Nonorientable manifolds being a boundaries

Question

I will say that a manifold (smooth, compact, without boundary) is itself boundary if there is some smooth compact manifold $W$ with boundary, such that $M=\partial W$. I'm interested in nontrivial example of nonorientable manifold $M$ which is connected and is a boundary. What is trivial: you can take any (nonorientable compact) manifold $M$ and consider the cylinder $W=M \times [0,1]$. Therefore the boundary would be $\partial W=M \sqcup M$ so it is easy to obtain nonorientable (but not connected) manifold as a boundary. If we admit noncompact manifolds then every manifold $M$ may be obtained as a boundary of $M \times [0,\infty)$. So I'm interested in such examples where $W$ is assumed to be compact and $M$ connected.

Answer 1

The simplest possible example is the Klein bottle. One way to think of the Klein bottle is the quotient $S^1 \times I /\sim$, where $(z,0) \sim (\bar z,1)$ (considering $S^1 \subset \Bbb C$). (This way of defining it comes essentially from the standard representation; equating the red arrows gives you $S^1 \times I$, and the relation of equating the blue arrows is our $\sim$.)

Then the exact same construction, replacing $S^1$ with $D^2$, gives us a manifold whose boundary is the Klein bottle: $D^2 \times I /\sim$, where again $(z,0) \sim (\bar z, 1)$.

That this is the simplest possible means, in particular, that $\Bbb{RP}^2$ does not bound any compact manifold. There are a variety of ways to see this. One is to use Lefschetz duality with $\Bbb Z/2\Bbb Z$ coefficients to prove that for odd-dimensional manifolds $M$, $\chi(\partial M) = 2\chi(M)$; but of course $\chi(\Bbb{RP}^2) = 1$.

One can actually show that the connected non-orientable surfaces that are boundaries are precisely those that are connected sums of Klein bottles. Proof: all connected non-orientable surfaces are of the form $\#_n \Bbb{RP}^2$ (that is, the connected sum of $n$ copies of $\Bbb{RP}^2$); and $\chi(\#_n \Bbb{RP}^2) = 2-n$, and as the Euler characteristic must be even, $n$ must be even, so that our surface is of the form $\#_k K = \#_{2k} \Bbb{RP}^2$.

Now note that the connected sum of surfaces that bound compact 3-folds also bounds a compact 3-fold (consider the boundary connected sum of the 3-folds the bound; this is the same as the usual connected sum, but instead of gluing around a neighborhood of an interior point, we do so on a neighborhood of a point on the boundary.)

More generally, a closed $n$-manifold is the boundary of a compact $(n+1)$-fold if and only if its Steifel-Whitney numbers vanish.