TITLE: A Compact Manifold with odd Euler characteristic whose tangent bundle admits a field of lines
QUESTION [12 upvotes]: I understand that the top Stiefel Whitney class is an obstruction for the tangent bundle of a manifold to have a trivial line sub-bundle. I am looking for a counterexample when removing the word "trivial", i.e: A compact manifold $M$ of dimention n such that $w_n(TM)\neq0$ (or equivalently $\chi(M)$ is odd) and there exists a sub-bundle $\xi\subset TM$ with $\operatorname{rank}(\xi)=1$.
After some thought I understand why one cannot find an example using surfaces, odd-dimensional manifolds or any of $S^n$,$T^n$,$\mathbb{R}P^n$,$\mathbb{C}P^n$.   I also understand why manifolds with $H^1(M;\mathbb{Z}_2)=0$ will not work, and why such a manifold (if it exists) cannot be null-cobordant.
My intuition on 4-manifolds (or god forbid anything higher dimensional) is not good enough to know where to look. Any insight as to where one should look for such a creature (or why the hell it should not exist) will be greatly appreciated.

REPLY [16 votes]: I believe there is no example satisfying all your constraints.  If I recall (my memory is a little foggy on this) the result likely goes back to Hopf, and one of his variations on the Poincare-Hopf index theorem. This question might be addressed in the Milnor and Stasheff text. Here is one way to argue the point. 
Say the tangent bundle of the manifold $N$ admits a field of lines.  Then (at worst) some 2:1 cover of $N$ admits an everywhere non-vanishing vector field.  Since the Euler characteristic is the obstruction to such a vector field existing (Poincare-Hopf index theorem) the Euler characteristic of this covering space is zero.  But $\chi N$ is a multiple of the euler characteristic of the cover.  
i.e. your assumption that the Euler characteristic is odd excludes the possibility of a 1-dimensional sub-bundle.

REPLY [2 votes]: Let $\gamma$ be the canonical (real) line bundle over $RP^2$. In other words, $w_1(\gamma)\neq 0$. The total Stiefel-Whitney class of the Whitney sum $\gamma\oplus\gamma$ is $(1+w_1(\gamma))^2=1+w_1(\gamma)^2$. So $w_2(\gamma\oplus\gamma)=w_1(\gamma)^2$ which is nonzero.