Subject: Re: question abt ratl homology spheres Date: Mon, 14 Oct 2002 15:33:42 -0500 From: Clarence Wilkerson If M is the manifold, then one always has the collapse map M --> M/M^{n-1} \approx S^n, so your condition is equivalent to having S^n as an unstable retract of M^n, rationally. But if x and y are rational cohomology classes in H^*(M) of complementary degree, then if xy = [M], then on pulling back to [S^n], one has 0 = [S^n], which is a contradiction. So there is no rational homology or cohomology to M below the top. dim. Clarence Wilkerson __________________________________________________ Subject: Re: question abt ratl homology spheres Date: Mon, 14 Oct 2002 17:00:40 -0400 (EDT) From: Walter Neumann Most rational homology spheres X in dimension 3 have contractible universal cover (this will be true whenever the fundamental group is infinite). Since a map from S^3 must lift to the universal cover, it will be forced to have degree zero in that case. To give an explicit example, the Brieskorn manifold $X(p,q,r)= \{ (x,y,z)\in C^3 : x^p+y^q+z^r=0 |(x,y,z)|=1 \}$ is an integral homology sphere for 1 Subject: Question about rational homology spheres > Date: Mon, 14 Oct 2002 12:09:31 -0400 (EDT) > From: Piotr Hajlasz > > I have a question about rational homology spheres. I work with analysis, > > and I know almost nothing about topology, however it turned out that the > > following question is very important for my recent paper. > > QUESTION 1. Let $X$ be an $n$-dimensional smooth compact orientable > manifold without boundary. What is a necessary and sufficient condition > > for the existence of a smooth map $f:S^n-> X$ of nonzero degree? > > I believe that the answer is the following: > > CONJECTURE. Such mapping exists if and only if $X$ is a rational > homology > sphere. > > I have some evidence for this conjecture. First of all it is a > necessary condition. $X$ has to be a rational homology sphere - it > easily > follows form the Poincare duality (as I work in analysis, it was easier > for me to see that the Poincare duality for de Rham cohomology implies > that the de Rham cohomologies of $X$ have to be the same as those for > the > sphere, which means $X$ is a rational homology sphere). > > As concerns the other implication I can only prove it for simply > connected > rational homology spheres, so assume now that $X$ is a simply connected > rational homology sphere. As I understand the Hurewicz theorem > modulo the Serre class of torsion groups implies that the Hurewicz > homomorphism in dimension $n$ is the isomorphism mod the class of > torsion > groups. Since the Hurewicz homomorphism is defined by the degree it > follows that one can even find the map of degree $1$ from $S^n$ onto > $X$. > > On the other hand the examples of non simply connected rational homology > > spheres that I know (lens spaces, Poincare homological dodakedral > sphere) > have the property that the mapping $f:S^n-> X$ of nonzero degree exists. > > If the answer to the question below is in the positive, then the above > proof for simply connected rational homological spheres implies the > nonsimply connected case as well. > > QUESTION 2. Is the universal cover of a rational homology sphere a > rational homology sphere? > > Piotr Hajlasz > Department of Mathrmatics > University of Michigan > ____________________________________________________ > Subject: Re: question abt ratl homology spheres Date: Mon, 14 Oct 2002 17:27:50 -0400 (EDT) From: Yuli Rudyak Just a comment to the question of Hajlasz. If a closed $n$-manifold (in particular, rational homology sphere) $M$ has an infinite fundamental group then there is no map of $n$-sphere to $M$ of non-zero degree. Indeed, every map from sphere to $M$ passes thru the universal cover of $M$ which is non-compact. Yuli Rudyak ________________________________________________ Subject: Re: question abt ratl homology spheres Date: Mon, 14 Oct 2002 14:40:18 PDT From: Kevin Iga I submitted a note to "Homotopy, Homology and Applications" as follows: \title{Manifolds that have degree 1 maps from spheres} \author{Kevin Iga} \date{January 16, 2001} \maketitle \begin{abstract} Let $M^n$ be a connected $n$-manifold. $M^n$ is a homotopy sphere if and only if there exists a continuous map $f:S^n\to M^n$ from a sphere to $M^n$ of degree 1. This strengthens the result of Sol Schwartzman in his recent paper \cite{S}. \end{abstract} Here is the proof: \begin{theorem} Let $M^n$ be a connected manifold. Then $M^n$ is a homotopy sphere if and only if there exists a continuous map $f:S^n\to M^n$ of degree 1 (meaning $f_*:H_n(S^n)\to H_n(M^n)$ is an isomorphism). \end{theorem} \begin{proof} If $M^n$ is a homotopy sphere, then by the definition of homotopy equivalence, the map $f$ exists and gives an isomorphism on all homology groups. Conversely, suppose $f$ exists with the above property. We will prove $M^n$ is a homotopy sphere. Now $H_n(M^n)\cong \zz$ only when $M^n$ is a closed orientable manifold, so for $f_*$ on $H_n$ to be an isomorphism, $M^n$ must be closed and orientable. Now if $n=0$ or $n=1$, then by the classification of compact manifolds in these dimensions, the theorem follows trivially. So we assume $n>1$. If $n>1$, we show that $M^n$ must be simply-connected. Indeed, let $\tilde{M}$ be the universal cover of $M$. Then since $n>1$, $f$ lifts to $\tilde{f}:S^n\to \tilde{M}$, in such a way that if $\pi:\tilde{M}\to M$ is the projection map for the covering space $\tilde{M}$, then $f=\pi\circ \tilde{f}$. Therefore, $f_*=\pi_*\circ \tilde{f}_*:H_n(S^n)\to H_n(M^n)$. The degree of $f_*$ is 1, and the degree is multiplicative under composition, so the degree of $\pi_*$ is $\pm 1$. Therefore, $\pi$ is the identity covering map, and thus $\tilde{M}=M$. Therefore, $M$ is simply connected. Next we prove $M^n$ is a homology sphere. Indeed, if there were non-zero homology in dimension $i$ for some $i$ between $1$ and $n-1$, it would arise either in homology with rational coefficients (if it were free) or homology with $\zz/p$ coefficients for some prime $p$ (if it were a torsion element). In any case, if we let $F$ be whichever field has non-zero $H_i(M,F)$, then Poincar\`e duality implies that there is an isomorphism $PD:H_i(M,F)\to H^{n-i}(M,F)$ such that $(\alpha\cup PD(c)) [M^n] = \alpha(c)$. Since $F$ is a field, this means $H^i(M,F)$ is isomorphic to $\hom(H_i(M,F),F)$ through the evaluation map. So (assuming $H_i(M,F)$ is non-zero) there exist $\alpha\in H^i(M,F)$ and $\beta=PD(c)\in H^{n-i}(M,F)$ so that $(\alpha\cup\beta)[M^n]=1$. Applying $f$ we get \begin{eqnarray*} f^*(\alpha\cup \beta)(f_*[M^n])&=&1\\ (f^*(\alpha)\cup f^*(\beta))(f_*[M^n])&=&1. \end{eqnarray*} We note that here $[M^n]$ is a homology class in $H_n(M,F)$ not $H_n(M)$ as before, but since $H_{n-1}(M)$ has no torsion, $H_n(M)\otimes F$ and $H_n(M,F)$ are isomorphic through the usual correspondence on chains. Furthermore, $f_*:H_n(S^n)\to H_n(M^n)$ being degree 1 implies that $f_*:H_n(S^n,F)\to H_n(M^n,F)$ is degree 1. Therefore $(f^*(\alpha)\cup f^*(\beta))[S^n]=1.$ On the other hand, $S^n$ has no cohomology classes except in dimension $0$ and $n$, and $\alpha$ and $\beta$ are not in these dimensions, so $f^*(\alpha)$ and $f^*(\beta)$ are zero. This gives us a contradiction, unless $M^n$ has no homology between dimensions $0$ and $n$. Therefore, $M^n$ is a homology sphere. Thus $f$ is a map between simply-connected spaces that induces an isomorphism on homology. By Whitehead's theorem \cite{Sp}, $f$ is a weak homotopy equivalence. Since manifolds are CW complexes, $f$ is a homotopy equivalence \cite{Sp} and thus $M^n$ is a homotopy sphere. \end{proof} \begin{thebibliography}{99} \bibitem{S}Sol Schwartzman, {\em Parallel Tangent Hyperplanes}, Proceedings of the AMS v. 130, (2002), pp. 1457--1458. \bibitem{Sp}Edwin Spanier, {\em Algebraic Topology}, Springer-Verlag, New York, 1966, pp. 399, 405. \end{thebibliography} \end{document} ________________________________________________ Subject: Re: question abt ratl homology spheres Date: Mon, 14 Oct 2002 19:48:25 -0400 From: Allen Hatcher Reply on rational homology spheres: There are lots of 3-dimensional integral homology spheres M whose universal cover is R^3, so every map S^3 ---> M is nullhomotopic. To construct such an M take two nontrivial knots K and K' in S^3, delete open tubular neighborhoods of them to produce compact 3-manifolds X and X' each having a torus boundary, then glue X and X' together by a diffeomorphism between their boundaries that interchanges longitude and meridian, just as S^3 is built from two solid tori by gluing their boundaries by a diffeomorphism interchanging longitude and meridian. The resulting manifold M is an integral homology sphere by a Mayer-Vietoris argument. Classical theorems in 3-manifold topology imply that the univeral cover of M is R^3. (The loop theorem implies that the fundamental group of the common torus of X and X' in M injects into the fundamental group of M, so M is a Haken manifold, and Waldhausen showed that the universal cover of a Haken manifold is R^3.) If Thurston's Geometrization Conjecture for 3-manifolds is true, then the only homology 3-sphere with finite fundamental group is the Poincare homology sphere. Allen Hatcher _________________________________________________ Subject: Re: question abt ratl homology spheres Date: Mon, 14 Oct 2002 17:04:07 -0700 (PDT) From: Igor Belegradek Could you post this to the Topology discussion list? The first part is the answer to Goodwillie, the second part is the answer to the today's question on rational homology spheres. Thanks, Igor ---------------------------------------------------------------- 1) In the response to my question to the list, Goodwillie asked whether there exists a finite 3-dimensional cell complex which has cohomological dimension < 3 and is not homotopy equivalent to a 2-dimensional cell complex. I just want to mention that in the recently published Handbook of Algebraic topology, page 1264, G. Mislin says that the answer to the question is still unknown. 2) Here is a response on question about rational homology spheres. >QUESTION 1. Let $X$ be an $n$-dimensional smooth compact orientable >manifold without boundary. What is a necessary and sufficient >condition for the existence of a smooth map $f:S^n-> X$ of nonzero >degree? > >I believe that the answer is the following: > >CONJECTURE. Such mapping exists if and only if $X$ is a rational >homology sphere. The conjecture is incorrect, I think. In the paper below there are examples of flat manifolds which are rational homology spheres. So if X is such a flat manifold, any map f:S^n-> X is null-homotopic, since X is aspherical (and n>1), so f has degree zero. Szczepanski, Andrzej Aspherical manifolds with the $Q$-homology of a sphere. Mathematika 30 (1983), no. 2, 291--294 (1984). Igor Belegradek