# Euler characteristic zero but non-parallelizable four-dimensional manifold

The title speaks for itself. Does there exist a four-dimensional smooth manifold which admits a non-vanishing continuous vector field (Euler characteristic zero) but not a global frame (non-parallelizable)? I've read about five-dimensional examples, but I am interested only in dimension four. Not a topologist - please, as simple as possible. References preferred. Thank you.

• I'm a student who knows a little so I might be wrong. What about $X = S^2 \times T^2$ ? The Euler characteristic is $\chi(X) = \chi(S^2).\chi(T^2)=0$. On the other side, $T(X) \cong T(S^2) \times T(T^2)$. I don't think this is trivial (my informal argument : if it is you can get a global frame for $S^2$ which is wrong).
– user171326
Commented Aug 22, 2016 at 15:56
• Sorry my idea doesn't work, I just saw that $S^2 \times T^2$ was parallelizable.
– user171326
Commented Aug 22, 2016 at 16:01
• Thanks for trying. Much better than ignorance. Commented Aug 22, 2016 at 16:34
• @N.H. Using a non-trivial $S^2$ bundle over $T^2$ does work though, see my answer below. Commented Aug 22, 2016 at 16:40
• @Bedovlat : Thanks for your kind words. Michael Albanese : really nice solution !
– user171326
Commented Aug 22, 2016 at 17:35

Let $L$ be a complex line bundle over $T^2 = S^1\times S^1$ such that $\langle c_1(L), [T^2]\rangle = 1$; such a line bundle exists because $c_1 : \operatorname{Vect}^1_{\mathbb{C}}(T^2) \to H^2(T^2; \mathbb{Z}) \cong \mathbb{Z}$ is an isomorphism. Then define $M$ to be the total space of the projectivised bundle $\mathbb{P}(L\oplus\varepsilon_{\mathbb{C}}^1)$. Note that $\pi : M \to T^2$ is a $\mathbb{CP}^1$-bundle. We can lift the two linearly independent vector fields on $T^2$ to $M$ via an Ehresmann connection, so $M$ admits at least two linearly independent vector fields; in particular, $\chi(M) = 0$.
As $L$ is an open subset of $M$, $M$ contains the image of the zero section of $L$ as a compact submanifold, call it $E$. As $\langle e(L), [T^2]\rangle = \langle c_1(L), [T^2]\rangle = 1$, $E$ has self-intersection $1$; in particular, it's mod $2$ intersection number $E\cdot E$ is $1$. On the other hand, the mod $2$ intersection number is also given by $E\cdot E = \langle \alpha\cup\alpha, [T^2]\rangle$ where $\alpha \in H^2(M; \mathbb{Z}_2)$ is the Poincaré dual of the $\mathbb{Z}_2$-fundamental class of $E$. Using the language of Steenrod squares and Wu classes, one can show that $x\cup x = (w_2 + w_1\cup w_1)\cup x$ for all $x \in H^2(M; \mathbb{Z}_2)$; see my answer here for example. As $M$ is orientable, we see that $$1 = E\cdot E = \langle\alpha\cup\alpha, [T^2]\rangle = \langle w_2\cup\alpha, [T^2]\rangle,$$ so $w_2 \neq 0$ (i.e. $M$ is not spin). As such, $M$ cannot admit three linearly independent vector fields; if it did, then $TM \cong \ell\oplus\varepsilon_{\mathbb{R}}^3$ for some real line bundle $\ell$ and hence $w_2 = 0$.