Two solid N_3 glued by its boundary - MathOverflow

Two solid N_3 glued by its boundary

2009-12-06T05:31:59Z

Let $N_3$ be the genus three non orientable surface. Do we have an analogous 3d manifold as the solid torus and the solid Klein bottle for $N_3$? I don't see how to extend the ideas related to the 3d lens spaces. Any feedback would be super-welcome

Answer by Sam Nead for Two solid N_3 glued by its boundary

2009-12-06T12:44:52Z

$N_k$ is the connect sum of $k$ copies of the real projective plane, so it has Euler characteristic $2 - k$. For $k$ even $N_k$ bounds a connect sum of $k/2$ copies of the solid Klein bottle, as in Henry's comment. When $k$ is odd the rank of $H_1$ (with $Z/2Z$ coefficients) is odd. Because of "half lives, half dies" the boundary of a three-manifold must have even rank in $H_1$. So $N_k$, for $k$ odd, does not bound.

Half lives, half dies can be found in Hatcher's three-manifold notes as Lemma 3.5, but you'll need to use $Z/2Z$ coefficients. My copy of the universal coefficient theorem is all rusty, so if I've made a mistake, it is here.

Answer by Oscar Randal-Williams for Two solid N_3 glued by its boundary

2009-12-06T13:10:35Z

Alternatively, the characteristic number $\langle w_2, [N_k] \rangle$ is just the mod 2 reduction of the Euler characteristic (as $w_2$ is the mod 2 reduction of the Euler class) so is -k modulo 2. Thus if k is odd this characteristic number is nontrivial, and $N_k$ cannot bound.

Answer by Allen Hatcher for Two solid N_3 glued by its boundary

2009-12-06T16:15:29Z

It is a general fact that a closed manifold of odd Euler characteristic cannot bound a compact manifold. This can be deduced pretty easily from the fact that a closed manifold of odd dimension has Euler characteristic zero (a consequence of Poincaré duality) as follows. Suppose N is the boundary of a compact manifold P. Let M be the double of P, the union of two copies of P glued along N. Then the Euler characteristics of M, N, and P are related by:

$\chi(M)=\chi(P)+\chi(P)-\chi(N)$

Thus $\chi(M)$ and $\chi(N)$ are congruent mod 2. If the dimension of N is even, then M is a closed manifold of odd dimension so $\chi(M)=0$, hence $\chi(N)$ is even. And if the dimension of N is odd then $\chi(N)=0$ anyhow.

I should have put this in my book!