Talk:Euler class

Vandermonde and Euler class
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)

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)

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)

construction of orientation class
An orientation of $$E$$ amounts to a continuous choice of generator of the cohomology
 * $$H^r(F, F \setminus F_0; \mathbb{Z})$$

of each fiber $$F$$ relative to the complement $$F \setminus F_0$$ to its zero element $$F_0$$.

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

You can get it from Thom Isomorphism


 * $$H^r(F, F \setminus F_0; \mathbb{Z})=H^0(X,\mathbb{Z})$$

--刻意(Kèyì) 17:37, 20 April 2009 (UTC)