# The Plonk permutation argument

## Case for $\beta = 1$

Let $\sigma$ be an identity permutation. Let us set evaluations of $f$ on $H$ as:
- $f(\mathbf{g}) = 4$, therefore: $f'(\mathbf{g}) = 5 + \gamma$
- $f(\mathbf{g}^2) = 4$, therefore: $f'(\mathbf{g}^2) = 6 + \gamma$
- $f(\mathbf{g}^3) = 4$, therefore: $f'(\mathbf{g}^3) = 7 + \gamma$
- $f(\mathbf{g}^4) = 4$, therefore: $f'(\mathbf{g}^4) = 8 + \gamma$

Similarly, let us set evaluations of $g$ on $H$ as:
- $g(\mathbf{g}) = 5$, therefore: $g'(\mathbf{g}) = 6 + \gamma$
- $g(\mathbf{g}^2) = 5$, therefore: $g'(\mathbf{g}^2) = 7 + \gamma$
- $g(\mathbf{g}^3) = 5$, therefore: $g'(\mathbf{g}^3) = 8 + \gamma$
- $g(\mathbf{g}^4) = 1$, therefore: $g'(\mathbf{g}^4) = 5 + \gamma$

Hence:
$Z(\mathbf{g}^4) = \frac{(5 + \gamma)(6 + \gamma)(7 + \gamma)(8 + \gamma)}{(6 + \gamma)(7 + \gamma)(8 + \gamma)(5 + \gamma)}=1$

However, clearly $f \neq \sigma(g)$.

The key thing here is that we could make $f' = \sigma(g')$.

## Case for $\gamma = 0$

Let $\sigma$ be an identity permutation. Let us set evaluations of $f$ on $H$ as:
- $f(\mathbf{g}) = 0$, therefore: $f'(\mathbf{g}) = \beta$
- $f(\mathbf{g}^2) = 4$, therefore: $f'(\mathbf{g}^2) = 4 + 2\beta$
- $f(\mathbf{g}^3) = 0$, therefore: $f'(\mathbf{g}^3) = 3\beta$
- $f(\mathbf{g}^4) = 2$, therefore: $f'(\mathbf{g}^4) = 2 + 4\beta$

Similarly, let us set evaluations of $g$ on $H$ as:
- $g(\mathbf{g}) = 0$, therefore: $g'(\mathbf{g}) = \beta$
- $g(\mathbf{g}^2) = 1$, therefore: $g'(\mathbf{g}^2) = 1 + 2\beta$
- $g(\mathbf{g}^3) = 0$, therefore: $g'(\mathbf{g}^3) = 3\beta$
- $g(\mathbf{g}^4) = 8$, therefore: $g'(\mathbf{g}^4) = 8 + 4\beta$

Hence:
$Z(\mathbf{g}^4) = \frac{\beta\cdot(4+2\beta)\cdot 3\beta\cdot (2 +4\beta)}{\beta\cdot(1+2\beta)\cdot3\beta\cdot(8+4\beta)}=\frac{2(2+\beta)\cdot 2(1+2\beta)}{(1+2\beta)\cdot4(2+\beta)}=1$

However, clearly $f \neq \sigma(g)$.

The key thing here is that even that $f' \neq \sigma(g')$, the products of their evaluations on $[4]$ are equal.