Counting parameters as an obstruction argument to write any polynomial representing a cubic curves in $\mathbb{C}P^2$ in a canonical equation

Question

I was reading a heuristic argument of why the all the cubic curves in $\mathbb{C} P^2$ cannot be transformed into a canonical one as the quadratic ones (e.g. if they are not degenerate they can be written as $x^2+y^2 + z^2$, otherwise as $x^2+y^2$ or $x^2$). The argument is as follows: the vector space of homogeneous polynomials of degree 3 is 10 dimensional, a cubic curve is defined up to a non-zero scalar by a homogeneous degree 3 polynomial, so the coefficients form a projective space of dimension 9. A projective transformation is determined uniquely by 4 points, (now comes the part that I don't understand) this gives 8 parameters and therefore we cannot expect that any cubic curve can be transformed to a standard one.

Answer 1

It may be easier to think about the linear algebra side of things. A projective transformation of $\Bbb CP^2$ is given by a matrix in $PGL_3(\Bbb C)$. The space of such matrices is 8-dimensional: the set of matrices with nonvanishing determinant inside $M_{3\times 3}(\Bbb C)$ is open, thus of dimension 9, and then we take the projectivization, which naturally lowers the dimension by 1 to 8.

Now let's think about what being able to transform an arbitrary polynomial to a single standard unique element under a projective transformation: this means that there is an element of $PGL_3$ so that when we act on our polynomial, we get the unique equation we want. By turning this around, this means that every element of our vector space is in the orbit of our unique element. But this is impossible for us: the orbit of a point under $PGL_3$ has dimension at most $\dim PGL_3=8$, which means it can't get to everything in a $9$-dimensional space.