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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?

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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. – algori Jun 8 '11 at 5:28
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. – Hal Sadofsky Jun 8 '11 at 7:09
What if you drop the word "rational" from condition 2? – Theo Johnson-Freyd Jun 8 '11 at 12:42
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. – algori Jun 9 '11 at 10:43
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. – algori Jun 9 '11 at 11:59


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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?

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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 $H^*(X;\mathbb{Z})$. – Mike Shulman Jun 8 '11 at 15:10
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. – John Baez Jun 9 '11 at 10:26

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