Analogue to covering space for higher homotopy groups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:45:31Z http://mathoverflow.net/feeds/question/953 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/953/analogue-to-covering-space-for-higher-homotopy-groups Analogue to covering space for higher homotopy groups? jc 2009-10-17T23:13:42Z 2009-11-21T15:40:58Z <p>The connection between the fundamental group and covering spaces is quite fundamental. Is there any analogue for higher homotopy groups? It doesn't make sense to me that one could make a branched cover over a set of codimension 3, since I guess, my intuition is all about 1-D loops, and not spheres.</p> http://mathoverflow.net/questions/953/analogue-to-covering-space-for-higher-homotopy-groups/957#957 Answer by Reid Barton for Analogue to covering space for higher homotopy groups? Reid Barton 2009-10-17T23:45:56Z 2009-10-17T23:45:56Z <p>There's certainly a homotopy-theoretic analogue. A universal cover of a connected space X is (up to homotopy) a simply connected space X' and a map X' -> X which is an isomorphism on &pi;<sub>n</sub> for n >= 2. We could next ask for a 2-connected cover X'' of X': a space X'' with &pi;<sub>k</sub>X'' = 0 for k &lt;= 2 and a map X'' -> X' which is an isomorphism on &pi;<sub>n</sub> for n >= 3. The homotopy fiber of such a map will have a single nonzero homotopy group, in dimension 1--it will be a K(&pi;<sub>2</sub>X, 1). (For the universal cover the fiber was the discrete space &pi;<sub>1</sub>X = K(&pi;<sub>1</sub>X, 0).)</p> <p>An example is the Hopf fibration K(Z, 1) = S<sup>1</sup> -> S<sup>3</sup> -> S<sup>2</sup>.</p> <p>Geometrically it's harder to see what's going on with the 2-connected cover than with the universal cover, because fibrations with fiber of the form K(G, 1) are harder to describe than fibrations with discrete fibers (covering spaces).</p> http://mathoverflow.net/questions/953/analogue-to-covering-space-for-higher-homotopy-groups/958#958 Answer by Eric Wofsey for Analogue to covering space for higher homotopy groups? Eric Wofsey 2009-10-17T23:50:57Z 2009-10-17T23:50:57Z <p>Just like there is a universal cover of every space, there is a natural n-connected space X_n that maps to any space X. To construct this space, you can add cells of dimension n+2 and higher to X to get a space Y together with a map X \to Y which is an isomorphism on \pi_i for i \leq n, but such that \pi_i(Y)=0 for i>n. The homotopy fiber X_n \to X of this map is then the "n-connected cover" of X; X_n is n-connected but has the same homotopy groups as X above n, as can easily be seen from the long exact sequence of the fibration. Details of this, as well as a proof of uniqueness of the n-connected cover, are in <a href="http://www.math.cornell.edu/~hatcher/AT/ATchapters.html" rel="nofollow">Hatcher</a> starting on page 410.</p> <p>More generally, if you started with an (n-1)-connected space, you could both kill the homotopy groups of X above n and kill a subgroup of \pi_n(X), and then the homotopy fiber would be an "n-cover" of X corresponding to that subgroup of \pi_n(X).</p> http://mathoverflow.net/questions/953/analogue-to-covering-space-for-higher-homotopy-groups/1020#1020 Answer by David Corfield for Analogue to covering space for higher homotopy groups? David Corfield 2009-10-18T09:36:29Z 2009-10-18T09:36:29Z <p>Rather than just taking homotopy groups for a single dimension, you can also think about the kind of algebraic entity that detects homotopy in two consecutive dimensions, or indeed any number of consecutive dimensions. In this <a href="http://golem.ph.utexas.edu/category/2008/06/fundamental%5F2groups%5Fand%5F2cover.html" rel="nofollow">discussion</a>, following on from an earlier <a href="http://golem.ph.utexas.edu/category/2008/04/questions%5Fon%5F2covers.html" rel="nofollow">post</a>, we're looking at the fundamental 2-group which picks up homotopy in dimensions 1 and 2.</p> http://mathoverflow.net/questions/953/analogue-to-covering-space-for-higher-homotopy-groups/1508#1508 Answer by Peter Arndt for Analogue to covering space for higher homotopy groups? Peter Arndt 2009-10-20T22:24:37Z 2009-10-20T22:24:37Z <p>The previous answers gave analogues to the universal covering space and looked at the homotopy groups of these analogues.</p> <p>However, the analogy to the n=1 case is not complete: While \pi_ 1(X) classifies the automorphisms over X of the universal covering space, the \pi_ n(X) don't classify the automorphisms over X of these n-connected analogues. In fact, the higher \pi_ n(X) don't seem to classify anything else than homotopy classes of maps S^n-->X.</p> <p>This is one of the motivations for using n-groupoids as invariants of spaces, see the discussion here, right before the references: <a href="http://ncatlab.org/nlab/show/fundamental+group+of+a+topos" rel="nofollow">http://ncatlab.org/nlab/show/fundamental+group+of+a+topos</a></p> <p>or, starting on page 17 with a nice story on the invention of the higher homotopy groups, and the desire for a non-abelian homology: <a href="http://www.intlpress.com/hha/v1/n1/a1/" rel="nofollow">http://www.intlpress.com/hha/v1/n1/a1/</a></p> http://mathoverflow.net/questions/953/analogue-to-covering-space-for-higher-homotopy-groups/6357#6357 Answer by Somnath Basu for Analogue to covering space for higher homotopy groups? Somnath Basu 2009-11-21T08:02:12Z 2009-11-21T15:40:58Z <p>There may be a geometric but partial answer to your question. This is an idea I learnt from Dennis Sullivan. As we know, passing from $X$ to its universal cover $\widetilde{X}$ kills the fundamental group. Now, by Hurewicz we can assume that $H_2(\widetilde{X})=\pi_2(\widetilde{X})$, whence killing $H_2(\widetilde{X})$ suffices. If we assume $H_2(\widetilde{X})$ is torsion free then each generator $\alpha_i\in H_2(\widetilde{X})$ corresponds to a circle bundle $E_i$ over $\widetilde{X}$, i.e., $H_2(E_i)=H_2(\widetilde{X})/\mathbb{Z}\alpha_i$. Thus, if $\widetilde{X}$ was a manifold of dimension $n$ and $H_2(\widetilde{X})$ was free of rank $k$ then taking successive circle bundles we get a manifold $E$ of dimension $n+k$. This has the same higher homotopy groups ($\pi_i$ for $i>2$) as that of $X$. The example given by Reid Barton is an illustration of this. However, for manifolds this is as far as you can go since killing even the free part of $\pi_3(\widetilde{X})$ (or, equivalently the free part of $H_3(E)$) requires bundles over $E$ with fibre $\mathbb{CP}^\infty$, which lands us outside the realm of finite dimensional manifolds. </p>