It is well known that three lines are said to be concurrent precisely when they all meet at a point, namely the point of concurrency.
In the paper On Sets Defining Few Ordinary Lines (v3) by Green & Tao, they write (page 21):
Finally, we give an example that lies on a cuspidal singular cubic curve, which after projective transform can be written as $$ \gamma := \big\{\,[x, y, z] : yz^{2} = x^{3}\,\big\}. $$ Removing the singular point at $[0, 1, 0]$, we may parameterise the smooth points $\gamma^{\ast}$ of this curve by $\big\{\,[t, t^{3}, 1] : t \in \mathbb{R}\,\big\}$. One can compute after a brief determinant computation that three distinct smooth points $[t_{1}, t^{3}_{1}, 1]$, $[t_{2}, t^{3}_{2}, 1]$ and $[t_{3}, t^{3}_{3}, 1]$ on the curve are concurrent precisely when $t_{1} + t_{2} + t_{3} = 0$.
I have added bold text for emphasis.
Question 1
• What do they mean by the three points are concurrent? They have described three points as concurrent on multiple occasions, not just this one.
I understand that any projective point $[a, b, c]$ is an equivalence class and that it is a line, in some sense, but I imagine this is not what they mean.
Question 2
In addition, is the discriminant they mention related to the discriminants described in the Conic sections of the Discriminant Wikipedia page? If so, which discriminant do they reference? I have never used discriminants like this before, so I do not understand the sketched proof of their claim.
Concurrent = Collinear
As Lord Shark the Unknown writes, they do mean to say collinear instead of concurrent.
Indeed, observe the following:
Proposition. Let $$ \big[t_{1}, t_{1}^{3}, 1\big],\qquad \big[t_{2}, t_{2}^{3}, 1\big],\qquad \big[t_{3}, t_{3}^{3}, 1\big] $$ be three distinct projective points. The points are collinear precisely when $t_{1} + t_{2} + t_{3} = 0$.
Proof. We may consider the affine points $P_{1} = \big(t_{1}, t_{1}^{3}\big), P_{2} = \big(t_{2}, t_{2}^{3}\big), P_{3} = \big(t_{3}, t_{3}^{3}\big)$ in place of the projective ones since each have a $Z$-coordinate of $1$. The points are collinear if and only if the line $L_{1}$ through $P_{1}$ and $P_{2}$ is the same as the line $L_{2}$ passing through $P_{2}$ and $P_{3}$. The gradient of $L_{1}$ is given by $$ \frac{t_{2}^{3} - t_{1}^{3}}{t_{2} - t_{1}} = t_{2}^{2} + t_{2}t_{1} + t_{1}^{2}\,. $$ Similarly, the gradient of $L_{2}$ is given by $$ \frac{t_{3}^{3} - t_{12}^{3}}{t_{3} - t_{2}} = t_{3}^{2} + t_{3}t_{2} + t_{2}^{2}\,. $$ The $y$-intercept of $L_{1}$ is $$ t_{2}^{3} - \big(t_{2}^{2} + t_{2}t_{1} + t_{1}^{2}\big)t_{2} = -t_{1}t_{2}^{2} - t_{1}^{2}t_{2} = -t_{1}t_{2}\big(t_{1} + t_{2}\big). $$ Similarly, the $y$-intercept of $L_{2}$ is $$ t_{3}^{3} - \big(t_{3}^{2} + t_{3}t_{2} + t_{2}^{2}\big)t_{2} = -t_{2}t_{3}^{2} - t_{2}^{2}t_{3} = -t_{2}t_{3}\big(t_{2} + t_{3}\big). $$ Now, the gradients of $L_{1}$ and $L_{2}$ are equal if and only if $$ t_{2}^{2} + t_{2}t_{1} + t_{1}^{2} = t_{3}^{2} + t_{3}t_{2} + t_{2}^{2} $$ which rearranges to $$ t_{1}(t_{1} + t_{2}) = t_{3}(t_{2} + t_{3}). $$ The $y$-intercepts of $L_{1}$ and $L_{2}$ are equal if and only if $$ -t_{1}t_{2}\big(t_{1} + t_{2}\big) = -t_{2}t_{3}\big(t_{2} + t_{3}\big) $$ which also rearranges to $$ t_{1}(t_{1} + t_{2}) = t_{3}(t_{2} + t_{3}). $$ The above equation rearranges to $$ t_{1}t_{2} - t_{2}t_{3} = t_{3}^{2} - t_{1}^{2} \qquad\text{so that}\qquad t_{2}(t_{1} - t_{3}) = (t_{3} - t_{1})(t_{3} + t_{1}). $$ By dividing out the $(t_{1} - t_{3})$ term we obtain $$ t_{2} = -(t_{3} + t_{1}) $$ and the result follows.$\qquad\square$