The space of homotopy classes of maps of products of spheres

Proposition 17.6.1 of "Differential form in Algebraic Topology" by Bott and Tu proves the following beautiful result:

$[S^{q}, X]\simeq \frac{\pi_{q}(X,x)}{\pi_{1}(X,x)}$

where $S^{q}$ is the $q$-sphere, $X$ is any path connected space and $[S^{q}, X]$ is the set of homotopy classes of continuous maps from $S^{q}$ to $X$. I was wondering if there was an analogous result for the product of spheres, namely:

$[S^{q}\times S^{p}, X] = ?$

I am mostly interested in the cases where $S^{q}$ and $S^{p}$ are both parallelizable and $X$ is the classifying space for $U(k)$, namely $X = BU(k)$. In other words, I am interested in the set of equivalence classes of complex vector bundles of rank $k$ on $S^{q}\times S^{p}$.

Thanks.

• If $K$ is well-pointed then there is a fibration $\text{map}(K,X) \rightarrow X$ given by evaluation at the basepoint and its fiber over a point in $X$ is the space of pointed maps. The result you're after is a special case of the corresponding exact sequence' $\pi_1X \rightarrow [K,X]_* \rightarrow [K,X]$ which is (by definition) the statement that $\pi_1$ acts on the path components of the fiber and the quotient is given by the path components of the total space (assuming that $X$ is connected). – Dylan Wilson Mar 23 '16 at 19:56
• Focusing on your specific situation, here's what's happening: Given a pointed map $K \rightarrow X$ and a path $\gamma$ in $X$, we can try to imagine shifting' this map along the path (think monodromy) and ending up with a new map where the basepoint of $K$ is sent to the endpoint of the path. Any map is basepoint-presrving for some basepoint of $X$ (trivially). Choosing a path from that one to $x$ gives a way of taking an arbitrary map and making it basepoint preserving. But there's ambiguity because there are different homotopy classes of paths between these two points. – Dylan Wilson Mar 23 '16 at 20:21
• The collection of choices, however, is acted on by $\pi_1$ and the quotient gets rid of the ambiguity. – Dylan Wilson Mar 23 '16 at 20:21
• You can describe the pointed maps $S^q \times S^p \to X$ in terms of homotopy groups, whitehead products and some obstruction theory. Unfortunately in your case this would require knowledge of the homotopy groups of your matrix groups. Often these can be difficult to compute, but you have enough information to do rational computations. – Ryan Budney Mar 23 '16 at 20:43
• There's a discussion related to Dylan Wilson's comments in Hatcher's algebraic topology book section 4.A. Note that the generalization of the result you are quoting from Bott and Tu only reduces the problem of computing the homotopy classes of maps between two spaces $K\rightarrow X$ to two things you need to know: (1) the set of homotopy classes of maps fixing a basepoint in the two spaces ("pointed" maps) and (2) the action of the fundamental group of $X$ on this set. When $K=S^q$, (1) is just the definition of $\pi_q(X)$, but for your $K$ it may be quite hard to compute. – j.c. Mar 24 '16 at 1:23

This is not a complete answer, but is intended to expand on some of the comments above.

As mentioned by j.c. and Dylan Wilson in the comments, the set of unbased homotopy classes $[S^p\times S^q,X]$ is a quotient of the set of based homotopy classes $\langle S^p\times S^q,X\rangle$ under an action of $\pi_1(X)$. Since you say you are mostly interested in the case $X=BU(k)$, which is simply-connected, we may as well work with pointed homotopy classes.

The way I would try to analyse this (and perhaps this is the approach Ryan Budney had in mind in his comment) is using the cofibration sequence $$S^{p+q-1}\to S^p\vee S^q\to S^p\times S^q \to S^{p+q}\to S^{p+1}\vee S^{q+1}\to \cdots$$ in which the first map is the attaching map of the top cell of $S^p\times S^q$. The fourth map is the suspension of this attaching map, therefore is null-homotopic. There results an exact sequence of pointed sets $$1 \to \langle S^{p+q},X\rangle \to \langle S^p\times S^q,X\rangle \to \langle S^p\vee S^q,X\rangle \to \langle S^{p+q-1},X\rangle$$ which written in terms of homotopy groups becomes $$0\to \pi_{p+q}(X) \to \langle S^p\times S^q,X\rangle \to \pi_p(X)\oplus\pi_q(X)\to \pi_{p+q-1}(X).$$ The last map is the Whitehead product. Hence if you know the homotopy groups of $X$ (which for $X=BU(k)$ are known in a stable range by Bott periodicity) and the Whitehead products, you have a good chance of describing the set $\langle S^p\times S^q, X\rangle$ (which by the way is not a group, since $S^p\times S^q$ is not a co-H-space).

Edit: In the comments, user DLIN asks about the case $X=Gl_N(\mathbb{C})$. We can say more in this case, since $X$ is a path-connected H-space. Therefore all Whitehead products in $\pi_*(X)$ are trivial, and the exact sequence of sets above becomes a short exact sequence of groups $$0\to \pi_{p+q}(X) \to \langle S^p\times S^q,X\rangle \to \pi_p(X)\oplus\pi_q(X)\to 0.$$ Note that the middle term is now a group (using an H-space multiplication $\mu:X\times X\to X$). Furthermore, this sequence splits; the last map, which is given by restriction to the two sphere factors, is split by sending $(f,g)$ to $\mu\circ (f\times g)$.

Now suppose we know that $\langle S^p\times S^q, X\rangle$ is abelian. Then we could conclude that $$\langle S^p\times S^q, X\rangle \cong \pi_{p+q}(X)\oplus\pi_p(X)\oplus\pi_q(X).$$ This would be the case, for example, if there are two commuting H-space structures on $X$ (see the Eckmann-Hilton argument).

• If $X=GL(N,\mathbb C)$, $p$ is even and $q$ is odd, can we say the homotopy group $[S^p\times S^q,X]$ splits? – DLIN Aug 8 '16 at 11:30
• @DLIN: Do you want to conclude that $\langle S^p\times S^q, X\rangle\cong\pi_p(X)\oplus\pi_q(X)$ when $X=GL_N(\mathbb{C})$? Since all Whitehead products are trivial in a path-connected H-space, the last map in my final exact sequence above will be trivial, but I'm not sure about the first map. – Mark Grant Aug 8 '16 at 20:08
• No, actually, I mean $[S^p\times S^q,X]=\pi_p(X)\oplus \pi_q(X)\oplus \pi_{p+q-1}(X)$. – DLIN Aug 9 '16 at 2:06
• @DLIN: Actually, it seems you are right (up to a shift in dimension), see my edit. – Mark Grant Aug 9 '16 at 6:56
• @PPR: For $p=q=1$ this is quite easy to see: suspension gives a group homomorphism $\pi_1(S^1\vee S^1)\to \pi_2(S^2\vee S^2)$ whose target is an abelian group, and the attaching map represents a commutator $aba^{-1}b^{-1}$. For the general case, see Whitehead's "Elements of homotopy theory", Theorem X.8.20. – Mark Grant Aug 22 '18 at 7:44