q i}\}_{i\in\Nat}$ is the standard filtration of the algebra $A_1$ associated with the generators $X$ and $Y$. For all $i\in\Z\setminus\{0\}$ and $f\in K[H]\setminus K$, $$\label{degsf}
\deg \s^i(f) = \deg f \text{ and } \deg (1 - \s^i)(f)  = \deg f - 1.$$

[**Proof of Theorem \[20Sep16\]**]{}: [*(i)*]{} [*If $P, Q\in A_{1,\leq 1}$ then $P = \tau(Y)$ and $Q=\tau(X)$ for some $\tau\in\Aut_K (A_1)$*]{}: Clearly, $P = aY + bX + \l$ and $Q = cY +dX + \mu$ for some $a, b, c, d, \l, \mu \in K$. Then $1 = [P,Q]=ad-bc$. So, the automorphism $\tau$ can be chosen of the form $\tau(Y) = aY + bX + \l$ and $\tau(X) = cY +dX + \mu$.

So, till the end of the proof we assume that at least one of the polynomials $P$ or $Q$ does not belong to the space $A_{1, \leq 1}$. In view of the relation $1 = [P,Q] = [-Q,P]$, we can assume that $P \notin A_{1, \leq 1}$. In view of Equation (\[xiaut1\]), we can assume that the highest homogeneous part of $P$, say $P_p \in A_{1,p}$, satisfies the condition that $p \geq 2$. Since $m(P)\leq 2$, either $P = P_p$ (if $m(P)=1$) or otherwise $P = P_r + P_p$ for some nonzero $P_r \in A_{1,r}$ where $r<p$.

[*(ii)*]{} $(m(P), m(Q)) \neq (1,1)$: Suppose that $m(P)= m(Q)=1$, we seek a contradiction. Then $P =\alpha X^p$ and $Q=\beta Y^p$ for some nonzero polynomials $\alpha, \beta\in K[H]$. Then $$1 = [P,Q] = \alpha \s^p(\beta) (p,-p) - \beta \s^{-p}(\alpha) (-p,p)$$ $$=
\alpha \s^p(\beta) (p,-p) - \beta \s^{-p}(\alpha) \s^{-p}((p,-p)) =
(1-\s^{-p})(\alpha \s^p(\beta) (p,-p)).$$ Since $p \geq 2$ (or $P\notin A_{1, \leq 1}$), $0 = \deg 1 = \deg\, (1-\s^{-p})(\alpha \s^p(\beta) (p,-p)) =
\deg \alpha + \deg \beta +\deg\, (p,-p) -1$ (by Equation (\[degsf\])) $\geq 0+0+p-1 \geq 2-1=1$, a contradiction.

[*(iii)*]{} $(m(P), m(Q)) \neq (1,2)$: Suppose that $m(P)=1$ and $m(Q)=2$. Then $P = \alpha X^p$ for some $p\geq 2$ and $Q=Q_s + Q_q$ where $Q_s \in A_{1,s}$, $Q_q \in A_{1,q}$ and $s<q$. By Lemma \[a20Sep16\], the equality $[P,Q]=1$ implies that $[P, Q_s]=1$ and $[P, Q_q]=0$. By the case (ii), this is not pos