Projective space and the differentiable structure. Examples in DoCarmo.

Question

I'm studying the DoCarmo's Riemannian Geometry more specifically, I'm working in the projective space $P^{n}(\mathbb{R})=\{\text{The set of straight lines of }\hspace{2mm}\mathbb{R}^{n+1}\hspace{2mm} \text{which pass through the origin } \}$, (Example 2.4). In this example Docarmo does the following

1. He established the equivalence relation $(x_{1}, x_{2}, \cdots, x_{n})\sim{(\lambda x_{1},\cdots,\lambda x_{n})}$ and considered $(\mathbb{R}^{n}\setminus\{0\})/\sim{}$.
2. He defined the neighborhoods $V_{i}=\{[x_{1},x_{2},\cdots,x_{n}]\vert\hspace{2mm}x_{i}\neq 0\}$.
3. Defined the following mappings $x_{i}:\mathbb{R}^{n}\to V_{i}$ given by $x_{i}(x_{1} ,x_{2},\cdots, x_{n})=[x_{1},\cdots, y_{i},1,y_{i-1}, \cdots, y_{n}]$.

With this steps DoCarmo says that $P^{n}(\mathbb{R})$ is a differentiable manifold. But my doubt is the following: For me with the steps 1,2,3 he just only giving structure of differentiable manifold for $(\mathbb{R}^{n}\setminus\{0\})/\sim{}$ and not for $P^{n}(\mathbb{R})$, so we need to do other step for have the structure for $P^{n}(\mathbb{R})$. It is true or not?

In similar way in the example 4.7 (another description of projective space) He established a equivalence relation on the sphere that identifies $p \in\mathbb{S}^{n}$ with its antipodal point , $A(p)=-p$ and consider $\mathbb{S}^{n}/\sim$. Later he considered the following function $\pi: \mathbb{S}^{n}\to P^{n}(\mathbb{R})$ the canonical projection, i.e., $\pi(p)=\{p,-p\}$ and here again I'm stuck since I'm not sure if $\pi: \mathbb{S}^{n}\to P^{n}(\mathbb{R})$ or $\pi: \mathbb{S}^{n}\to \mathbb{S}^{n}/\sim$.

Thanks in advance!

Answer 1

I think your confusion can be solved by explaining what is sometimes called the transfer of structure.

Abstractly.

Suppose $X$ and $Y$ are two sets, $f\colon X\to Y$ is a bijection, and assume furthermore that $X$ is endowed with some structure. Then, through $f$, we can consider that $Y$ is endowed with the same structure, such that $f$ becomes an isomorphism of this structure.

This is a vague notion: the way we transfer the structure depends on the context.

Concrete example: if $X$ is a group.

Suppose that $X$ is a group, with neutral element $e_X$ and with group law $\cdot_X$. Then one can define $$ \forall a,b\in Y,\quad a\cdot_Y b = f\left( f^{-1}(a)\cdot_Xf^{-1}(b)\right), $$ and check that $(Y,\cdot_Y)$ is a group with neutral element $e_Y=f(e_X)$. Furthermore, $f\colon X\to Y$ is now a group isomorphism.

What has been done is the following: to multiply two elements in $Y$, just look at them in $X$ through $f$, multiply them in $X$ (since we know how to do that here), and send the result in $Y$ thanks to $f$ again.

What does it have to do with your question?

Assume from now that $X$ is endowed with a topology $\mathcal{O}_X = \{U \subset X \mid U \text{ is open}\}$. Then $\{f(U) \mid U \subset X \text{ is open}\}$ is now a topology on $Y$, and $f\colon X\to Y$ is an homeomorphism.

Assume now that $X$ is endowed with a differentiable structure $\{(U_i,\varphi_i)\}$. Then you can check that $\{(f(U_i),\varphi_i\circ f^{-1}\}$ is now a differentiable structure on $Y$, making $f$ into a diffeomorphism. In fact, when looking at the transition functions, $f$ "magically" disappears, since $$ (\varphi_i\circ f^{-1}) \circ \left( \varphi_j\circ f^{-1}\right)^{-1} = \varphi_i\circ f^{-1} \circ (f^{-1})^{-1} \circ \varphi_j^{-1} = \varphi_i\circ f^{-1}\circ f \circ \varphi_j^{-1} = \varphi_i \circ \varphi_j^{-1}, $$ and this is the key ingredient in the proof.

So what Do Carmo is in fact doing is defining a differentiable structure on $\left(\Bbb R^{n+1}\setminus\{0\} \right)/\sim$, and transferring it to $\Bbb RP^n$ via the bijection $$ [x] \in \left(\Bbb R^{n+1}\setminus\{0\} \right)/\sim \quad \longmapsto \mathrm{span}\left(\{x\}\right) \in \Bbb RP^n. $$ This uniquely determines a differentiable structure on the projective space.

Answer 2

Lets proceed geometrically starting in $\mathbb{R}^{n+1}\setminus\{0\}$. Two points are in the same equivalence class if and only if they're on the same line through the (missing) origin so the first condition tells us we're working in a projective space. This lets us identify each line with it's equivalence class to ensure uniqueness. The reason we want $x_i \neq 0$ for our chart is that it allows us to identify the point of intersection between a line and the hyperplane $x_i=1$ to represent the line uniquely. This chart misses all the lines parallel to $x_i=1$ but since $0$ is not in the set we can find some $i$ that works. This is also where we lose a dimension since the hyperplane will look like $\mathbb{R}^n$ rather than $\mathbb{R}^{n+1}$. This is the representation the OpenGL standard uses for computer graphics since it's useful when calculating vanishings points of a perspective projection. Historically this is where the projective transform got its name.

Lets look at $\mathbb{R}^3\setminus \{0\}$ and $i=3$ so that every point $(x_1,x_2,x_3)$ in question is represented by $(x_1,x_2,1)$. While it's not the usual vector normalization you can see the main idea here, divide by $x_i\neq 0$ and we have the point of intersection. However points like $(1,2,0)$ require a different chart because there is no scalar $\lambda$ such that $\lambda (1,2,0)=(x_1,x_2,1)$. You can see that it's in the plane $x_3=0$ which is parallel to $x_3=1$ and so needs a different chart.

This faux normalization motivates looking at vector normalization and the map $x/\vert x \vert$ that takes $x$ to a point on the unit sphere in $\mathbb{R}^{n+1}$. This map gives us two points to represent each line and we solve it in the same way we did previously, by defining and equivalence relation so that we talk about the class $\{x,-x\}$ instead. this ensure we have a unique representative for each line in projective space. Your projection operation $\pi$ will be from $x$ and $-x$ to $\{x,-x\}$. In this case picking a representative is trivial since the distinction between $x$ and $-x$ is arbitrary so we don't have to be as picky when choosing one.

Many of the most interesting properties of projective space relate to how the antipodal gluing doing strange things. Like if you physically try to glue all the antipodal points together on a sphere it doesn't take long to figure out that you're going to get some twisting and self-intersection and you can start to see why it's not the same as the half sphere which merely looks like $\mathbb{R}^n$.

In both cases we can see the loss of dimension, either as points in a hyperplane or as points on the surface of a hypersphere. There is a geometric intuition as to why it's only one dimension as well that I would encourage to think on.