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.

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

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