Spaces that are both homotopically and cohomologically finite - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:41:00Z http://mathoverflow.net/feeds/question/67209 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67209/spaces-that-are-both-homotopically-and-cohomologically-finite Spaces that are both homotopically and cohomologically finite John Baez 2011-06-08T05:16:48Z 2011-06-10T18:18:56Z <p>Is it true that every connected space with</p> <p>1) just finitely many nontrivial homotopy groups, all finite, </p> <p>and </p> <p>2) just finitely many nontrivial rational cohomology groups, all finite rank, </p> <p>is weakly homotopy equivalent to a point? </p> <p>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. </p> <p>In 1998, Carles Casacuberta wrote:</p> <blockquote> 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. </blockquote> <p>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.</p> <p>For some further discussion of these issues see the comments starting here:</p> <p><a href="http://golem.ph.utexas.edu/category/2011/05/mbius_inversion_for_categories.html#c038299" rel="nofollow">http://golem.ph.utexas.edu/category/2011/05/mbius_inversion_for_categories.html#c038299</a></p> <p>and also these slides and references:</p> <p><a href="http://math.ucr.edu/home/baez/counting/" rel="nofollow">http://math.ucr.edu/home/baez/counting/</a></p> <p><b>Edit</b>: 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 <i>rational</i> cohomology.</p> <p>So let me try again. I can think of two ways:</p> <p><b>Fix #1:</b> Is it true that every connected space with</p> <p>1) just finitely many nontrivial homotopy groups, all finite, </p> <p>and </p> <p>2) just finitely many nontrivial integral cohomology groups, all finitely generated,</p> <p>is weakly homotopy equivalent to a point? </p> <p><b>Fix #2:</b> Is it true that every connected finite CW complex with just finitely many nontrivial homotopy groups, all finite, is homotopy equivalent to a point?</p> http://mathoverflow.net/questions/67209/spaces-that-are-both-homotopically-and-cohomologically-finite/67211#67211 Answer by Tom Church for Spaces that are both homotopically and cohomologically finite Tom Church 2011-06-08T05:34:01Z 2011-06-08T05:34:01Z <p>$\mathbb{R}P^\infty$.</p> http://mathoverflow.net/questions/67209/spaces-that-are-both-homotopically-and-cohomologically-finite/67246#67246 Answer by Neil Strickland for Spaces that are both homotopically and cohomologically finite Neil Strickland 2011-06-08T12:05:01Z 2011-06-10T18:18:56Z <p>This is really a comment rather than an answer. </p> <p>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?</p>