Bousfield Classes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:06:28Z http://mathoverflow.net/feeds/question/88155 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88155/bousfield-classes Bousfield Classes Jon Beardsley 2012-02-10T22:43:41Z 2012-02-11T04:04:04Z <p>This question has a few parts:</p> <p>1) Is the Bousfield class of <code>$\langle E\rangle$</code> the class of $E$-acyclics, i.e. <code>$\langle E\rangle=\left\{ X:E\wedge X=0\right\}$</code> or is it the class of spectra which are Bousfield equivalent to $E$? It seems that both of these definitions are used. In either case, which is then the correct characterization of the partial ordering on the Bousfield lattice, as the former definition of Bousfield class seems amenable to reverse inclusion, but the latter does not.</p> <p>2) Secondly, the following statement is made in Hovey and Palmieri's "Structure of the Bousfield Lattice" and I feel that it is probably really obvious, but that I must be missing something:</p> <p>Given an element of the Bousfield lattice <code>$\langle E\rangle$</code> such that <code>$\langle E\rangle &lt;\langle H\mathbb{F}_p\rangle$</code> we have that <code>$E\wedge H\mathbb{F}_p=0$</code> else <code>$\langle E\rangle\geq \langle H\mathbb{F}_p\rangle$</code>. I cannot see why this is necessarily true in general and assume it must have something to do with the nature of <code>$H\mathbb{F}_p$</code> that I am missing. Can anyone help me out? I do not believe any other suppositions are made of $E$ than it is a spectrum. </p> <p>So perhaps (2) depends on (1) in some way. H&amp;P say that <code>$\langle E\rangle$</code> is the class of acyclics, but then how is (2) true in that case? I also believe that, definitions aside, (2) would follow if it was the case that <code>$H\mathbb{F}_p\wedge H\mathbb{F}_p=0$</code> but I don't think that's true....is it?</p> <p>Please excuse my ignorance!</p> http://mathoverflow.net/questions/88155/bousfield-classes/88162#88162 Answer by Tom Goodwillie for Bousfield Classes Tom Goodwillie 2012-02-10T23:16:43Z 2012-02-10T23:16:43Z <p>I don't know the preferred notation, so let's be noncommittal and denote the class of $E$-acyclic spectra by $(E)$. Now let's call $E$ and $E'$ equivalent if and only if $(E)=(E')$. And now let's denote by $[E]$ the class of all spectra equivalent to $E$. </p> <p>You can make a lattice whose elements are the classes $(E)$, ordered by inclusion. Since $(E)=(E')$ if and only if $[E]=[E']$, you get an isomorphic lattice whose elements are the $[E]$. </p> <p>I think the special thing about $H\mathbb F_p$ that is being used is that the smash product of any spectrum with $H\mathbb F_p$ is (an $H\mathbb F_p$-module and therefore) always a coproduct of copies of $\Sigma^n H\mathbb F_p$ for various integers $n$, so that if $E\wedge H\mathbb F_p$ is not contractible then $E\wedge H\mathbb F_p$ has some $\Sigma^n H\mathbb F_p$ as a retract. This implies that any spectrum $X$ that is not $H\mathbb F_p$-acyclic is not $E$-acyclic, either, since $(X\wedge E)\wedge H\mathbb F_p$ has a suspension of $X\wedge H\mathbb F_p$ as a retract.</p> http://mathoverflow.net/questions/88155/bousfield-classes/88163#88163 Answer by Charles Rezk for Bousfield Classes Charles Rezk 2012-02-10T23:21:24Z 2012-02-10T23:21:24Z <p>1) The best convention is probably this. Identify $\langle E\rangle$ with the class of $X$ for which $E\wedge X\neq 0$. The the "Bousfield class of $E$" is the class $\langle E\rangle$, and we say that "$E$ and $F$ have the same Bousfield class" if $\langle E\rangle = \langle F\rangle$. </p> <p>With this convention, the partial ordering on Bousfield classes is just class containment: $\langle E\rangle \leq \langle F\rangle$ iff $\langle E\rangle \subseteq \langle F\rangle$.</p> <p>2) The new ingredient comes from the Kunneth theorem for ordinary homology, which tells you that $$H\mathbb{F}_p \wedge X \wedge Y = 0 \quad\text{iff either}\quad H\mathbb{F}_p\wedge X=0 \quad\text{or}\quad H\mathbb{F}_p\wedge Y=0.$$ If $\langle E\rangle&lt; \langle H\mathbb{F}_p \rangle$, then there is an $X$ such that $H\mathbb{F}_p\wedge X\neq0$ but $E\wedge X=0$. This would satisfy $H\mathbb{F}_p\wedge E\wedge X=0$, and to be consistent with Kunneth, we must have $H\mathbb{F}_p\wedge E=0$.</p> http://mathoverflow.net/questions/88155/bousfield-classes/88176#88176 Answer by John Palmieri for Bousfield Classes John Palmieri 2012-02-11T04:04:04Z 2012-02-11T04:04:04Z <p>Charles and Tom have answered question (2) very nicely.</p> <p>For question (1), for discussing inequalities like $\langle F \rangle \leq \langle E \rangle$, it doesn't matter too much which point of view you take. With Charles's version, $\leq$ means "is contained in", i.e., $\subseteq$; if you take $\langle E \rangle$ to mean $E$-acyclics instead, then $\leq$ means "contains", i.e., $\supseteq$. If you take $\langle E \rangle$ to be the equivalence class of $E$, then you define inequality explicitly: $$\left(\langle F \rangle \leq \langle E \rangle\right) \Leftrightarrow \left(E \wedge X = 0 \Rightarrow F \wedge X = 0\right).$$ <em>However</em>, if someone wants to use notation like "$F \in \langle E \rangle$", well, I will insist that this is a terrible idea, since $\langle E \rangle$ can mean such three very different classes of spectra, depending on the author. Writing "$F \in \langle E \rangle$" can only lead to confusion. Instead, just say "$F$ is $E$-acyclic" or "$E \wedge F \neq 0$" or "$\langle F \rangle = \langle E \rangle$", depending on which one you mean.</p>