Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I was wondering if the sum $TS^{2n}\oplus TS^{2n}$ is a trivial bundle? The same is true for spheres of odd dimension (one can find a nowhere zero section of the second bundle, add it to the first, the first becomes trivial and the rest of second bundle plus trivial bundle of rk 2 is trivial too).

It seems that one should take $2n$ sections $v=(v_1,\dots,v_{2n})$ of $TS^{2n}$ (for example projections of coordinate vector fields from $\mathbb R^{2n+1}$ to $S^{2n}$), $u=(u_1,\dots,u_{2n})$ for the second part and than perturb a little u=u+av, v=v+bu.

Nevertheless I can not prove that it works. From the other point of view I see no reasons for this bundle to be non-trivial.

share|cite|improve this question

2 Answers 2

up vote 15 down vote accepted

Yes. Let $V$ be a real vector bundle whose base is a $d$-dimensional manifold or cell complex, and whose fibers are $r$-dimensional. Then (1) if $r>d$ then $V=W\oplus \epsilon$ where $\epsilon$ is a trivial rank one bundle, and (2) if $r>d+1$ then the rank $r-1$ bundle $W$ is determined up to isomorphism by $V$. In particular stably trivial bundles of rank greater than $d$ are trivial.

share|cite|improve this answer

[[Edit]]: I hastily interpreted "sum" as the "direct product", although $\oplus$ is almost surely (or surely) referring to Whitney sum:

No. $\chi(S^{2n}\times S^{2n})=\chi(S^{2n})^2=4$ and so the Euler class of $S^{2n}\times S^{2n}$ is nontrivial (it pairs to the Euler characteristic under Poincare-duality). But that means there cannot exist a nonvanishing section of $TS^{2n}\oplus TS^{2n}$ (such a section would split the bundle with a trivial summand and the Euler class of a trivial bundle is zero), i.e. it is not trivial.

share|cite|improve this answer
(I think he meant a bundle with base $S^{2n}$.) – Tom Goodwillie Mar 26 '13 at 22:01
@Chris, but there is a non-vanishing section of $T(S^{2n}) \oplus T(S^{2n})$. For example, with $n=2$ consider the vector fields that are the derivatives of "rotation about the x-axis" and "rotation about the y-axis". Those two vector fields on $S^2$ have no common zeros, so their direct sum is everywhere non-zero. – Ryan Budney Mar 26 '13 at 22:06
Ah I took "sum" as direct product, although $\oplus$ is standard notation for Whitney sum. – Chris Gerig Mar 26 '13 at 22:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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