manifolds with no plane fields

[jairob]

I'm looking for examples of compact manifolds M (without boundary)
whose tangent bundle admits no non-trivial invariant sub-bundles. That
is, M has no continuous field of k-planes, for all k with 0<k<dim(M).

(In dim.2, that is easy: M is any surface with non-zero Euler
characteristic.)

Are there other simple examples?
Do all even-dimensional spheres satisfy the property?

Thanks in advance!

[Jeffrey Rolland]

Actually, any smooth manifold of any dimension with non-zero Euler
characteristic will not admit a nowhere-vanishing vector field, and
therefore its tangent bundle will not admit a non-trivial invariant sub-
bundles.

For manifolds with zero Euler characteristic, look at the Stiefel-Whitney
classes. It's been a long time, but I believe there is a theorem to the
effect that a smooth manifold's tangent bundle admits k linearly
independent vector field if and only if its last k Stiefel-Whitney
classes vanish. So, look for manifolds without vanishing Stiefel-Whitney
classes, and you're in business.

Hope this helps.

Sincerely,
--
Jeffrey Rolland
<wilds...@hotmail.com>

[jairob]

> Actually, any smooth manifold of any dimension with non-zero Euler
> characteristic will not admit a nowhere-vanishing vector field, and
> therefore its tangent bundle will not admit a non-trivial invariant sub-
> bundles.

Wait, is this true? For example, S^2 \times S^2 has no nowhere-
vanishing vector field (and the Euler characteristic is 4), but
nevertheless it admits a continuous field of planes.

> For manifolds with zero Euler characteristic, look at the Stiefel-Whitney
> classes. It's been a long time, but I believe there is a theorem to the
> effect that a smooth manifold's tangent bundle admits k linearly
> independent vector field if and only if its last k Stiefel-Whitney
> classes vanish. So, look for manifolds without vanishing Stiefel-Whitney
> classes, and you're in business.

Oh, I didn't know about these SW-classes! (My algebraic topology
course didn't go that far.) Thanks!

According to mathworld.wolfram.com/Stiefel-WhitneyClass.html, the
Stiefel-Whitney classes are not always the obstruction. But since I
only desire to obtain a few examples of manifolds with no plane
fields, they will probably help. I guess I got to visit the library.

Thanks again.
Jairo

[John M. Lee]

Actually, nonzero Euler characteristic isn't enough. For example, S2
x S2 has Euler characteristic 4, but it does admit 2-plane field (the
tangent planes to either copy of S2). A manifold can admit a k-plane
field without having a nonzero vector field.

I don't know the answer to the original question.

Jack Lee
_______________________________________________
John M Lee, Professor of Mathematics
University of Washington Mathematics Department, Box 354350
Seattle, WA 98195-4350
l...@math.washington.edu, 206-543-1735
http://www.math.washington.edu/~lee

> <wildstar...@hotmail.com>

[Robert]

On Dec 4 2009, 7:58�am, "John M. Lee" <l...@math.washington.edu>
wrote:

> Actually, nonzero Euler characteristic isn't enough. �For example, S2
> x S2 has Euler characteristic 4, but it does admit 2-plane field (the
> tangent planes to either copy of S2). �A manifold can admit a k-plane
> field without having a nonzero vector field.
>
> I don't know the answer to the original question.
>
> Jack Lee
> _______________________________________________
> John M Lee, Professor of Mathematics
> University of Washington Mathematics Department, Box 354350
> Seattle, WA 98195-4350

> l...@math.washington.edu, �206-543-1735http://www.math.washington.edu/~lee

I just saw this and thought I'd comment: The original question about
whether the even-dimensional spheres are examples is a good one. The
answer is 'yes': The tangent bundle of S^{2n} is irreducible, i.e.,
not the sum of two subbundles of positive dimension. The reason is
the Euler
class. Every vector bundle over S^{2n} is orientable because S^{2n}
is
simply connected. If TS^{2n} = V_1 \oplus V_2, then we'd have that
the Euler classes satisfy e(TS^{2n}) = e(V_1)*e(V_2), but, since e(TS^
{2n})
is not zero, but there are no nonzero cohomology classes on S^{2n}
between
0 and 2n, this equation has no solutions except V_1 or V_2 have rank
0.

-- RLB