# Is $(\#^k \Bbb{RP}^2) \times I$ an $\mathbb{RP}^2$-irreducible 3-manifold?

Consider $S$ a surface homeomorphic to a connected sum of $n$ projective planes, $n \geq 2$. Can there be a two sided projective plane embedded in $[-\epsilon,\epsilon]\times S$?

• And absolutely no personal input either. – Did Jan 7 '16 at 15:57
• what is the meaning of two sided protective plane?? and yes I agree with @MikeMiller – Anubhav Mukherjee Jan 7 '16 at 16:21
• @Anubhav.K: It's an embedding of $\Bbb{RP}^2$ that extends to an embedding of $\Bbb{RP}^2 \times [-1,1]$; that is to say, it's an embedding of $\Bbb{RP}^2$ with trivial normal bundle. One can find this definition in any book about 3-manifolds, including Hatcher's online notes, especially in the context of the sphere theorem. – user98602 Jan 7 '16 at 16:23
• @Anubhav.K So, you agreed with "there is absolutely nothing unclear" while asking for clarification... – user147263 Jan 7 '16 at 21:47

Call your surface $\Sigma$ so as to avoid confusion with spheres. Our first course of action is to note that $\Sigma$ has no 2-torsion in its fundamental group. (Actually, there's a much stronger true fact: a finite CW complex with contractible universal cover has no torsion in its fundamental group. I will not need or prove this.) To see this, note that if it did, it has as a covering space a noncompact surface with fundamental group $\Bbb Z/2$; but the only simply connected surfaces are $S^2$ and $\Bbb R^2$, and $\Bbb R^2$ does not have any continuous involutions with no fixed point. (It's easier to see that neither $\Bbb R^2$ or the unit disc with hyperbolic metric have isometric involutions with no fixed point; we only need to work in the case of isometric quotients by the uniformization theorem.)
Something slightly stronger than your question is true: there's not even a 2-sided $\Bbb{RP}^2$ in $\Sigma \times S^1$. (Indeed, there's no embedded $\Bbb{RP}^2$ at all, since a manifold with a 1-sided $\Bbb{RP}^2$ has $\Bbb{RP}^3$ as a connected summand; and we have no 2-torsion in our fundamental group, so that's not possible.)
To see this, note that a 2-sided $\Bbb{RP}^2$ cannot possibly disconnect the manifold: this would imply that $\Bbb{RP}^2$ bounds a compact 3-manifold, which it does not. The fact that it does not disconnect implies that it is a homologically nontrivial submanifold (meaning it represents a nonzero class in $H_2$): you can find a loop that intersects with $\Bbb{RP}^2$ in precisely one point, and mod 2 intersection numbers are defined on the level of homology classes.
Now recall that $\pi_1(S^1 \times \Sigma)$ had no 2-torsion, so the map $\pi_1(\Bbb{RP}^2) \to \pi_1(S^1 \times \Sigma)$ is trivial, and the map from $\Bbb{RP}^2$ factors through the universal cover of $S^1 \times \Sigma$: that is, through $\Bbb R^3$. But $\Bbb R^3$ is contractible, which implies your 2-sided embedding of $\Bbb{RP}^2$ was null-homotopic. This contradicts the fact above that it was a homologically nontrivial submanifold.