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?

  • 1
    My conjecture: someone set their students an assignment on characteristic classes, and wrote a bunch of superficially plausible garbage on wikipedia in order to catch out plagiarists. – Johannes Nordström Oct 16 '14 at 10:04
  • 1
    Why not ask the author directly? Btw, this question was already discussed on the talk page but apparently without an answer. – ThiKu Oct 16 '14 at 17:06
  • FWIW the same mistake is to be found in – ThiKu Oct 18 '14 at 6:23
  • 1
    Apart from that one should replace "Vandermonde determinant and discriminant" by "Pfaffian and determinant", there is also the question how this follows from a Splitting principle. – ThiKu Oct 18 '14 at 6:35
  • 1
    Here is a weak relationship I know of, which of course doesn't relate to Wiki's claim: You can compute the total Chern class of the flag manifold, written as a product of 2nd degree cohomology classes. Then you use Vandermonde determinants to see that its Todd genus equals 1. (Chapter III.14 of Hirzebruch's Topoloical Methods in Algebraic Topology) – Chris Gerig Oct 19 '14 at 6:18
up vote 3 down vote accepted

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.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.