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The real projective space has been defined in my notes as ${\mathbb{P}}^n={\mathbb{R}}^{n+1}\setminus \{0\}/\sim $ where the equivalence relation identifies points of the line passing through the origin. I am trying to make sense of the charts that cover ${\mathbb{P}}^n$

It says that we should observe that for i=1,...n+1. the subsets $U_i=\{[a]=[a_{0}:\cdots :a_{n}],a_{i}\neq 0\}$ which are well defined are open.... (why is that? I haven't been able to prove it......(1))

...and they cover the space (.... (2) why isn't one of these maps enough to cover the whole projective space?

Assuming as local coordinates of a point [a] of $U_i$ the non-homogeneous coordinates $a_1/a_i,..., a_{i-1}/a_i, a_{i+1}/a_i,...,a_{n+1}/a_i ,$ we get the continuous function $x_i:U_i \to {\mathbb{R}}^n : x_i[a]= (a_1/a_i,..., a_{i-1}/a_i, a_{i+1}/a_i,...,a_{n+1}/a_i)$ whose inverse is $b \mapsto [b_1,...,b_{i-1},1,b_{i},..., b_{n}]$

(....(3). How come the definition of homogeneous coordinates is not being respected here? When I learned about homogeneous coordinates only the last coordinate was the one that had to be non-zero to allow division by it. Instead the maps they are defining here allow to pass from homogeneous to non homogeneous coordinates dividing by the i-th coordinate. How do I make sense of that?)

Can someone shed some light?

some_math_guy
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  • What did you try? Do you know how to define the topology on $\mathbb P^n$? Do you know what is the notation $[a_1, \cdots, a_{n+1}]$? – Arctic Char Nov 28 '21 at 22:03
  • @Artic Char These are all new things to me, so I have been reading about those things in several sites: projective spaces, homogeneous coordinates, and charts... still I haven't been able to make sense of these points. The topology should be the quotient topology. I have been thinking about this for several days – some_math_guy Nov 28 '21 at 22:17
  • Yes it is the quotient topology. You can use the definition of quotient topology directly to check that $U_i$ are open. – Arctic Char Nov 28 '21 at 22:18
  • @ArcticChar So $U_i$ is open if and only if its pre-image is open. So it suffices to see that the $\pi^{-1} (U_i)$ where $\pi$ is the projection map onto the quotient is open. If I an not mistaken $\pi^{-1} (U_i)= \mathbb{R}^n$ which is open. Makes sense? – some_math_guy Nov 28 '21 at 22:46
  • I meant $\mathbb{R}^{n+1}$ – some_math_guy Nov 28 '21 at 22:53
  • No. At the very least it does not contain the origin.... – Arctic Char Nov 28 '21 at 23:05
  • @ArcticChar Do you mean that since $\pi: {\mathbb{R}}^{n+1}\setminus {0} \to {\mathbb{P}}^n={\mathbb{R}}^{n+1}\setminus {0}/\sim $, then $\pi^{-1}(U_i )= {\mathbb{R}}^{n+1}\setminus {0}$? – some_math_guy Nov 28 '21 at 23:13
  • Your equality is wrong. How do you get to that? Did you use the condition $a_i\neq 0$? – Arctic Char Nov 28 '21 at 23:18
  • Or can you write down explicitly $\pi$? – Arctic Char Nov 28 '21 at 23:19
  • $\pi( (a_0,...,a_n)) = [(\lambda a_0, ... ,\lambda a_n)]=[( a_0, ... , a_n)]$ – some_math_guy Nov 28 '21 at 23:26
  • What is $\lambda$ and how do you get to $\mathbb R^{n+1}$ from this? – Arctic Char Nov 28 '21 at 23:28
  • lambda is any non-zero real number – some_math_guy Nov 28 '21 at 23:29
  • So how do you calculate? – Arctic Char Nov 28 '21 at 23:30
  • $\pi( (a_0,...,a_n)) = [( a_0a_i, ...,a_{i-1}a_i, a_i ,a_{i+1}a_i,... , a_na_i)]$. I have to add one component that is never zero(for points not at infinity) in the i-th position – some_math_guy Nov 28 '21 at 23:33
  • I am not sure if homogeneous coordinates , which have one more component are being used only in the codomain or in the donain as well of the function pi – some_math_guy Nov 28 '21 at 23:41
  • Why do you have to do that? Just use the definition – Arctic Char Nov 28 '21 at 23:41
  • $\pi(a)=[a]$, but I don't know if a and [a] have n, n+1 or n+2 components since homogeneous coordinates have should add a component. So a should have n+1 components and [a] should have n+2 components – some_math_guy Nov 28 '21 at 23:43

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Your question does not deal with charts on $\mathbb P^n$, but with open sets covering $\mathbb P^n$. These sets may be taken as the domains of charts, but that is another question.

The simplest way to answer your question is to observe that the quotient map $\pi : \mathbb R^{n+1} \setminus \{0\} \to \mathbb P^n$ restricts to a quotient map $\phi = \pi \mid_{S^n} : S^n \to \mathbb P^n$. This is easily proved; see also When is the restriction of a quotient map $p : X \to Y$ to a retract of $X$ again a quotient map?

$\phi$ identifies pairs of antipodal points $x, -x$. It is an open map: Let $U \subset S^n$ be open. We have $\phi^{-1}(\phi(U)) = U \cup (-U)$, where $-U = \{-x \mid x \in U\}$. The antipodal map $a : S^n \to S^n,a (x) = -x$, is a homeomorphism, thus $-U = a(U)$ is open. We conclude that $\phi^{-1}(\phi(U))$ is open, and therefore $\phi(U)$ is open because $\phi$ is a quotient map.

For $j = 1,\ldots,n+1$ define $U_j^\pm = \{(x_1,\ldots,x_{n+1}) \in S^n \mid (-1)^{\pm 1} x_j > 0 \}$; these are $2(n+1)$ open sets which cover $S^n$. We have $\phi(U_j^+) = \phi(U_j^-) = \{ [x_1 : \dots : x_{n+1}] \mid x_j \ne 0 \}$ and these sets (which we denote by $U_j$) are open in $\mathbb P^n$. Clearly they cover $\mathbb P^n$. This answers $(1)$ and $(2)$. Note that all $n+1$ of these sets are needed to cover $\mathbb P^n$ because the points $[x_1 : \dots : x_{n+1}]$ with $x_i = 0$ for $i \ne j$ and $x_j = 1$ are contained only in $U_j$ and in no other $U_i$.

Concerning $(3)$: You say "When I learned about homogeneous coordinates only the last coordinate was the one that had to be non-zero to allow division by it". This is not true, the points having this property lie in $U_{n+1}$. But the points $[x_1 : \dots : x_n:0]$ do not lie in $U_{n+1}$. In fact, division by $x_j$ makes sense precisely when $[x_1 : \dots : x_{n+1}] \in U_j$.

Paul Frost
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  • What I meant concerning (3) is that as you said $[x_1 : \dots : x_n:0]$ are the points at infinity , with $x_1 : \dots : x_n$ not all zero and $[x_1 : \dots : x_n: x_{n+1}]$ with$ x_{n+1} \neq 0$ are the other ones. So aren't these all of the points in $\mathbb{P}^n$? Why do we need to considered n+1 sets where each component is not zero?Those should already be considered in these two sets – some_math_guy Nov 29 '21 at 00:00
  • Another thing, $[x_1 : \dots : x_n:0]$ is a point in the projective space, but shouldn't it have n+2 components? I saw some exercises of conversion from non-homogeneous to homogeneous coordinates, and a coordinate is added So when passing from R^n+1 where points are represented with non-homogeneous coordinates to the quotient where homogeneous coordinates are used, I was expecting that classes in the quotient have n+2 components. How do I make sense of it? – some_math_guy Nov 29 '21 at 00:08
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    The points of $\mathbb P^n$ are equivalence classes of points of $\mathbb R^{n+1} \setminus {0}$. These have $n+1$ (non-homogeneous) coordinates, thus the points of $\mathbb P^n$ have $n+1$ homogeneous coordinates, not $n+2$. Just count in $[x_1 : \dots : x_{n+1}]$. In fact the points $[x_1 : \dots : x_{n+1}]$ with $x_{n+1} \ne 0$ (i.e. the points in $U_{n+1}$) can be identified with the points of $\mathbb R^n$; just use the function in $(3)$ of your question. The set of points $[x_1 : \dots : x_n:0]$ at infinity form a set $N_{j+1}$ which is the complement of $U_{n+1}$ in $\mathbb P^n$. – Paul Frost Nov 29 '21 at 00:36
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    But $N_{j+1}$ is closed and not open. Thus it cannot be the domain of a chart. In fact, you need all of the open set $U_1,\ldots, U_n$ to cover $N_{j+1}$. – Paul Frost Nov 29 '21 at 00:39