6
$\begingroup$

There's an operation $\Lambda^2: KO^0(X) \to KO^0(X)$ such that $\Lambda^2(x+y)-\Lambda^2(x)-\Lambda^2(y)=xy$, which comes from the antisymmetric power. Similarly, there's a $\Lambda^2: KO^4(X) \to KO^8(X)$ with the same property, since the antisymmetric square of an $Sp$ bundle is an $O$ bundle.

Do they extend to the entire $KO^*$ so that $\Lambda^2: KO^d(X) \to KO^{2d}(X)$ and $\Lambda^2(x+y)-\Lambda^2(x)-\Lambda^2(y)=xy$?

If so, is there a corresponding operation in $KO(-,\mathbb{Z}/n\mathbb{Z})$?

$\endgroup$

0

You must log in to answer this question.

Browse other questions tagged .