Puiseux Series and Tropical Varieties

by ayushkhaitan3437

Puiseux series- This field is denoted by \Bbb{C}[[t]]. Note that we have a double brace “[[ ]]” instead of “[]”. This implies that we have infinite series instead of finite ones (which would be polynomials). The Puiseux laurent series is denoted as \Bbb{C}((t)). This means that t is also allowed to have negative powers. Now \mathcal{K}=\bigcup\limits_{n\geq 1} \Bbb{C}((t^{1/n})), which just means that \mathcal{K} contains all rational powers of t now, and not just integral powers, as in the Laurent series. We seem to be generalizing in every successive step.

Now we define a valuation: v:\mathcal{K}^{\times}\to\Bbb{Q}: v(\sum a_it^{i/N})=\min(i/N), for a_i\neq 0. So what we’ve essentially done is that we’ve written all rational powers (including integral ones) as fractions with denominator N. Clearly, amongst all the denominators in the rational powers of t, N has to be the largest denominator.

If the Puiseux series converges, then we have v(f)=\lim\limits_{t\to 0^+}\frac{\log (f(t))}{\log t}

Why is that? It seems to me that \lim\limits_{t\to 0^+}\frac{\log (f(t))}{\log t} would give us the sum of all rational powers of t, which could possibly be infinite. Then why do we just get the lowest one?

Now for X\in (\mathcal{K}^\times)^n, set Trop(X) to be v(f)\subset\Bbb{Q}^n. Take X=x+y+1=0 for example. One should think about this variety as x(t)+y(t)+1=0 instead, where x and y are power series in t (with rational powers). Then there are three possibilities:

i) v(x)> 0 and v(t)=0.

ii) v(x)=0 and v(y)> 0

iii) v(x)=v(y)\leq 0.

These cases can be easily deduced to contain all possibilities. For instance, if v(x)<0, then v(y)<0 too. This is because v(x)<0 implies that x has negative powers of t. This implies that y(t) too has to have negative powers of t, as x(t)+y(t)=-1. When one of them contains strictly positive powers, the other has to contain -1 and no negative powers of t, which implies that if x(t)>0, then y(t)=0.