# Thom isomorphism from the ABGHR perspective

In ABGHR Thom spectra are described in the following way: we start with a morphism of Kan complexes $X\to \mathbb{S}\text{-line}$, where $\mathbb{S}\text{-line}$ is an $\infty$-groupoid which is equivalent to the well known $BGL_1(\mathbb{S})$. Then the Thom spectrum $Mf$ is defined to be the colimit of the composition of functors (of $\infty$-categories) $X\to\mathbb{S}\text{-line}\to\mathbb{S}Mod$. If we have a commutative ring spectrum $R$ there is a functor $\mathbb{S}\text{-line}\to R\text{-line}$ and it is a theorem that the colimit of the composition $X\to\mathbb{S}\text{-line}\to R\text{-line}\to RMod$ is equivalent to $Mf\wedge R$. There is a fibration of $\infty$-groupoids $R\text{-triv}\to R\text{-line}$ that we should think of as being our model of $EGL_1(R)\to BGL_1(R)$. If the composition $X\to R\text{-line}$ lifts to $R\text{-triv}$ then we say that $Mf$ is $R$-oriented and we have a Thom isomorphism $Mf\wedge R\simeq X_+\wedge R$.

My question is the following: we know classically that $Mf\wedge Mf\simeq X_+\wedge Mf$ (perhaps there are some conditions I need to put on $f$ to make this true?). So, if we assume $X$ is an infinite loop space and $X\to \mathbb{S}\text{-line}$ is a morphism of infinite loop spaces then we should have that $X\to\mathbb{S}\text{-line}\to Mf\text{-line}$ lifts to $Mf\text{-triv}$, yielding the classical Thom isomorphism described above. Is there an easy way to see that there is such a lift in this framework?

A more general question is: we know that such lifts always give such Thom isomorphisms, but are there conditions under which we can deduce a lift from such an isomorphism?

Let's try to see what a lift $\tilde f : X\to R\textrm{-triv}$ is. Recall that $R\textrm{-triv}$ is the $\infty$-groupoid of $R$-lines with a specified isomorphism with $R$, so a lift $\tilde f$ corresponds to giving a natural equivalence of $f$ with the constant functor $X\to R\textrm{-line}$ with value $R$. This is the same as giving an $R$-module map from the colimit of $f$ to $R$ which is an equivalence after precomposing with all the inclusions $f(x)\to \textrm{colim }f$. But $\textrm{colim }f=Mf\wedge R$, so a lift is effectively giving an $R$-linear map $Mf\wedge R \to R$ such that for all $x$ the composition $f(x)\wedge R\to R$ is an equivalence. Moreover an $R$-module map $Mf\wedge R\to R$ is the same thing as a spectrum map $Mf\to R$.
Specializing in your case it is easy to see that the identity $Mf\to Mf$ gives exactly the Thom class you need (because the map $f$ is multiplicative, that is a loop space map), so you're done.
• Does that spectrum map you describe, $Mf\to R$ need to be multiplicative? – Jonathan Beardsley Nov 30 '14 at 17:18
• No, why should it be? You certainly do not need it in the proof of the Thom isomorphism. In fact the Thom isomorphism is true even when $f$ is not a loop map (so $Mf$ has no ring structure) – Denis Nardin Nov 30 '14 at 17:20