I am working through the book "introduction to tropical algebra" and page 70 and 71 example 2.2.10 states
"An automorphism of the torus is an invertible map specified by Laurent monomials. Thus the automorphism group of $\mathbb{T}^n$ isomorphic to GL(n, Z). Here the matrix entries are the exponents of the monomials.
The invertible integer map $U = \begin{pmatrix} 1 & -1\\ 1 & -2 \end{pmatrix} $ represents the automorphism $(x, y) \mapsto (xy, x^{−1}y^{−2})$ of the torus $\mathbb{T}^2$ and of its coordinate ring $\mathbb{C}[x^{±1}, y^{±1}]$. The image of the curve $X = \{(x, y) ∈ \mathbb{T}^2 : f(x, y) = 0\}$ in Example 1.8.1 under the automorphism U is the curve defined by $(U ◦ f)(x, y) = c_2 + c_5x + c_1y + c_3xy + c_4x^2y^2$ "
Where example 1.8.1 states
"Let X be the curve in $(\mathbb{C}^∗)^2$ defined by the polynomial $f(x, y) = c_1 + c_2xy + c_3x^2y+c_4x^3y + c_5x^3y^2$ Here $c_1, c_2, c_3, c_4, c_5$ are any complex numbers..."
My question is, how is the curve in example 2.2.10 obtained from the given automorphism? For example after applying the given automorphism why does $c_1$ gain a $y$ factor and why is the $c_2$ term $c_2 xy$ not mapped to $c_2 (xy)(x^{−1}y^{−2})=c_2y^{-1}$ ?
I think there must be something I am missing here, so any help you could provide would be greatly appreciated.
Well, it indeed contains some inaccuracy.
Normally, when we apply a transformation $U$ to a subset $X=\{(x,y):f(x,y)=0\}$, then we have to apply $U^{-1}$ on the defining equation because $$(a,b)\in U(X)\ \iff\ U^{-1}(a,b)\in X\ \iff\ f\left(U^{-1}(a,b)\right)=0\,.$$
The notation $U\circ f$ especially with the stated strict equality is not too clean.
What the received equation instead describes, is thus the $U^{-1}$ transform $U^{-1}(X)$ of the curve $X$, but of course, the $U$ transform can be obtained the same way.
So, $(x,y)\in U^{-1}(X)\iff f(U(x,y))=0\iff$ $$\matrix{c_1+c_2y^{-1}+c_3x+c_4x^2y+c_5xy^{-1}=0& \iff & |\,\cdot y\\ c_1y+c_2+c_3xy+c_4x^2y^2+c_5x=0\,.}$$