# $4$-manifold with fundamental group $\Bbb Z/4\Bbb Z$

Before I write my question, I want to write some thoughts.

Let $$M$$ be a connected topological manifold such that $$\pi_1(M)=\Bbb Z/3\Bbb Z$$. Then, considering its orientation $$2$$-fold cover, which is connected, I can say $$M$$ is orientable. Now, an example of such a closed $$3$$-manifold is $$L(3,1)$$.

Now, this type of argument can not be given if I consider $$\pi_1(M)=\Bbb Z/4\Bbb Z$$ to conclude $$M$$ is orientable. But Euler characteristic of an odd-dimensional closed manifold is always zero, so we cannot say $$\Bbb Z/4\Bbb Z$$ is the fundamental group of any closed connected non-orientable $$3$$-manifold, as $$H_1(M,\Bbb Z)$$ is infinite when $$M$$ is closed non-orientable connected $$3$$-manifold.

Again this logic can not be given for $$4$$-dimensional closed connected manifold. So, I am wondering if the following fact. I assume closed means compact without boundary.

Does there exist closed connected $$4$$-manifolds both orientable and non-orientable type having fundamental group $$\Bbb Z/4\Bbb Z$$?

Any help will be appreciated.

• Every finitely presented group is the fundamental group of a compact $4$- dimensional manifold: take the standard 2-dimensional CW complex with that fundamental group, immerse it into $\Bbb R^4$ and take a regular neighborhood. Sep 15 '20 at 19:50
• Does this argument gives both orientable and non-orientable, can you elaborate more. I will really appreciate if you write this as an answer.
– User
Sep 15 '20 at 19:57
• @JCAA: that construction seems to me to produce either an open manifold or a manifold with boundary. I think you want to immerse into $\mathbb{R}^5$ and take the boundary of a tubular neighborhood, as Somnath Basu does here: mathoverflow.net/a/15421/290 (I don't know whether the result is orientable or not though.) Sep 15 '20 at 20:01

As pointed out, every finitely presented group arises as the fundamental group of a closed smooth orientable four-manifold.

The same is not true of non-orientable manifolds as $$\mathbb{Z}/3\mathbb{Z}$$ illustrates. A necessary condition is that the group must have an index two subgroup (i.e. the fundamental group of the orientable double cover). This turns out to be sufficient. That is, a finitely presented group is the fundamental group of a closed smooth non-orientable four-manifold if and only if it has an index two subgroup; see this question.

• My bad! I should check this website before asking this question. Thanks. +1 for your help.
– User
Sep 16 '20 at 6:22

A non-orientable example: consider the automorphism $$f : S^2 \times S^2$$ given by $$(x, y) \mapsto (y, -x)$$ where $$-$$ denotes the antipode map. This map has order $$4$$ and gives a free action of $$\mathbb{Z}/4$$ on $$S^2 \times S^2$$, so its quotient is a closed $$4$$-manifold $$X$$ with $$\pi_1(X) \cong \mathbb{Z}/4$$. Since $$\chi(S^2 \times S^2) = 4$$ we have $$\chi(X) = 1$$ so $$X$$ is non-orientable; alternatively, we can check that $$f$$ acts by $$-1$$ on $$H^4(S^2 \times S^2)$$.

• But, here we can not take orientation two-cover to get another orientable manifold. Am I right? Thanks for your answer.
– User
Sep 15 '20 at 19:55
• @Math: why not? The orientation double cover is the quotient of $S^2 \times S^2$ by $f^2$, which sends $(x, y)$ to $(-x, -y)$. It turns out to be the real Grassmannian $\text{Gr}_2(\mathbb{R}^4)$. Sep 15 '20 at 19:57
• Sorry, I was thinking that $\pi$ induced an index $2$-subgroup of $\Bbb Z/4\Bbb Z$ as covering induced map is injective.
– User
Sep 15 '20 at 20:03