most recent 30 from http://mathoverflow.net 2013-05-19T08:23:53Z http://mathoverflow.net/feeds/question/108416 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108416/mathbbz-2-action-on-spectra-given-by-inversion Akhil Mathew 2012-09-29T17:34:36Z 2012-09-29T19:58:22Z

<p>This is a somewhat naive question to which I don't know the answer. There is a map of spectra $S \to S$, defined up to homotopy, given by multiplication by $-1$, and it satisfies the relation $(-1)^2 \simeq 1$. Can this be refined to a strict $\mathbb{Z}/2$-action on the sphere $S$ (and thus on all spectra)? </p> <p>My feeling is that the answer is probably not, but I haven't been able to prove it. I do know that when one inverts $2$, there is a $\mathbb{Z}/2$-action: if I understand correctly, the obstructions to rigidifying a homotopy $\mathbb{Z}/2$-action to an actual $\mathbb{Z}/2$-action on a spectrum $X$ live in the $\mathbb{Z}/2$-cohomology groups of $B Aut(X)$ (where $Aut(X)$ is the monoid of homotopy self-equivalences of $X$), which should be zero when $2$ acts invertibly on $\pi_*X $. </p>

http://mathoverflow.net/questions/108416/mathbbz-2-action-on-spectra-given-by-inversion/108417#108417 Tyler Lawson 2012-09-29T18:36:14Z 2012-09-29T19:58:22Z

<p>There's often no self-map of $S$ which induces multiplication by $(-1)$, because it's usually not cofibrant-fibrant. There's no <em>homotopical</em> obstruction to it existing, because the map $E \mapsto E \wedge S^1$ is an autoequivalence of the stable homotopy category, and $\mathbb{Z}/2$ acts on $S^1$. Or, if you prefer, we a map of based $\mathbb{Z}/2$-spaces <code>$$ \mathbb{Z}/2_+ \to S^0 $$</code> and taking the homotopy fiber of the induced map of suspension spectra, we get a homotopy fiber sequence of spectra with $\mathbb{Z}/2$-action <code>$$ S \to S \wedge \mathbb{Z}/2_+ \to S $$</code> with the first term having a negation action.</p> <p>One can see visibly that in, e.g., the category of symmetric spectra, the sphere has no $-1$ self-map, but the equivalent free spectrum <code>$F_1(S^1)$</code> does (for appropriate definitions of "S^1"). In the framework of Adams' "Stable homotopy and generalized homology", because maps are only defined "eventually", the analogue of <code>$F_1(S^1)$</code> and $S$ are isomorphic objects, so you do have this map.</p> <p>So if you want it on-the-nose, it depends on your framework. If you want it homotopically, it always exists (roughly because the map <code>$BGL_1(S) \to B\mathbb{Z}/2$</code> splits).</p>