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Let R be the 2-periodic complex K-theory spectrum, or any other naturally occuring 2-periodic E-infty ring spectrum. The suspend-once functor gives an autoequivalence of the category of R-module spectra, and since R is 2-periodic applying it twice is isomorphic to the identity functor. In some coarse sense this defines an action of Z/2 on the category of R-modules.

There is a better, more complicated and more correct notion of Z/2-action on a collection of module spectra. Module spectra form an infty-category, and we can ask if Z/2 acts on this infty-category. One way to put it: the suspend-once functor and the homotopy between suspend-twice and the identity give us a map from RP^2 into the classifying space of automorphisms of this infty-category. Does this extend to a map from BZ/2 = RP^infty?

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4 Answers 4

up vote 14 down vote accepted

At least for complex K-theory, I believe the answer is yes. If you want an explicit construction, you can use the fact that there's a symmetric monoidal functor from the 2-groupoid of Clifford algebras and Morita equivalences to the groupoid of invertible K module spectra. So it suffices to construct a monoidal functor from Z/2 into this 2-groupoid, which is a much more concrete problem.

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As Jacob says, to produce a map $RP^\infty\to B\mathrm{Pic}(KU)$, we should use the fact that there is an infinite loop splitting $\mathrm{Pic}(KU)=\mathrm{Cliff}_{\mathbb{C}} \times Y$, where $Y$ is the 3-connected cover.

So, when I got to do the exercise of computing homotopy classes of maps $RP^\infty\to B\mathrm{Cliff}_{\mathbb{C}}$ (using knowledge of the $\infty$-loop k-invariants, as decribed in the other answers), I get $[RP^\infty, B\mathrm{Cliff}_{\mathbb{C}}] = \mathbb{Z}/8$: each stage in the postnikov tower of $B\mathrm{Cliff}$ adds a factor of $2$, and all the extensions are non-trivial. In particular, the elements which are congruent to $1$ mod 2 in $\mathbb{Z}/8$ are non-trivial on the bottom cell. So, if I've done the calculation right (and that's a big if), there are four distinct ways to make $\mathbb{Z}/2$ act so that the generator acts by suspension.

I find it interesting that I get $\mathbb{Z}/8$, which reminds me of real Bott periodicity. In fact, the calculations make it look like $$\mathrm{map}_*(RP^\infty, B\mathrm{Cliff}_{\mathbb{C}}) \approx \mathrm{Cliff}_{\mathbb{R}}$$ as infinite loop spaces.

Is there a way to take a real Clifford algebra $A$, and functorially produce from it a monoidal $\mathbb{Z}/2$-action on (the symmetric monodial Morita $2$-groupoid of) complex Clifford algebras?

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Let $\mathcal{P}$ denote the topological category of $R$-modules that are invertible under the smash product over $R$, with $R$-linear weak equivalences as morphisms. The map $n\mapsto[\Sigma^nR]$ gives a map $\mathbb{Z}\to\pi_0(B\mathcal{P})$, and if $R$ is $2$-periodic this factors through $\mathbb{Z}/2$. Now $\mathcal{P}$ is symmetric monoidal and $\pi_0(B\mathcal{P})$ is already a group (not just a monoid) so there is a spectrum $K(\mathcal{P})$ with $\Omega^\infty K(\mathcal{P})=B\mathcal{P}$. The loop space $\Omega B\mathcal{P}$ is the space $GL_1(R)$ of invertible components in $\Omega^\infty R$. I think that what you are asking for is equivalent to a map $\mathbb{R}P^\infty\to BB\mathcal{P}$, or to a loop map $\mathbb{Z}/2\to \Omega^\infty K(\mathcal{P})$. It would probably be more natural to ask for an infinite loop map $\mathbb{Z}/2\to \Omega^\infty K(\mathcal{P})$, or in other words a map of spectra $H\mathbb{Z}/2\to K(\mathcal{P})$. In the case $R=KU$, some things about the spectrum $K(\mathcal{P})$ are well understood. The key point is that if you complete at any prime, the loop space known as $BSU_{\otimes}$ becomes equivalent to $BSU$. (The proof involves choosing a topological generator of $\mathbb{Z}^\times_p$, so the results for different primes cannot be patched together in any obvious way.) If I have the numbers straight, this should mean that the $4$-connected cover of $K(\mathcal{P})$ becomes equivalent (after $p$-completion) to $\Sigma^5kU$. Unfortunately, by passing to the connective cover we lose the information that is relevant for the immediate question. However, this at least suggests that the problem may be solvable by quite classical methods.

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7  
I believe that this essentially never exists as an infinite loop map: for any commutative ring spectrum R, multiplication by η carries the class of R[1] in pi_0 of Pic(R) to -1 in (pi_0 R)^* = pi_1( Pic(R) ), so R[1] cannot be obtained from an infinite loop map even from Z into Pic(R) unless R has characteristic 2. –  Jacob Lurie Nov 18 '10 at 0:53

EDIT: This argument claimed to prove something it did not, based on a stupid mistake (pointed out by Jacob Lurie in the comments).

Let's let ${\rm Pic}(KU)$ be the category of invertible $KU$-modules and weak equivalences, and $G$ be the subcategory of objects weakly equivalent to $KU$ or its suspension. These are symmetric monoidal with $\pi_0$ a group, so they are infinite loop spaces associated to spectra ${\rm pic}(KU)$ and $g$ respectively. $G$ acts on the category of $KU$-modules, and so your question is roughly: Can the map $\mathbb{Z}/2 = \pi_0 G \subset \pi_0 {\rm Pic}(KU)$ be lifted to a splitting of the map $BG \to B\mathbb{Z}/2$?

First we note that there is a fiber sequence $BGL_1(KU) \to {\rm Pic}(KU) \to \pi_0 {\rm Pic}(KU)$, and similarly for $G$. The map $BG \to B\pi_0 G$ is a principal bundle with fiber $BBGL_1(KU)$, obtained by delooping a fiber sequence of spectra $\Sigma {\rm gl}_1(KU) \to g \to H\pi_0(g)$; thus splitting is equivalent to asking for the triviality of the classifying map $B\pi_0(g) \to BBBGL_1(KU)$. If this were true, then the composite classifying map $B\pi_0(G) \to B^3 \pi_0(GL_1(KU))$ would also be trivial. This is a map $K(\mathbb{Z/2},1) \to K(\mathbb{Z/2},3)$, classifying the first stage in the Postnikov tower, and it is either trivial or the nontrivial stable map classifying the Steenrod operation $Sq^2$ trivial.

We can use naturality to show that the stable map must be $Sq^2$. Let $\Sigma$ be the category of finite sets, which is the free symmetric monoidal category on one object; it has a symmetric monoidal map to the subcategory $G$ of $KU$-modules by sending the set with $n$ objects to $\Sigma^n KU$. Taking the associated spectra (K-theory objects) gives us a map from the sphere to $g$ which, on $\pi_0$ and $\pi_1$, are the projection map $\mathbb{Z} \to \mathbb{Z/2}$ and an isomorphism $\mathbb{Z/2} \to \mathbb{Z/2}$ (the twist map switches signs!) respectively. The k-invariants in the Postnikov tower are maps $H{\mathbb{Z}} \to \Sigma^2 H\mathbb{Z}/2$ and $H{\mathbb{Z}/2} \to \Sigma^2 H\mathbb{Z}/2$. The former is nontrivial (from Jacob's $\eta$-multiplication argument) and so naturality implies that the latter is nontrivial as well.

$KU$ could be any 2-periodic commutative ring spectrum with $2 \neq 0$ in $\pi_0(R)$ here.

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2  
Things swept under the rug here include: Derived Morita theory establishes Pic(KU) as a full subcategory of the category of automorphisms of KU-mod. –  Tyler Lawson Nov 18 '10 at 3:42
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Sq^2: K(Z/2,1) -> K(Z/2,3) is nullhomotopic. So your argument actually shows that the first obstruction vanishes. The next obstruction lives in H^5(BZ/2, Z) and therefore automatically vanishes. Higher obstructions vanish because the map from Pic(KU) to its 3-truncation splits (the splitting coming from the map from the Clifford category into Pic(KU) ). –  Jacob Lurie Nov 18 '10 at 4:05
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@Jacob: In my defense, I can only say: Sometimes, I am just a blockhead. –  Tyler Lawson Nov 18 '10 at 5:06
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However, the argument you just gave then seems to show that for any 2n-periodic commutative ring spectrum with 2n-torsion-free homotopy groups concentrated in even degrees, Z/2n acts (nonuniquely) on the module category - correct? –  Tyler Lawson Nov 18 '10 at 5:08
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Yes, I believe so. –  Jacob Lurie Nov 19 '10 at 19:06

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