Which homology groups of a closed orientable 6-manifold can be isomorphic to $\mathbb{Z}^3$? 
List all $i$ for which there is a closed orientable $6$-manifold $M$ with $H_i(M) =\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$.

I am working on an old exam problem and this one stumped me. Progress so far: 
I can take $Y=\mathbb{T}^3 \times \mathbb{S}^3$ and get the desired group for $i =1,5,2,4$ I think. But this construction I believe gives $H_3(Y)=\mathbb{Z}^2$. What to do in this case?  
 A: First of all, $Y = T^3\times S^3$ is indeed a closed orientable six-dimensional manifold with $H_i(Y; \mathbb{Z}) \cong \mathbb{Z}^3$ for $i = 1, 2, 4, 5$. Furthermore, $H_0(Y; \mathbb{Z}) \cong \mathbb{Z}$, $H_3(Y; \mathbb{Z}) \cong \mathbb{Z}^2 $, and $H_6(Y; \mathbb{Z}) \cong \mathbb{Z}$.
It is also possible to find a closed orientable six-dimensional manifold $M$ with $H_i(M; \mathbb{Z}) \cong \mathbb{Z}^3$ for $i = 0, 6$. For example, $M = S^6\sqcup S^6\sqcup S^6$. However, if we also require $M$ to be connected, no such example exists.
So the only question which remains is whether there exists a closed orientable six-dimensional manifold $M$ with $H_3(M; \mathbb{Z}) \cong \mathbb{Z}^3$. Suppose $M$ is such a manifold and fix an orientation. By Poincaré duality, $H^3(M; \mathbb{Z}) \cong H_3(M; \mathbb{Z})$ and cup product defines a non-degenerate bilinear form
\begin{align*}
H^3(M; \mathbb{Z})\times H^3(M; \mathbb{Z}) &\to \mathbb{Z}\\
(\alpha, \beta) &\mapsto \langle \alpha\cup\beta, [M]\rangle
\end{align*}
where $[M] \in H_6(M; \mathbb{Z})$ is the fundamental class of $M$ and the angled brackets on the right denote the pairing between cohomology and homology - note, in general the cup product is non-degenerate on $H^3(M; \mathbb{Z})_{\text{free}}$, but for our given $M$, $H^3(M; \mathbb{Z})$ has no torsion. Note that 
$$\alpha\cup\beta = (-1)^{3\times 3}\beta\cup\alpha = -\beta\cup\alpha,$$ 
so the above form is skew-symmetric. 
Choose a basis for $H^3(M; \mathbb{Z})$. The corresponding matrix representation of the form is a $3\times 3$ integer matrix $A$. As the form is non-degenerate, $A$ is non-singular (i.e. $\det A \neq 0$), and as the form is skew-symmetric, $A$ is skew-symmetric (i.e. $A^T = -A$). Note that
$$\det A = \det A^T = \det (-A) = (-1)^3\det A = -\det A,$$
but this is impossible as $\det A \neq 0$. Therefore, no such $A$, and hence $M$, can exist.

More generally, the proof above can be modified to give the following result:

Let $M$ be a closed orientable manifold of dimension $4n + 2$. Then the rank of $H^{2n+1}(M; \mathbb{Z})$ is even, i.e. $b_{2n+1}(M)$ is even.

Together with Poincaré duality, this gives rise to the following corollary:

Let $M$ be a closed orientable manifold of dimension $4n + 2$. Then $\chi(M)$ is even.

In contrast, for any integer $k$, there exists a closed orientable manifold $M$ of dimension $4n$ with $\chi(M) = k$. In particular, such manifolds can have odd Euler characteristic, e.g. $M = \mathbb{CP}^{2n}$ which has Euler characteristic $\chi(\mathbb{CP}^{2n}) = 2n + 1$.
