5
$\begingroup$

I'm not sure this is research level so if this is not appropriate, feel free to move the question to StackExchange. However, I post it here since my "fake proof" is based on a (recent) paper and I'm not sure it would be fit for SE.

Everything is done $\infty$-categorically.

I'm using the paper A simple universal property of Thom ring spectra, by Antonin-Camarena and Barthel (see here for the arXiv version), and specifically using proposition 4.9 and corollary 5.9 of that paper.

For clarity, let me quote them here :

Prop 4.9 : if $f: S^{k+1}\to BGL_1(R)$ is a based map and $\overline f : \Omega^n\Sigma^n S^{k+1}\to BGL_1(R)$ is the corresponding $n$-fold loop map, then for any $A\in Alg_R^{\mathbb E_n}$, there is an equivalence of spaces $Map_{Alg_{R}^{\mathbb E_n}}(M\overline f,A) \simeq \Omega^{\infty +k+1}A$, if $A$ has characteristic $\chi(f)$, otherwise the space is empty.

Where $M\overline f$ denotes the Thom ring spectrum associated to $\overline f$, that is, the colimit of $\Omega^n\Sigma^nS^{k+1} \to BGL_1(R)\to Mod_R$

And

Corollary 5.9 : There is a map $\Omega^2\Sigma^2S^1\to BGL_1S^0$ whose associated Thom spectrum is equivalent to $H\mathbb Z$ as $\mathbb E_2$-ring spectra.

I changed the phrasing : in the paper it was written $\Omega^2S^3$, but I wrote it that way to emphasize the connection between the two results, and to emphasize the fact that a map is obtained as a $2$-fold loop map from $S^1$. So corollary 5.9 puts us in the situation of proposition 4.9 with $k+1=1$.

We get

Let $A\in Alg_\mathbb S^{\mathbb E_2}$, then if that space is nonempty, there is an equivalence of spaces $Map_{Alg_\mathbb S^{\mathbb E_2}}(H\mathbb Z,A) \simeq \Omega^{\infty +1}A$

My problem is the following : consider $K= H\mathbb Z\otimes_{\mathbb S}H\mathbb Z$. This is an $\mathbb E_\infty$-algebra (which is the coproduct of $H\mathbb Z$ with itself as $\mathbb E_\infty$-algebras), and in particular there are two inclusion map $H\mathbb Z\overset{in_i}\to K$, $i=0,1$ that are $\mathbb E_\infty$-maps.

Forgetting some structure, they are also $\mathbb E_2$-maps. But note that $Map_{Alg^{\mathbb E_2}}(H\mathbb Z, K) \simeq \Omega^{\infty +1}K$, so its $\pi_0$ is $\pi_1K = H_1(H\mathbb Z; \mathbb Z) = 0$ (indeed, one can easily prove that $H_{n+1}(K(\mathbb Z,n);\mathbb Z) = 0$ for all $n$, for instance by providing a cell-structure with no $n+1$-cells).

It follows that $Map_{Alg^{\mathbb E_2}}(H\mathbb Z, K)$ is connected, so $in_0\simeq in_1$ as $\mathbb E_2$-maps.

What this means is that $in_0\otimes in_1 : H\mathbb Z\otimes H\mathbb Z\to K\otimes K$ (tensor products will be over $\mathbb S$, unless explicitly stated) is equivalent to $i\otimes i$ as $\mathbb E_2$-maps, or even as maps of spectra, where $i : H\mathbb Z\to K$ is any $\mathbb E_2$-map (for instance $in_0$ or $in_1$)

But the identity of $K$ factors as $H\mathbb Z\otimes H\mathbb Z\overset{in_0\otimes in_1}\to K\otimes K\to K$ as a map of spectra, and so, as a map of spectra, it factors as $H\mathbb Z\otimes H\mathbb Z\overset{i\otimes i}\to K\otimes K\to K$.

But now this latter map can be upgraded to a map of $\mathbb E_\infty$-algebras : indeed if we chose $i=in_0$ for instance, we can see this latter map as $H\mathbb Z\otimes H\mathbb Z \to H\mathbb Z \overset{in_0}\to K$ where the first map is obtained from the coproduct property, applied to the identity two times.

This means that $id_K$ factors, as a map of spectra, through $H\mathbb Z$. Looking at homotopy groups, this is absurd (it would mean $\pi_iK = 0$ for $i\neq 0$, which is known to be false)

Of course there must be a mistake somewhere but I don't see where it is, so my question is :

Where is the mistake ?

$\endgroup$
4
  • 6
    $\begingroup$ the first mistake is a typo in the paper: $\mathbb{Z}$ is a Thom spectrum on $\Omega^2(S^3\langle 3\rangle)$ not $\Omega^2S^3$, so proposition 4.9 does not apply. $\endgroup$ Mar 24, 2020 at 15:15
  • 5
    $\begingroup$ the next question one might ask is "ok, then what goes wrong with the same argument for $\mathbb{F}_p$"? and the answer is that $\pi_1(\mathbb{F}_p \wedge \mathbb{F}_p) \ne 0$. So assuming everything else you wrote is correct (I didn't check) you would be showing that $\tau_0$ detects the difference between the left and right unit as $\mathbb{E}_2$-maps, which would be a fun fact $\endgroup$ Mar 24, 2020 at 15:18
  • 2
    $\begingroup$ @DylanWilson : ah that's what that was ! it did feel weird when reading the appendix, but since they referred to some other paper for how to globalize their maps (from $\mathbb Z_p$ to $\mathbb Z$), I figured that the globalization process somehow made this go from $S^3\langle 3\rangle$ to $S^3$ (although I should have known that it wasn't reasonable). Thanks ! And yeah; I had figured that out for $\mathbb F_p$ - I guess what you call $\tau_0$ is the generator of the $\pi_1$ ? $\endgroup$ Mar 24, 2020 at 15:35
  • 4
    $\begingroup$ @DylanWilson : and I guess your comment completely settles the question (unless someone comes along and finds another mistake), so if you could write it as an answer, I could accept it and close the question $\endgroup$ Mar 24, 2020 at 15:37

0

Your Answer

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