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)?
Best Answer
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 $H^\ast(M;\mathbb{Q})\cong H^\ast(S^m;\mathbb{Q})$.