# Two solid N_3 glued by its boundary

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

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I don't understand this question. Are you asking whether there's a non-orientable handlebody bounded by N_3? If by N_3 you mean the non-orientable surface of Euler characteristic -2, the answer is yes. (Just take a solid Klein bottle and attach a 1-handle.) Please clarify. –  HJRW Dec 6 '09 at 6:23
The title of this question should be: "Does $N_3$ bound?". Then you could mention the application you have in mind (which I guess is gluing together copies of the bounded manifold?). Also, could you explain how lens spaces play a role in this question? –  Sam Nead Dec 6 '09 at 12:47
Sam, if N_3 would bound we should try to develop a similar theory of lenspaces where the the building block were solid N_3's with respective boundaries a pair of 2-torus with one point $\mathbb{R}$-blowups but, not now... anyway thanks a lot for your interest –  janmarqz Dec 7 '09 at 0:18

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!

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yeah, I completely agree, perhaps in the next update... Prof thanks so much! –  janmarqz Dec 7 '09 at 0:27
Ok, it is perfectly clear that N<sub>3</sub> can not bound an ORIENTABLE 3-manifold... but, what about a NON ORIENTABLE 3d-orbifold? Could the cone $N_3\times I/(N_3\times 1\sim *)$ possibly? Is there a similar half live or half die for 3d-orbifolds? –  janmarqz Dec 7 '09 at 19:28
You'll have to settle on a precise definition of orbifold cobordism to make sense of this question but if you allow any kind of orbifold locus you get all compact manifolds as boundaries of orbifolds by considering $M \times [-1,1] / \mathbb Z_2$ where the action of $\mathbb Z_2$ on $[-1,1]$ is reflection about the origin. $\mathbb Z_2$ acting trivially on the $M$ factor. –  Ryan Budney Dec 11 '09 at 5:13

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.

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thanks Oscar, you are a real savant on the subject... –  janmarqz Dec 7 '09 at 0:24

$N_k$ 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.

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