Analogue to covering space for higher homotopy groups? - MathOverflow

Analogue to covering space for higher homotopy groups?

2009-10-17T23:13:42Z

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.

Answer by Reid Barton for Analogue to covering space for higher homotopy groups?

2009-10-17T23:45:56Z

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 π_n for n >= 2. We could next ask for a 2-connected cover X'' of X': a space X'' with π_kX'' = 0 for k <= 2 and a map X'' -> X' which is an isomorphism on π_n 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(π₂X, 1). (For the universal cover the fiber was the discrete space π₁X = K(π₁X, 0).)

An example is the Hopf fibration K(Z, 1) = S¹ -> S³ -> S².

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).

Answer by Eric Wofsey for Analogue to covering space for higher homotopy groups?

2009-10-17T23:50:57Z

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 Hatcher starting on page 410.

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).

Answer by David Corfield for Analogue to covering space for higher homotopy groups?

2009-10-18T09:36:29Z

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 discussion, following on from an earlier post, we're looking at the fundamental 2-group which picks up homotopy in dimensions 1 and 2.

Answer by Peter Arndt for Analogue to covering space for higher homotopy groups?

2009-10-20T22:24:37Z

The previous answers gave analogues to the universal covering space and looked at the homotopy groups of these analogues.

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.

This is one of the motivations for using n-groupoids as invariants of spaces, see the discussion here, right before the references: http://ncatlab.org/nlab/show/fundamental+group+of+a+topos

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: http://www.intlpress.com/hha/v1/n1/a1/

Answer by Somnath Basu for Analogue to covering space for higher homotopy groups?

2009-11-21T08:02:12Z

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.