I'm learning projective geometry and need help to understand the following statement :
We give the quadratic form in $\mathbb{R}^3$ (where $\mathbf{x} = (x_1, x_2, x_3)$) :
$$q(\mathbf{x}) = 2x_1^2 - 2x_1x_2 + ax_2^2 + 2x_1x_3 - 4x_2x_3 + 3x_3^2.$$
The quadratic equation $q = 0$ is well defined in the real projective plane.
Given the title of my post I think you can guess my next question: why is $q = 0$ well defined in $\mathbb{RP}^2$? In general, what does it mean for a quadratic equation to be well defined in $\mathbb{RP}^2$? Since the real projective plane is the collection of all lines passing through the origin in $\mathbb{R}^3$, is $q = 0$ a line in $\mathbb{RP}^2$? I think the answer to the last question is no but it's still unclear to me what $q = 0$ represents in $\mathbb{RP}^2$.
Note : if anyone asks for the parameter $a$, this is part of an exercise where we first need to find the values of $a$ for which $q(\mathbf{x})$ is positive definite. I'm only interested for answers to the questions related to the real projective plane.
The statement "[t]he quadratic equation $q = 0$ is well defined in the real projective plane" means that the set $\{\ell \in \mathbb{RP}^2 \mid q(\ell) = 0\}$ (which is a collection of lines) is a well-defined set; that is, there is an unambiguous definition of when $\ell \in \mathbb{RP}^2$ satisfies $q(\ell) = 0$.
For this to be well-defined, we need to make sure that whether $q(\ell) = 0$ or not depends only upon the line itself, not the point we chose on it; i.e. that the property $q(\ell) = 0$ is a property of the line $\ell$, not a property of the point $(x_1, x_2, x_3)$. In order to do this, we need to show that if $(y_1, y_2, y_3)$ is another point in $\ell$ with $(y_1, y_2, y_3) \neq (0, 0, 0)$, then $q(y_1, y_2, y_3) = 0$ if and only if $q(x_1, x_2, x_3) = 0$ so that it doesn't matter which non-origin point on $\ell$ we chose.
As $\ell$ is a one-dimensional vector space and $(x_1, x_2, x_3) \in \ell$ is not the zero vector, it forms a basis. In particular, for any other vector in $\ell$, it must be a multiple of $(x_1, x_2, x_3)$. In particular, $(y_1, y_2, y_3) = \lambda(x_1, x_2, x_3) = (\lambda x_1, \lambda x_2, \lambda x_3)$ for some real number $\lambda$; as $(y_1, y_2, y_3)$ is not the zero vector, $\lambda$ is non-zero. Now note that
\begin{align*} q(y_1, y_2, y_3) &= q(\lambda x_1, \lambda x_2, \lambda x_3)\\ &= 2(\lambda x_1)^2 - 2(\lambda x_1)(\lambda x_2) + a(\lambda x_2)^2 + 2(\lambda x_1)(\lambda x_3) - 4(\lambda x_2)(\lambda x_3) + 3(\lambda x_3)^2\\ &= 2\lambda^2x_1^2 - 2\lambda^2x_1x_2 + a\lambda^2x_2^2 + 2\lambda^2x_1x_3 - 4\lambda^2x_2x_3 + 3\lambda^2x_3^2\\ &= \lambda^2(2x_1^2 - 2x_1x_2 + ax_2^2 + 2x_1x_3 - 4x_2x_3 + 3x_3^2)\\ &= \lambda^2q(x_1, x_2, x_3). \end{align*}
As $\lambda \neq 0$, $\lambda^2 \neq 0$ and hence $q(y_1, y_2, y_3) = 0$ if and only if $q(x_1, x_2, x_3) = 0$. Therefore, the notion $q(\ell) = 0$ is well-defined (i.e. only depends on the line $\ell$), and hence so is the set $\{\ell \in \mathbb{RP}^2 \mid q(\ell) = 0\}$.
More generally, a polynomial $p$ is called homogeneous if there is $k \in \mathbb{R}$ such that $p(\lambda{\bf x}) = \lambda^kf({\bf x})$ for all $\lambda \in \mathbb{R}$. Note, the calculation above shows that the polynomial you asked about is homogeneous with $k = 2$. It's not hard to show that if $p$ is a homogeneous polynomial, then $k = \deg p$ and every monomial in $p$ has top degree (as is the case for your polynomial). If $p$ is a homogeneous polynomial of $n + 1$ variables, then the notion $p(\ell) = 0$ is well-defined in $\mathbb{RP}^n$.