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I want to ask the problem 4-C in the characteristic classes written by John W. Milnor.

Problem [4-C]. A manifold $M$ is said to admit a field of tangent $k$-planes if its tangent bundle admits a sub-bundle of dimension $k$. Show that $P^n$ admits a field of tangent $1$-planes if and only if $n$ is odd.

From right to left, I can prove it by using the fact that $S^n$ has a non vanishing vector field when $n$ is odd. But from left to right, I try to use the Whitney product theorem to prove.

Can anybody give me a hint or a proof? Thank you.

user26857
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ljh8372
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  • I wrote this up a few weeks ago. (given a nonzero family of 1 planes) $TP^n=e\times V$, and $V$ has a zero $n$th SW-class and $e$ has a total SW class either equal to $1$ or $1+a$. Now just see that for even n, that $(1+a)^{n+1}$ cannot be written as a product of two such things. – PVAL-inactive Oct 13 '14 at 02:57
  • I check what PVAL said, Thank you. – ljh8372 Oct 13 '14 at 06:49
  • It's worth noting everything in site is a unit, so $(a+1)f=(a+1)^{n+1}$ implies $f=(a+1)^n$. – PVAL-inactive Oct 13 '14 at 13:57

2 Answers2

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An odd dimensional projective space admits of a nowhere zero vector field
The trick is to embed the corresponding sphere into complex numerical space: $S^{2n-1}\subset \mathbb C^n $.
We then have a nowhere vanishing vector field $S^{2n-1}\to TS^{2n-1}: v\mapsto iv$ which, being invariant under the antipodal map $S^{2n-1}\to S^{2n-1}:v\mapsto -v$, descends to the quotient manifold $\mathbb P^{2n-1} $ as a nowhere zero vector field $s:\mathbb P^{2n-1}\to T\mathbb P^{2n-1}$
The existence of such a nowhere tangent field is equivalent to the existence of a trivial rank one subbundle of the tangent bundle: here the subbundle $L\subset T\mathbb P^{2n-1}$ is given by $L(v)=\mathbb R\cdot s(v)$

An even dimensional projective space does not admit of a nowhere zero vector field

A nowhere vanishing vector field on $\mathbb P^{2n}$ would lead to the existence of a trivial rank one subbundle $L\subset T\mathbb P^{2n}$ and thus to a (non-canonical) splitting $T\mathbb P^{2n}=L\oplus V$ where $L\cong \underline {\mathbb R}$
But then the top Stiefel-Whitney class of $T\mathbb P^{2n}$ should vanish: $w_{2n}(T\mathbb P^{2n})=0$
However this is false: the total Stiefel-Whitney class of $T\mathbb P^{2n}$ is $(1+h)^{2n+1}$ so that $$w_{2n}(T\mathbb P^{2n})=\binom {2n+1}{2n}h^{2n}=(2n+1)\cdot h^{2n}=h^{2n}\in H^{2n}(\mathbb P^{2n},\mathbb Z/2)=\mathbb Z/2\cdot h^{2n}$$ This contradiction shows that there is no rank one trivial subbundle of $T\mathbb P^{2n}$

Edit
As commented by PVAL, in the even dimensional case I have proved that $T\mathbb P^{2n}$ doesn't contain a trivial line bundle but haven't proved that it doesn't contain a non trivial one . Here is a proof of that last fact:
If $T\mathbb P^{2n}$ contained a non-trivial line bundle, it would be the tautological line bundle $\gamma$ (the only non trivial line bundle on $\mathbb P^{2n}$) and we would have $(1+h)^{2n+1}=(1+h)\cdot w(V)$ since $T\mathbb P^{2n}=\gamma\oplus V$.
Hence we would deduce that $w(V)=(1+h)^{2n}$ because, as PVAL judiciously comments, $1+h\in H^\star (\mathbb P^{2n},\mathbb Z/2)$ is an invertible element .
But then we deduce $w_{2n}(V)=h^{2n}\neq0\in H^{2n}(\mathbb P^{2n},\mathbb Z/2)$: this is absurd because $V$ has rank $2n-1$ and cannot have a non-zero Stiefel-Whitney class of degree $2n$.

Georges Elencwajg
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    I don't think a field of tangent $1$-planes is supposed to be equivalent to a nowhere zero section. Rather I think Milnor means $L$ (in your notation) could be any rank 1 subbundle (i.e. not necessairly an orientable one). So one also has to deal with the case where the total SW class of $L$ is $1+h$ (not just when it is $1$) In that case $w(V)=(1+h)^n$ necessairly as $w(L)$ and $w(TP^n)$ are both units, but then $w_n(V)\ne 0$ which is a contradiction. – PVAL-inactive Jan 10 '15 at 11:03
  • @PVAL: everything I wrote is correct (I think) but you are right that it only answers the implication: odd dimensional $\implies $ subbundle. My other implication is correct but weaker than what the exercise requires. – Georges Elencwajg Jan 10 '15 at 12:08
  • Dear @PVAL: I have now edited the text, completing it in the direction you suggest. Thanks a lot for your vigilance. – Georges Elencwajg Jan 10 '15 at 12:43
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$\mathbb{P}^n(\mathbb{R}) \simeq S^n/{\pm 1}$, that is $S^n$ is a two-fold cover of $\mathbb{P}^n$. Fields of tangent $1$-planes are $1$-dimensional sub-bundles of the tangent bundle. Consider such a field on $\mathbb{P}^n$. Pull back to $S^n$ and get a field on $S^n$ ( invariant under $x\mapsto -x $). Since $S^n$ is simply connected ($n\ge 2$) this $1$- bundle is orientable and from it we get a nonzero vector field on $S^n$ (once the direction is chosen, just take the vector of unit length).

So this is possible only for odd $n$.

orangeskid
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