# When is Thom isomorphism a ring map?

Let $$R$$ be an $$E_{\infty}$$-ring spectrum and $$B$$ be an $$E_\infty$$-space. Suppose we have an $$E_\infty$$-map $$f: B \to BGL_1(S^0)$$ such that the composite $$f_R: B \to BGL_1(S^0) \to BGL_1(R)$$ is null, then a choice of null-homotopy produces Thom isomorphism which is a weak-equivalence $$u: Mf \wedge R \simeq B_+ \wedge R,$$

where $$Mf$$ is the Thom spectrum associated to $$f$$.

Note that, both sides of the Thom isomorphism are ring spectra.

Q: I wonder when $$u$$ is a ring-map?

A possible guess is that if $$f_R$$ is homotoped to null via infinite-loop space maps, then maybe $$u$$ is a ring map. I am not quite sure if this is true or how to see this.

## 1 Answer

One comment is that you have to be careful, because to be an $$E_{\infty}$$-map is not a property but rather additional structure.

You are exactly right in that what is needed is a null-homotopy of $$B \rightarrow BGL_{1}(R)$$ as a map of $$E_{\infty}$$-spaces. This is Proposition 3.16 in Omar and Toby's "A simple universal property of Thom ring spectra" when you set $$A = R$$.

• Cool, this is exactly what I was looking for! – Prasit Jan 4 '20 at 22:23