Classification of maps $S^{1}\times S^{3} \to S^{3}$ up to homotopy?

Question

What is the classification of maps $S^{1}\times S^{3} \to S^{3}$ up to homotopy?

(I am firstly not sure my terminology is correct - I am looking for the suitable interpretation of "up to continuous deformation" which includes and generalizes the usual homotopy maps from an $n$-sphere into some space.)

I know that maps $ S^{3} \to S^{3}$ are characterized by $\pi_3(S^3)=\mathbb{Z}$, and maps $S^{4} \to S^{3}$ by $\pi_4(S^3)=\mathbb{Z}_2$. Also, any map $S^{1}\times S^{3} \to S^{3}$ induces a map $S^{3} \to S^{3}$, as well as a map $S^{1}\to S^{3}$, which has trivial homotopy, so my guess is that maps $S^{1}\times S^{3} \to S^{3}$ fall into equivalence classes isomorphic to $\mathbb{Z}$, though I am unsure how to show this rigorously.

Answer 1

Here is yet another answer. You can never have too many :)

There is additional structure present, because the space $S^3$ carries a topological group structure as the unit quaternions. Thus, the homotopy set $[S^1\times S^3,S^3]$ carries a natural group structure and we can compute this group. To this end, note that there exists a delooping $BS^3$, i.e. $S^3\simeq\Omega BS^3$. This allows us to compute \begin{align} [S^1\times S^3,S^3]&\cong[S^1\times S^3,S^3]_{\ast}\cong[S^1\times S^3,\Omega BS^3]_{\ast}\cong[\Sigma(S^1\times S^3),BS^3]_{\ast}\\ &\cong[\Sigma S^1\vee\Sigma S^3\vee\Sigma S^4,BS^3]_{\ast}\cong[S^2,BS^3]_{\ast}\times[S^4,BS^3]_{\ast}\times[S^5,BS^3]_{\ast}\\ &\cong\pi_2(BS^3)\times\pi_4(BS^3)\times\pi_5(BS^3)\cong\pi_1(S^3)\times\pi_3(S^3)\times\pi_4(S^3)\\ &\cong\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}. \end{align} Here, we have used that $S^3$ is simply connected (so there is no difference between pointed and unpointed homotopy classes), the adjunction $\Sigma\dashv\Omega$, the splitting formula $\Sigma(X\times Y)\cong\Sigma X\vee\Sigma Y\vee\Sigma(X\land Y)$ and the relationship $\pi_{n+1}(BG)\cong\pi_n(G)$.

Answer 2

Your guess is rather naïve: There is no good reason that homotopy classes of maps out of a product should correspond to products of homotopy classes out of the factors in general, and I'd think that generally determining the former in terms of the latter would be quite difficult.

For your particular example, however, you're in luck: You do not get $\mathbb{Z}$ as a result, but from this answer we have the following theorem:

Theorem (Steenrod): Let $X$ be a complex of dimension $n+1$, where $n \geq 3$. Then there is a bijection $$\newcommand{\Z}{\mathbb{Z}}\newcommand{\Sq}{\mathrm{Sq}} [X, S^n] \cong H^n(X; \Z) \times \operatorname{Coker} \phi, $$ where $\phi$ is the following composition: $$ H^{n-1}(X; \Z) \xrightarrow{\mod 2} H^{n-1}(X; \Z / 2) \xrightarrow{ \Sq^2 } H^{n+1}(X; \Z / 2) $$

Now $S^1 \times S^3$ has dimension 4, so the computation becomes a matter of plugging in: $[S^1 \times S^3, S^3] \cong H^3(S^1 \times S^3) \times \operatorname{Coker} \phi \cong \Z \oplus \Z / 2$ seeing as $H^2(S^1 \times S^3) = 0$ and $H^4(S^1 \times S^3; \Z / 2) \cong \Z / 2$.

Answer 3

Here is an alternative argument to the one Ben gave.

Consider the fiber sequence $\Omega S^3 \to \mathcal{L} S^3 \to S^3$ where $\mathcal{L}S^3 = \mathsf{Map}(S^1, S^3)$ is the free loop space. It induces a long exact sequence on homotopy groups which we will use in two ways.

The first consequence is that there is an exact sequence $\pi_1 \Omega S^3 \to \pi_1 \mathcal{L} S^3 \to \pi_1 S^3$, so $\pi_1 \mathcal{L} S^3$ is sandwiched between $0$'s, meaning it is itself $0$. Similarly about $\pi_0$. Thus, $\mathcal{L} S^3$ is simply-connected. Thus, $$ [S^1 \times S^3, S^3] \cong [S^3, \mathcal L S^3] \cong [S^3, \mathcal L S^3]_* \cong \pi_3 \mathcal{L} S^3 $$ where the second equivalence uses the simply-connectedness to pass to based homotopy classes. For this, we consider the portion in the LES reading $\pi_3 \Omega S^3 \to \pi_3 \mathcal{L} S^3 \to \pi_3 S^3$. Since $\mathcal{L}S^3 \to S^3$ admits a section (by sending $x \in S^3$ to the constant loop at $x$), this is a split exact SES. Therefore, $$ \pi_3 \mathcal L S^3 \cong \pi_3 \Omega S^3 \oplus \pi_3 S^3 \cong \pi_4 S^3 \oplus \pi_3 S^3 \cong \mathbb{Z}/2 \oplus \mathbb{Z}. $$ A bit of work that is slightly hidden in this argument is $\pi_4 S^3 \cong \mathbb{Z}/2$ which does require some computational techniques, e.g. some Serre spectral sequence argument.

Answer 4

As others have alluded to, you are asking to find the cohomotopy group $\pi^3(S^1\times S^3)$, which is $\mathbb{Z}\oplus\mathbb{Z}/2$. More generally, $\pi^3(X)$ is known for any closed connected oriented 4-manifold $X$ by Theorem 1 of the paper Kirby, Melvin, & Teichner - Cohomotopy sets of 4-manifolds. Namely, if $X$ has odd intersection form, then $\pi^3(X) \cong H_1(X)$, while if $X$ is spin, then $\pi^3(X) \cong H_1(X)\oplus\mathbb{Z}/2$. This result uses the Theorem of Steenrod mentioned in Ben Steffan's answer, and has a careful analysis of the splitting of the exact sequence mentioned in the comment by Derived Cats.