Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $SX$ be the suspension of CW complex. What are some results available to determine the homotopy groups of $\pi_n(SX)$?

share|improve this question

3 Answers 3

up vote 19 down vote accepted

This, in general, an incredibly difficult problem. Even we just want to compute the rational homotopy groups of the suspension of $X$ and $X$ is simply connected, where we can do everything using rational homotopy theory to reduce things to commutative DGAs, this starts to involve things like free Lie algebras and the like.

Locally at the prime 2, there is actually a famous long exact sequence when $X$ is a sphere called the EHP long exact sequence. It relates the suspension homomorphism, the Hopf map, and a "whitehead product" map. This gives rise to the EHP spectral sequence that, funnily enough, starts with the 2-local homotopy groups of odd-dimensional spheres and computes the 2-local homotopy groups of spheres. Miller and Ravenel have a paper titled "Mark Mahowald's work on the homotopy groups of spheres" that covers some of this material in detail.

Another approach is to say: The "stable" homotopy groups of $X$ are a first-order approximation using the Freudenthal suspension theorem that Andrea mentioned. There is then a "quadratic" correction term that you can try to use to get an approximation of the homotopy groups that is correct out to roughly three times the connectivity, and so on. These lead into the subject of Goodwillie calculus.

For a classifying space $K(G,1)$, neither of these approaches work very well, because the higher homotopy groups are going to depend pretty intricately on your group itself. For instance, $\pi_3$ of the suspension of the classifying space of a free group is the set of symmetric elements in $G_{\text{ab}}\otimes G_{\text{ab}}$ where $G_{\text{ab}}$ is the abelianization of $G$, and for a general group it lives in an exact sequence between something involving such symmetric elements and the second group homology of $G$. I don't know a closed form for it but maybe someone else knows better.

EDIT: Let me at least be precise, there's an exact sequence $$\pi _4 (\Sigma BG)\rightarrow H_3 G \rightarrow (G_{\text{ab}}\otimes G_{\text{ab}})^{\mathbb Z/2} \rightarrow \pi_3 (\Sigma BG)\rightarrow H_2 G\rightarrow 0$$ Note that for $R$ a ring, an element of $(G_{\text{ab}}\otimes G_{\text{ab}})^{\mathbb Z/2}$ gives rise to an $R$-valued symmetric bilinear pairing on $\mathsf{Hom}(G_{\text{ab}},R)$.

EDIT FOR THE FINAL TIME: sorry for the multiple revisions, switching back and forth between homology and cohomology gave me errors. the exact sequence above should be correct now.

share|improve this answer
Excellent answer! Thanks –  Skye Oct 24 '09 at 1:39
The functor on abelian groups A |-> (symmetric elements in A tensor A) is not right exact, and has a 0-th right derived functor &Gamma;<sup>2</sup>. This functor is also the "2nd divided square" construction. Is &Gamma;<sup>2</sup> G<sub>ab</sub> the closed form you want? –  Charles Rezk Oct 28 '09 at 16:36
You are right, this "symmetric elements" functor preserves surjections but not coequalizers, but I think the functor I want is its 0'th left (nonabelian) derived functor. –  Tyler Lawson Oct 29 '09 at 12:27

Since suspension of a $k$-sphere is a $k+1$-sphere, you have a natural suspension homomorphism $\pi_k(X)\rightarrow \pi_{k+1}(SX)$.

Freudenthal suspension theorem says that if $X$ is $n-1$ connected, $n\geq 2$, then the suspension homomorphism is actually an isomorphism for $k<2n-1$ and surjective for $k=2n-1$.

This is called the stable range. It implies that $\pi_{k+n}(S^n(X))$ eventually stabilizes to a group, called the $k^{\text{th}}$ stable homotopy group of $X$. Here $S^n$ is the $n^{\text{th}}$ iterated suspension.

Outside this range the relation is not so clear, even when you take $X$ to be a sphere (although certainly there is some result unknown to me). Have a look to a table of low-dimensional homotopy groups of spheres to see this wild behaviour.

share|improve this answer
Thanks for your information. If X has good connectness, that will be good by the suspension theorem. But if X is not, eg X is the classifying space, are there some approaches? –  Skye Oct 23 '09 at 13:45

I can't help but mention a cute application of the suspension isomorphism of (co)homology combined with Pioncare duality to show that for a compact closed manifold $M^n$, if the suspension $\sum M$ is homotopic to a closed orientable manifold, then $M$ is a homology sphere. This actually concerns the homology of the suspension, rather than homotopy groups. I apologize if it doesn't help your question.

In general, for a CW complex $M$, we have the suspension isomorphism $\tilde H_i(M)\cong \tilde H_{i+1}(\sum M)$, the reduced homology. And $\tilde H^i(M)\cong \tilde H^{i+1} (\sum M)$ for reduced cohomology. Now if we have Poincare duality on both sides, we would have $\tilde H^i(M)\cong \tilde H^{i+1}(M)$. passing from reduced and non reduced (co)homology tells us $M$ is a homology sphere.

There is a relevent exercise in "Elements of homology theory" by Viktor Vasilʹevich Prasolov on P45.

Edit: $\tilde H^i(M)\cong \tilde H^{i+1}(M)$ for suitable dimensions. e.g. you can start the induction from $H^{n-1}(M)=0$ and then $H^{k-1}(M)=H^{k-2}(M)$ etc.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.