# Signs For Products Defined Homotopy-Theoretically

Let E be an $$E_\infty$$ ring spectrum and let's work in your favorite model for spectra with a symmetric monoidal product. Let $$X$$ be some suspension spectrum (or anything with a nice diagonal?). Then (I think) $$E^*X=[X,\Sigma^*E]$$ has a graded commutative product that should obey the Koszul sign rules.

Explicitly given $$\alpha: X\to \Sigma^n E$$ and $$\beta: X\to \Sigma^m E$$ we form their product:

$$X\xrightarrow{\Delta}X\wedge X \xrightarrow{\alpha \wedge \beta} \Sigma^{n}E\wedge\Sigma^mE\xrightarrow{\cong}\Sigma^{n+m}E\wedge E\to \Sigma^{n+m}E$$

I have heard it said that the sign difference between $$\alpha\beta$$ and $$\beta\alpha$$ comes from swapping the smash-terms in those last maps, but I am not sure why this is the case. Is there an intuitive or obvious reason why swapping (only sphereical?) smash terms should correspond to changing a sign?

In this answer I use the convention that $$S^n = \Bbb R^n_c$$, that is, we identify the sphere with the one-point compactification of $$\Bbb R^n$$. Then the homeomorphism $$W_{p,q}: S^p \wedge S^q \to S^{p+q}$$ is the identity on the subspace $$\Bbb R^p \times \Bbb R^q \to \Bbb R^{p+q}$$ (and sends $$\infty \times S^q$$ and $$S^p \times \infty$$ to $$\infty$$).

Set $$p + q = n$$.

Now the point is that the map $$S^n \xrightarrow{W_{p,q}^{-1}} S^p \wedge S^q \xrightarrow{\tau_{p,q}} S^q \wedge S^p \xrightarrow{W_{q,p}} S^n$$

is is given away from infinity as the map $$\Bbb R^n \to \Bbb R^p \times \Bbb R^q \to \Bbb R^q \times \Bbb R^p \to \Bbb R^n$$ where the first and last map are the identity but the middle term is the swap map $$\tau_{p,q}(v,w) = (w,v).$$

Now there are two homotopy equivalences $$S^n \to S^n$$ up to homotopy: the one of degree 1 and degree -1. Because the map described above is a homeomorphism (which is smooth on $$\Bbb R^n$$) we can compute its degree by seeing whether it is orientation-preserving or reversing. So what you need to know is that the swap map has sign $$(-1)^{pq}$$.

Now apply this to what you're interested in. In the end you have two maps $$f_1, f_2: X \to S^{n+m} \wedge E$$ with $$f_2 = \tau_{n,m} f_1$$. Then use the argument above --- which implies that that $$\tau_{n,m} = (-1)^{nm}$$ on the sphere spectrum $$\Bbb S^{n+m}$$ --- to see that $$f_2 = (-1)^{nm} f_1.$$

I think BTW that $$`$$thing with nice diagonal' is not much different than a suspension spectrum. But I could be wrong, I didn't think about it.

• See John's answer here about the diagonal. Jun 6 at 5:22