What is the state of knowledge about the dimension of homotopy types? By the latter I mean the minimal number which is the dimension of a topological space representing the homotopy type. The open Eilenberg–Ganea conjecture concerns 1-types. What is known for n-types?

By the dimension of a CW complex I mean the dimension of its highest cell. For general spaces one can take Lebesgue covering dimension, however the primary interest is in CW complexes.

  • $\begingroup$ What notion of dimension of a topological space do you have in mind? $\endgroup$ Mar 2, 2016 at 2:16
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    $\begingroup$ By the dimension one can understand the Lebesgue covering dimension. I am mostly interested in spaces with a CW homotopy type. At least for the finite CW complexes the covering dimension coincides with the dimension of the highest cell. $\endgroup$ Mar 2, 2016 at 2:37

1 Answer 1


If you are willing to make "niceness" assumptions, assuming that the n-type is, say, nilpotent with finite fundamental group, then the dimension, in any reasonable sense, will always be infinite, since it will have homology in infinitely many dimensions.

The first result along these lines is due to Serre:

J.-P. Serre, Cohomologie modulo 2 des complexes d’Eilenberg–Mac Lane, Comment. Math. Helv. 27 (1953), 198–232.

I wrote a couple of papers about this back in the days:



where I prove that the cohomology, also as a ring, will be huge.

If you don't want to make assumptions on the fundamental group, then it is an open problem.

There are no known examples of n-types that are not $K(G,1)$s but that are homotopy equivalent to finite dimensional CW-complexes. And that there are no such examples has been stated as a conjecture by Casacuberta-Castellet in "Mislin's book of conjectures":


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    $\begingroup$ In your first sentence, do you mean to say that "it will have nontrivial homology in infinitely many dimensions"? As it stands, the statement does not appear to imply what has been concluded from it. $\endgroup$ Mar 2, 2016 at 14:03
  • $\begingroup$ @ViditNanda yes, that was obviously a confusing typo, now corrected. $\endgroup$ Mar 2, 2016 at 16:17

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