# No irreducible parallelizable manifold of a given dimension

What is an example of a closed 4-manifold $M$ such that $M$ is parallelizable and $M$ is topologically (or at least smoothly) irreducible?

Recall, $M$ is said to be topologically/smoothly irreducible if it is not homeomorphic/diffeomorphic to a product of two topological/smooth manifolds.

This question is a particular case of a more general question:

For which integers $n$ is every closed, parallelizable $n$-manifold necessarily reducible?

For example, this is the case when $n = 2$.

• Due to <Vogel, Existence of Engel structures, Ann. of Math. v. 169> for parallelizable manifolds M_1 and M_2, their connected sum connected sum with with S^2 x S^2 also parallelizable. E.g., (n+1)(S^3 x S^1) connected sum with n(S^2 x S^2) parallelizable. Some of which, I believe, are not products. – valeri Aug 23 '15 at 15:21

I don't know the answer for general dimensions, but here is an argument that shows that the four-dimensional manifolds suggested by valeri fit your criteria (provided $n > 0$).

The Dold-Whitney Theorem states that if $E_1$ and $E_2$ are orientable rank four vector bundles over an oriented four-dimensional manifold $M$, then $E_1$ and $E_2$ are isomorphic if and only if they have the same second Stiefel-Whitney class, first Pontryagin class, and Euler class. In particular, if $M$ is orientable and we take $E_1 = TM$ and $E_2 = \varepsilon^4$, we arrive at the following corollary:

A smooth closed four-manifold $M$ is parallelisable if and only if it is spin with $\tau(M) = 0$ and $\chi(M) = 0$.

Note, this immediately implies the statement in valeri's comment. In fact, $S^2\times S^2$ can be replaced by any orientable spin manifold with signature zero and Euler characteristic 4.

Now consider the parallelisable manifolds $M_n := (n+1)(S^1\times S^3)\# n(S^2\times S^2)$ suggested by valeri and note that $\pi_1(M_n) = F_{n+1}$, the free group on $n + 1$ generators. If $n = 0$, then $M_0 = S^1\times S^3$ is reducible, so let's only consider $n > 0$ from now on.

If $M_n$ were homeomorphic to a product of surfaces, one of them would have to be a torus as $\chi(M_n) = 0$. Therefore $\mathbb{Z}\oplus\mathbb{Z}$ would be isomorphic to a subgroup of $F_{n+1}$, but this is impossible as every subgroup of a free group is free.

Suppose now that $M_n$ is homeomorphic to $S^1\times X$ for some three-manifold $X$. Then $F_{n+1} \cong \mathbb{Z}\oplus\pi_1(X)$. As $\pi_1(X)$ is finitely generated and isomorphic to a subgroup of a free group, $\pi_1(X) \cong F_k$ for some non-negative integer $k$. Note that

$$H_2(F_{n+1}) \cong H_2(K(F_{n+1}, 1)) \cong H_2(B_{n+1}) \cong 0$$

where $B_{n+1} = \bigvee_{i=1}^{n+1}S^1$ is a bouquet of $n+1$ circles. On the other hand,

$$H_2(\mathbb{Z}\oplus F_k) \cong H_2(K(\mathbb{Z}\oplus F_k, 1)) \cong H_2(S^1\times B_k) \cong H_2(B_k)\oplus H_1(B_k) \cong \mathbb{Z}^k.$$

So we must have $k = 0$, in which case $F_{n+1} \cong \mathbb{Z}\oplus\pi_1(X) \cong \mathbb{Z}$. But this is impossible as $\mathbb{Z}$ is abelian and $F_{n+1}$ is not (because $n + 1 > 1$).

Therefore, for $n > 0$, the manifolds $M_n = (n + 1)(S^1\times S^3)\# n(S^2\times S^2)$ are irreducible (smoothly, topologically, and homotopically).

• Note, the argument used here can be used to show that $F_n$ cannot be written as a non-trivial direct sum. – Michael Albanese Mar 21 '18 at 23:01