Verifying triplet collinearity in $\mathbb{C}^2$ using determinants I am asked to find triplets of collinear points in $\mathbb{C}^2$, and I am asked to use the determinants. we are taught that in the complex projective plane, we normally assume $z=1$.
My question is: is it true that there is a corresponding point from $\mathbb{C}^2 \to \mathbb{P}^2_{\mathbb{C}}$ such that, if we are given points $(a'_1, a'_2), (b'_1, b'_2), (c'_1, c'_2) \in \mathbb{C}^2$, then they are collinear if and only if
$$det
\begin{bmatrix}
    a_1 & a_2 & 1 \\
    b_1 & b_2 & 1 \\
    c_1 & c_2 & 1 \\
\end{bmatrix} = 0
$$
where $a_1 = \frac{a'_1}{z}, a_2 = \frac{a'_2}{z}$ and so on? Many thanks, it's our first taste in this area, apologies if any of the above sounds weird!
 A: That is correct.
Suppose $p_i = (a_i, b_i), i = 1,2,3$ are colinear. Then the points $(a_i, b_i, 1), i = 1,2,3$ are colinear within the plane $z = 1$.
But remember that we want to think of a point $[a : b : c] \in \mathbb{P}^2$ as the line $(ta, tb, tc) : t \in \mathbb{C}$. So actually, what we want are the three lines $\bar{p_i} = \{(ta_i, tb_i, t) : t \in \mathbb{C}\}$.
Let $\ell = \{p_1 + s(p_1 - p_2) : s \in \mathbb{C}\}$ be the line in $\mathbb{C}^2$ containing $p_1$ and $p_2$ (and also $p_3$). Then in $\mathbb{P}^2$, $\ell$ becomes a "line of lines" (i.e. a plane). That is:
\begin{align}
\bar \ell &= \{(t_1a_1, t_1b_1, t_1) + s((t_1a_1, t_1b_1, t_1) - (t_2a_2, t_2b_2, t_2)) : t_1, t_2, s \in \mathbb{C}\} \\
&= \{(1+s)t_1(a_1, b_1, 1) - st_2(a_2, b_2, 1) : t_1, t_2, s \in \mathbb{C}\} \\
&= \{t_1'(a_1, b_1, 1) + t_2'(a_2, b_2, 1) : t_1', t_2' \in \mathbb{C}\} \\
&= \operatorname{span}\{(a_i,b_i,1): i = 1,2\}.
\end{align}
And in general, the correspondence is:




$\mathbb{P}^n$
$\mathbb{C}^{n+1}$




point
line


line
plane


$k$-plane
$(k + 1)$-plane




So if $p_1,p_2,p_3$ lie on a common line in $\mathbb{P^2}$ (or in $\mathbb{C}^2$) then they lie in a common plane in $\mathbb{C}^3$.
If $(a_i, b_i, 1), i = 1,2,3$ lie in a common plane, then the paralellepiped they form has volume $0$, i.e. the determinant you wrote down is $0$.
In general, you could look not just where $z = 1$ but take any non-zero point on the three lines you get in $\mathbb{C}^3$ and then the relationship is
$$\det\begin{pmatrix} t_1a_1 & t_1b_1 & t_1 \\
t_2a_2 & t_2b_2 & t_2 \\
t_3a_3 & t_3b_3 & t_3 \\
\end{pmatrix}
= t_1t_2t_3 \det \begin{pmatrix} a_1 & b_1 & 1 \\
a_2 & b_2 & 1 \\
a_3 & b_3 & 1 \\
\end{pmatrix}$$
In projective space we only ask whether a polynomial is zero or non-zero rather than what it's value is since you can always scale all of the coordinates and get a new value.
