How to show that a map which relates to Donaldson–Thomas invariants is an automorphism?

Question

I am reading the lecture notes INTRODUCTION TO DONALDSON–THOMAS INVARIANTS. I have a question in the end of page 1 about the proof of a map is an automorphism.

Let $m>0$ be an integer. Let $\overline{A} = Q[x_1, x_2]$ be an algebra with multiplication: \begin{align} x_1^ax_2^b \cdot x_1^c x_2^d = (-1)^{m(ad-bc)} x_1^{a+c} x_2^{b+d}. \end{align}

For any $a,b \in \mathbb{N}^2\backslash \{0\}$, define a homomorphism $T_{a,b}: \overline{A} \to \overline{A}$ by \begin{align} x_1 \mapsto x_1 \cdot (1-x_1^a x_2^b)^{-mb}, \\ x_2 \mapsto x_2 \cdot (1-x_1^a x_2^b)^{ma}. \end{align} It is said that this map is an automorphism. How to show that $T_{a,b}$ is a bijection?

It is said that the algebra $\overline{A}$ is commutative. But according to the definition of the multiplication, this algebra is not commutative. Am I correct?

Thank you very much.

Answer 1

The fact that the algebra is commutative has been discussed in the comments. So, I will address the bijectivity of $T_{a,b}$. I will assume we believe that we have a well defined algebra homomorphism since it is only asked why $T_{a,b}$ is a bijection. Here is a sketch of bijectivity.

Our multiplication respects the usual (multi)degree of monomials since $x_1^ax_2^b \cdot x_1^cx_2^d = \pm x_1^{a+c}x_2^{b+d}$. We will consider some term order. Notice that $T_{a,b}(x_1^cx_2^d)$ is a power series with lowest term $x_1^cx_2^d$. More generally given any formal power series $f$ with lowest term $x_1^cx_2^d$, the power series $T_{a,b}(f)$ has lowest term $x_1^cx_2^d$. It follows that $T_{a,b}(f) = 0$ only if $f = 0$ and so $T_{a,b}$ is injective. To show $T_{a,b}$ is surjective we will show any monomial $x_1^cx_2^d$ is in the image. To do this start with $f_0 = x_1^cx_2^d$, then $T_{a,b}(f_0)$ is a power series with lowest term $x_1^cx_2^d$. Now define $f_1$ so that $T_{a,b}(f_1)$ cancels the second lowest term of $T_{a,b}(f_0)$. Consider $T_{a,b}(f_0 + f_1)$ and repeat.