The 2-Dimensional Projective Space $\mathbb{P}^2$ Has the Basis with *Three* (?) Vectors $\mathbf{e}_i$, $i = 1, \dots, 3$?

Question

My textbook says the following:

Consider a set of basis vector $\mathbf{e}_i$, $i = 1, \dots, 3$ for a 2-dimensional projective space $\mathbb{P}^2$. For reasons to become clear, we will write the indices as subscripts. With respect to this basis, a point in $\mathbb{P}^2$ is represented by a set of coordinates $x^i$, which represents the point $\sum_{i = 1}^3 \mathbf{e}_i$. We write the coordinates with an upper index, as shown. Let $\mathrm{\mathbf{x}}$ represent the triple of coordinates, $\mathrm{\mathbf{x}} = (x^1, x^2, x^3)^T$.

My understanding from linear algebra is that a basis is a linearly independent spanning set. We then define the dimension of a vector space to be equal to the number of vectors in a basis for said vector space. Therefore, a vector space can have multiple different bases, but each basis must contain the same number of vectors, since this defines the dimension of the vector space.

So how is it that in the above excerpt the author says that the 2-dimensional projective space $\mathbb{P}^2$ has the basis with three vectors $\mathbf{e}_i$, $i = 1, \dots, 3$? Shouldn't the basis consist of only 2 vectors, since this is a 2-dimensional vector space?

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

Projective space is not a vector space, so the linear algebraic notions of dimension and basis do not apply in this case. However, one calls it two-dimensional because it is when we consider alternative notions of dimension, e.g. when we look at $\mathbb P^2$ over $\mathbb R$ then this will be a two-dimensional manifold, or over an arbitrary field we can still formulate an algebro-geometric notion of dimension.

Intuitively it should not be a surprise that $\mathbb P^2$ is two-dimensional: Although you use three coordinates to describe a point, these are not independent from each other: Any point $x=(x_0:x_1:x_2)$ has a non-zero entry, say $x_0\neq 0$. Then dividing by $x_0$ gives an alternative representation of the same point $x=(1:x_1/x_0:x_2/x_0)$, so in fact "locally" you only need two numbers to describe a point, i.e. you only have two degrees of freedom.

Answer 2

What is the projective plane $\Bbb P^2$ and indeed the projective space $\Bbb P^n$ for every $n$?

Start with a vector space $V$ of dimension $n+1$ and let $V^\prime=V\setminus\{0\}$. Then the projective space associated to the vector space $V$ is, by definition, $$ \Bbb P(V)=V^\prime/\sim $$ where $\sim$ denotes the homothetic equivalence, namely $$ v_1\sim v_2\Longleftrightarrow\text{$v_1=\lambda v_2$ for some $\lambda\in F^\times$} $$ (here $F$ denotes the field of coefficients of the vector space $V$). In geometric terms $\Bbb P(V)$ is the set of lines through $0$ in $V$.

Thus any point $P\in\Bbb P(V)$ is represented by a vector $v\in V$ and if we have fixed a basis $(e_1,...,e_{n+1})$ of $V$ we can associate to $P$ the coordinates of $(x_1,...,x_{n+1})$ of $v$ with respect to the chosen basis.

Of course this is not fully well-defined: if we choose a different representant $v^\prime$ of $P$, we will have $v^\prime=\lambda v$ for some $\lambda\in F^\times$ and so $$ x_i^\prime=\lambda x_i\qquad i=1,...,n+1 $$ where $(x_1^\prime,...,x_{n+1}^\prime)$ are the coordinates of $v^\prime$. Thus, the coordinates of $P$ are defined only up to a common non-zero factor: one speaks of homogeneous coordinates of $P$ (classically they're numbered from $0$ to $n$).

To see that $\Bbb P(V)$ has dimension $n$, notice that since $0\notin V^\prime$ no point $P$ has coordinates all $0$'s. Thus a point $P$ will have some coordinate not zero and we may assume it is the last. Thus, up to rescaling coordinates, $P$ will have coordinates of the form $$ P\equiv(y_1,...,y_n,1). $$ Moving $P$ around a little will leave the last coordinate fixed (again, up to rescaling) and the other $y$'s changing arbitrarily. But that is describing an affine space of dimension $n$, so $\Bbb P(V)$ is locally of dimension $n$.