Spaces with same homotopy and homology groups that are not homotopy equivalent? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:18:48Z http://mathoverflow.net/feeds/question/53399 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent Spaces with same homotopy and homology groups that are not homotopy equivalent? Dylan Thurston 2011-01-26T20:32:47Z 2011-01-27T13:55:22Z <p>A common caution about Whitehead's theorem is that you need the map between the spaces; it's easy to give examples of spaces with isomorphic homotopy groups that are not homotopy equivalent. (See <a href="http://mathoverflow.net/questions/3540/are-there-two-non-homotopy-equivalent-spaces-with-equal-homotopy-groups" rel="nofollow">http://mathoverflow.net/questions/3540/are-there-two-non-homotopy-equivalent-spaces-with-equal-homotopy-groups</a>). It's surely also true that the pair (homotopy groups, homology groups) is not a complete invariant, but can anyone give examples? That is, I'm looking for spaces $X$ and $Y$ so that $\pi_n(X) \simeq \pi_n(Y)$ and $H_n(X;\mathbb{Z}) \simeq H_n(Y; \mathbb{Z})$ but $X$ and $Y$ are still not (weakly) homotopy equivalent.</p> <p>(Easier examples are preferred, of course.)</p> http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53408#53408 Answer by Ben Wieland for Spaces with same homotopy and homology groups that are not homotopy equivalent? Ben Wieland 2011-01-26T21:29:53Z 2011-01-26T21:56:06Z <p>A fairly easy obstruction is the action of the fundamental group on the homotopy groups. Here is an example where the homotopy groups are isomorphic as abstract groups, but not preserving the action of the fundamental group.</p> <p>Form a fibration over a circle with fiber a wedge of spheres $S^2\vee S^2$. That is, take the wedge, cross with an interval, and identify the ends by a homotopy equivalence of the wedge. Homotopy equivalences are parameterized by $GL_2(\mathbb Z)$, the action on homology. The homotopy type of the space remembers the monodromy as the action of the fundamental group on the homotopy groups. The homotopy groups are that of the universal cover, which does not depend on the choice of monodromy. Compute the homology by the Serre spectral sequence. This involves the homology of $\mathbb Z$ acting on $\mathbb Z^2$ by the monodromy. If the monodromy is hyperbolic, the homology vanishes and the space has the homology of the circle. Thus two different hyperbolic matrices give spaces with isomorphic homotopy and homology groups.</p> <p>In a completely different direction, there are examples of pairs of simply connected spaces such that the Postnikov truncations are equivalent, but the inverse limits of their Postnikov towers are not, but I think such examples have to be pretty large.</p> http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53409#53409 Answer by Johannes Ebert for Spaces with same homotopy and homology groups that are not homotopy equivalent? Johannes Ebert 2011-01-26T21:31:22Z 2011-01-27T13:55:22Z <p>Take a finite group $G$, a finite-dimensional $\mathbb{Q}$-vector space $V$ and two representations actions of $G$ that are inequivalent but the spaces of coinvariants both have the same dimension, say zero. </p> <p>The first concrete example that comes to my mind is $G=Z/4$, $V=Q[i]$, with the two actions where the generator acts by $-1$ or $i$. These two actions are not even equivalent under outer automorphisms of $G$.</p> <p>Let $n \geq 3$ be an odd integer and consider the Eilenberg-Mac-Lane space $K(V,n)$, which inherits two $G$-actions. The two Borel-constructions $EG \times_G K(V,n)$ have the same homotopy groups. But the $n$th homotopy groups are not isomorphic when considered as a $\pi_1$-module; so these two spaces are not homotopy equivalent.</p> <p>The homology can be computed from the Leray-Serre spectral sequence of the fibration $EG \times_G K(V,n) \to BG$. Recall $E^{2}_{pq}= H_p (G; H_p (K(V;n)))$.</p> <p>To begin with, $\tilde{H}_* (K(V,n), \mathbb{Z}) =V$ if $*=n$ and $0$ otherwise. Thus $E^{2}_{pq}=0$ unless $q=0$ (then it is the group homology $H_p(G;Z)$ or $(p,q)=(0,n)$, in which case it is $V_G=0$. Thus the projections $EG \times K(V,n) \to BG$ are homology equivalences.</p> <p>Finally note that the Eilenberg Mac Lane spaces can be realized as abelian topological groups and $G$ acts fixing the basepoint. Thus the maps $EG \times_G K(V;n)) \to BG$ have sections, which are homology equivalences as well. So the construction even produces a homology equivalence between two spaces with abstractly isomorphic homotopy groups.</p> <p>EDIT: since there is a homology equivalence between these two spaces, it follows that the homology with coefficients, the cohomology rings and even the actions of the Steenrod algebra for all primes are isomorphic.</p> http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53410#53410 Answer by Somnath Basu for Spaces with same homotopy and homology groups that are not homotopy equivalent? Somnath Basu 2011-01-26T21:52:34Z 2011-01-26T21:52:34Z <p>Following up on John's comment, one can consider $S^2$-fibrations over $S^2$. There are two of them since such fibrations are classified by $\pi_1(\textrm{Diff}^{+}(S^2))=\mathbb{Z}_2$. One of them is $S^2\times S^2$ while the other can be shown to be the connected sum of $\mathbb{CP}^2$ and $\overline{\mathbb{CP}}^2$. These two spaces have the same homology. They have the same homotopy groups since they both form the base of a $S^1$-fibration with total space $S^2 \times S^3$. However, the intersection forms are not equivalent and hence they are not homotopy equivalent.</p>