Splitting of tangent bundle

Question

Is it possible to give an example of $n$ dimensional manifold with the property that the tangent bundle $TM$ cannot be expressed as Whitney sum of two subbundles? It is certain true for two sphere; it is certainly not true for three dimensional manifold since every three manifold is parallelizable. You can always split $TM$ as $\gamma \oplus Q$ where $\gamma$ is one dimensional if you can find non vanishing vector field (i.e. where the Euler class is zero which is the case for odd dimension) but what about examples with $n$ even, larger than $2$?

Answer 1

To expand on my comment above, suppose that $TS^{2k}\cong\xi\oplus\eta$ for some non-trivial vector bundles $\xi$ and $\eta$ over $S^{2k}$ of dimensions $m$ and $\ell$, respectively. Hence $0<m,\ell<2k$. Since $S^{2k}$ is simply-connected, both $\xi$ and $\eta$ are oriented, hence possess Euler classes $$ e(\xi)\in H^m(S^{2k};\mathbb{Z}),\qquad e(\eta)\in H^{\ell}(S^{2k};\mathbb{Z}). $$ Since $S^{2k}$ is $(2k-1)$-connected, both are zero.

However, Property 9.6 of Euler classes in Milnor and Stasheff then gives that that $$e(TS^{2k})=e(\xi)\cup e(\eta)=0.$$ This is a contradiction, since $$ 0\neq 2=\chi(S^{2k}) = \langle e(TS^{2k}),[S^{2k}]\rangle $$ (this is Corollary 11.12 in Milnor and Stasheff).

Answer 2

I like to play around with projective spaces for these type of questions. The tangent bundles of real projective spaces have stiefel whitney class $(1+a)^{n+1}$. So for example for $n=4$, we have stiefel whitney class $1+a+a^4$. This polynomial is irreducible hence the tangent bundle does not split.

EDIT:

The comment below made me realise I was not clear. If the tangent bundle splitted, let's say $T\mathbb{RP}^4=E\oplus F$, we would find for the stiefel Whitney class that $$ sw(T\mathbb{RP}^4)=sw(E)sw(F) $$ But $\dim E+ \dim F=4$ and the right hand side maximally has terms of the form $a^4$. The Stiefel Whitney class $1+a+a^4$ would factor then also in $\mathbb{Z}_2[a]$ (not only in $\mathbb{Z}_2[a]/a^5$). It is true that $1+a+a^4$ is reducible in $\mathbb{Z}_2[a]/a^5$, and in fact stably the tangent bundle does split as $T\mathbb{RP}^4\oplus \mathbb{R}\cong\tau^5$, where $\tau$ is the dual of the tautological bundle.

Answer 3

A condition that prevents a vector bundle from splitting a Whitney sum is that the projection of the associated sphere bundle induces a non-surjective map on some homotopy group. This is proved by L.Guijarro and G.Walschap in Transitive holonomy groups and rigidity in nonnegative curvature who show that the above condition forces the normal holonomy of the bundle to be transitive (which is clearly impossible if the bundle splits as a Whitney sum).

For example, they explain in the paper that the tangent bundle to $CP^n$ with $n>1$ splits as a Whitney sum if and only if $n$ is odd.