The well-known identity for complex points forming an equilateral triangle reads
$$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$$
I have a doubt concerning the uniqueness of this identity
Is $z_1^2 + z_2^2 + z_3^2 – z_1 z_2 – z_2 z_3 – z_3 z_1$ the only quadratic form in 3 variables that when replacing with the complex points of an equilateral triangle is equal to 0?
Another related question, in case that's not unique,
Is $z_1^2 + z_2^2 + z_3^2 – z_1 z_2 – z_2 z_3 – z_3 z_1$ the smallest quadratic form, in the sense of the smallest absolute determinant of its associated matrix?
A quadratic form in three variables has 6 parameters, and I've recognized 3 equations: 1) the quadratic form is equal to $0$, 2) the translated quadratic form when $z_i \mapsto z_i+\zeta$ for each $i$ is equal to 0 (translation invariance), and 3) the rotated quadratic form when $z_i \mapsto z_ie^{i\pi/3}$ is equal to $0$ (rotational invariance). How can I narrow down the other parameters?
Best Answer
Let $\omega = e^{i\frac{2\pi}{3}}$ be a cubic root of unity. For any $z_1,z_2,z_3 \in \mathbb{C}$, let
$$\begin{align} a &= \frac13 (z_1 + z_2 + z_3)\\ b &= \frac13 (z_1 + z_2\omega + z_3\omega^2)\\ c &= \frac13 (z_1 + z_2\omega^2 + z_3\omega) \end{align}$$ It is easy to check $$\begin{align} z_1 &= a+b+c\\ z_2 &= a+b\omega^2+c\omega\\ z_3 &= a+b\omega+c\omega^2 \end{align}$$ Let $Q(z_1,z_2,z_3)$ be the quadratic form $$z_1^2+z_2^2+z_3^2-z_1z_2 - z_2z_3 -z_3z_1$$ The reason that $Q(z_1,z_2,z_3) = 0$ iff $\triangle z_1z_2z_3$ is equilateral comes down to the fact $$Q(z_1,z_2,z_3) = (z_1 + z_2\omega + z_3\omega^2)(z_1 + z_2\omega^2+z_3\omega) = 9bc$$ So $$Q(z_1,z_2,z_3) = 0 \iff bc = 0 \iff b = 0 \lor c = 0$$ When $b = 0$, $(z_1,z_2,z_3) = (a+c,a+c\omega,a+c\omega^2)$ and we can obtain $z_2,z_3$ by rotating $z_1$ with respect to $a$ counter-clockwisely for angles $120^\circ$ and $240^\circ$ respectively. This implies $\triangle z_1,z_2,z_3$ is equilateral and $z_1,z_2,z_3$ are ordered counter-clockwisely with respect to center $a$.
By a similar argument, if $c = 0$, $\triangle z_1,z_2,z_3$ is again equilateral but $z_1,z_2,z_3$ are ordered clockwisely with respect to $a$.
To show the quadratic form $Q(z_1,z_2,z_3)$ is essentially unique. Let $Q'(z_1,z_2,z_3)$ be another quadratic form which vanishes on and only on equilateral triangles. Expand $Q'(z_1,z_2,z_3)$ in terms of $a,b,c$, there are coefficients $\alpha,\beta_1,\beta_2, \gamma_1,\gamma_2,\gamma_3$ (not al zero) such that
$$Q'(z_1,z_2,z_3) = \alpha a^2 + a (\beta_1 b + \beta_2 c) + (\gamma_1 b^2 + \gamma_2 bc + \gamma_3 c^2)\tag{*1}$$
Pick any $z_1,z_2,z_3$ such that $\triangle z_1z_2z_3$ is equilateral. If one translate all vertices by same amount, $z_k \mapsto z_k + t$, the resulting triangle is also equilateral. However, $a \mapsto a + t$ but $b,c$ remains the same. This means for that specific $b,c$. $(*1) = 0$ for all $a$. This forces
$$\alpha = \beta_1 b + \beta_2 c = \gamma_1 b^2 + \gamma_2 bc + \gamma_3 c^2 = 0$$
It is trivial to find equilateral triangles with $b = 0, c\ne 0$ or triangles with $b \ne 0, c = 0$. This forces $\beta_1 = \beta_2 = \gamma_1 = \gamma_3 = 0$. As a result, $\gamma_2 \ne 0$ and
$$Q'(z_1,z_2,z_3) = \gamma_2 bc = \frac{\gamma_2}{9} Q(z_1,z_2,z_3)$$
From this, we can deduce the quadratic form $Q(z_1,z_2,z_3)$ is unique up to non-zero scaling constants.