Example of closed orientable manifold with a nonzero vector field, but not parallelizable

Question

Can you give a simple example of a closed orientable manifold with an everywhere nonzero section of its tangent bundle, but where the tangent bundle is not trivial?

If the manifold is not orientable, it is not too hard to come up with examples (e.g. Klein bottle). But all the obvious constructions I can think of fail when you require orientability. For example, a natural thing to do is to take a product of a parallelizable manifold (say $S^1$) with a manifold with no everywhere nonzero section of its tangent bundle. But there is no guarantee that this will have nontrivial tangent bundle: for example, $S^1\times S^n$ is parallelizable for all $n$.

Answer 1

The sphere $S^5$ is closed, orientable, and there is an explicit nonvanishing section of its tangent bundle: $(x_2, -x_1, x_4, -x_3, x_6, -x_5)$ (if you view $S^5 \subset \mathbb{R}^6$ in the standard way). But Adams' theorem on the Hopf invariant says that the only parallelizable spheres are $S^0$, $S^1$, $S^3$, and $S^7$. (I chose $S^5$, but any odd sphere other than these four examples works.)

Answer 2

By the Poincare-Hopf theorem and its converse, if $X$ is a closed connected smooth orientable manifold, then $X$ admits a nonvanishing vector field iff the Euler characteristic of $X$ is zero. So it suffices to find an example of such a manifold whose Euler characteristic is zero but such that at least one of its Stiefel-Whitney or Pontryagin classes does not vanish.

Such an example does not exist in low dimensions (and note that the Euler characteristic condition automatically holds in all odd dimensions by Poincare duality):

• If $\dim X = 1$, then $X$ must be $S^1$, which is parallelizable.
• If $\dim X = 2$, then $X$ must be $S^1 \times S^1$, which is parallelizable.
• If $\dim X = 3$, then any closed oriented smooth $3$-manifold is parallelizable.

So $\dim X \ge 4$. Unfortunately I'm having some trouble writing down a $4$-dimensional counterexample, and I'm not sure if one exists.

But here are a bunch of $5$-dimensional counterexamples: let $X = S^1 \times Y$ where $Y$ is a closed connected smooth orientable $4$-manifold with nonzero signature, for example $\mathbb{CP}^2$. Then $X$ is closed, connected, smooth, orientable, and has Euler characteristic $0$, but by the signature theorem $p_1(Y) \neq 0$ and so $p_1(X) \neq 0$, so $X$ is not stably parallelizable.

More generally you can take $X = S^1 \times Y$ where $Y$ is closed, connected, smooth, orientable, but not stably parallelizable, e.g. because at least one of its Stiefel-Whitney or Pontryagin classes does not vanish. This construction requires that $\dim Y \ge 4$.

Edit: An easier example is to take $X = \mathbb{RP}^5$. Like $S^5$, this is closed, connected, smooth, orientable, and has Euler characteristic $0$, hence has a nonvanishing vector field (which I think can be constructed from a corresponding nonvanishing vector field on $S^5$). But its Stiefel-Whitney classes don't all vanish; for example, $w_2(\mathbb{RP}^5) \neq 0$. So $X$ is not stably parallelizable. This doesn't require any big theorems.

Answer 3

Take the connected sum of two copies of circle x three sphere and two copies of the complex projective plane. Then the Euler characteristic is zero and the signature is nonzero. This gives a closed oriented 4D example to a manifold having a nowhere zero vector field and not paralízale.