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:



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> For which integers $n$ is every closed, parallelizable $n$-manifold necessarily reducible?
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For example, this is the case when $n = 2$.



No irreducible parallelizable manifold of a given dimension
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:



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


