= Solution
Work over $\mathbb Q$ and put $b_i=\dim H^i(M;\mathbb Q)$. For a closed <orientable> six-dimensional <manifold>, <Poincare duality> gives $b_i=b_{6-i}$, so
$$
\chi(M)=2(b_0-b_1+b_2)-b_3.
$$
The remaining <Betti number> is even. Indeed, the <cup product> pairing
$$
H^3(M;\mathbb Q)\times H^3(M;\mathbb Q)\longrightarrow\mathbb Q,
\qquad (u,v)\longmapsto\langle u\smile v,[M]\rangle
$$
is nondegenerate by duality and skew-symmetric by graded commutativity, since $(-1)^{3\cdot3}=-1$. 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 and
$$
\boxed{\chi(M)\equiv0\pmod2.}
$$
This 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, giving
$$
\chi(\mathbb {RP}^6)=1-1+1-1+1-1+1=\boxed{1}.
$$
It 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 gives
$$
\chi(M\mathbin{\#}N)=\chi(M)+\chi(N)-2.
$$
A <connected sum> of $r$ copies of $\mathbb {CP}^3$ and $s$ copies of $S^3\times S^3$ has
$$
\chi=2+2r-2s,
$$
with the empty sum interpreted as $S^6$. Given any even integer $n$, take
$$
r=\max\{n/2-1,0\},\qquad
s=\max\{1-n/2,0\}.
$$
Both are nonnegative integers and the displayed <Euler characteristic> equals $n$. Hence \b[<every even integer is the Euler characteristic of a closed oriented six-manifold>], including zero and negative integers; the construction is connected.
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