# Thread: Topological Manifolds - Charts on Real Projective Spaces

1. I am reading John M. Lee's book: Introduction to Smooth Manifolds ...

I am focused on Chapter 1: Smooth Manifolds ...

I need some help in fully understanding Example 1.3: Projective Spaces ... ...

My questions are as follows:

Question 1

In the above example, we read:

" ... ... define a map $\displaystyle \phi_i \ : \ U_i \longrightarrow \mathbb{R}^n$ by

$\displaystyle \phi_i [ x^1, \ ... \ ... \ , x^{n+1} ] = ( \frac{x^1}{x^i} , \ ... \ , \frac{x^{i-1}}{x^i} , \frac{x^{i+1}}{x^i}, \ ... \ , \frac{x^{n+1}}{x^i} )$

This map is well defined because its value is unchanged by multiplying x by a nonzero constant. ... ... "

Now, in the above, the domain of $\displaystyle \phi_i$ is shown as an $\displaystyle (n+1)$-dimensional point ... ... BUT ... ... $\displaystyle \phi_i$ is a map with a domain consisting of lines in $\displaystyle \mathbb{R}^{n + 1}$, so shouldn't the dimension of the domain be $\displaystyle n$ ... ?

Maybe we have to regard the equivalence classes of the quotient topology involved as $\displaystyle (n+1)$-dimensional points and recognise that points $\displaystyle x = \lambda x$ where $\displaystyle \lambda \in \mathbb{R}$ ... ... is that right?

(The statement about the map being well defined is presumably about recognising equivalence classes as on point in the projective space ... ... is that right?

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Question 2

In the above text from Lee's book we read:

"... ... Because $\displaystyle \phi_i \circ \pi$ is continuous ... ... "

How do we know that $\displaystyle \phi_i \circ \pi$ is continuous ... ?

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Question 3

In the above text from Lee's book we read:

"... ... In fact $\displaystyle \phi_i$ is a homeomorphism, because its inverse is given by

$\displaystyle {\phi_i}^{-1} [ u^1, \ ... \ ... \ , u^{n} ] = [ u^1, \ ... u^{i-1}, 1, u^i, \ ... \ , u^{n} ]$

as you can easily check ... ... "

I cannot see how Lee determined this expression to be the inverse ... why is the inverse of $\displaystyle \phi_i$ of the form shown ... how do we get this expression ... and why is it continuous (as it must be since Lee declares $\displaystyle \phi_i$ to be a homeomorphism ... ...

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Question 4

Just a general question ... in seeking a set of charts to cover $\displaystyle \mathbb{RP}^n$, why does Lee bother with the $\displaystyle \tilde{U_i}$ and $\displaystyle \pi$ ... why not just define the $\displaystyle U_i$ as an open set of $\displaystyle \mathbb{RP}^n$ and define the $\displaystyle \phi_i$ ... ... ?

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Hope someone can help with the above three questions ...

Help will be appreciated ... ...

Peter

2. Originally Posted by Peter
Question 1

In the above example, we read:

" ... ... define a map $\displaystyle \phi_i \ : \ U_i \longrightarrow \mathbb{R}^n$ by

$\displaystyle \phi_i [ x^1, \ ... \ ... \ , x^{n+1} ] = ( \frac{x^1}{x^i} , \ ... \ , \frac{x^{i-1}}{x^i} , \frac{x^{i+1}}{x^i}, \ ... \ , \frac{x^{n+1}}{x^i} )$

This map is well defined because its value is unchanged by multiplying x by a nonzero constant. ... ... "

Now, in the above, the domain of $\displaystyle \phi_i$ is shown as an $\displaystyle (n+1)$-dimensional point ... ... BUT ... ... $\displaystyle \phi_i$ is a map with a domain consisting of lines in $\displaystyle \mathbb{R}^{n + 1}$, so shouldn't the dimension of the domain be $\displaystyle n$ ... ?
What is an $(n+1)$-dimensional point, and what is the dimension of a domain? It appears you are using the term 'dimension' very loosely here. The only concern that could arise here is that the maps $\phi_i$ may not be well-defined.

Originally Posted by Peter
(The statement about the map being well defined is presumably about recognising equivalence classes as on point in the projective space ... ... is that right?
It's about making sure that the definition of $\phi_i$ does not depend on the choice of a representative of an equivalence class.

Originally Posted by Peter
Question 2

In the above text from Lee's book we read:

"... ... Because $\displaystyle \phi_i \circ \pi$ is continuous ... ... "

How do we know that $\displaystyle \phi_i \circ \pi$ is continuous ... ?
Since each component of $\phi_i \circ \pi$ is a continuous function, then $\phi_i \circ \pi$ is continuous.

Originally Posted by Peter
Question 3

In the above text from Lee's book we read:

"... ... In fact $\displaystyle \phi_i$ is a homeomorphism, because its inverse is given by

$\displaystyle {\phi_i}^{-1} [ u^1, \ ... \ ... \ , u^{n} ] = [ u^1, \ ... u^{i-1}, 1, u^i, \ ... \ , u^{n} ]$

as you can easily check ... ... "

I cannot see how Lee determined this expression to be the inverse ... why is the inverse of $\displaystyle \phi_i$ of the form shown ... how do we get this expression ... and why is it continuous (as it must be since Lee declares $\displaystyle \phi_i$ to be a homeomorphism ... ...
Let $[x^1,\ldots, x^{n+1}]\in U_i$ and set $(x^1/x^i,\cdots,x^{i-1}/x^i,x^{i+1}/x^i,\ldots, x^{n+1}/x^i) = (u^1,\ldots, u^n)$. Then $$x^1 = x^iu^1,\ldots, x^{i-1} = x^iu^{i-1}, x^{i+1} = x^iu^i,\ldots, x^{n+1} = x^iu^n.$$ So $$[x^1,\ldots, x^{n+1}] = [x^iu^1,\ldots, x^{i-1}u^{i-1},x^i,x^{i}u^{i+1},\ldots, x^iu^n] = [u^1,\ldots, u^{i-1},1,u^i,\ldots, u^n].$$ Hence, the solution of the equation $\phi_i[x^1,\ldots, x^{n+1}] = (u^1,\ldots, u^n)$ is $[x^1,\ldots, x^{n+1}] = [u^1,\ldots,u^{i-1},1,u^i,\ldots, u^n]$. This tells us how to define $\phi_i^{-1}$.

Now $\phi_i^{-1}$ is the composition of continuous functions: $\phi_i^{-1} = \pi \circ f_i$, where $f_i : (u^1,\ldots, u^n) \to (u^1,\ldots, u^{i-1},1,u^i,\ldots, u^n)$. As the composition of continuous functions is continuous, $\phi_i^{-1}$ is continuous.

Originally Posted by Peter
Question 4

Just a general question ... in seeking a set of charts to cover $\displaystyle \mathbb{RP}^n$, why does Lee bother with the $\displaystyle \tilde{U_i}$ and $\displaystyle \pi$ ... why not just define the $\displaystyle U_i$ as an open set of $\displaystyle \mathbb{RP}^n$ and define the $\displaystyle \phi_i$ ... ... ?
That's probably a question to ask him about, but I think the kind of conciseness you're proposing would be lost to several readers of his book. The projective space is given the quotient topology, so having the quotient map $\pi$ is a necessity.