I have two very concrete and simple question. Just in case I write downwards what led me into this.

My questions: Let $\mathbf{SH}(X)$ be the stable homotopy category of Voevodsky. Denote $S^n$ the simplicial $n$-sphere and also the spectrum it defines by infinite suspension.

1.- Is the following diagram commutative? $$ \begin{matrix} S^a\wedge S^{b}&\buildrel{\sim}\over{\to} & S^{a+b}\\ \sigma \downarrow & &\downarrow ^{(-1)^{a+b}}\\ S^b\wedge S^a &\buildrel{\sim}\over{\to} & S^{a+b} \end{matrix} $$ where $\sigma$ denotes the permutation.

2.- If so, can you give a proof or a concrete reference?

Thanks in advance.

What led me to this question? Recall from Cisinki and Deglise that the algebraic de Rham cohomology is represented by a commutative ring spectrum $E_{\mathrm{dR}}$, in other words: $$ H^p(X,\Omega_X^\bullet)=\mathrm{Hom}_{\mathbf{SH(X)_\mathbb{Q}}}(1_X,E_{\mathrm{dR}}[p]). $$ Note that commutative spectrum means that there is a product $\mu \colon E\wedge E\to E$ satisfying the commutative diagram if we switch the source. We know that the algebraic de Rham cohomology is anticommutative or graded commutative, (because the product comes from the wedge product on differential forms) In other words, $\alpha \in H^p(X,\Omega_X^\bullet)$ and $\beta \in H^q(X,\Omega_X^\bullet)$ then $\alpha \cdot \beta =(-1)^{p+q}\beta\cdot \alpha$.

Let me recall the definition of product in the $E$-cohomology. Let $\alpha\in \mathrm{Hom}_{\mathbf{SH(X)_\mathbb{Q}}}(1_X,E_{\mathrm{dR}}[p])$ and $\beta \in \mathrm{Hom}_{\mathbf{SH(X)_\mathbb{Q}}}(1_X,E_{\mathrm{dR}}[p'])$, then $$ \alpha\cdot \beta \colon 1_X\to E_{\mathrm{dR}}[p]\wedge E_{\mathrm{dR}}[p']\simeq E_{\mathrm{dR}}\wedge E_{\mathrm{dR}}\wedge 1_X[p+p']\buildrel{\mu}\over{\longrightarrow}E_{\mathrm{dR}}\wedge 1_X[p+p']\simeq E_{\mathrm{dR}}[p+p'] $$ Recall that $1_X[p]=S^p\wedge 1_X$, in order for this product to be anticommutative or graded commutative the diagram $$ \begin{matrix} 1_X[p]\wedge 1_X[p']&\buildrel{\sim}\over{\to} & 1_X[p+p']\\ \sigma \downarrow & &\downarrow (-1)^{p+p'}\\ 1_X[p']\wedge 1_X[p]&\buildrel{\sim}\over{\to} & 1_X[p+p'], \end{matrix} $$ where $\sigma$ denotes the permutation, must commute. Right?

  • $\begingroup$ I know this is probably frustrating, but I'm not convinced that the kind of commutativity you are asking for is natural. I mean, yes, it'd be nice, but I don't think $S^{a+b}$ knows enough about $a$ and $b$ and $\wedge$ to have useful opinions on that square. $\endgroup$ – Jesse C. McKeown Sep 24 '15 at 22:48
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    $\begingroup$ The sign should be $(-1)^{ab}$. This square is a functorial image of the analogous square in the ordinary stable homotopy category, where the commutativity boils down to computing the degree of reflections on spheres. $\endgroup$ – Marc Hoyois Sep 25 '15 at 0:51
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    $\begingroup$ @Tintin Being the homotopy category of a stable model category, $SH(X)$ is tensored over $SH$. The functor $SH\to SH(X)$ I'm talking about is $E\mapsto E\otimes 1_X$. You may define spheres so that the horizontal maps are equalities, why is that a problem? $\endgroup$ – Marc Hoyois Sep 25 '15 at 13:19
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    $\begingroup$ It's impossible to answer 1) unless you specify what the horizontal arrows are. $\endgroup$ – Fernando Muro Sep 25 '15 at 15:47
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    $\begingroup$ @Tintin This follows from the universal property of $SH$, a reference for which is Higher Algebra, Cor. More concretely, if you take $T$-$S^1$-prespectra in simplicial presheaves as a model for $SH(X)$, then the functor sending a simplicial spectrum to $\Sigma^\infty_T$ of the corresponding constant presheaf is left Quillen for appropriate model structures. $\endgroup$ – Marc Hoyois Sep 26 '15 at 1:30

Thanks to Marc Hoyois which pointed the answer in a comment above. I am just writting that with more details:

1.- Yes, the diagram is commutative.

2.- This already happens in classic stable homotopy category. One may find the concrete statement in Adam's "Stable homotopy and generalised homology" at the beggining of section III.4 It is not anything particular from the product of spheres $S^n$ but it is a general fact from the smash product of spectra. Since this reference is hard to find in electronic format I copy the page here enter image description here

(One obtains our diagram of 1 for classic spectra out of the lower right diagram of Adam's page. Note that there is a missprint on the upper right corner, it should be $Y\wedge X$ and not $X\wedge Y$). Thanks everyone for the comments and help.

  • $\begingroup$ Je dirais même plus, oui, le diagramme est commutatif. Je vois que vous n'avez pas quitté la rue du Labrador depuis votre enquête sur le fétiche arumbaya à l'oreille cassée... $\endgroup$ – Georges Elencwajg May 20 '16 at 21:10

This is actually an answer to a question you did not ask; sorry.

The corresponding diagram $T\wedge T\to T\wedge T$ neither commutes nor anticommutes. An observation of Morel (that was studied in detail in section 16.2. of http://arxiv.org/abs/0912.2110) is that $SH(X)[1/2]$ splits into two summands $SH_+(X)\bigoplus SH_-(X)$ such that this diagram is commutative on $SH_+(X)$ and is anticommutative on $SH_-(X)$. The second summand is "often" non-zero and non-torsion. The De Rham cohomology (as well as $-\wedge E_{\mathrm{dR}}$) kills $SH_-(X)$ (since $E_{\mathrm{dR}}$ is orientable); yet $SH_-(X)$ can be detected by certain Witt cohomology theories.

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    $\begingroup$ The diagram for $a=b=0$ does commute by the axioms for a symmetric monoidal category. $\endgroup$ – Marc Hoyois Sep 25 '15 at 0:09
  • $\begingroup$ Sorry; I was thinking about the diagram $T\wedge T\to T\wedge T$. That was stupid of me! $\endgroup$ – Mikhail Bondarko Sep 25 '15 at 6:32

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