Hausdorff property of projective space

  • A
  • Thread starter Korybut
  • Start date
  • #1
Korybut
48
1
Hello!

I am reading "Differential Geometry and Mathematical Physics" by Rudolph and Schmidt. And they have and example of manifold (projective space). I believe that there is a typo in the book, but perhaps I miss something deep.
Definitions are the following
$$\mathbb{K}^n_\ast=\{\mathbf{x}\in \mathbb{K}^n : \mathbf{x}\neq 0\},\;\; \mathbb{K}^n_1=\{\mathbf{x}\in \mathbb{K} : \| x \|=1\}$$

They pick up two representatives ##\mathbf{x}\neq \mathbf{y}## both with unit norm ##\| \mathbf{x}\|=\|\mathbf{y}\|=1## in ##\mathbf{K}^{n+1}_\ast##

To proceed with the neighborhoods they define function
$$l(\mathbf{x},\mathbf{y}):=\min_{\lambda \in \mathbb{K}_1} \|\mathbf{x} \lambda-\mathbf{y}\|$$
Then the pre-images of the neighborhoods are defined as follows
$$K_{\mathbf{x}}:=\{ \mathbf{z}\in \mathbb{K}^{n+1}_\ast : \max_{\lambda \in \mathbb{K}_1} \| \frac{\mathbf{z}}{\|\mathbf{z}\|}\lambda-\mathbf{x}\|<\frac{1}{2} l(\mathbf{x},\mathbf{y})\} \in \mathbb{K}^{n+1}_\ast$$

I've considered the case ##\mathbb{K}=\mathbb{R}## and ##\mathbb{R}P^1##. Then there are only two possibilities for ##\lambda=\pm 1##. And I found that set ##K_{\mathbf{x}}## is simply empty. This ##\max## in definition of ##K_{\mathbf{x}}## looks like a typo. Am I right?
 

Answers and Replies

  • #2
martinbn
Science Advisor
3,051
1,391
For ##\mathbb RP^1## you need ##\mathbb K^2=\mathbb R^2##.
 
  • #3
Korybut
48
1
For ##\mathbb RP^1## you need ##\mathbb K^2=\mathbb R^2##.
I do take ##\mathbb{R}^2##.

Suppose
$$\mathbf{x}=\left(\begin{matrix} 1 \\ 0 \end{matrix}\right),\;\; \mathbf{y}=\left(\begin{matrix} 0 \\ 1 \end{matrix} \right)$$
then this ##l## function is equal to ##\sqrt{2}##. On the other hand ##\max## in the definition of ##K_{\mathbf{x}}## takes values in ##[\sqrt{2},2]## for all possible choices of ##\mathbf{z}##. There is no way it can be less then ##\frac{1}{\sqrt{2}}##.
 
  • #4
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
5,469
1,402
That doesn't seem wrong to me. I think the neighborhood you are describing around ##x## is all lines that make at most a 90 degree angle with it. But every line makes at most a 90 degree angle (since both sides of the line map to the same point in protective space)

Edit: whoops, you said the set is empty, not everything. Ignore this
 
  • #5
martinbn
Science Advisor
3,051
1,391
I do take ##\mathbb{R}^2##.

Suppose
$$\mathbf{x}=\left(\begin{matrix} 1 \\ 0 \end{matrix}\right),\;\; \mathbf{y}=\left(\begin{matrix} 0 \\ 1 \end{matrix} \right)$$
then this ##l## function is equal to ##\sqrt{2}##. On the other hand ##\max## in the definition of ##K_{\mathbf{x}}## takes values in ##[\sqrt{2},2]## for all possible choices of ##\mathbf{z}##. There is no way it can be less then ##\frac{1}{\sqrt{2}}##.
I see, you meant the the field is the reals.

It does look suspicious. What do they mean by ##\mathbb K_1##? Is it ##\mathbb K^n_1## for ##n=1## or something else?
 
  • #6
Korybut
48
1
Authors did not provide any specific definition for ##\mathbb{K}_1## so I think ##\mathbb{K}_1=\mathbb{K}_1^1##.
I believe it should be ##\min## in the definition of ##K_{\mathbf{x}}##.
 
  • #7
Leobardo
1
0
Hello!

I am reading "Differential Geometry and Mathematical Physics" by Rudolph and Schmidt. And they have and example of manifold (projective space). I believe that there is a typo in the book, but perhaps I miss something deep.
Definitions are the following
$$\mathbb{K}^n_\ast=\{\mathbf{x}\in \mathbb{K}^n : \mathbf{x}\neq 0\},\;\; \mathbb{K}^n_1=\{\mathbf{x}\in \mathbb{K} : \| x \|=1\}$$

They pick up two representatives ##\mathbf{x}\neq \mathbf{y}## both with unit norm ##\| \mathbf{x}\|=\|\mathbf{y}\|=1## in ##\mathbf{K}^{n+1}_\ast##

To proceed with the neighborhoods they define function
$$l(\mathbf{x},\mathbf{y}):=\min_{\lambda \in \mathbb{K}_1} \|\mathbf{x} \lambda-\mathbf{y}\|$$
Then the pre-images of the neighborhoods are defined as follows
$$K_{\mathbf{x}}:=\{ \mathbf{z}\in \mathbb{K}^{n+1}_\ast : \max_{\lambda \in \mathbb{K}_1} \| \frac{\mathbf{z}}{\|\mathbf{z}\|}\lambda-\mathbf{x}\|<\frac{1}{2} l(\mathbf{x},\mathbf{y})\} \in \mathbb{K}^{n+1}_\ast$$

I've considered the case ##\mathbb{K}=\mathbb{R}## and ##\mathbb{R}P^1##. Then there are only two possibilities for ##\lambda=\pm 1##. And I found that set ##K_{\mathbf{x}}## is simply empty. This ##\max## in definition of ##K_{\mathbf{x}}## looks like a typo. Am I right?
 
  • #8
mathwonk
Science Advisor
Homework Helper
11,392
1,631
yes "max" seems to be a typo, and should have been "min" as you say. the mysterious K1 is apparently meant to be the non zero scalars, or even just plus and minus 1 will do, I think. The point is to take a nbhd consisting of points represented by vectors on a line which is near the given representative. There are many representatives of each "point", so measure nearness by the nearest one. They have essentially said, given two unit length representatives x,y, of points p,q, define their distance d as the closest distance between y and any representative of the point p. (Odd that they choose an asymmetric looking, but not actually asymmetric, definition). Then take as neighborhood of p, any point having a representative closer to x than half of that distance. Or, since they have stated it in terms of representatives, take as neighborhood preimage, any representative z such that the closest point of the line through z, is closer to x than d/2.

This is about as unintuitive and cloaked a description as I have ever seen. Since points are lines, they could have just defined the distance between two points p,q as the minimum distance d between any two unit length representatives of these two lines, then defined the neighborhood of one point p, to be all lines closer to p, in this sense, than d/2.

I think this is right, a least without spending any more time on it, and that you have already understood it correctly.

Annoying typos like that are made more likely when the writing is so unclear and unfriendly to the reader. I.e. then even the proof reader, and the authors themselves, cannot get it right. On the other hand, they wopuld probably appreciate a remark from you correcting their error, as writing a readable book is extremely challenging.
 
  • #9
WWGD
Science Advisor
Gold Member
6,291
8,174
Maybe there is a point-set theorem out there on when quotients respect the Hausdorff property.
 
  • #10
mathwonk
Science Advisor
Homework Helper
11,392
1,631
I suspect the OP does not need any help, but since this has been on my mind, I wish to get it off. The simplest way to visualize the projective space as a manifold is to represent each point of projective space as a pair of antipodal points of the unit sphere. As WWGD suggests, this represents projective space as a quotient of the unit sphere by an action of the group Z/2Z. An epsilon ball around a point of projective space is then given by a pair of epsilon balls, one around each antipodal point. Then since it is obvious that any finite number of points of the sphere possesses disjoint epsilon balls, one sees that given any two pairs of antipodal points, there is an epsilon such that all 4 epsilon balls centered at these points are disjoint. Hence it is obvious that projective space is hausdorff, and a manifold too.

edit: To consider the idea of WWGD, indeed it seems plausible that any quotient of a hausdorff manifold by a finite group of automorphisms (homeomorphisms) is a hausdorff space, and if the action is without fixed points, is also locally euclidean, hence a hausdorff manifold.
 
Last edited:

Suggested for: Hausdorff property of projective space

Replies
27
Views
2K
  • Last Post
Replies
1
Views
976
Replies
1
Views
2K
Replies
3
Views
672
  • Last Post
Replies
0
Views
432
Replies
7
Views
935
Replies
22
Views
2K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
0
Views
293
Top