Hausdorff property of projective space

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  • Thread starter Korybut
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  • #1
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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
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For ##\mathbb RP^1## you need ##\mathbb K^2=\mathbb R^2##.
 
  • #3
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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
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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
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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
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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}}##.
 

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