agol —Preceding unsigned comment added by Agol (talk • contribs) 02:28, 9 July 2010 (UTC)Reply

Agreed. I think that the splitting principle relation of the Euler class to the top Pontryagin class is more that the square of a pfaffian is the determinant. Mike Stone (talk) 21:13, 28 July 2013 (UTC)Reply

http://mathoverflow.net/questions/184527/is-there-any-relationship-between-the-euler-class-and-the-vandermonde-determinan --Kamsa Hapnida (talk) 17:47, 16 October 2014 (UTC)Reply

An orientation of ${\displaystyle E}$ amounts to a continuous choice of generator of the cohomology

${\displaystyle H^{r}(F,F\setminus F_{0};\mathbb {Z} )}$

of each fiber ${\displaystyle F}$ relative to the complement ${\displaystyle F\setminus F_{0}}$ to its zero element ${\displaystyle F_{0}}$.

Would somebody mind elaborating, how to get this relative cohomology class? Thanks, Jack

You can get it from Thom Isomorphism

${\displaystyle H^{r}(F,F\setminus F_{0};\mathbb {Z} )=H^{0}(X,\mathbb {Z} )}$

--刻意(Kèyì) 17:37, 20 April 2009 (UTC)Reply