# Non-orientable 3-manifolds

I am reading Non-orientable 3-manifolds of small complexity (Topology and its Applications 133 (2003) pp 157-178, arXiv:math/0211092), by Amendola and Martinelli. In this work $\mathbb P^2$-irreducible complexity-6 manifolds are listed. There are five of them. I wonder about the following non-orientable manifolds.

1. Take $S^2\times I$ and glue its top sphere to its bottom sphere with the antipodal homeomorphism or with a reflection in plane homeomorphism. Let's denote it by $S^2\widetilde\times S^1$.
2. $\mathbb P^2\times S^1$

I assume that those two manifolds are not $\mathbb P^2$-irreducible. I don't know how to embed $\mathbb P^2$ into the first one. The preposition 1.3 on page 5 of the abovementioned work says that a Stiefel-Whitney surface cannot be a sphere. It seems to me that it is sphere for the first manifold.

Both of them have double cover $S^2\times S^1$, and the fundamental groups are $\mathbb Z$ and $\mathbb Z + \mathbb Z_2$. What are the fundamental groups of the five manifolds of complexity 6 in the above work?

At the same time I have the following additional questions about non-orientable 3-manifolds.

A. In case of surfaces we obtain every non-orientable one by the connected sum of an orientable one with $\mathbb RP^2$. Is there an analog in 3-manifolds? I.e. Is every non-orientable closed 3-manifold being connected sum of orientable one and $S^2\widetilde\times S^1$ ? (clarified on 2018-09-03)

B. A non-orientable surface with a removed disk is embeddable in $\mathbb R^3$. Can we embed a non-orientable $M^3$ with removed ball into $\mathbb R^4$ ?

C. Is the regular neighborhood of a loop changing orientation in a 3-manifold homeomorphic to a solid Klein bottle ?

EDIT 2018-07-08 I add following new question.

D. Each non-orientable surface is double covered by orientable one. We can ask whether every 3-manifold with infinite fundamental group is double cover of some non-orientable one.

The answer to my question A is negative but still it seems that we can somehow convert orientable manifold into non-orientable by attaching handle which change orientation. When Stiefel-Whitney surface is sphere then it is the case.

• (A) No. What you see in dimension 2 is a very special case. (B) No again. There are obstructions to embedding punctured 3-manifolds in $\mathbb R^4$, for example there is one using the torsion linking form. (C) The boundary of a regular neighbourhood, yes. The neighbourhood itself is the associated disc bundle to the Klein bottle. – Ryan Budney Jul 5 '18 at 14:55
• Thank you again. At least we can sort irreducible non-orientable 3-manifolds by genus of Stiefel-Whitney surface. – Marek Mitros Jul 6 '18 at 6:00
• The notion of P2-irreducible contains in particular irreducible. So there are no non-trivial spheres. – Bruno Martelli Jul 6 '18 at 16:22
• by a full Kleinbottle you mean solid Kleinbottle? – janmarqz Jul 9 '18 at 2:13
• Sorry for my language. I mean $D^1\times D^2$ with top and bottom $D^2$ glued with reflection homeomorphism. I guess this is called "solid Klein bottle". – Marek Mitros Jul 9 '18 at 4:43

For question D, I think the answer is negative. Consider the semidirect product $\Lambda=\mathbf{Z}^2\rtimes_A\mathbf{Z}$, with $A=\begin{pmatrix}25 & 7\\ 7 & 2\end{pmatrix}$. This is the $\pi_1$ of some SOL-type closed 3-manifold.
This matrix is not a square in $\mathrm{GL}_2(\mathbf{Z})$, by a naive computation (use that the centralizer consists of symmetric matrices, and there exist no integers $a,b,c$ such that $(a^2+b^2,b^2+c^2)=(25,2)$). In addition one checks that every element of order 2 in the normalizer of $\langle A\rangle$ in $\mathrm{GL}_2(\mathbf{Z})$ is contained in $\mathrm{SL}_2(\mathbf{Z})$. Therefore, any subgroup of $\mathrm{GL}_2(\mathbf{Z})$ containing $\langle A\rangle$ as subgroup of index 2 is contained in $\mathrm{SL}_2(\mathbf{Z})$. In addition, since the determinant of $A-I$ is $-25$, every subgroup of index $\le 5$ in $\Lambda$ contains $\mathbf{Z}^2$. Hence, if $\Gamma$ is a group containing $\Lambda$ as subgroup of index 2, then the intersection of subgroups of index $\le 4$ contains $\mathbf{Z}^2$, and taking the centralizer of its derived subgroup retrieves $\mathbf{Z}^2$, which is thus normal in the larger group $\Gamma$. The previous applies to the image of $\Gamma$ in $\Gamma/\mathbf{Z}^2$. So in case $\Gamma$ is torsion-free, the corresponding manifold is orientable.