# Topology of a cone of RP^2

I had already posted this on mathoverflow and was advised to post the same here. So here it goes:

X={x,y,z|$x^2+y^2+z^2≤1$ and $z≥0$} i.e. X is the top half of a 3-Disk.

Z=X/E, where E is the equivalence relation on the the plane z = 0 which is as follows:

(x,y,0)∼(−x,−y,0).

I was told that this space is equivalent to a cone of RP2 (Real Projective Plane).

I want to know the following facts about "Z"

1) Is this a manifold with a boundary?

2) If it is a manifold with a boundary, what are the points of Z that make the boundary?

3) Is it simply connected?

4) What is the minimum Euclidean dimension in which Z can be embedded in?

Thank you very much for your help. I am new to topology and this problem came up as a part of my project. Any help is appreciated.

Thank you. Will.

-

1. It's not a manifold either with or without boundary. For the proof, look at the cone point c. If U is a small open set around c, then $U-c$ is homeomorphic to $\mathbb{R}P^2 \times (0,1]$. However, this is not homeomorphic to either $\mathbb{R}^3-pt$ nor a half space minus a point.

2. In fact, it's contractible (the cone over anything is contracitble). Thus, it's simply connected.

3. I'm not sure, but it can definitely be topological embedded in R^5. To see this, let $f$ be an embedding of $\mathbb{R}P^2$ into $\mathbb{R}^4$ (which exists by Whitney's embedding theorem). Define $G:Z\rightarrow\mathbb{R}^5$ by $G(p, t) = (tf(p),t)$. This will be a topological embedding.

The minimal possible answer is 4 since that's the smallest in which $\mathbb{R}P^2$ can be embedded into.

-
Is the space $Z$ described in the questyion really a cone on $P^2$? – Mariano Suárez-Alvarez Oct 19 '10 at 16:55
@Mariano: I didn't think about it too hard because of the way the question is stated. I must confess that I haven't convinced myself one way or the other. In the mean time, one can interpret the occurances of "it" in my answer to mean "the cone over RP^2". Then, I think it's at least mathematically correct even if it doesn't at all pertain to the problem asked! – Jason DeVito Oct 19 '10 at 17:19
@Mariano: I now think that $Z$ is $\mathbb{R}P^2$. Here is my proposed homeomorphism. For each $0<r\leq 1$, let $Y_r$ denote the points in the upper half disc with distance $r$ from the origin. Then, the equivalence relation restricted to $Y_r$ gives a quotient $T_r$ which is homeomorphic to $\mathbb{R}P^2$. Then, thinking of the disc as the cone over $Y_1$, we get a map from the disc to the cone over $\mathbb{R}P^2$ sending a point $(p,r)$ to $(\pi(p),r)$ where $\pi:Y_1\rightarrow T_1$ is the projection map. It's easy to check that $f$ is continuous and surjective. – Jason DeVito Oct 19 '10 at 18:21
Further, $f$ is not 1-1, but one has f(x) = f(y) iff x~y. This means that $f$ descends to a map of $D/~$ which is 1-1. Since $D/~$ is compact, the inverse map is automatically continuous, so the descended map is a homeomorphism. – Jason DeVito Oct 19 '10 at 18:23
I guess it doesn't make sense to talk about the boundary of $Z$ in terms of a manifold, since $Z$ is not a manifold. What about talking about boundary of $Z$ as a topological space (as a subspace of $\mathbb{R}^3$). Would the points on the surface of the hemisphere ($x^2 + y^2 + z^2 = 1$) belong to the boundary of Z (as a subspace of $\mathbb{R}^3$) – Will Oct 19 '10 at 20:01

First consider just the hemisphere $X'\subseteq X$, and define $Z'= X'/E$. This is the real projective plane, in one of its many incarnations. You may have seen $\mathbb{RP}^2$ presented as the space of lines through the origin in $\mathbb{R}^3$. To get from that presentation to this one, note that each such line intersects $S^2 \subseteq \mathbb{R}^3$ in exactly two points, and for those lines where one point is above the xy-plane and the other is below, we've just chosen the $z>0$ representative. This is a manifold without boundary, and it is not simply-connected. Indeed, its universal cover is $S^2$, which is a two-fold cover. Hence its fundamental group has two elements.

Now, note that in $Z$, we're literally just taking the cone over $Z'$ -- those rays from the origin that end at points in $X'$ that were identified in $Z'$ are themselves identified. It is a non-trivial (?) fact that $\mathbb{RP}^2$ is not the boundary of any 3-manifold. I'm pretty sure that if the cone over a space is a manifold, then the boundary is the manifold itself, although I can't instantly tell you why this should be true. (In fact, $\mathbb{RP}^2$ generates the unoriented cobordism ring in dimension 3.) Note that the cone over any space is simply-connected, since it is contractible (to the cone point). I don't know the answer to 4, though I'd suspect it's either 4 or 5, since 4 is the minimum dimension you need in order to embed $\mathbb{RP}^2$. In general, the question of minimum dimension for embedding a particular manifold is not so easy, beyond the a priori bounds given by the Whitney embedding theorem. And this isn't even a manifold...

-
Actually I think that the question of whether the cone over a manifold is a manifold depends on your definitions. For example, $I=[0,1]$ is a 1-manifold (with boundary), and the cone $CI$ is homeomorphic to a disk. So is this a manifold? It doesn't look like it in the usual setup, of course. On the other hand, a natural qualification for your question would be, "Does it inherit the structure of a differentiable manifold from its definition as a quotient of a subset of $\mathbb{R}^3$? The answer in that case is no, since (for starters) the origin doesn't have enough 'stuff' around it. – Aaron Mazel-Gee Oct 19 '10 at 6:32
Whether the cone over a manifold is a manifold depends mostly on the manifold! Consider a torus, for example... – Mariano Suárez-Alvarez Oct 19 '10 at 6:34
On the other hand, there's no way that the cone on a cylinder could be called a manifold (with or without boundary), because a neighborhood of the cone point wouldn't be homeomorphic to a 3-ball or a piece of 3-d halfspace. I think the anomaly I pointed out might be specific to 1-dimensional manifolds, and that to have any hope of having your cone be (homeomorphic to) a manifold in higher dimensions you need to be boundaryless. Which still doesn't answer the question of why the boundary would have to be just $\mathbb{RP}^2$... – Aaron Mazel-Gee Oct 19 '10 at 6:47
@ Mariano: Yes of course it depends mostly on the manifold! I was just pointing out a weird example I noticed. – Aaron Mazel-Gee Oct 19 '10 at 6:48
Thank you very much for your comments and answers. Just want to clarify first that this is not some homework problem in a class and this space came up as a part of my research. 1) So $Z$ is not a manifold? A cone of 2-sphere is homeomorphic to a 3-Disk which is a manifold with a boundary. Similarly, a cone of a 2-Disk is homeomorphic to upper half of a 3-Disk (again a manifold with a boundary). On the other hand I think a cone of a torus is not a manifold. So whether the cone space is a manifold or not depends entirely on the manifold. – Will Oct 19 '10 at 17:23

This should be a comment somehwere, but got to long.

Will, the cone on $P^2_\mathbb R$ is not a manifold. Consider the long exact sequence for integral reduced homology of the pair $(C,C')$ where $C=C(P^2_\mathbb R)$ is the cone over $P^2_\mathbb R$ and $C'=C\setminus\{a\}$ is the complement of the apex of the cone. Since $C$ is contractible and $C'$ deformation-retracts onto $P^2_\mathbb R$, you get isomorphisms $H^\sharp_2(C,C')\cong H^\sharp_{1}(P^2_\mathbb R)\cong\mathbb Z/2\mathbb Z$.

If follows that $C$ is not a manifold: in a manifold $M$, for every point $p\in M$ we have that the integral reduced homology $H^\#_\bullet(M,M\setminus\{p\})$ is that of a sphere.

(Generalizing this reasoning, you get a rather strong condition on a manifold for its cone to be also a manifold. I'm sure the topologists among our fellow M.SEers know of a precise characterization.)

In the same way, we see that $C$ is not a manifold with boundary, because in such a space $H^\#_\bullet(M,M\setminus\{p\})$ is, for every $p$, either identically zero or that of a sphere.

-
I'm not a topologist, but I think the question of whether or not the cone of a manifold is a manifold is a very difficult one. For example, let $M$ be the Poincare dodecahedral space. (This a a closed 3-manifold with pi_1 a perfect group and homology groups the same as S^3). Then, it's known that the suspension $\Sigma M$ is NOT a manifold, but the suspension $\Sigma \Sigma M$ is homeomorphic to $S^5$. In thinking of a suspension as two cones stuck base to base, I think this provides evidence that figuring out when the cone is a manifold is tough. – Jason DeVito Oct 19 '10 at 22:24
@Jason: $\mathbb RP^2$ does not bound any manifold because its Euler characteristic is odd. A cone on it, if it was a manifold, would have $\mathbb RP^2$ as its boundary. – Ryan Budney Oct 19 '10 at 23:56
@Ryan: I understand that the cone on $\mathbb{R}P^2$ is not a manifold (in fact, see my own proof in the accepted answer). I was more remarking on Mariano's comment about general conditions for the cone of a manifold to be a manifold. Of course, there are some easy very strong conditions (as Mariano pointed out), but I was trying to show evidence that sufficient conditions would have to be somewhat harder to work out. – Jason DeVito Oct 20 '10 at 0:05