Exercises:Saul - Algebraic Topology - 9/Exercise 9.7

Exercises

Exercise 9.7

Show that the cone on the real projective plane, i.e. $C(\mathbb{RP}^2)$, is not a topological 3-manifold with boundary.

I might have two solutions for this, I accidentally answered the wrong question initially. $\newcommand{\crp}{C(\mathbb{RP}^2)}$

Solution 1

I struggled to focus writing this up, however the idea should be clear, if the presentation isn't up to my usual standard please forgive me. Aditionally:

• $\langle G\rangle$ means the "normal subgroup generated by $G$", fortunately where it is used here $G$ is trivial, thus is a normal subgroup itself so $\langle 0\rangle\eq 0$, thus the question of "normal subgroup closure" never pops up in this solution.

Before we start, I want to note that we make heavy use of the fundamental group, and I may write this without a base-point, this is because for a path-connected space the fundamental group based anywhere is isomorphic to any other fundamental group based elsewhere of the space. Please know that I know this.

First we consider $\crp$ as a topological space and work out some fundamental group isomorphisms.

• Immediately, as any topological cone is contractible space we see:
  • $\pi_1(\crp)\cong 0$ (where $0$ denotes the trivial group, we do not use $1$ even though fundamental groups are usually not Abelian)

Let $p\in\crp$ be the apex of the cone, i.e. the image of $\mathbb{RP}^2\times\{1\}$ under the the cone's quotient map, some people may identify $X\times\{0\}$ as the apex, we do not.

Consider $\crp-\{p\}$, this is easily seen to be homeomorphic to $\mathbb{RP}^2\times[0,1)$

• Note that $\mathbb{RP}^2\times[0,1)$ has $\mathbb{RP}^2$ as a strong deformation retraction (by slowly "projecting down" along $[0,1)$), and thus:
  • $\pi_1(\crp-\{p\})\cong\pi_1(\mathbb{RP}^2)\cong\frac{\mathbb{Z} }{2\mathbb{Z} }$

We summarise:

• $\pi_1(\crp)\cong 0$ and
• $\pi_1(\crp-\{p\})\cong\pi_1(\mathbb{RP}^2)\cong\frac{\mathbb{Z} }{2\mathbb{Z} }$

We now use the "fact" it is a manifold to tackle the claim: $\pi_1(\crp)\cong\pi_1(\crp-\{p\})$ - this will make use of the Seifert-Van Kampen theorem.

• Let $x\in\crp$ be an arbitrary point that is not the implicit basepoint for $\pi_1$
  1. suppose that $x$ is in the interior of the manifold, then:
     • there exists a homeomorphism: $\varphi:U\rightarrow\mathbb{B}^3$ which takes an open neighbourhood of $x$ onto the open unit ball at the origin.
       • Suppose WLOG that $\varphi(x)\eq 0$ (we can do this as if this wasn't the case we can use that translations are homeomorphisms in Euclidean space to get a new chart, $\varphi':z\mapsto \varphi(z)-\varphi(x)$ instead, this is also a homeomorphism and has the property that $\varphi'(x)\eq 0$)
         • Note that homeomorphic spaces have isomorphic fundamental groups so $\pi_1(U)\cong\pi_1(\mathbb{B}^3)$
         • Note also given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself so
           • $\varphi\big\vert_{U-\{x\} }:(U-\{x\})\rightarrow(\mathbb{B}^3-\{0\})$ is a homeomorphism
         • We now use the Seifert-Van kampen theorem to see:
           • $$\pi_1(\crp)\cong\frac{\pi_1(\crp-\{x\})\ast\pi_1(U)}{\langle \pi_1\big((\crp-\{x\})\cap U\big)\rangle}$$ which by homeomorphic spaces have isomorphic fundamental groups:

             $$\cong\frac{\pi_1(\crp-\{x\})\ast\pi_1(\mathbb{B}^3)}{\langle\pi_1(U-\{x\})\rangle}$$ (by replacing spaces with homeomorphic ones)

             $$\cong\frac{\pi_1(\crp-\{x\})\ast\pi_1(\mathbb{B}^3)}{\langle\pi_1(\mathbb{B}^3-\{0\})\rangle}$$ (by replacing spaces with homeomorphic ones)

             $$\cong\frac{\pi_1(\crp-\{x\})\ast 0}{0}$$ as $\mathbb{B}^3$ is clearly contractible, and $\mathbb{B}^3-\{0\}$ strongly deformation retracts to the 2-sphere which has trivial fundamental group (easily seen as any loop can be "pulled" to its base point in $\mathbb{S}^2$)

             $\cong\pi_1(\crp-\{x\})$

         • We conclude:
           • $\pi_1(\crp)\cong\pi_1(\crp-\{x\})$ - provided $x$ is an interior point of the $3$-manifold $\crp$
  2. now suppose $x\in\partial(\crp)$ - that is $x$ is a boundary point of our 3-manifold with boundary
     • We see that there exists a homeomorphism: $\varphi:U\rightarrow\mathbb{H}^3$ where $U$ is an open neighbourhood of $x$ and $\mathbb{H}^3$ denotes the 3-dimensional half space:
       • $\mathbb{H}^3:\eq\{(x_1,x_2,x_3)\in\mathbb{R}^3\ \big\vert\ \Vert (x_1,x_2,x_3)\Vert < 1 \wedge x_3\ge 0\} \eq \mathbb{B}^3\cap\{(x_1,x_2,x_3)\in\mathbb{R}^3\ \big\vert\ x_3\ge 0\}$
     • Such that $\big(\varphi(x)\big)_3\eq 0$ (i.e. on the $x_3\eq 0$ plane in $\mathbb{R}^3$)
       • Again WLOG we can suppose $\varphi(x)\eq 0$ - for the same reasoning as the previous case
         • As before we note that $(U-\{x\})\cong(\mathbb{H}^3-\{0\})$ by the restriction of $\varphi$ to the left hand side of the $\cong$
         • We use the Seifert-Van Kapmen theorem again:
           • $$\pi_1(\crp)\cong\frac{\pi_1(\crp-\{x\})\ast\pi_1(U)}{\langle \pi_1\big((\crp-\{x\})\cap U\big)\rangle}$$ which by homeomorphic spaces have isomorphic fundamental groups:

             $$\cong\frac{\pi_1(\crp-\{x\})\ast\pi_1(\mathbb{H}^3)}{\langle\pi_1(U-\{x\})\rangle}$$ (by replacing spaces with homeomorphic ones)

             $$\cong\frac{\pi_1(\crp-\{x\})\ast\pi_1(\mathbb{H}^3)}{\langle\pi_1(\mathbb{H}^3-\{0\})\rangle}$$ (by replacing spaces with homeomorphic ones)

             $$\cong\frac{\pi_1(\crp-\{x\})\ast 0}{0}$$ as $\mathbb{H}^3$ and $\mathbb{H}^3-\{0\}$ are both clearly contractible. (To convince yourself of this take the origin for $\mathbb{H}^3$, you can just "straight-line" pull everything onto it. For $\mathbb{H}^3-\{0\}$ pick the point $(0,0,1)$ to pull back on, again straight line onto this point)

             $\cong\pi_1(\crp-\{x\})$

         • We conclude:
           • $\pi_1(\crp)\cong\pi_1(\crp-\{x\})$ - provided $x$ is a boundary point of the $3$-manifold with boundary$\crp$
  • Thus in either case:
    • $\pi_1(\crp)\cong\pi_1(\crp-\{x\})$
• Since $x\in\crp$ was arbitrary (and not the implicit base-point) we see $\pi_1(\crp)\cong\pi_1(\crp-\{x\})$ holds for all $x$
• Next we combine everything to see:
  • $0\cong\pi_1(\crp)\cong \pi_1(\crp-\{p\})\cong\pi_1(\mathbb{RP}^2)\cong\frac{\mathbb{Z} }{2\mathbb{Z} }$
    • Thus: $0\cong\frac{\mathbb{Z} }{2\mathbb{Z} }$ - obviously a contradiction!
• Thus $\crp$ cannot be a $3$-manifold with boundary
  • In fact it cannot be a $3$-manifold without boundary either, as per the first half where we assumed $x$ was a point interior to the manifold.

This completes the proof.