# Euler class and Vandermonde polynomial

«If the rank $r$ is even, then this cohomology class $e(E) \cup e(E)$ equals the top Pontryagin class $p_{r/2}(E)$. Under the splitting principle, this corresponds to the square of the Vandermonde polynomial equaling the discriminant: the Euler class corresponds to the Vandermonde polynomial, the basic alternating polynomial, while the top Pontryagin class corresponds to the discriminant, a symmetric polynomial. More formally, the Euler class of a direct sum of line bundles is the Vandermonde polynomial (orientation determines the order of the line bundles up to sign), while top Pontryagin class is the discriminant.»

Can anybody clarify me this? In particular I don't understand the sense of "the Euler class of a direct sum of line bundles is the Vandermonde polynomial". I checked the references but there is nothing about that.

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