Projective Space vs. Vector Space

Question

In my geometry course we defined the projective space $\mathbb{P}(V)$ of a vector space $V$ as

$\mathbb{P}(V) := \{ U \subseteq V \, | \, U \text{ is a vector subspace of dimension } 1 \}$.

Let's observe $\mathbb{P}(\mathbb{R}^2)$. This is the set of all lines passing through the origin. In my imagination these lines cover $\mathbb{R^2}$. So instead of thinking about $(x, y)^T \in \mathbb{R}^2$ as a point only, I can think about them as a generator, of one of the lines in $\mathbb{P}(\mathbb{R}^2)$. [Beside the origin itself...]

Why can I not just say that $\mathbb{R}^2$ "is the same" as $\mathbb{P}(\mathbb{R^2})$? [Or at least a subset/superset]

I would like to understand the differences between the projective space and the vector space itself. First of all, what is the difference?

Bonus-Question:

Does the answer to the above question depends on the vector space? What happens if $V$ would be $\mathbb{C}^n$, or $\mathbb{F}_q^n$ [Field with $q$ elements], or a set of linear functions, etc. ?

Answer 1

You can't identify $\mathbb{P}(\mathbb{R}^2)$ with $\mathbb{R}^2$, but there's something like you'd like to think. $\def\vv#1#2{\left(\begin{smallmatrix}#1\\#2\end{smallmatrix}\right)} \def\ww#1#2{\begin{pmatrix}#1\\#2\end{pmatrix}}$

Consider, over the set $\mathbb{R}^2\setminus\{\vv{0}{0}\}$ the relation $$ \ww{a}{b}\sim\ww{c}{d} \text{ if and only if there exists $\rho\ne0$ such that } \ww{c}{d}=\rho\ww{a}{b} $$ This is an equivalence relation over the set. Note that each equivalence class corresponds to a single vector subspace of $\mathbb{R}^2$, the one generate by any element in the equivalence class.

Then you can “identify” $\vv{x}{y}$ with a point in the projective space, provided you “don't distinguish” between, say, $\vv{1}{2}$ and $\vv{\smash-1}{\smash-2}$ or $\vv{3}{6}$.

These are indeed the homogeneous coordinates, often written $(1:2)$ to mean the equivalence class of $\vv{1}{2}$. Thus $(1:2)=(-1:-2)=(3:6)$.

You can do similarly when dealing with any vector space. If you fix a basis of $V$, then a one-dimensional subspace is identified by any of its nonzero vectors; if you take the coordinates of such a nonzero vector with respect to the chosen basis, you get the homogeneous coordinates with respect to the coordinate system induced by the chosen basis.

What you get is actually a bijection from $\mathbb{P}(V)$ to the quotient set $(\mathbb{R}^n\setminus\{0\})/{\sim}$, where the equivalence relation is defined as before.

In the simple case of $\mathbb{P}(\mathbb{R}^2)$, there is a basis involved, namely the canonical basis. Choosing a different basis will induce a different coordinate system on the projective line.

Is $\mathbb{R}$ relevant? No, everything works the same over any base field.

Answer 2

Why can I not just say that $\mathbb{R}^2$ "is the same" as $\mathbb{P}(\mathbb{R^2})$?

Simple, because it is not the same. For one, $\mathbb R^2$ has the zero vector which doesn't generate any line (or rather, it generates too many of them). Also, $(1,1)$ and $(2,2)$ are two points in $\mathbb R^2$, but they generate only one line.

Answer 3

Well, the affine space ${\Bbb R}^2$ is a subset of $P^2({\Bbb R})$. If you are familiar with homogeneous coordinates, $P^2({\Bbb R})$ has the following subsets:

$$\{(a:b:1)\mid a,b\in {\Bbb R}\}$$ which corresponds bijectively to ${\Bbb R}^2$ by matching $(a:b:1)$ with $(a,b)$,

$$\{(a:1:0)\mid a\in {\Bbb R}\}$$ which is a line at infinity, and

$$\{(1:0:0)\}$$ which is a point at infinity.

Notice the normalization of the last coordinate to $1$.

The above definition of the projective plane works over any field.