Timeline for Non-unique splittings of homotopy idempotents
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 9, 2015 at 23:02 | answer | added | Charles Rezk | timeline score: 4 | |
| Apr 13, 2015 at 17:17 | comment | added | Mike Shulman | For instance, here: github.com/HoTT/HoTT/blob/master/theories/Idempotents.v#L815 is a formalized proof that the identity idempotence homotopy of the identity map always has a unique splitting. That doesn't rule out that a nonidentity idempotence homotopy might have more than one splitting, but it's not immediately obvious to me how to find an example. | |
| Apr 13, 2015 at 13:19 | comment | added | Mike Shulman | @TylerLawson Ah, but even if $r,s,r',s'$ are all also the identity, then $K\neq K'$ or $H\neq H'$ isn't enough to ensure that the splittings are different as splittings, because an "equivalence of splittings" consists of an equivalence between the retract types (the domains of $s$ and $s'$) that may not be the identity. | |
| Apr 13, 2015 at 12:59 | comment | added | Tyler Lawson | Right, sorry, the second comment I wrote was supposed to be about that. In the case where f is the identity, the data of the splitting is $(K, H)$ and the output idempotence homotopy is $K^{-1}H$. | |
| Apr 13, 2015 at 6:40 | comment | added | Mike Shulman | @TylerLawson I'm not asking whether a given $f$ can admit different homotopies $f\circ f \sim f$; I know that that's true (see the penultimate paragraph of the question). I'm asking whether one homotopy $f\circ f \sim f$ can be obtained from more one distinct splitting. Or am I misunderstanding what you wrote? | |
| Apr 11, 2015 at 19:03 | comment | added | Tyler Lawson | (What worries me is that perhaps one needs to be more careful about "composing" homotopies, since that involves a higher cell.) | |
| Apr 11, 2015 at 19:01 | comment | added | Tyler Lawson | So if you did want to take $f$ to be the identity, you might be looking for a space $X$ and elements $K, H, K', H'$ loops at the identity in $Map(X,X)$ (which has abelian fundamental group) so that $K^{-1} H = (K')^{-1} H'$, but $K \neq K'$ or $H \neq H'$. Does that sound reasonable? | |
| Apr 11, 2015 at 18:58 | comment | added | Tyler Lawson | Mike, unless I'm mistaken the construction of a splitting involves the choice of a section $s$, a retraction $r$, and choices of homotopy $H: r \circ s \sim 1$, $K: s\circ r \sim f$. This produces your choice of homotopy $(K^{-1} \circ K^{-1}) \cdot (1 \circ H \circ 1) \cdot K: f \circ f \sim f$. For example, even if (as Will suggests) you have $f = id$, you can get any nontrivial elements in $\pi_1$ of the self-mapping space by e.g. picking $K$ to be the trivial homotopy and $H$ arbitrary. | |
| Dec 13, 2014 at 0:27 | comment | added | Mike Shulman | @WłodzimierzHolsztyński, certainly in a 1-category, if the inverse or direct limit exists, then it splits the idempotent. But the question is about homotopy idempotents, and in that case the proof that the section-retraction composite is the identity of the (co)limit does not work unless the witness of idempotence is at least partially coherent. There is a counterexample in Warning 1.2.4.8 of Lurie's Higher Algebra showing that not every homotopy idempotent in spaces splits, even though of course all sequential limits and colimits of spaces exist. | |
| Dec 12, 2014 at 17:58 | comment | added | Włodzimierz Holsztyński | If there is still an idempotent splitting controversy (as oppossed to a clarification :-) then you, @MikeShulman, may present a short statement about the seeming contradiction, and each of us may present their component of the not resolved yet problem; then I am sure that we should obtain a clear picture soon. | |
| Dec 12, 2014 at 17:01 | comment | added | Włodzimierz Holsztyński | My full theorem about the idempotent was: Let $\ f\ $ be an idempotent. Then the following $\ 3\ $ conditions are equivalent: 1.there is the direct limit; 2.there is the inverse limit; 3.$\ f\ $ splits. | |
| Dec 12, 2014 at 16:35 | comment | added | Włodzimierz Holsztyński | But it does when a limit exists. I've already provided a pretty good outline of the proof, and I can fill it up with details (most anybody can). | |
| Dec 12, 2014 at 16:12 | comment | added | Mike Shulman | @WłodzimierzHolsztyński, if you're claiming that that always works, then how do you explain the fact that not every homotopy idempotent splits? | |
| Dec 12, 2014 at 6:33 | comment | added | Włodzimierz Holsztyński | BTW, I was credited for this construction by DAE & RG. I used it for obtaining a shape version of the Wall's example. However, in that paper (published in Ann of Math, 1975) by DAE & RG they blatantly and shamelessly have stolen my result (on shape variation of Wall's example), in broad daylight. (There are many cases of stealing but perhaps never like that one :-) | |
| Dec 12, 2014 at 6:14 | comment | added | Włodzimierz Holsztyński | If $\ X^*\ $ is the inverse limit (when it exists) then the canonical projection morphisms $\ f^* : X^*\rightarrow X\ $ is the $\ell$-morphism, and the $r$-morphism $\ r:X\rightarrow X^*\ $ is induced by the bunch of copies of $\ f\ $ where $\ X\ $ is mapped into $n$-th term $X$ in the inverse sequence. This provides the needed split. The direct limit case is dual, and gives split again when the limit exists. | |
| Dec 12, 2014 at 5:25 | comment | added | Mike Shulman | @WłodzimierzHolsztyński, in what context do you mean that? (Homotopy) inverse limits and direct limits always exist in spaces, but not every homotopy idempotent splits. | |
| Dec 12, 2014 at 5:17 | comment | added | Włodzimierz Holsztyński | An arbitrary idempotent morphism $\ f\ $ splits if any one (or both :-) of the two conditions hold: there exists the countable inverse limit $\ \ldots\rightarrow_f X\rightarrow_f X\ $ or there exists the countable direct limit $\ X\rightarrow_f X\rightarrow_f\ldots\ $ (I proved it in 1969 but most likely it was known some years before). | |
| Dec 12, 2014 at 4:38 | answer | added | Will Sawin | timeline score: 0 | |
| Dec 12, 2014 at 4:03 | comment | added | Will Sawin | I think it's more complicated even than that: two of nontrivial faces are not the $2$-homotopy, but rather the $2$-homotopy composed with $f$. In my case $f$ coming first gives the same thing but afterwards is trivial. Maybe the simplical perspective is better? I think my idea of a constant function must be wrong, because any homotopy splitting of it must be contractible which makes all the maps and homotopies unique. | |
| Dec 11, 2014 at 18:26 | comment | added | Mike Shulman | Yes, you can formulate it using cubes. Lurie does it in HTT 4.4.5 using simplices (of course). I'm not convinced that opposite faces will cancel: if you write out the cube, I think you'll see that the 2-homotopy only appears on five of its faces; the sixth is a naturality square for the 1-homotopy. Thus, if we fill in the cube, we assert $x+x=x+x+x$ where $x$ is our chosen element of $\pi_2(X)$, hence $x=0$. This is more obvious in the simplicial version where a 4-simplex has five 3-simplices as faces. But what about starting with $\pi_3$ instead of $\pi_2$? | |
| Dec 10, 2014 at 19:34 | comment | added | Will Sawin | because opposite faces cancel. The only problem is if the nontrivial element of $\pi_2(X)$ pulls back to a trivial element of $\pi_2( Map(X,X))$, which I don't know how to rule out. | |
| Dec 10, 2014 at 19:33 | comment | added | Will Sawin | If this is right, I think you might be able to construct two different coherentizations of the homotopy $X \to pt \to X$ for some space $X$, maybe $X = \mathbb CP^\infty$. All the functions involved in the homotopy will be constant functions. Start with the obvious homotopies, but choose the 2-homotopy between the two homotopies $f^3 \to f$ to be a sphere representing a nontrivial element of $\pi_2(X)$. Then you can glue on a $3$-cell, $4$-cell, etc. to ensure the higher homotopies exist. The only cell that can trivialize that element of $\pi_2(X)$ is the $3$-cell, and I think it doesn't | |
| Dec 10, 2014 at 19:28 | comment | added | Will Sawin | Let me see if I understand what a fully coherent idempotent is: You have a homotopy $f \circ f \sim f$, which gives you two homotopies $ f\circ f \circ f \sim f\circ f \sim f$, and then you demand a 2-homotopy between them? And then you furthermore have a cube of ways to get from $f^4$ to $f$, and the aforementioned 2-homotopy gives you the faces of the cube, and you demand a 3-homotopy filling the cube, and so on. | |
| Dec 10, 2014 at 18:18 | history | asked | Mike Shulman | CC BY-SA 3.0 |