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For any positive integer $k$, is there a smooth, closed, non-parallelisable manifold $M$ such that the maximum number of linearly independent vector fields on $M$ is $k$?

Note that any such $M$, for any $k$, must have Euler characteristic zero by the Poincaré-Hopf Theorem.

Without the non-parallelisable hypothesis, the $k$-dimensional torus is an easy example.

One might think that the product of a $k$-dimensional torus with a manifold which admits no nowhere-zero vector fields provides an example, but this doesn't necessarily work. For example, the product of a torus and an even dimensional-sphere is actually parallelisable. The reason this approach fails with the even-dimensional sphere is that it is has stably trivial tangent bundle; maybe using a manifold with non-stably trivial tangent bundle might work.

Determining the maximal number of linearly independent vector fields on a manifold is not easy in general. Even for spheres, the answer is complicated: if $n + 1 = 2^{4a+b}u$ where $a \geq 0$, $0 \leq b \leq 3$ and $u$ is odd, then the maximal number of linearly independent vector fields on $S^n$ is $8a + 2^b - 1$. So for the first few odd values of $n$ we have

$$ \begin{array}{c|c} & \text{maximum number of linearly}\\ n &\text{independent vector fields on}\ S^n\\ \hline 1 & 1\\ 3 & 3\\ 5 & 1\\ 7 & 7\\ 9 & 1\\ 11 & 3\\ 13 & 1\\ 15 & 8 \end{array} $$

The above result shows that the answer to the initial question is yes for $k \equiv 0, 1, 3, 7 \bmod 8$.

Michael Albanese
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2 Answers2

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Consider $M$, the product of the Klein bottle $K^2$ with $k-1$-torus $T^{k−1}$. This manifold is nonorientable, hence, nonparallelizable. On the other hand, it is the total space of a circle bundle over $T^k$, since $K^2$ is a circle bundle over the circle. Let $H$ be a (smooth) connection on this bundle. Take $k$ independent vector fields $X_1,...X_k$ on $T^k$ and lift them to $M$ via the connection $H$. Hence, $M$ admits $k$ independent vector fields. It cannot have $k+1$ vector fields since it is not parallelizable. I am sure there are orientable examples as well, but this requires more work.

Moishe Kohan
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  • I was trying to see how to obtain an orientable example by looking at three manifolds $M$ which were $S^1\times S^1$ bundles over $S^1$. One way to construct such a manifold is by taking the mapping torus $M_f$ of a homeomorphism $f : S^1\times S^1 \to S^1\times S^1$. The product $(S^1)^{k-2}\times M_f$ will always have at least $k$ linearly independent vector fields by the same argument as in your answer. Maybe there is an $f$ such that $(S^1)^{k-2}\times M_f$ will have exactly $k$ linearly indepenedent vector fields. I haven't been able to establish this though. – Michael Albanese May 05 '16 at 03:16
  • Also note that this approach can be seen as a generalisation of the one you gave in your answer as the Klein bottle is the mapping torus associated to the antipodal map on $S^1$. – Michael Albanese May 05 '16 at 03:17
  • This approach won't work. Note that $T((S^1)^{k-2}\times M_f) \cong \varepsilon^k\times L$ for some line bundle $L$. But $L$ is orientable, and hence trivial, so $(S^1)^{k-2}\times M_f$ is parallelisable. The same problem will occur if $M_f$ is replaced by the total space of any fibre bundle over $S^1$ with toric fibres. – Michael Albanese May 05 '16 at 16:02
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My advisor gave me an example of a closed orientable four-manifold which admits two linearly independent vector fields but not three.

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

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 $w_2 = 0$.

From this construction we obtain, for any $k \geq 2$, an example of a closed, orientable, non-parallelisable manifold such that the maximum number of linearly independent vector fields it admits is $k$, namely $M\times T^{k-2}$. It admits at least $k$ linearly independent vector fields as it is a $\mathbb{CP}^1$-bundle over $T^k$, and it does not admit any more as it is not spin (a product of manifolds is spin if and only if the individual factors are spin). For $k = 1$, we can take $S^5$.

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Michael Albanese
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