Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

By a "homotopy commutative diagram," I mean a functor $F: \mathcal{I} \to \mathrm{Ho}(\mathrm{Top})$ to the homotopy category of spaces. By a "strictification," I mean a lifting of such a functor to the category $\mathrm{Top}$ of topological spaces. I am curious about simple instances where such a "strictification" does not exist: that is, there are obstacles to making a diagram that commutes only up to homotopy strictly commute.

For example, it is known that a homotopy coherent diagram in topological spaces can always be strictified. In other words, if one imposes homotopy commutativity, but also keeps track of all the homotopies and requires that they satisfy compatibility conditions with one another, then one can strictify. More precisely, a homotopy coherent diagram $\mathcal{I} \to \mathrm{Top}$ can be described as a simplicial functor $\mathfrak{C} \mathcal{I} \to \mathcal{Top}$, where $\mathfrak{C}\mathcal{I}$ (notation of HTT) is a "thickened" version of $\mathcal{I}$ where the usual relations in $\mathcal{I}$ only hold up to coherent homotopy. Then it is a theorem of Dwyer and Kan that the projective model structures on homotopy coherent diagrams (i.e. $\mathrm{Fun}(\mathfrak{C} \mathcal{I}, \mathrm{SSet})$) and on commutative diagrams (i.e. $\mathrm{Fun}(\mathcal{I}, \mathrm{SSet})$ are Quillen equivalent under some form of Kan extension.

Part of this question is to help myself understand why homotopy coherence is more natural than homotopy commutativity. One example I have in mind is the following: let $X$ be a connected H space, and suppose $X $ is not weakly equivalent to a loop space. Then one can construct a simplicial diagram in the homotopy category given by the nerve construction applied to $X$ (as a group object in the homotopy category). This cannot be lifted to the category of spaces, because then Segal's de-looping machinery would enable us to show that $X$ was the loop space of the geometric realization of this simplicial space.

However, this example is rather large. Is there a simple, toy example (preferably involving a finite category) of a homotopy commutative diagram that cannot be strictified?

share|improve this question
5  
See the discussion starting on p.401 of: Cooke, George Replacing homotopy actions by topological actions. Trans. Amer. Math. Soc. 237 (1978), 391–406. –  Oscar Randal-Williams Nov 23 '11 at 17:30
6  
The obstructions to which you allude in your first paragraph are introduced in Dwyer-Kan, "Realizing diagrams in the homotopy category by means of diagrams of simplicial sets," MR0744648, and "An obstruction theory for diagrams of simplicial sets," MR0749527. Your example can be adapted in the following way: any H-space that is not an A_n-space defines a functor $\Delta_{\leq n}\to\mathrm{Ho}\mathcal{S}$ that cannot be rectified. –  Clark Barwick Nov 23 '11 at 19:28
5  
A naturally occurring example of a functor $\mathcal I\to Ho(Top)$ is the one where $\mathcal I$ is the category of abelian groups and the functor takes $G$ to a Moore space: a simply connected space whose only nontrivial homology group is $H_n\cong G$ for some fixed $n\ge 2$. People have studied the obstructions to strictifying this. There are probably pretty small diagrams of abelian groups for which it can't be done. Google "equivariant Moore space". –  Tom Goodwillie Nov 23 '11 at 22:14
    
Thanks for all the references! –  Akhil Mathew Nov 26 '11 at 16:14
    
On the other hand, for which categories I is Top^I -> Ho(Top)^I dense? I.e. what is the condition on I that ensures that there is never an obstruction against rectifying a diagram of shape I? Cf. also mathoverflow.net/questions/39431/… . –  Matthias Künzer Dec 14 '11 at 7:10
add comment

5 Answers

up vote 10 down vote accepted

EDIT: Tom Goodwillie, in the comments, points out (interpreted in a quite charitable way) that there are two mistakes with the following argument. The $\pi_1$-obstruction does exist. However:

  • It misreads the question and assumes that there is some portion of an actual diagram which is homotopy commutative.

  • Even given that, it makes a mistake: it asserts that we can ignore the indeterminacy. This is wrong. For any space in the diagram, the space of maps to the terminal object $S^1$ is not simply-connected, but instead is homotopy equivalent to $S^1$. This allows you to simply erase the obstruction in $\pi_1$ by simply making different choices of homotopies.

As such, it's advisable that what's written below be demoted and I'll try to get a correct version later.


One of the classical obstructions to realizability is for cubical diagrams. Here, the category $I$ is the poset of subsets of $\{0,1,2\}$. Given such a diagram which is homotopy commutative, you can get twelve maps (one per edge in the cube) and six homotopies (one per face, which are well-defined up to "multiplication by an element in $\pi_1$") and the collection of all possible ways to compose maps, or compose maps with homotopies, gives rise to a hexagon in the space $Map(X,Z)$. Here $X$ is the image of initial object and $Z$ is the image of the terminal object in the cube. If the diagram is actually commutative then you can choose your homotopies to be trivial, and get a trivial hexagon; if it is equivalent to an honestly commutative diagram then you can choose your six homotopies on faces so that the hexagon in $Map(X,Z)$ can be filled in with a disc, or equivalently "represents the trivial element in $\pi_1$".

All scare quotes in the above are places where I'm neglecting basepoints. You have to be more careful in a real-world situation.

Here's an example where all the spaces in the diagram are contractible, except for the terminal one. This makes it easy to ignore the indeterminacy.

Let $Z$ be $S^1$, viewed as a quotient of $[0,1]$. We have six subsets of $Z$, which are the images of $$ [0,1/3], [1/3, 2/3], [2/3, 1], \{0\}, \{1/3\}, \{2/3\} $$ We get a corresponding cubical diagram in the homotopy category as follows. The space $S^1$ and its six subspaces define an honestly commutative diagram which is almost all of a commutative cube, except that it's missing the initial vertex. Let the initial vertex be a point $\ast$, which maps isomorphically to all three objects $\{0\}, \{1/3\}, \{2/3\}$.

(Sorry, I'm not really up to TeXing up the commutative diagram on MO today, it would be much easier to grasp.)

This diagram in the homotopy category is homotopy commutative. In fact, all the spaces are connected and the sources of nontrivial morphisms are contractible, so the diagram has no choice but to commute. The six homotopies all occur in contractible mapping spaces, so there is no $\pi_1$-indeterminacy. The hexagon maps to the mapping space $Map(\ast,S^1) \cong S^1$, and if you use the most obvious choices of homotopies then the hexagon maps to $S^1$ by a homotopy equivalence.

In this instance you can choose a witness to each commutativity diagram. The failure to rectify homotopy commutativity occurs here because your witnesses aren't telling stories that are compatible.

share|improve this answer
    
I ignored the "finite category" restriction. A useful exercise might be to write down a chain-level version of this diagram. –  Tyler Lawson Nov 23 '11 at 20:03
1  
I guess I was thinking that the range category (topological spaces) is not particularly finite in any sense. –  Tyler Lawson Nov 25 '11 at 6:44
4  
I think that this example answers a slightly different question form what was asked. I think the question is: "What is an example of a functor $X:I\to Ho(Top)$ such that there is no functor $I\to Top$ such that the composed functor $I\to Top\to Ho(Top)$ is isomorphic to $X$?" –  Tom Goodwillie Nov 26 '11 at 19:49
4  
In Tyler's example, a lifting is given after restriction to a certain subcategory of $I$, and what does not exist is a lifting on all of $I$ such that on the subcategory it is equivalent (in some homotopy-coherence sense) to the given lifting. –  Tom Goodwillie Nov 26 '11 at 19:52
2  
@Tyler: Dihedral of order $10$. –  Tom Goodwillie Dec 9 '11 at 3:15
show 6 more comments

One thing to be aware of is that, even when liftings to strict diagrams exist, their non-uniqueness is a serious matter. For example, consider the square pushout diagram in which $S^n$ maps to a point, and to $D^{n+1}$ by inclusion; the fourth space is $S^{n+1}$. Compare this with a different diagram in which the four spaces are the same and three of the maps are the same but the map $S^n\to D^{n+1}$ is replaced by a constant map (to some point in the boundary). This is again strictly commutative, and it yields an isomorphic diagram in $Ho(Top)$, but the one square is "highly connected" while the other is not: the first square gives a map from the homotopy fiber of $S^n\to \star$ (which is $S^n$) to the homotopy fiber of $D^{n+1}\to S^{n+1}$ (which is homotopy equivalent to $\Omega S^{n+1}$) that induces an isomorphism of $H_n$, while the second square induces a map between precisely the same two homotopy fibers which, being a constant map, cannot possibly induce an isomorphism.

EDIT

Of course there are similar examples involving chain complexes instead of spaces.

Note that in the case of the category $Ch(K)$ of chain complexes over a field $K$ the existence of a lifting is automatic. In fact, let $Ch'(K)$ be the full subcategory consisting of chain complexes whose boundary operators are zero: the composed functor $Ch'(K)\to Ch(K)\to Ho(Ch(K))$ is an equivalence of categories, so every diagram (of any shape) in the homotopy category can be lifted.

But of course this does not mean that we are silly to bother with chain complexes that have nonzero boundary operators. The point is that a commutative diagram in the homotopy category is not good for much. For example, Mayer-Vietoris sequences arise from certain square diagrams of chain complexes (say, those which are both pushouts and pullbacks), but two of these may be isomorphic as diagrams in the homotopy category and give different results.

share|improve this answer
    
Thanks. I guess your example also points out the difference between homotopy coherence and homotopy commutativity: to give a map from a space into a homotopy fiber product $A \times_{h B} C$ is not only the same as giving maps $X \to A, X \to C$ such that the push-forwards to $B$ are homotopic, but also to give a choice of homotopy. –  Akhil Mathew Nov 28 '11 at 5:54
add comment

Generically speaking, the obstructions to rigidification are generalized Toda brackets.

Let C be the partial order {0,1}x{0,1}x{0,1}. (C for cube) Let f:A->B, g:B->C and h:C->D be pointed maps such that gf and hg are null homotopic. Lay f, g and h along three edges of the cube with A at 000, B at 001, C at 011, D at 111. Put * (a point) at all other vertices of C and fill in the rest of the arrows in the only way possible. This gives a hunky dory diagram in the homotopy category. The diagram can be rigidified if and only if the Toda bracket contains 0. Jeff Smith

share|improve this answer
6  
Welcome to MO, Mr. Smith. –  Ryan Budney Dec 2 '11 at 22:59
1  
Thanks for this response. (As a student of your work on combinatorial model categories, I second the welcome!) –  Akhil Mathew Dec 3 '11 at 4:19
add comment

A good place to look for finite toy examples is in the theory of group extensions with non-abelian kernel.

A homomorphism $G \to Out(H)$ is the same as a functor $G \to Ho(Top)$ sending $*$ `to $BH$, where $G$ is viewed as a category with one object $*$ and $G$ as morphisms. By the Dwyer-Kan obstruction theory (noting that the diagram is centric) the obstructions to lifting this to a diagram in Top lies in $H^3(G;Z(H))$.

A closer inspection shows that this obstruction class is the same as the obstruction in $H^3(G;Z(H))$ to constructing an extension

$$1 \to H \to ? \to G \to 1$$

It is possible to calculate this obstruction group in a number of cases:

Eilenberg-MacLane show in the 1947 Annals paper "Cohomology Theory in Abstract Groups II" that given a group $G$ and an abelian group $C$ with an action of $C$, and a class $v$ in $H^3(G;C)$ then there exist a group $H$ with center $C$ and a morphism $G \to Out(H)$ realizing the class $v$. (A special case is proven in MacLane's book Homology, and there is an exposition on some of this in Brown's book on group cohomology.) Note that in this construction $H$ may be infinite even if $G$ and $C$ are not.

It is also possible to give completely finite examples: Given a group H there is a universal obstruction in $H^3(Out(H);Z(H))$ associated to this group, determining whether the extension

$$1 \to H \to ? \to Out(H) \to 1$$

exists. Other obstruction classes will be a pull-back of this one (in particular it has the best chance of being non-zero).

The topological interpretation of this obstruction class is that it is the unique k-invariant of the space $BAut(BH)$, where Aut means the space of self-homotopy equivalences. (Recall that group extensions correspond to fibrations, and fibrations with fiber $F$ are classified by maps into $BAut(F)$, and this is a good way to understand group extensions with non-abelian kernel.)

Now, it is possible to calculate this class explicitly for small groups $H$. It of course vanishes when H is abelian. It also turns out to vanish for $H = D_8$ or $Q_8$, but is actually non-zero for $D_{16}$ and $Q_{16}$.

[As an aside I can say that the H^3 class always vanishes for compact connected Lie groups, which is a theorem of de Siebenthal from the 1950's. Kasper Andersen and I proved that this generalizes to p-compact groups (in a G&T paper from 2008), which was why Kasper and I examined the case where $H$ finite.]

[As a further aside I can say that there are a lot of similar obstruction questions over relatively small categories such as the orbit category and corresponding centric diagrams, which occur in e.g., in the theory of p-compact groups and p-local finite groups, when addressing certain uniqueness questions -- here lucily the obstructions are usually zero, although this is usually not so easy to prove, and in particular don't seem to follow from "formal" arguments.]

share|improve this answer
add comment

I'm not sure if this is what Tom was referring to, but here's an explicit example that Adams gave of a homotopy-associative H-space that couldn't be strictified, this is quoted from Stasheff's book on H-spaces:

Let $Y$ be a Moore space, $Y(G, 2n-1)$ where divisibility by $2$ and $3$ but no other primes is possible. Thus $Y$ has cohomology only in dimensions $2n-1$ and for $i \ne n$, $\pi_iY \cong \sum_{p>3} \pi_i(S^{2n-1})_p$. Consider $Y \subset \Omega^2\Sigma^2Y$ which is an associative H-space. The obstructions to forming $Y \times Y \rightarrow \Omega^2\Sigma^2Y \times \Omega^2\Sigma^2Y \rightarrow \Omega^2Y$ into $Y$ lie in $H^i(Y \wedge Y; \pi_i(\Omega^2\Sigma^2Y,Y))$. The relative groups are trivial to at least $5(2n)-3$, so the obstructions vanish. Similarly, there are no obstructions to deforming an associating homotopy within $\Omega^2\Sigma^2Y$ to one in $Y$. That $Y$ is not a loop space follows from the decomposability of $P^nu_{2n} \cdot u_{2n}P$ for $p>3$ and $n \not\vert p-1$.

I assume that he's referring to Steenrod operations at the end there... Anyway, I know there are probably fancier examples using Zabrodsky mixing, but I like this one because we see an explicit obstruction to straightening the homotopy associativity, via the Steenrod operations.

share|improve this answer
    
I realize you mentioned something about this in your question, but I figured an explicit example would give you the "toy" you were looking for. –  Dylan Wilson Nov 24 '11 at 1:02
    
Thanks. This is pretty interesting; I'm going to have a look at Stasheff's book on H spaces. –  Akhil Mathew Nov 27 '11 at 14:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.