0
$\begingroup$

Following Spanier's book on algebraic topology chapter $1$, section $6$ about suspensions, I'm wondering about the following questions:

1) We know that $S^n$ is an $H$-cogroup for all $n\geq1$ because $S^n = S(S^{n-1})$ where $S$ is the "reduced suspension" functor, which has been proven to take values in the category of $H$-cogroups. Is it also true that $T^n = S^1 \times \dots \times S^1$ is an $H$-cogroup? If so, how does one show this?

2) Suppose the answer to the above question is yes (which I intuitively think it is), then is there an $H$-cogroup isomorphism $S^n \to T^n$? Or at least an $H$-cogroup homomorphism? If so, then there would be a natural transformation between the functors $\pi_{S^n}$ and $\pi_{T^n}$, or even a natural equivalence if the map $S^n \to T^n$ is isomorphic (but it cannot be a homeomorphism. I'm a bit confused here, can it they be $H$-cogroup isomorphic but not homeomorphic?)

3) What I'm really after is a correspondence between $\pi_{S^n}(X)$ and $\pi_{T^n}(X)$ for given $X$ and $n\geq1$. The two are group-isomorphic, right? Is the above the right way to show this? If not, what is?

Note on notation: $\pi_Q$ is a covariant functor from $\mathsf{hTop}_{\bullet}$ to $\mathsf{Grp}$ which sends $X\in \operatorname{Obj}(\mathsf{hTop}_{\bullet})$ to the homotopy classes of basepoint-preserving maps $\operatorname{hom}_{\mathsf{hTop}_{\bullet}}(Q,X)$ and a morphism $[f]\in \operatorname{hom}_{\mathsf{hTop}_{\bullet}}(X,Y)$ to $$([g]\mapsto[f\circ g])\in \operatorname{hom}_{\mathsf{hTop}_{\bullet}}(\operatorname{hom}_{\mathsf{hTop}_{\bullet}}(Q,X), \operatorname{hom}_{\mathsf{hTop}_{\bullet}}(Q,Y)).$$

$\endgroup$
  • 1
    $\begingroup$ $\pi_k T^n$ is trivial for all $k>1$. There are lots of homotopically nontrivial maps between tori. $\endgroup$ – user98602 Aug 29 '15 at 23:38
  • 1
    $\begingroup$ Wait, why is $\pi_{S^n}(X)$ isomorphic to $\pi_{T^n}(X)$? Last I remember, $\pi_{S^2}(S^1)$ was trivial, but $\pi_{T^2}(S^1)$ is non-trivial. $\endgroup$ – Dan Rust Aug 29 '15 at 23:39
  • $\begingroup$ I guess my pretense was wrong, and I'm glad to have found out! $\endgroup$ – PPR Aug 29 '15 at 23:42
  • 1
    $\begingroup$ I believe that the only connected manifold cogroups are spheres, and that this is proved somewhere on this site. $\endgroup$ – user98602 Aug 29 '15 at 23:46
  • 2
    $\begingroup$ $S^n$ and $T^n$ are not even homotopy equivalent (except when $s = 1$; they can be distinguished by $H_1$), so there's no hope for them to be equivalent as anything more structured. $\endgroup$ – Qiaochu Yuan Aug 30 '15 at 0:00
2
$\begingroup$

1) As is shown here, if $M$ is a compact connected $n$-dimensional manifold which is a co-$H$-space, then $M$ is simply connected and has the same integral homology as $S^n$ (i.e. $M$ is a $\mathbb{Z}$-homology sphere). By a version of Whitehead's Theorem (see Hatcher's Algebraic Topology, Corollary $4.33$), $M$ has the same homotopy groups as $S^n$ (i.e. $M$ is a homotopy sphere), so by the generalized Poincaré conjecture, $M$ is homeomorphic to $S^n$; note, as $M$ was assumed to be connected, $n > 0$.

As every $H$-cogroup is a co-$H$-space (because every group is a monoid), we see that the only compact connected orientable manifolds which could possibly be $H$-cogroups are the spheres $S^n$ for $n > 0$. As $n > 0$, $S^n = \Sigma S^{n-1}$, so we see that they are in fact $H$-cogroups (and hence co-$H$-spaces).

In summary, if $M$ is a compact connected manifold, then:

  • $M$ is a co-$H$-space if and only if $M$ is homeomorphic to $S^n$ for some $n > 0$.
  • $M$ is an $H$-cogroup if and only if $M$ is homeomorphic to $S^n$ for some $n > 0$.

In particular, there is no compact connected manifold which is a co-$H$-space but not an $H$-cogroup.

By the above, we see that $T^n$ is not an $H$-cogroup for $n > 1$. For $n = 1$, we see that $T^1 = S^1$ is an $H$-cogroup.

2) As $T^n$ is not an $H$-cogroup for $n > 1$, there is no $H$-cogroup isomorphism or homomorphism $S^n \to T^n$. For $n = 1$, $S^1$ and $T^1$ are homeomorphic (via the identity).

3) As $T^n$ is not an $H$-cogroup for $n > 1$, $\pi_{T^n}(X)$ is not naturally a group in general; in particular, it cannot be isomorphic as a group to $\pi_{S^n}(X)$.

Although $\pi_{T^2}(X)$ is not naturally a group is general, the set $[T^2, X]_*$ of pointed homotopy classes of maps $T^2 \to X$ can be computed in terms of $\pi_1(X)$ and Whitehead products, see this MathOverflow question which asks about maps from the product of two arbitrary spheres. I don't know if the argument there can be generalised to $T^n$.

$\endgroup$
  • $\begingroup$ Would you agree that if someone has difficulties distinguishing $S^n$ and $T^n$ then this answer might not prove that useful? (It is good for reference, however.) I think one should address the fact the OP is failing to distinguish $T^n$ and $S^n$. $\endgroup$ – Pedro Tamaroff Jul 18 '16 at 3:40
  • 1
    $\begingroup$ @PedroTamaroff: I don't think the OP has difficulties distinguishing these two spaces, I'm not sure why you think they do. $\endgroup$ – Michael Albanese Jul 18 '16 at 5:16
  • $\begingroup$ It seems you haven't read the question with enough care, then. $\endgroup$ – Pedro Tamaroff Jul 18 '16 at 5:31
  • $\begingroup$ @MichaelAlbanese, thank you very much, now I have some order in my confusion about 1) and 2). What I was really after is 3) which I will now look into. $\endgroup$ – PPR Jul 18 '16 at 10:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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