Projective Geometry interpretation of a set

Question

I have the following example: Lets see the projective straight line $l \subset \mathbb{P^2}$ that goes through the points $[1:1:0]$ and $[1:0:1]$ (first is this any diferent that $(1,1,0)$ and $(1,0,1)$?)

The example starts by finding the subspace generated by the vectors $(1,1,0)$ and $(1,0,1)$, and they reach the equation: $-x_0+x_1+x_2=0$

Then, it concludes that: $l=\{[x_0:x_1:x_2]\in\mathbb{P^2}:-x_0+x_1+x_2=0\}$

My question is: what exactly is this last set? (And a intuition for it). Isn't this a plane instead of a line?

I'm having some trouble finding out the meaning of apparently elementary definitions of Projective Geometry. If there are some notes out there that could help me getting the first intuition about it, I would be grateful if you could share them to me.

Answer 1

Notation 1:1:0 is better suited for projective geometry than 1,1,0 because 1:1:0 is equivalent to 2:2:0, and to 99:99:0, etc. One can say that ':' stands for a ratio.

A line on a projective plane can be represented by a plane in $\mathbb R^3$ which contains the origin (0,0,0), i.e. $ax_1+bx_2+cx_3=0$, and $a,b,c$ are not uniquely defined but can be multiplied by any non-zero scalar, i.e., for any $\alpha != 0$ the set $(\alpha a,\alpha b, \alpha c)$ defines the same plane, and thus the same projective line.

If you plug in the values of your two points into the plane equation you'll get two equations for $a,b,c$, and $(-1,1,1)$ will be one of the possible (non-unique) solutions, produce the plane equation $-1*x_1+1*x_2+1*x_3=0$

Answer 2

The points of $\mathbb{P}^{n-1}$ are the 1D subspaces of the vector space $\mathbb{F}^n$ (where $\mathbb{F}$ is your field, or skew field if we're feeling adventerous). The lines of $\mathbb{P}^{n-1}$ are the 2D subspaces of $\mathbb{F}^n$, and so on. If it helps, imagine the unit sphere $S^{n-1}\subset\mathbb{R}^n$ as a double cover of $\mathbb{RP}^{n-1}$: the 1D subspaces of $\mathbb{R}^n$ are "collapsed" to antipodal pairs of points, the 2D subspaces (planes through the origin) are "collapsed" to great circles (which are one-dimensional), and so on.

So, $(1,1,0)$ and $(1,0,1)$ are two vectors in $\mathbb{F}^3$. If $\lambda$ is a scalar from $\mathbb{F}$ other than $1$, then $\lambda(1,1,0)$ is a different vector of $\mathbb{F}^3$ than $(1,1,0)$ is. However, $[1:1:0]$ and $[\lambda:\lambda:0]$ represent the same point in projective space for every nonzero scalar $\lambda$. The notations $(1,1,0)$ and $[1:1:0]$ refer to elements of two different spaces, although obviously they're related (the nonzero vectors of $\mathbb{F}^n$ turn into points of $\mathbb{P}^{n-1}$, with a $\mathbb{F}^\times$-worth of vectors going to every point).

So $[1:1:0]$ and $[1:0:1]$ are points of $\mathbb{P}^2$. The line they span in $\mathbb{P}^2$ by definition comes from the plane they span in $\mathbb{F}^3$. If we use $\mathbb{F}=\mathbb{R}$ and imagine the double cover $S^2$, we have two points on a sphere, which span a great circle, and we can find the great circle will be the intersection of the sphere with the plane spanned by any two nonzero vectors representing the two points on the sphere. The real projective plane $\mathbb{RP}^2$ works the same way, except points of $\mathbb{RP}^2$ correspond to antipodal pairs of points on $S^2$ and not just single points.