Commutativity of naive smash product

Question

I am reading Switzer right now, and I came upon this line in the definition of naive smash product of spectra (p.256):

Some of the naive smash products are commutative ($E\wedge_{BC}F \simeq F\wedge_{CB} E $ provided $\beta(a) \gamma(a)$ is always even).

Why? I am at loss, it seems to me that if I were to write out the products just by definition, $(E\wedge_{BC} F)_{\alpha(a)} = E_{\beta(a)} \wedge F_{\gamma(a)}$, the smash product is always commutative. Where does the oddity condition come in?

Thanks!

Answer 1

You also want the bonding maps $\Sigma(E \wedge_{BC} F)_{\alpha(a)}) \to (E \wedge_{BC} F)_{\alpha(a)+1}$ to be the same. If you examine the formulas on page 255, you'd notice that in one direction there is a twist by a power of the antipode map on $S^1$. The power is either $\beta(a)$ or $\gamma(a)$, and you can show that a necessary and sufficient condition for the bonding maps to be equal is that $\beta(a) \gamma(a)$ is even for all $a$.