Non-orientable 6-manifold with certain Betti numbers My question is about an aside that came about when answering the following question Closed oriented manifold with middle Betti is one with odd degree..
Is there any sequence $(1,a_1,a_2,a_3,a_4,a_5,0)$ $a_i \geq 0$ integers which cannot be the Betti number sequence of a compact non-orientable $6$-manifold?
I can think of examples (from products of real projective spaces, non-orientable surfaces etc.) with $(1,0,0,0,0,0)$, $(1,1,1,1,1,1,0)$, $(1,0,1,0,0,0,0)$, $(1,1,0,0,0,0)$ , $(1,0,0,0,1,0,0)$ then connect sum gives almost every possibility. But for some sequences I not able to think of an example, for example $(1,0,0,1,0,0,0)$.
Sorry if this is a naïve question I haven't thought about non-orientable manifolds for a long time. My expectation is that probably everything can be realised but I don't know enough examples (perhaps fibre bundles are enough to give everything)?
 A: I claim that in every even dimension, any sequence of rational Betti numbers can occur.
As you noted, in each dimension $k$, it's sufficient to find examples $N^k_m$ with $0 < m < k$ with $H^\ast(N^k_m;\mathbb{Q})\cong H^\ast(S^m;\mathbb{Q})$, for then connect summing gives everything.  EDIT:  This is wrong.  connect summing two non-orientable manifolds adds rational cohomology in degree $k-1$.  End Edit
Now, if $m$ is even, one can simply use $N^k_m = S^m\times \mathbb{R}P^{k-m}$.  (Since $m$ and $k$ are both even, so is $k-m$, and $\mathbb{R}P^{even}$ has the rational cohomology of a point.)
So, we need only handle the case where $m$ is odd.  Set $n:=k-m$, which is also odd since $k$ is even and $m$ is odd.  Consider the $\mathbb{Z}_2$ action on $S^m \times S^n$ with $-1\ast(x,y) = (-x, r(y))$ where $r:S^n\rightarrow S^n$ is any reflection in a hyperplane.
This action is free since it is free on the first factor.  Call the resulting quotient manifold $M$.  I claim that $N^k_m = M$ works.
Because $m$ is odd, the antipodal map is orientation preserving.  Since $r$ reverses orientation, it follows that the $\mathbb{Z}_2$ action reverses orientation.  In particular, $M$ is non-orientable.
I claim that $H^\ast(M;\mathbb{Q}) \cong H^\ast(S^m;\mathbb{Q})$.  To this this, first note that since $\mathbb{Z}_2$ is finite, we have a transfer homomorphism $f:H^\ast(M;\mathbb{Q})\rightarrow H^\ast(S^m\times S^n;\mathbb{Q})$ for which the composition $f\circ \pi^\ast$ is multiplication by $2$ (which is an isomorphism with $\mathbb{Q}$ coefficients).  In particular, $\pi^\ast$ is an injection, so the rational Betti numbers of $M$ are zero except possibly in degrees $0,m,n,m+n$.  Further, the Betti numbers are $a_0, a_m, a_n, a_{m+n}$ are all bounded by $1$ (except when $m = n$, in which case $a_{m}$ is bounded by $2$).
Now the antipodal action on $S^m$ is free with quotient $\mathbb{R}P^m$.  The associated bundle construction now shows that $M$ is the total space of an $S^n$ bundle over $\mathbb{R}P^m$.  Pulling back this bundle along the double cover $S^m\rightarrow \mathbb{R}P^m$, we obtain a commutative diagram $$\begin{array} AS^n & \longrightarrow & S^n \\
\downarrow & & \downarrow \\
S^m\times S^n & \longrightarrow & M\\
\downarrow &  & \downarrow \\
S^m & \longrightarrow & \mathbb{R}P^m
\end{array}$$
The induced map on cohomolgoy $H^m(\mathbb{R}P^m;\mathbb{Q})\rightarrow H^m(S^m;\mathbb{Q}))\rightarrow H^m(S^m\times S^n;\mathbb{Q})$ is easily seen to be non-trivial.  It follows from commutativity that $H^m(\mathbb{R}P^m;\mathbb{Q})\rightarrow H^m(M;\mathbb{Q})$ must be non-zero, so $H^m(M;\mathbb{Q})$ is non-trivial.
In addition, if $H^n(M;\mathbb{Q}) \neq 0$ (or $H^m(M;\mathbb{Q})$ has dimension $2$ when $m=n$), then we have a problem:  If $x\in H^m(M;\mathbb{Q})$ is non-zero and $y\in H^n(M;\mathbb{Q})$ is non-zero (and $x$ and $y$ are independent if $m = n$), then $\pi^\ast(xy) = \pi^\ast(x)\pi^\ast(y)\neq 0$, which implies $H^{m+n}(M;\mathbb{Q}) = 0$.  Since we already know $M$ is non-orientable, this is absurd.  Thus, we conclude that $H^\ast(M;\mathbb{Q})\cong H^\ast(S^m;\mathbb{Q})$.
