TITLE: Is there any relationship between the Euler class and the Vandermonde determinant?
QUESTION [12 upvotes]: Several Wikipedia articles claim that the relationship between the Euler class $e(V)$ and the top Pontryagin class $p_k(V)$ of an oriented $2k$-dimensional real vector bundle $V$ corresponds, via the splitting principle, to the relationship between the Vandermonde determinant and the discriminant (in particular, in each case the former is the square of the latter). 
As far as I can tell, this can't possibly be true. Most obviously, the Vandermonde determinant and the discriminant have degrees that are much too large: they grow quadratically rather than linearly in the dimension / number of variables. More importantly, the splitting principle for oriented $2k$-dimensional real vector bundles and, say, rational characteristic classes involves invariance under the Weyl group of $\text{SO}(2k)$, which is an index $2$ subgroup of the group of signed permutations on $k$ letters rather than an alternating group as the appearance of the Vandermonde determinant would suggest. In terms of invariant polynomials on the Lie algebra we should instead think of the relationship between the Pfaffian and the determinant. 
So, should the Wikipedia articles be corrected? Or am I missing some less obvious connection?

REPLY [3 votes]: I've taken the liberty of removing this claim about the Vandermonde determinant from all of the relevant Wikipedia articles I could find, listed below for convenience. 

Euler class
Pontryagin class
Vandermonde polynomial
Characteristic class
Splitting principle
Alternating polynomial