Okay, so I found the result that the tangent-bundle of any product of spheres is parallizable, given that some element of the product is either $S^1$, $S^3$, or $S^7$. I prove this as follows, first noting that since the higher-spheres are simply-connected, they $\Phi^{\mathbb{R}}_1(S^i)=1$, since rank-1 vectorbundles are in correspondence with double-covers (over a paracompact space). Thus for $j\in \{1, 3, 7\}$ have (omitting the pull-back map of the projection which occurs on most terms, and switching $1$ with the normal bundle by above) that $T(S^i\times S^k)=T(S^i)\oplus T(S^k)$$=i\oplus T(S^k)=$$(i-1)\oplus N(S^k)\oplus T(S^k)=(i-1)\oplus (k+1)$$=k+i$ if $k>1$ and if $k=1$ it is trivial, now factoring and using the trivial bundles to parallize the other spheres in this way, we get the conclusion.

My question is does this extend? I want to say that it is true for all products, due to the hope that since the tangent-bundle is associated to the clutching-map using the upper and lower hemispheres of the rotation matrices, that the product will be the matrix-direct-sum of the rotation matrices which I want to say is null-homotopic, but I don't know enough to try to go deeper into this aurugment.

Does anyone have a solution to this problem?

  • $\begingroup$ I don't think clutching functions are helpful here, at least when used naively. That is because the contractible neighborhoods for $S^i \times S^k$ will be products of hemispheres, so there's four of them in total with mutual intersections being homotopic to either $S^{i-1}$, $S^{k-1}$ or $S^{i-1} \times S^{k-1}$. Correspondingly, there will be multiple clutching functions depending on which overlap we're talking about -- the one you mention is for passing from southern to northern hemispheres in both spheres simultaneously. $\endgroup$ – Marek Feb 6 '14 at 12:41
  • $\begingroup$ But otherwise this is a very nice and probably classical question that hopefully some local expert will answer soon. To add my gut feeling -- you can't expect parallelizability in general, tangent bundles of spheres are quite complicated and there's no a priori reason for their direct sums to be trivial besides the trivial reason of adding the normal bundle. $\endgroup$ – Marek Feb 6 '14 at 12:45
  • $\begingroup$ This is false for $S^{2n} \times S^{2n}$ with $n>0$, because the Euler class of the tangent bundle is nonzero. $\endgroup$ – Dylan Wilson Feb 11 '14 at 18:48
  • $\begingroup$ Ah, so it is. I have removed it $\endgroup$ – Pax Kivimae Feb 11 '14 at 21:10

The result is:

The product of spheres $X=S^{n_1}\times S^{n_2}\times \ldots \times S^{n_k}$ (containing at least two factors) is parallelizable iff at least one of the $n_i$ is odd.

First, if all the $n_i$ are even, the $\chi(S^{n_1}) = 2$ for all $i$, and hence $\chi(X) = 2^k \neq 0$. This implies $X$ is not parallizable. In fact, it implies that every vector field on $X$ has a $0$, so there is no hope of constructing $\sum n_i$ independent vector fields.

Conversely, let's assume for simplicity that $n_1$ is odd. Since there is a nonvanishing vector field $V$ on $S^{n_1}$, we may decompose $TS^{n_1}$ as the Whitney sum $\operatorname{span}\{V\}\oplus N$ where $N$ is the normal space to $V$. Note that $\operatorname{span}\{V\}$ clearly has a section, so is isomorphic to a trivial rank 1 vector bundle $\epsilon$ over $S^{n_1}$. That is, $TS^{n_1} \cong \epsilon \oplus N$.

If $p_i$ denotes the $i$th projection map $p_i:X\rightarrow S^{n_i}$, then clearly we have $TX \cong \bigoplus p_i^\ast(TS^{n_i})$ (which is true for any product, regardless of whether or not the factors are spheres.)

Since the pullback of a trivial bundle is trivial, we have $TX \cong \epsilon\oplus p_{n_1}^\ast(N)\oplus \bigoplus p_{n_i}^\ast (TS^{n_i})$.

Now, since $TS^n \oplus \epsilon$ is trivial (because it's isomorphic to $TS^n \oplus \nu$ where $\nu$ is the normal bundle of $S^n\subseteq \mathbb{R}^{n+1}$), it follows that, for example, $p_2^\ast(TS^{n_2}\oplus \epsilon) \cong p_2^\ast(TS^{n_2})\oplus \epsilon$ is also trivial. At this point, one can use induction to show that $TX \cong p_1^\ast(N)\oplus \epsilon^{K}$ for some large $K$. In particular, $K\geq 2$. But $p_1^\ast(N)\oplus \epsilon^2 \oplus \epsilon^{K-2}\cong p_1^\ast(N\oplus \epsilon^2)\oplus \epsilon^{K-2} \cong p_1^\ast(TS^{n_1}\oplus \nu)\oplus \epsilon^{K-2}$ and therefore, is trival, because $TS^{n_1}\oplus \nu$ is trivial.


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