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By a homotopy idempotent I mean a map $f:X\to X$, where $X$ is a space, equipped with a homotopy $f\circ f \sim f$. In contrast to the situation in stable homotopy theory (where $X$ would be a spectrum or a chain complex), not every homotopy idempotent splits; there is a counterexample in Warning 1.2.4.8 of Higher Algebra.

I'm interested in the dual question of uniqueness: supposing that a homotopy idempotent splits (up to homotopy), can it have multiple inequivalent splittings? Put differently, can a space have multiple inequivalent retracts that induce the same homotopy idempotent?

The meaning of "equivalent" here is somewhat subtle. A homotopy idempotent is, of course, in particular an ordinary idempotent in the homotopy category, and splittings of ordinary idempotents are unique (up to isomorphism). Therefore, any two splittings of the same homotopy idempotent must be incoherently homotopy equivalent. That is, if the first splitting consists of $s:A\to X$ and $r:X\to A$ such that $r\circ s \sim 1$ and $s\circ r \sim f$, and similarly the second consists of $A'$, $s'$, and $r'$, then we must have an equivalence $e:A\simeq A'$ such that $s' \circ e \sim s$ and $e \circ r \sim r'$. However, these homotopies might not be coherently related to the given ones.

Note that Corollary 4.4.5.7 and Proposition 4.4.5.12 of Higher Topos Theory imply that to give a splitting of a homotopy idempotent is equivalent to extending it to a "fully coherent idempotent". So the question could equivalently be phrased: can a given homotopy idempotent admit multiple inequivalent coherentifications?

One last note: I do mean, as I said, that a homotopy idempotent is equipped with a homotopy $f\circ f \sim f$. If the homotopy is allowed to vary, then certainly the same underlying map $f$ can admit inequivalent splittings/coherentifications. For instance, if $f$ is the identity on $S^1 = B\mathbb{Z}$, then in addition to the trivial homotopy $f\circ f \sim f$ there is a nontrivial one, which also admits a splitting, and any splitting or coherentification will remember the difference between these two homotopies. But if we fix a given homotopy $f\circ f \sim f$, can it be split/coherentified in multiple ways?

I expect the answer is yes, but I would like to see an explicit example.

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  • $\begingroup$ 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. $\endgroup$ – Will Sawin Dec 10 '14 at 19:28
  • $\begingroup$ 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 $\endgroup$ – Will Sawin Dec 10 '14 at 19:33
  • $\begingroup$ 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. $\endgroup$ – Will Sawin Dec 10 '14 at 19:34
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    $\begingroup$ @WłodzimierzHolsztyński, if you're claiming that that always works, then how do you explain the fact that not every homotopy idempotent splits? $\endgroup$ – Mike Shulman Dec 12 '14 at 16:12
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    $\begingroup$ @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. $\endgroup$ – Mike Shulman Dec 13 '14 at 0:27
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The question "can a given homotopy idempotent admit multiple inequivalent coherentifications" ought to be approachable by the standard spectral sequence machinery, so let me try to do that. I'll more or less follow the Dwyer-Kan approach.

Let $M=\langle f\,|\, f^2=f \rangle = \{1,f\}$, the walking idempotent as a monoid. Let $\mathcal{S}=$ the simplicially enriched category of CW-complexes, and $h\mathcal{S}$ its homotopy category.

Clearly, a "homotopy idempotent" gives rise to a functor $M\to h\mathcal{S}$. (But not quite conversely, as you've built a choice of homotopy into the definition.) A "coherentification" of it should amount to a commutative diagram $$ \begin{array}{ccc} {\widetilde{M}} & \to & \mathcal{S} \\ \downarrow &&\downarrow \\ M & \to & h\mathcal{S}, \end{array} $$ where $\widetilde{M}\to M$ is a cofibrant approximation to $M$ in simplicial monoids.

Let's fix a space $X$ and $f\colon X\to X$ such that $ff=f$, thus determining a functor $\gamma\colon \widetilde{M}\to M\to \mathcal{S}$. We want to compute something about a full subspace of $$\newcommand{\Map}{\mathrm{Map}} \Map_{s\text{Monoids}}( \widetilde{M}, \Map(X,X) ) $$ whose points are $\phi\colon \widetilde{M}\to \Map(X,X)$ such that $h(\phi)\colon \pi_0\widetilde{M}\to \pi_0\Map(X,X)$ coincides with our chosen $\gamma$.

There's a spectral sequence for this: $$ E_2^{s,t}= H_Q^s(M, A_t) \Longrightarrow \pi_{t-s}\Map_{s\text{Monoids}}(\widetilde{M},\Map(X,X))_{\gamma}. $$ The corner $E_2^{0,0}$ is anomalous. It is not given by cohomology; rather, it corresponds to a choice of $\gamma\colon M\to \pi_0\Map(X,X)$, which we have fixed here. We are going to be interested in the groups $E^{t,t}_2=H_Q^t(M,A_t)$ for $t\geq 1$, which potentially contribute to $\pi_0$.

The cohomology is "Quillen cohomology" of the monoid $M$. The coefficients are in an abelian group object in $\text{Monoids}_{/M}$. Such an abelian group object amounts to:

  • For each $x\in M$, an abelian group $A(x)$.
  • For each $x,y\in M$, group homomorphisms $x\cdot A(y)\to A(xy)$ and $\cdot y\colon A(x)\to A(xy)$ which are natural (e.g., $x\cdot(y\cdot a)=(xy)\cdot a$ and $1\cdot a=a$, and similarly on the right), and which commute: $(x\cdot a)\cdot y= x\cdot (a\cdot y)$.

From this you can build the monoid $A := \coprod A(x)$ with product defined by $$ a\cdot b := (a\cdot y)+ (x\cdot b) \qquad\text{for $a\in A(x), b\in A(y)$.} $$

For a monoid $M$ acting on a space $X$, we use the coefficient systems defined by $$ A_t(f) := \pi_t \Map(X,X)_f \qquad \text{for $f\in M$.} $$ The "$M$-actions" are defined by pointwise pre-or-post composition of a map $a\colon S^t\to \Map(X,X)$ with the constant map $S^t\to *\xrightarrow{f} M\to \Map(X,X)$ where $f\in M$.

Quillen cohomology is in principle hard to compute, because you need to choose a cofibrant resolution of $M$. However, there is a theorem: $$ H^t_Q(M;A) \approx H^{t+1}_{HM}(M,\{1\}; A), $$ where the other side is (relative) "Hochschild-Mitchell" cohomology. Basically, Hochschild-Mitchell cohomology is a Quillen cohomology of $M$, regarded as an object in the category of spaces equipped with a two-sided action by $M$. There is a nice bar complex for computing this.

Anyway, when I work through all this for $M=\{1,f\}$, I think I get the following. The Quillen cohomology $H_Q^t(M,A)$ for $t\geq0$ is equal to the cohomology of a complex: $$ B \to B \to B \to B \to \cdots. $$ The group $B=A(f)$. The first differential is $a\mapsto f\cdot a + a\cdot f - a$. The second differential is $a\mapsto f\cdot a-a\cdot f$. Then they alternate.

The two actions $f\cdot, \cdot f$ are commuting idempotents on the abelian group $B$. Thus, you can write elements of $B$ as $2\times 2$-matrices, so that the actions by $f$ amount to left and right multiplication by $\begin{bmatrix} 1&0\\0&0 \end{bmatrix}$. It is then easy to show that $$ H^t_Q(M, A) =0 \qquad \text{if $t\geq 1$.} $$ Note that $H^0_Q(M,A)$ is not generally $0$. Thus we ought to conclude that the space $\Map_{s\text{Monoid}}(\widetilde{M}, \Map(X,X))_\gamma$ is connected, with higher homotopy groups $\pi_t= H^0_Q(M,A_t)$.

Of course, this is not exactly right: the coefficient system $A_1$ is a group object, but may not be an abelian group object. So I need to contemplate the non-abelian cohomology $$ E_2^{1,1} = H^1_Q(M, A_1),\qquad A_1(f) = \pi_1\Map(X,X)_f. $$ That's more than I want to do right now. In any case, it seems to me the answer to your question comes down to this single group. [Roughly, this group seems to have something to do with "adjusting" a choice of homotopy $\alpha\colon f\sim ff=f$ by a choice of homotopy $\beta\colon f\sim f$; it's something like the orbits of the action $\alpha\mapsto (\beta\beta)\alpha(\beta^{-1})$ on the set $\{\alpha\,|\,\alpha\cdot f=f\cdot \alpha\}$.]

(It's possible, of course, that I've misinterpreted what you mean by "inequivalent" coherentifications. The approach I described presupposes one particular notion of equivalence, but there are others.)

(Or I may have screwed up some other way.)

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I think a counterexample in the form of a nontrivial coherentification of the identity exists.

Essentially the $n$th coherence isomorphism involves filling in a $S^{n-1}$ in $X^X$. So the obstruction to doing so is $\pi_{n-1}(X^X)$, and the choice when doing so is $\pi_n(X^X)$. To make our life as simple as possible, we want $X^X$ (or, more precisely, a connected component of it) to be an Eilenberg-MacLane space, so that we get to make one choice and then only have to check one obstruction.

This is presumably standard (or else I have made some trivial error), but if $X =K(G,n)$ then each connected component of $X$ is $K(G,n)$. Indeed:

$[S^k, X^X]= [S^k \times X, X]= H^n(S^k \times X, G)=H^n(S^k, G) \oplus H^n(X,G)$ (by the Kunneth formula and the fact that $X$ has no cohomology in degrees between $0$ and $n$.)

$H^n(X,G)$ tells you what connected component you're in, and $H^n(S^k, G)$ tells you $\pi_k$ of that component, which is $G$ if $k=n$ and $0$ otherwise.

In particular take $X= K(\mathbb Z/2, 3)$. Let's take the identity morphism as our idempotent and choose either a trivial or a nontrivial homotopy class for the $3$rd coherence isomorphism. We need to check that it's possible to find higher coherence isomorphisms. The only nontrivial condition is the fourth coherence isomorphism, which requires filling a five-simplex where all six faces come from the third coherence isomorphism (in two of the faces it must be composed with $f$, but $f$ is the identity so this doesn't change anything. I guess because $f$ is idempotent it will never change anything.) Whatever the signs on these faces are, adding six copies of something mod $2$ will produce $0$, so the obstruction vanishes.

However this all seems strange because a splitting of the identity idempotent is just a homotopy equivalence, so this seems to suggest there are non-equivalent equivalences of $X$ with something else.

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    $\begingroup$ Hmm. I'm not sure about the argument, but as in your last paragraph I'm pretty sure this can't happen with an identity map. Per HTT, the space of coherent idempotents is equivalent to the space of retractions, so the space of coherentifications of a homotopy idempotent should be equivalent to the space of its splittings (that induce the specified witness of idempotence $f\circ f \sim f$). But if $f$ is the identity with its canonical witness, then the latter space is equivalent to the space of "spaces equipped with an equivalence to $X$", and hence is contractible. $\endgroup$ – Mike Shulman Dec 12 '14 at 6:02
  • $\begingroup$ @MikeShulman What part of the argument are you most skeptical of? Perhaps precisely identifying the problem here will help prove or disprove the claim in general. $\endgroup$ – Will Sawin Dec 12 '14 at 23:13
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    $\begingroup$ If I had to pick where to look for a problem, I'd probably start with the assertion that we only need to fill one 5-simplex and all the other coherence comes for free. I'd probably go back to the definition of coherent idempotent in HTT and see how many simplices will actually need to be added. $\endgroup$ – Mike Shulman Dec 16 '14 at 11:31

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