What is the intuition behind the Freudenthal suspension theorem?

Question

The Freudenthal suspension theorem states in particular that the map $$ \pi_{n+k}(S^n)\to\pi_{n+k+1}(S^{n+1}) $$ is an isomorphism for $n\geq k+2$.

My question is: What is the intuition behind the proof of the Freudenthal suspension theorem?

Answer 1

There are two proofs I particularly like:

1. A Morse-theoretic proof, probably due to Bott, can be found in Milnors book. Idea: consider the space of all paths on $S^n$ from the north to the south pole, which is homotopy equivalent to $\Omega s^n$. There is the energy function on this space. One does Morse theory: critical points correspond to geodesics (they are not non-degenerate, but Bott proved that a good deal of Morse theory works nevertheless). The set of absolute minima is the space of minimal geodesics connecting the poles. This is homeomorphic to $S^{n-1}$. All other critical points have index at least (rouhgly) $2n$. Therefore, the inclusion of the set of absolute minima into the whole space is $2n$-connected.

2. A spectral sequence proof (see Kirk, Davis, Lectures on Algebraic topology). Consider the homology Leray-Serre spectral sequence of the path-loop fibration $\Omega S^n \to PS^n \to S^n$. The total space $P S^n$ is contractible. Look at the shape of the spectral sequence; you'll see that for $k \leq 2n$ (again, only a rough estimate), all differentials out of the $E_{k,0}$-slot are zero, except of the last one, which gives an isomorphism $H_k (S^n)=E_{k,0}^{2} \to E_{0,k-1}^{2} = H_{k-1} (\Omega S^n)$. It is credible (but nontrivial to prove, this uses the transgression theorem) that this isomorphism is the same as the composition $H_k (S^n) \cong H_{k-1} (S^{n-1}) \to H_{k-1} (\Omega S^n)$ of the suspension isomorphism and the natural map $S^{n-1} \to \Omega S^n$. Thus the natural map is a homology isomorphism in a range of degrees, and by the Hurewicz theorem, this holds for homotopy groups as well.

Answer 2

Maybe this differential topologic way of thinking the Freudenthal suspension is much more intuitive. By Pontrjagin's contruction you can identify $\pi_{n+k}(S^n)$ with equivalence classes of framed submanifolds $(N,\nu)$ of $S^{n+k}$. The image of this class under the Freudenthal suspension homomorphism is just the framed submanifold $(N,\tilde{\nu})$, where we identify $S^{n+k}$ with the equator of $S^{n+k+1}$ and the frame $\tilde{\nu}$ is obtained by $\nu$ just "adding" to $\nu$ the canonical normal frame of $S^{n+k}$ inside $S^{n+k+1}$. The fact that the map is an isomorphism for $n>k+1$ can now be achieved by general position arguments.

Answer 3

The Freudenthal theorem is really a special case of the phenomenon called "homotopy excision" aka the Blakers-Massey triad theorem. The idea is that one has an inclusion $$ (C_-X,X) \to (\Sigma X,C_+X) $$ given by gluing in $C_+X$, where $C_\pm X$ are copies of the cone on $X$. The Blakers-Massey theorem tells us that this map is $(2r+1)$-connected, where $r$ is the connectivity of $X$. Furthermore, the inclusion induces the suspension homomorphism on homotopy groups.

There are a variety of proofs of the homotopy excision theorem. In the case at hand, possibly the easiest proof is given in the paper cited by Jeff Strom in answering my question: https://mathoverflow.net/questions/54169.

By the way, in the dual case one has the map $\Sigma \Omega X \to X$. It is somewhat easier to check that the latter is $(2r+1)$-connected. Its homotopy fiber can be identified with $\Sigma (\Omega X) \wedge (\Omega X)$ which is $(2r)$-connected. So the map $\Sigma \Omega X \to X$ is $(2r+1)$-connected.

Answer 4

The suspension map $\Sigma: [A, X] \to [\Sigma A, \Sigma X]$ (of which yours is a special case) is adjoint to a map $[A, X] \to [A, \Omega\Sigma X]$ which is induced by the map $\sigma: X\to \Omega\Sigma X$ (adjoint to $\mathrm{id}_{\Sigma X}$). Thus determining the behavior of the suspension operation is reduced to the study of the map $\sigma$.

It can be proved that if $X$ is $(n-1)$-connected, then the map $\sigma$ is a $(2n-1)$-equivalence, and it follows that $\Sigma: [A, X] \to [\Sigma A, \Sigma X]$ is a bijection for all CW complexes $A$ with $\dim(A)< 2n-1$.

How is it proved? One method is to look at the James construction; another is to examine the comparison map $\xi: A\to F$ that arises when a cofiber sequence $A\to B\to C$ has been converted to a fibration sequence $F\to B\to C$.

Answer 5

Edit October 2013: I have given more information on the Blakers-Massey triad theorem on the ncatlab here, referring specifically to the algebraic determination of the critical, i.e. first non vanishing, group. This is described as a tensor product of relative homotopy groups, and of course a tensor product is trivial if one of the factors is trivial. So we get an intuitive view of the connectivity result and so for the Freudenthal Suspension Theorem. Indeed, they are both seen as a consequence of a higher order homotopy Seifert-van Kampen Theorem.

Original answer: John Klein's answer refers to the Blakers-Massey triad theorem; this was generalised to an $n$-ad connectivity theorem by Barratt and Whitehead, and of which Goodwillie has given a geometric proof, using general position arguments.

In these results there is interest in describing the critical group, an $n$-ad homotopy group, which may be nonabelian in general. Some abelian cases have been described by Barratt and Whitehead, following Blakers-Massey, with simple connectivity assumptions.

This problem was solved in Theorem 3.7 of

Ellis, G.J. and Steiner, R. "Higher-dimensional crossed modules and the homotopy groups of $(n+1)$-ads." J. Pure Appl. Algebra 46 (2-3) (1987) 117--136.

which uses crucially the van Kampen Theorem for $n$-cubes of spaces in

Brown, R. and Loday, J.-L. "Van Kampen theorems for diagrams of spaces", Topology 26~(3) (1987) 311--335. With an appendix by M. Zisman.

Of course connectivity results follow from the algebraic results, but the proof in this last paper is by induction with connectivity in one dimension implying surjectivity in the next.

One intuitive point is that the description of the critical group of an $n$-ad will involve generalised Whitehead products coming from sub $r$- and $s$-ads of the given $n$-ad, and an elaborate algebra (crossed $n$-cubes of groups, as defined by Ellis-Steiner) is needed to handle all this. The proofs do not involve general position at all.

For a fairly recent application of these ideas, see

Ellis, G.~J. and Mikhailov, R. "A colimit of classifying spaces". Advances in Math. (2010) arXiv: [math.GR] 0804.3581v1.