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Maybe every algebraic topology student, at some moment, will ask himself/herself the question: why are $\pi_\*$ so difficult and mysterious, especially when compared with (co)homology? Think about the weird connections between $\pi_n(S^k)$ and number theory... it is insane!

But one day I realize this idea may be wrong and biased.

Homology, is probably not really easier than homotopy.

One tends to think $H_\*$ is easy, only because most of us only care about finite-dimensional manifolds in our daily life. And then one have that nice vanishing result for higher-dimensional $H_\*$.

Consider the infinite-dimensional Eilenberg−MacLane space $K(\mathbb{Z}, n)$ when $n>2$. Its $\pi_\*$ is surely as easy as one could hope, but how about its $H_\*$, especially the torsions?

It appears to me, the better (?) statement might be

"Spaces with simple $\pi_\*$ tends to have complicated $H_\*$.

Spaces with simple $H_\*$ tends to have complicated $\pi_\*$."

This gives one some strange feeling. It is almost like some kind of "Fourier transformation", some kind of duality.

And this makes some spaces particularly interesting: spaces with both simple $H_\*$ and simple $\pi_\*$ at the same time.

Let me start with some simple examples:

1) $S^1 \simeq K(\mathbb{Z}, 1)$.

2) $\Sigma^g$, that is, Riemann surfaces.

3) $K(\mathbb{Z}, 2) \simeq CP^\infty$.

4) $K(G, 1)$ when $G$ is a finite group.

And I am looking forward to your examples of such spaces, especially comments from AT experts.

Thank you very much.

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Rational homotopy theory (for simply connected or just nilpotent spaces) shows exactly this dichotomy. –  Torsten Ekedahl Jun 29 '10 at 11:00
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The 'duality' between cohmology and homotopy goes by the name Eckmann-Hilton duality - ncatlab.org/nlab/show/Eckmann-Hilton+duality . –  David Corfield Jun 29 '10 at 11:23
    
The title is confusing in light of "simple homotopy" theory being an object of study. en.wikipedia.org/wiki/Simple-homotopy_equivalence –  jd.r Jun 30 '10 at 15:26

1 Answer 1

There are many $K(G,1)$ spaces with nicely computable homology groups. Besides the ones you listed, there are many compact 3-manifolds of this type. Also various configuration spaces such as the space of ordered $n$-tuples of distinct points in $\mathbb{R}^2$ which is a $K(G,1)$ for $G$ the pure braid group $P_n$. I'm sure a long list of such $K(G,1)$ spaces could be assembled.

For simply-connected spaces, examples are harder to find. There is no known simply-connected noncontractible finite CW complex whose homotopy groups are all known. This probably holds also with "finite" replaced by "finite-dimensional", so one needs to look at infinite-dimensional spaces. The first that comes to mind is certainly $\mathbb{C}P^\infty$. The spaces $K(\mathbb{Q},n)$ for arbitrary $n$ have nice homotopy and homology, though with the disadvantage that the homotopy and homology groups are vector spaces over $\mathbb{Q}$, so aren't finitely generated as groups. Perhaps the next simplest examples after these are the infinite unitary group $U=\cup_n U(n)$ and its classifying space $BU$, the infinite-dimensional Grassmannian, but for these it takes a deep theorem, Bott periodicity, to compute the homotopy groups. The real analogs $O$ and $BO$ aren't quite as nice, but should still qualify for having both simple homotopy groups and simple homology. If one is willing to go from spaces to spectra, there are the spectra arising in cobordism theories.

It might be interesting to compile a list of spaces whose homotopy and homology groups are all known, even if they're not exactly simple. For $K(G,1)$'s the list would be quite long, but for simply-connected spaces it must be much shorter.

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We should also point out that the (co)homology of $K(Z,n)$ and $K(Z/p,n)$ is completely known, and that this is the basis for powerful general methods of extracting some information about homotopy groups from homology. Bo Peng needs to learn about the Steenrod algebra and the Adams spectral sequence! –  Tom Goodwillie Jun 29 '10 at 14:52
    
You can't replace "finite" by "finite-dimensional." For odd $n$, the identification $S^n_{\mathbb Q}\cong K(\mathbb Q,n)$ gives an example of a space whose Hurewicz map is an isomorphism. (all examples?) –  Ben Wieland Jun 29 '10 at 19:44

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