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

$X=\{(x,y,z)|x^2+y^2+z^2\le 1$ 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 $\mathbb R\mathbb P^2$ (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.

ett
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3 Answers3

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  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.

Jason DeVito - on hiatus
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  • Is the space $Z$ described in the questyion really a cone on $P^2$? – Mariano Suárez-Álvarez 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 - on hiatus 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 - on hiatus 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 - on hiatus 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 $\mathbb{R}^3$) –  Oct 19 '10 at 20:01
  • @Jason: Z is definitely not $P^2_\mathhb R$: there is are no identifications happening in the interior of the half closed ball from which we start, so in $Z$ there are points with a neighborhood looking like an open $3$-ball. This does not happen in $P^2_\mathbb R$. – Mariano Suárez-Álvarez Oct 19 '10 at 21:39
  • @Mariano: I agree completely! What I meant to say a few comments above is that $Z$ is the CONE OVER $\mathbb{R}P^2$. In fact, I think the proof I wrote shows this fact. Sorry for the confusion! – Jason DeVito - on hiatus Oct 19 '10 at 22:18
  • @Will: Well, $Z$ is not naturally a subspace of $\mathbb{R}^3$, so I'm not sure how to compute the boundary as you say. Also, the image of the embedding into $\mathhbb{R}^5$ I wrote will have every point a boundary point, since it will be a smaller dimensional compact set. – Jason DeVito - on hiatus Oct 19 '10 at 22:20
  • There's a gap in your argument in point 1. For example, every point in a 2-manifold has a neighborhood basis of open sets homeomorphic to an open annulus, which is not homeomorphic to an open ball. You have to use something else, such as the local homology argument in the answer of @MarianoSuárez-Álvarez. – Lee Mosher Aug 17 '17 at 20:05
  • @Lee: I completely agree, and even know about using local homology to deal with this. I'm not sure what I was thinking back then (perhaps I didn't know of using local homology that way at the time). – Jason DeVito - on hiatus Aug 18 '17 at 03:08
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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...

Aaron Mazel-Gee
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  • 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 06:32
  • Whether the cone over a manifold is a manifold depends mostly on the manifold! Consider a torus, for example... – Mariano Suárez-Álvarez Oct 19 '10 at 06:34
  • 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 06:47
  • @ 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 06:48
  • 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. –  Oct 19 '10 at 17:23
  • It could be true that a cone of $RP^2$ is not a manifold. Would that mean that talking about a boundary for the space $Z$ does not make sense ? @Aaron: Can you please elaborate on " since (for starters) the origin doesn't have enough 'stuff' around it." ?? –  Oct 19 '10 at 17:25
  • No, Z is not a manifold. As some others have mentioned, one way to prove this is by "local homology". You take a neighborhood $U$ of a point $x$, and the local homology at $x$ is the homology of $U \backslash { x}$ (so long as you have chosen a sufficiently small neighborhood $U$). Up to homotopy (and hence in the eyes of homology), at the interior of an n-manifold this will be an (n-1)-sphere, and on a boundary point this will be an (n-1)-disk. Since the local homology of the cone point is neither of these, this cone cannot be a manifold. (Local homology is a homeomorphism invariant.) – Aaron Mazel-Gee Oct 20 '10 at 05:34
  • The precise way to measure "stuff" around a point depends on your perspective. If you're working with smooth manifolds, every point needs to have a neighborhood which is diffeomorphic to an n-ball. If you're working with topological manifolds, replace "diffeo" with "homeo". (These are practically the definitions of manifolds.) – Aaron Mazel-Gee Oct 20 '10 at 05:37
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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.

Mariano Suárez-Álvarez
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    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 - on hiatus 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 - on hiatus Oct 20 '10 at 00:05
  • I realise that this is an old question, but regarding your last sentence: for which $p$ would the homology be identically zero? – Irregular User Mar 09 '17 at 22:01
  • Suppose that $p$ is in the boundary of $M$. By excision and an homeomorphism, $H_(M,M-p)$ is the same thing as $H_(D,D-0)$, with $D={(x_1,\dots,x_n)\in\mathbb R^n:x_1^2+\cdots+x_n^2\leq1, x_1\geq0}$, the closed half ball, and $0$ the origin. Consider now the long exact sequence for the pair $(D,D-0)$ in homology. The inclusion $D-0\to D$ is an homotopy equivalence, so it induces isomorphisms in homology, and it follows that the relative homology is zero. – Mariano Suárez-Álvarez Mar 09 '17 at 22:52