Projective Basis Explanation, Visualisation

Question

I have trouble understanding the geometric idea behind a projective basis. So for example in $\mathbb{P}^2(K)$ we have $[1:0:0],[0:1:0],[0:0:1], [1:1:1]$ but why? I know that $$ [1:0:0] = k\cdot \vec{e_1} $$ then we have four lines. Why is there $[1:1:1]$? Isnt this a linear combination of the other ones? A very brief explanation would be enough. I really have no clue why there is always an extra vector or "line".

Answer 1

In projective geometry you cannot talk of a linear combination. Also, the idea that dimension corresponds to the number of elements in the basis is something you should get rid of when talking about projective geometry. So why do we have four points in a basis for a two-dimensional projective space? (Or $n + 2$ points in a basis for $\mathbb{P}^n K$?)

This idea arises when we're trying to fix a projective transformation. Let $[x], [y]$ and $[z]$ be three points in the projective space, and $x, y, z$ linearly independent in $K^3$ and suppose that $\varphi = [A]$ is a projective transformation on $\mathbb{P}^2 K$ (with $A$ a $3 \times 3$ matrix -- or linear transformation, if you want -- over $K$) such that $\varphi([1:0:0]) = [x]$, $\varphi([0:1:0]) = [y]$ and $\varphi([0:0:1]) = [z]$.

To make this example easier, I'll suppose $x = (0,1,0), y = (0,0,1), z = (1,0,0)$, but notice that the following argument would work for any three independent vectors. Then we have $A(1,0,0) = \lambda (0,1,0)$ for some $\lambda \in K_{\neq 0}$. Similarly we find $\mu, \nu \neq 0$ such that $$\varphi = [A] = [\begin{pmatrix} 0 & 0 & \nu \\ \lambda & 0 & 0 \\ 0 & \mu & 0 \end{pmatrix} ] = [\begin{pmatrix} 0 & 0 & 1 \\ \lambda/\nu & 0 & 0 \\ 0 & \mu/\nu & 0 \end{pmatrix} ].$$ Notice that this "basis" DOES NOT uniquely define $\varphi$ (we still have two "degrees of freedom" left!). This is not what we want, so we throw in another vector $(1,1,1)$ so that any three vectors are lineairly independent, and the same for our basis of images. Say, $\varphi([1:1:1]) = [2:3:4].$ (For the sake of example, but this fourth vector should be such that any three vectors in our basis are LI.) Then, $$A(1,1,1) = \xi (2,3,4) \iff \nu = 2\xi, \lambda = 3\xi, \mu = 4 \xi$$ for some $\xi \neq 0$, and this DOES fix our projective transformation: $$\varphi = [A] = [\begin{pmatrix} 0 & 0 & 2 \\ 3 & 0 & 0 \\ 0 & 4 & 0 \end{pmatrix} ],$$ and this is exactly what we want from our basis: it's the smallest number of points such that a transformation is uniquely determined on them.