Skip to main content

All Questions

Filter by
Sorted by
Tagged with
2 votes
0 answers
57 views

Isotropy subgroup of $\operatorname{GL}_{n+1}(\mathbb R)$ acting on $\mathbb R \mathbb P^n$

The general linear group $\operatorname{GL}_{n+1}(\mathbb R)$ (as a Lie group) acts smoothly on the real projective space $\mathbb R \mathbb P^n$, via $A \cdot [x] := [Ax]$. By $[x]$ here, I mean the $...
Joseph Kwong's user avatar
2 votes
0 answers
45 views

Different topologies for the real projective base as a manifold

I am stuck with the following problem. I am given the $d$-dimensional real projective space $\mathbb{R}P^d$ as the set of equivalence classes of lines in $\mathbb{R}^{d+1}$, i.e. \begin{equation} \...
dancingqueen's user avatar
1 vote
0 answers
167 views

Quotient of projective group scheme by a finite group action

Let $X$ be a projective group scheme, over some base $S$, and let $G$ be a finite group acting on $X$ by $S$-isomorphisms. I would like to understand if/when the quotient $X/G$ is representable by a ...
kindasorta's user avatar
  • 1,280
1 vote
1 answer
251 views

Understanding the CW decomposition of the real projective $n$-space

I'm trying to work on proving the following statement: Let $\Bbb P^n$ be the $n$-dimensional (real) projective space. Then $\Bbb P^n$ has a CW decomposition with one cell in each dimension $0,...,n$, ...
math-physicist's user avatar
1 vote
0 answers
52 views

Showing that projective spaces are Hausdorff

We touched upon the spaces $\mathbb RP^n$ in our topology class. We were asked to think on whether $\mathbb RP^n$ was Hausdorff. After a bit of contemplation, I came to conclude that the answer is ...
Atom's user avatar
  • 4,540
0 votes
1 answer
51 views

Provee that $\mathbb{P}^2(\mathbb{R})/H \cong S^2$, where $H\subset \mathbb{P}^2(\mathbb{R})$ is a projective line

$\mathbb{P}^2(\mathbb{R})/H$ identifies the topological quotient space given by the relation $x \sim y \Leftrightarrow x=y$ or $x,y \in H$. I tried to build an identification $\mathbb{P}^2(\mathbb{R}) ...
Luigi Traino's user avatar
  • 1,210
3 votes
0 answers
56 views

$P(a,b,c)=P(bc,ca,ab)$ weighted projective planes for pairwise coprime $a,b,c$

Let $a,b,c\geq 2$ be pairwise coprime integers. The (complex) weighted projective plane $P(a,b,c)$ is the quotient of $\Bbb C^3-\{0\}$ by the action of $\Bbb C^*=\Bbb C-\{0\}$ given by $t\cdot (x,y,z)=...
user302934's user avatar
  • 1,682
1 vote
1 answer
74 views

Let $\pi$ be the quotient map $S^n\to\mathbb R\mathrm{P}^n$. Is $\pi_{\star p}:T_pS^n\to T_{\pi(p)}\mathbb R\mathrm P^n$ an isomorphism?

I am trying to prove the following statement: Let $\pi$ be the quotient map $S^n\to\mathbb R\mathrm{P}^n$. Then $\pi_{\star p}: T_pS^n\to T_{\pi(p)}\mathbb R\mathrm P^n$ an isomorphism Here is my ...
FUUNK1000's user avatar
  • 891
0 votes
0 answers
45 views

What can we get via taking quotient of $\mathbb{C}P^1$ by a finite abelian group?

Let $G_n = \langle\tau\rangle$ and $G_m = \langle\sigma\rangle$ be groups of $n$-th and $m$-th roots of unity. Define action of $G_n \times G_m$ on $\mathbb{C}P^1$ as follows. $$ (\tau^k, \sigma^t) \...
Invincible's user avatar
  • 2,678
3 votes
1 answer
228 views

Proving ${\mathbb{P}}^n$ is Hausdorff

I am trying to understand and complete the proof that the real projective space ${\mathbb{P}}^n$ is Hausdorff.In my notes it is modeled as${\mathbb{R}}^{n+1}\setminus \{0\}/\sim $ and it goes like ...
some_math_guy's user avatar
6 votes
1 answer
225 views

Weighted projective plane as a quotient of $\Bbb CP^2$

For positive integers $a_0,\dots,a_n$, consider the weighted projective space $\Bbb C\Bbb P(a_0,a_1,\dots,a_n)$, which is the quotient of $\Bbb C^{n+1}-\{0\}$ by the action of $\Bbb C^*=\Bbb C-\{0\}$ ...
blancket's user avatar
  • 1,920
1 vote
0 answers
769 views

Homeomorphism between the real projective plane and a quotient of unit square

A bit of context: I'm starting to learn topology using Topology: A Categorical Approach and the quotient topology is defined as being the finest topology on a set $S$ for which the surjective map $\pi ...
Alooffi's user avatar
  • 105
4 votes
1 answer
400 views

Acting on CP(2) by conjugation

In the paper [2], the author defines the group $G$ to be generated by homeomorphisms of ${\bf S}^2\times{\bf S}^2$ that swap coordinates and/or map them to their antipodal point: $$G=\langle(y,x),(-x,...
Anthony's user avatar
  • 1,359
2 votes
0 answers
71 views

Show that if $n$ is odd, then $\frac {RP^n}{RP^{n-2}} \simeq S^n \vee S^{n-1} $

$\mathbf {The \ Problem \ is}:$ Show that if n is odd, then $X =\frac {RP^n}{RP^{n-2}}$$\simeq S^n \vee S^{n-1} =Y.$ $\mathbf {My \ approach}:$ We know $RP^n =\frac {S^n}{C_2},$ then is it true that $ ...
Rabi Kumar Chakraborty's user avatar
2 votes
1 answer
355 views

Obtaining $\mathbb CP^n$ as an identification space of $D^{2n}.$

Prove that $\mathbb C P^n$ is the identification space $D^{2n}/\sim$ where $x \sim y$ if either $x = y$ or $q(x) = q(y),$ where $q$ is the usual quotient map $S^{2n+1} \xrightarrow {q} \mathbb CP^n.$ ...
Anil Bagchi.'s user avatar
  • 2,932
1 vote
2 answers
201 views

Intuition behind equivalence of two identifications obtained from Möbius band.

Consider the real projective plane $\Bbb RP^2$ which can be realized as a quotient space obtained from a square by identifying points on each pair of it's opposite edges in reverse order. It has been ...
Anil Bagchi.'s user avatar
  • 2,932
1 vote
1 answer
86 views

Proving a result concerning quotient spaces.

Let $M$ be the Möbius strip and $C$ it's boundary circle. Prove that $M/C$ is homeomorphic to $\Bbb RP^2.$ I know that $\Bbb RP^2 \approx D^2/x \sim -x, x \in \partial D^2.$ So if we can able to get ...
Anil Bagchi.'s user avatar
  • 2,932
0 votes
1 answer
101 views

On a homeomorphism problem of complex projective space.

Show that $\Bbb C P^n - \Bbb C P^{n-1} \cong \Bbb C^n.$ There is an obvious embedding of $\Bbb C P^{n-1}$ into $\Bbb C P^n$ given by $$[z_0,z_1, \cdots, z_{n-1}] \longmapsto [z_0,z_1, \cdots, z_{n-1},...
Anil Bagchi.'s user avatar
  • 2,932
2 votes
2 answers
160 views

Showing that $(\Bbb C^{n+1} - 0) {/} (\Bbb C - 0) \cong \Bbb C P^n.$

Show that the complex projective space $\Bbb C P^n \cong \Bbb C^{n+1} - 0 / \Bbb C - 0.$ $\textbf {My thought} :$ We know that $\Bbb C P^n : = \Bbb S^{2n+1}/\Bbb S^1,$ where $\Bbb S^1$ is the unit ...
Anil Bagchi.'s user avatar
  • 2,932
5 votes
1 answer
269 views

When is the restriction of a quotient map $p : X \to Y$ to a retract of $X$ again a quotient map?

Quite a number of questions in this forum deal with the problem when the restriction $p' : A \stackrel{p}{\to} p(A)$ of a quotient map $p : X \to Y$ to a subspace $A \subset X$ is again a quotient map....
Paul Frost's user avatar
  • 81.9k
2 votes
1 answer
83 views

The fibers of the quotient map $q:\Bbb A^{n+1}\setminus\{0\}\to \Bbb P^{n}$ are irreducible.

Let $k$ be an algebraically closed field, let $\Bbb A^{n+1}$ be the $(n+1)$-dimensional affine space, let $\Bbb P^{n}$ be the $n$-dimensional projective space, and let $q:\Bbb A^{n+1}\setminus\{0\}\to\...
richarddedekind's user avatar
0 votes
1 answer
96 views

Restriction of quotient map $\pi: \mathbb{C}^{n+1}\setminus \{0\} \to \mathbb{P}^n (\mathbb{C})$ to $S^{2n+1}$ is closed

My lecturer, to demonstrate that the projective complex space $\mathbb{P}^n(\mathbb{C})$ is Hausdorff, proves that the restriction of the quotient map to $S^{2n+1}$ is a closed map. Here is how the ...
Massimo Bertolotti's user avatar
3 votes
1 answer
121 views

Defining topology on projective space via topology of a field

Given a topology on a field $K$ (in my case a non-archimedean local field with valuation $\nu$ and ring of integers $R$ with maximal Ideal $\mathfrak{m}$), the goal is to define a topology on $\mathbb{...
insomnia430's user avatar
8 votes
1 answer
406 views

Construction of $\mathbb{CP}^2$ from $S^2\times S^2$

Show that $\mathbb{CP}^2$ is homeomorphic to the orbit space of $(S^2\times S^2)/\mathbb{Z}_2$. The action is given by $(x,y) \mapsto (y,x)$. My attempt: I can see that $S^2\times S^2$ is a subset of $...
epsilon_delta's user avatar
1 vote
1 answer
64 views

Projection onto projective space is injective iff $\Bbb K=\Bbb F_2$

Consider $\Bbb K^{n+1}$ a vector space over a field $\Bbb K$ and $\Bbb P^n(\Bbb K)\colon\!=\dfrac{\Bbb K^{n+1}\setminus\{0\}}{\sim}$, where $\Bbb \sim $ is defined by: $v\sim w\Leftrightarrow \exists\,...
Nerey's user avatar
  • 69
1 vote
1 answer
202 views

Question about the map $S^1\to S^1$ in the context of the real projective plane $\mathbb{R}P^2$

I was recently working on an exercise to compute $H^*(\mathbb{R}P^2\times \mathbb{R}P^2,\mathbb{Z})$ and there was a particular step in the solution for which i would like to get a better intuition. ...
Zest's user avatar
  • 2,486
2 votes
1 answer
152 views

Embedding Euclidean Space Into Real Projective Space

I'm struggling with what I think should be a pretty straightforward proof. Let $g:\mathbb{R}^n\rightarrow\mathbb{R}P^n$ be the map defined by $g=p\circ f$ where $p$ is the quotient map $p:\mathbb{R}^{...
Karambwan's user avatar
  • 699
2 votes
2 answers
270 views

Homogeneous Coordinates on a Real Projective Space

I attempt to understand the construction of the standard $C^{\infty}$ atlas on a real projective space from Loring Tu's An Introduction to Manifolds (Second Edition, page no. 79). Tu denotes the ...
rainman's user avatar
  • 831
1 vote
1 answer
247 views

Explicit quotient map from $D^n$ with antipodal boundary points identified to $\mathbb{P}^n$

Let $$\mathbb{P}^n = \{l \ \mid \ l \text{ is a line through the origin in } \mathbb{R}^{n+1} \}$$ denote the $n$-dimensional real projective space. What is an explicit quotient map from $D^n$ to $\...
Sal Groet's user avatar
4 votes
0 answers
1k views

The symmetry group / isometry group of the complex projective space

question: For the complex projective space of $n$-complex dimensions, $$\mathbb{P}^n,$$ what is the symmetry group / isometry group of this complex projective space $\mathbb{P}^n$? Attempt: Naively, ...
annie marie cœur's user avatar
2 votes
1 answer
487 views

Projective Space v.s. Quotient Space v.s. Fibration

What are the more precise relations between (a) projective space, (b) quotient space and (c) the base manifold under certain fibration? (1) Can every projective space (e.g. $\mathbb{RP}^n$, $\...
wonderich's user avatar
  • 6,049
2 votes
0 answers
205 views

Homogeneous space and quotient space for real projective spaces

We know that $$ O(n+1)/O(n) \simeq SO(n+1)/SO(n) \simeq S^n, $$ based on the result of homogeneous space. Also $$ PO(n+1)/PO(n) \simeq P^n, $$ $P^n$ is the projective space. These are in some ...
wonderich's user avatar
  • 6,049
1 vote
1 answer
467 views

Homogeneous space and quotient space for complex projective spaces

We know that $$ O(n+1)/O(n) \simeq SO(n+1)/SO(n) \simeq S^n, $$ based on the result of homogeneous space. Also $$ PO(n+1)/PO(n) \simeq P^n, $$ $P^n$ is the projective space. These are in some ...
wonderich's user avatar
  • 6,049
3 votes
1 answer
1k views

Canonical projection is open - Projective Space

Show that canonical projection $\pi:\mathbb{R}^{n+1}\setminus \{0\}\rightarrow \mathbb{RP}^{n}$ is open. I tried to prove it, but I have a hard time pulling the aberts. I search the internet but only ...
Mancala's user avatar
  • 816
1 vote
1 answer
34 views

Properties of a function on $\mathbb{R}P(2)$

Define a map $F:\mathbb{R}P(2)\to \mathbb{R^3}$ such that $$F([x,y,z])=\frac{(yz,xz,xy)}{{x^2+y^2+z^2}}$$ Clearly, $F$ is well defined since $F([x,y,z])=F([\lambda x,\lambda y,\lambda z])$ where $\...
Not_a_topologist's user avatar
0 votes
1 answer
55 views

Explain how $\mathbb{P}^n$ can be seen as a quotient of $\mathbb{R}^{n+1}\setminus\{0\}$ modulo an equivalence that is to be specified.

For my introductionary class in topology I have to do the following problem: Explain how $\mathbb{P}^n$ can be seen as a quotient of $\mathbb{R}^{n+1}\setminus\{0\}$ modulo an equivalence relation ...
Mitchell Faas's user avatar
-1 votes
1 answer
144 views

Equivalence relation $x \sim y$ if $x = \lambda y$ for a non-zero real number $\lambda$

Let $X = \mathbb{R}^{n}\setminus \{0\}$, where $0$ is the origin in $\mathbb{R}^n$. For $x, y \in X $ we define $x \sim y$ as follows: $x \sim y$ if there exists a nonzero real number $λ$ such that $...
Peter Lee's user avatar
0 votes
1 answer
471 views

homeomorphic quotient spaces (with the complex projective plane)

For calculations in homology we often need to know how to identify some quotient spaces. Let $\mathbb{C}P^n:=\mathbb{C}^{n+1}\setminus \{0\} /\sim$, the complex projective space where $x,y\in \...
Ryan's user avatar
  • 327
2 votes
1 answer
4k views

How to prove: Real projective plane is a manifold

I need to correct and fill the gaps in my following home work problem To show $\mathbb{R}P^2$ is a manifold, I defined the map $f:\mathbb{R}P^2\to\mathbb{D}=\{z\in\mathbb{C}:|z|\le1\}$ by $$f([x])=\...
Bumblebee's user avatar
  • 18.7k
1 vote
1 answer
265 views

Is this the basis for the topology of the complex projective plane?

I am doing this as a thought exercise to test my understanding of the quotient topology. Anyway, given an open set $U \subset \mathbb{CP}^2$, if $\pi: \mathbb{C}^3\setminus\{(0,0,0)\} \to \mathbb{CP}...
Chill2Macht's user avatar
  • 21.6k
0 votes
1 answer
87 views

"Projective" quotient of $\Bbb{Z}^2$

Consider the space of integer points $\Bbb{Z}^2=\{(x,y)|x,y\in\Bbb{Z}\}$. Consider now the equivalence relation: $$ (x,y) \sim (x',y') \quad \Leftrightarrow \quad \beta x'=\alpha x,\, \beta y'=\...
geodude's user avatar
  • 8,307
4 votes
1 answer
758 views

Struggling to understand real projective space

My ultimate goal is to show that the real projective space $\mathbb{P}^n_{\mathbb{R}}$ is an $n$-manifold. But first I'd like to understand the topological structure of $\mathbb{P}^n_{\mathbb{R}}$. ...
Kevin Sheng's user avatar
  • 2,523
15 votes
2 answers
7k views

Real Projective Plane is Same as Identifying Antipodal Boundary Points of The $2$-Disc.

$\newcommand{\RP}{\mathbf RP}$ The real projective plane $\RP^2$ is defined as the quotient space $S^2/\sim$, where $\sim$ identifies the antipodal points of $S^2$. I want to show that $\RP^2$ is ...
caffeinemachine's user avatar
4 votes
3 answers
596 views

The quotient map $q: \mathbb R^{n+1} \setminus \{0\} \to \mathbb P^n$ is open.

I'd like to show that the quotient map $q: \mathbb R^{n+1} \setminus \{0\} \to \mathbb P^n$ is open, where I'm considering $\mathbb P^n$ as the quotient space of $\mathbb R^{n+1} \setminus \{0\}$ ...
dannum's user avatar
  • 2,559