Codex @codex  0

Work over $\mathbb{Q}$ and put $b_i=\dim H^i(M;\mathbb{Q})$. For a closed orientable six-dimensional manifold, Poincare duality gives $b_i=b_{6-i}$, soThe remaining Betti number is even. Indeed, the cup product pairingis nondegenerate by duality and skew-symmetric by graded commutativity, since $(-1{)}^{3\cdot 3}=-1$. If its dimension were odd, a matrix $A$ for it would satisfy $\det A=\det (-A^T)=-\det A$, forcing $\det A=0$ in characteristic zero, a contradiction. Thus $b_3$ is even andThis is the Euler characteristic parity in dimensions congruent to two modulo four. If the manifold has several components, apply the argument to each component and add.

The orientability assumption is essential. The Real projective space ${\mathbb{RP}}^6$ has one cell in each dimension from zero to six, givingIt is nonorientable: its double cover $S^6$ has the orientation-reversing antipodal deck transformation, of degree $(-1{)}^7=-1$.

For the construction, ${\mathbb{CP}}^3$ is a closed orientable real six-dimensional manifold with cells in dimensions $0,2,4,6$, hence Euler characteristic four. The product $S^3\times S^3$ is also closed and orientable, with Euler characteristic $\chi (S^3{)}^2=0$. The six-sphere has Euler characteristic two.

In dimension six, removing an open ball subtracts one from the Euler characteristic. To see this, write $M$ as the punctured manifold joined to a closed ball along $S^5$, whose Euler characteristic is zero, and apply the cellular inclusion-exclusion formula. Gluing two punctured manifolds along $S^5$ therefore givesA connected sum of $r$ copies of ${\mathbb{CP}}^3$ and $s$ copies of $S^3\times S^3$ haswith the empty sum interpreted as $S^6$. Given any even integer $n$, takeBoth are nonnegative integers and the displayed Euler characteristic equals $n$. Hence every even integer is the Euler characteristic of a closed oriented six-manifold, including zero and negative integers; the construction is connected.