6
$\begingroup$

Given a map of spaces $f:X\to BGL_1(R)$ for $R$ an $E_\infty$-ring spectrum (of course this can be done more generally) one can produce a Thom spectrum $Mf$ by a number of methods. Let's denote such a datum by $(X,f)$ and let $(X,\ast)$ denote the datum $X\to \ast\to BGL_1(R)$, whose associated Thom spectrum is $R\wedge\Sigma^\infty_+X$, which we will denote by $R[X]$. There is a morphism in $Top_{BGL_1(R)}$, which we might denote $(\Delta,-\times\ast):(X,f)\to (X\times X,f\times\ast)$ which Thomifies to the morphism of spectra $Mf\to Mf\wedge R[X]$ (note that the target of this map is an object of $Top_{BGL_1(R)}$ because we can multiply the two maps together using the multiplicative structure on $BGL_1(R)$). It is not hard to show that the map $(X,\ast)\to (X,\ast\times\ast)$ induces a co-$E_\infty$ comultiplication on $R[X]$. However, it seems much less clear to me that the Thom diagonal in general exhibits $Mf$ as an $E_\infty$-comodule over $R[X]$. This would follow from showing that the above diagonal map exhibits $(X,f)$ as a co-$E_\infty$-comodule in $Top_{BGL_1(R)}$. Intuitively, I believe this to be true because all of the necessary coherences exist on the diagonal map, and on the morphism $f$ we're simply crossing with the trivial map. Note that the above is not the trivial coaction, which would be given by $X\simeq X\times\ast\overset{id_X\times\ast}\to X\times X$.

For me, being a co-$E_\infty$ comodule simply means being an $E_\infty$-module over the $E_\infty$-algebra $R[X]$ in the opposite of the category of interest. Note also that in the category $Top_{BGL_1(R)}$ the symmetric monoidal structure is given by the infinite loop space structure of $BGL_1(R)$.

I have been playing with this for some time but have not been able to prove it with the tools I have.

$\endgroup$
3
  • $\begingroup$ Oh my. It just occurred to me that I may be making my life very difficult. Is this actually just a strictly cocommutative coaction? $\endgroup$ Feb 10, 2015 at 3:24
  • $\begingroup$ Is the map $f$ an $E_{\infty}$ map? $\endgroup$
    – Prasit
    Feb 10, 2015 at 3:56
  • $\begingroup$ @Prasit Nope. I'm only using the structure of $X$ as an $E_\infty$-coalgebra, which it gets from the diagonal map. $\endgroup$ Feb 10, 2015 at 4:44

1 Answer 1

4
$\begingroup$

The answer to this is yes, it is co-$E_\infty$. Let $\iota\colon BGL_1(R)\to Mod_R$ be the inclusion. Since colimit is left adjoint to the strong monoidal diagonal functor, it's oplax monoidal. Note that the constant functor $\kappa_R\colon X\to Mod_R$ is the monoidal unit for the pointwise monoidal structure in $Mod_R^X$ (hence a coalgebra over which every other functor is a comodule). So there is an equivalence $\iota\circ f\xrightarrow{\sim} (\iota\circ f)\otimes_{pw} \kappa_R$ in $Mod_R^X$, where $\otimes_{pw}$ is the pointwise tensor product. Taking the colimit of this, and using that colimit is oplax monoidal gives us a coaction of $colim(\kappa_R)\simeq R[X]$ on $colim(\iota\circ f)\simeq Mf$. This is all as monoidal (e.g. $\mathbb{E}_k$-monoidal) as its various moving parts allow it to be. The only thing one needs to check is that the resulting morphism $Mf\to Mf\otimes_R R[X]\simeq Mf\otimes X$ is the "Thom diagonal" in the literature. This follows from Theorem 4.15 in my preprint.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.