Is it true that every connected space with

1) just finitely many nontrivial homotopy groups, all finite,


2) just finitely many nontrivial rational cohomology groups, all finite rank,

is weakly homotopy equivalent to a point?

In 1953 Serre proved that any noncontractible simply-connected finite CW-complex has infinitely many nontrivial homotopy groups. That kills off a lot of possible counterexamples.

In 1998, Carles Casacuberta wrote:

However, we do not know any example of a finite CW-complex with finitely many nonzero homotopy groups which is not a $K(G, 1)$, and the results of this paper suggest that it is unlikely that there exist any.

I'm interested in my question because the spaces it asks about are the connected spaces whose homotopy cardinality and Euler characteristic are both well-defined. These concepts are morally 'the same', but it seems the spaces on which they're both defined are in very short supply, unless we stretch the rules of the game and use tricks for calculating divergent alternating products or sums.

For some further discussion of these issues see the comments starting here:


and also these slides and references:


Edit: Condition 1) was supposed to say our space is "cohomologically finite", while 2) was supposed to say it's "homotopically finite". It's been pointed out that condition 1) is too weak: spaces like $\mathbb{R}P^\infty = K(\mathbb{Z}/2,1)$ exploit this weakness and serve as easy counterexamples to my question. They are cohomologically infinite in some sense, but not in a way detected by rational cohomology.

So let me try again. I can think of two ways:

Fix #1: Is it true that every connected space with

1) just finitely many nontrivial homotopy groups, all finite,


2) just finitely many nontrivial integral cohomology groups, all finitely generated,

is weakly homotopy equivalent to a point?

Fix #2: Is it true that every connected finite CW complex with just finitely many nontrivial homotopy groups, all finite, is homotopy equivalent to a point?

  • 10
    $\begingroup$ If one allows infinite CW complexes, then the answer is no: take a $K(G,1)$ with $G$ finite. If one does not, the answer is yes: take the universal cover and then apply Serre's theorem. $\endgroup$ – algori Jun 8 '11 at 5:28
  • 7
    $\begingroup$ Just to point out that the fundamental group is not important here, you can generalize Tom and algori's answers. If G is finite (and abelian) then K(G,n) for n>1 also gives a space with one non-zero homotopy group (which is finite) and with no rational cohomology. $\endgroup$ – Hal Sadofsky Jun 8 '11 at 7:09
  • 1
    $\begingroup$ What if you drop the word "rational" from condition 2? $\endgroup$ – Theo Johnson-Freyd Jun 8 '11 at 12:42
  • 3
    $\begingroup$ Dear John -- re fix no 2: isn't it answered in my first comment? Re fix no 1: I don't know the answer, but if it is positive, then this would probably be not so easy to prove, since a positive answer implies a positive answer to Casacuberto's question for spaces with finite fundamental group. $\endgroup$ – algori Jun 9 '11 at 10:43
  • 1
    $\begingroup$ I'm not sure what the homotopy cardinality is, but let me try to explain the identity $e(\mathbb{R} P^\infty) =\frac{1}{2}$. Suppose $X$ is a CW complex which is not of finite type but which has a finite $d$-sheeted cover $\tilde X$ that is homotopy equivalent to a finite CW complex. Then one can define the Wall characteristic $\chi(X)$ of $X$ as $\frac{1}{d}e(\tilde X)$. This is a rational number which is invariant under homotopy equivalences and does not depend on the choice of $\tilde X$. If $X$ is itself homotopy equivalent to a finite CW complex, we get the Euler characteristic. $\endgroup$ – algori Jun 9 '11 at 11:59



Yes to Fix #2.

A space with finitely many homotopy groups, all finite, is in the Bousfield class generated by the Eilenberg-MacLane spaces for those groups in those dimensions. The Sullivan Conjecture---Miller's Theorem---implies that any such space has no nontrivial maps into any finite-dimensional space. Thus if the space is finite (-dimensional) its identity function is homotopic to the constant function.

There's a lot more to say about this point of view.

The fundamental relation is that the space of pointed maps $\mathrm{map}_*(X,Y)$ is weakly contractible. Thus $X$ can't 'see' $Y$ at all from a homotopy-theoretical perspective.

Sullivan Conjecture/Miller's theorem: If $G$ is a (locally) finite group, then $BG$ cannot see any finite-dimensional space.

Then we have

Zabrodsky Lemma: If $F \to E \to B$ is a fibration sequence (with $B$ path-connected) and if $F$ cannot see $Y$, then $B$ can see $Y$ if and only if $E$ can see $Y$.

Now a simple induction shows that for any $n\geq 1$, $K(G,n)$ cannot see any finite-dimensional space; and then that any space $X$ with finitely many nonzero homotopy groups, all finite, also cannot see finite-dimensional spaces.

  • 1
    $\begingroup$ This is very cool! Thanks for posting. $\endgroup$ – David White Jul 31 '16 at 13:37
  • $\begingroup$ Nice. One of my homotopy theory pals says he's used to Bousfield classes being used stably, giving a stable version of what you said: "any finite complex with only finitely many non-zero stable homotopy groups is stably contractible." He's wondering how the unstable version works. $\endgroup$ – John Baez Aug 2 '16 at 6:56
  • 1
    $\begingroup$ @JohnBaez I'm definitely using the unstable version, defined by $X$ and $Y$ are in the same class if (for every space $Q$) the space of maps from $X$ into $Q$ is weakly contractible if and only if the same is true of $Y$. To show that a spectrum maps trivially into all finite complexes, it suffices to show that it maps trivially into spheres. Surprisingly, this is true unstably too! For a finite-type CW complex $X$ with abelian fundamental group, $\mathrm{map}_*(X,S^n) \sim *$ for all $n\geq 1$ if and only if $\mathrm{map}_*(X,K) \sim *$ for all finite-dimensional spaces $K$. $\endgroup$ – Jeff Strom Aug 2 '16 at 16:24

This is really a comment rather than an answer.

I am dubious that the Euler characteristic of $X$ can be considered well-defined if $H^\ast(X;\mathbb{Q})$ is finitely-generated but $H^\ast(X;\mathbb{Z})$ is not. If $H^*(X;\mathbb{Z})$ is finitely generated then the Euler characteristic of $H^\ast(X;K)$ is constant for all fields $K$. If $X=\mathbb{R}P^\infty$ then we have an Euler characteristic of $1$ for any field whose characteristic is odd or zero. In characteristic two the Poincare series can be regarded as the rational function $f(t)=1/(1-t)$ and by putting $t=-1$ you get an Euler characteristic of $1/2$. Perhaps there is some context in which the Euler characteristic can be defined as an adele?

  • 5
    $\begingroup$ I tend to think of the "real" definition of Euler characteristic as being the trace of the identity map in the stable homotopy category. In that case the condition one needs for well-definedness is dualizability of $\Sigma_+^\infty(X)$, which implies finite generation of $H^*(X;\mathbb{Z})$. $\endgroup$ – Mike Shulman Jun 8 '11 at 15:10
  • $\begingroup$ Thanks, Neil. I think you're right, and I've attempted to restate my conjecture in a way that saves it by taking this idea into account. $\endgroup$ – John Baez Jun 9 '11 at 10:26

Let me call a connected space $X$ with just finitely many non-trivial homotopy groups all of which are finite, a "$\pi$-finite space", and a space with just finitely many non-trivial integral homology groups all of which are finitely generated, a "space of finite-type" (e.g a finite CW complex is of finite type). The following comes close to answering the question, but there is still a small gap...

First, finite homotopy groups actually imply finite integral homology groups for example by Allen Hatcher's answer to this question. The main idea is to use the serre spectral sequence for the fiber sequence $\tilde{X} \to X \to B(\pi_1X)$ to reduce to the simply connected case and then use the standard technique of Serre classes (note that finitely presented homotopy groups do not imply finitely generated homology groups). Thus, for a connected $\pi$-finite space, the only extra condition in being of finite type is the vanishing of homology in high dimensions.

Second, Serre's theorem quoted in the question (which appears in this paper as theorem 10 of section 24) implies that if X of finite type and has just finitely many non-zero homotopy groups, then $X$ is (weakly) contractible (the finite CW complex is just the most famous special case). This shows that a $\pi$-finite simply-connected space of finite type is contractible.

For a connected but not simply connected $\pi$-finite space of finite type $X$, we would like to apply the argument for the universal cover $\tilde{X}$. Note that it is still $\pi$-finite (obviously) and has finite (hence finitely generated) homology groups (by Hatcher's answer), thus the only question is the vanishing of homology in high dimensions. If $X$ is a finite CW complex then so is $\tilde{X}$ and then we're good. This also holds if $X$ is a retract up to homotopy of a finite CW complex (I think this is equivalent to being a compact object in the $\infty$-category of spaces). More generally, if $X$ is of finite cohomological dimension in the sense that all cohomology with local coefficients vanish above a certain dimension, then so is $\tilde{X}$ and again we're fine. So in all these special cases, $\tilde{X}$ must be contractible. I don't know if this is the case in general though.

Finally, if $\tilde{X}$ is contractible then $X = BG$ for a finite group $G$, and if it is not trivial, then it always has non-trivial integral homology in infinitely many dimensions (e.g. look at this question and answers). Thus, $X$ itself must be contractible.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

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