10
$\begingroup$

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.

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
21
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Prof. Goodwillie, may be an alternative proof is that; it is well known that the complexification of the tangent bundle of $S^{n}$ is trivial. I read this statement in a paper of Swan, but I do not know what is the reason of this triviality?is there an obvious reason for this? $\endgroup$
    – Ali Taghavi
    Commented Dec 4, 2015 at 14:04
  • $\begingroup$ I don't know offhand. $\endgroup$
    – Tom Goodwillie
    Commented Dec 4, 2015 at 17:02
0
$\begingroup$

[[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.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ (I think he meant a bundle with base $S^{2n}$.) $\endgroup$
    – Tom Goodwillie
    Commented Mar 26, 2013 at 22:01
  • 2
    $\begingroup$ @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. $\endgroup$
    – Ryan Budney
    Commented Mar 26, 2013 at 22:06
  • $\begingroup$ Ah I took "sum" as direct product, although $\oplus$ is standard notation for Whitney sum. $\endgroup$
    – Chris Gerig
    Commented Mar 26, 2013 at 22:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .