K-theory product trivial on $K^1$ due to stability?

Question

Let $K^0$ be the topological complex vector bundle $K$-theory functor. On the one hand, we have that the product on $K^0(\Sigma X)$ vanishes by a Mayer-Vietoris argument. On the other hand, we define $K^1(X) = K^{-1}(X)=K^0(\Sigma X)$, hence is it correct to conclude the graded product for the topological $K$-theory ring vanishes in all odd degrees?

Answer 1

No, it is not. Recall that by the Chern character isomorphism, rational K-theory is periodic rational cohomology, so the cup product on odd rational K-theory doesn't vanish in any situation where the cup product on odd rational cohomology doesn't vanish; the simplest example is $S^1 \times S^1$.

The stability argument is irrelevant because the cup product is not a stable operation. The cup product on $K^0(\Sigma X)$ is an operation $K^0(\Sigma X) \times K^0(\Sigma X) \to K^0(\Sigma X)$ but the cup product on $K^1(X)$ is an operation $K^1(X) \times K^1(X) \to K^2(X) \cong K^0(X)$; these operations have nothing to do with each other. Note that your argument applies equally well, and is equally irrelevant, to ordinary cohomology.