I'm fairly new to the whole topic of tropicalization and I'm having trouble to even understand the following introductory definition:

[General setting: Let $K$ be a field which is complete with respect to a non-archimedean valuation $v: K*=K \setminus \{0\} \rightarrow \mathbb{R}$ which induces an absolute value $|.|= \exp(-v(.))$ on $K$.]

"For every $K$-split torus $\mathbb{G}_m^n$ and every basis $\{ \chi_1, \dotsc, \chi_n\}$ of its character lattice $\Lambda$, we define a tropicalization map

$$ trop: (\mathbb{G}_m^n)^{an} \rightarrow \mathbb{R}^n, ~~ \gamma \mapsto (\log |\chi_1(\gamma)|, \dotsc, \log |\chi_n(\gamma)|)"$$

My ideas/questions:

  1. I gathered that $\mathbb{G}_m$ denotes an algebraic group over a field $K$ given by the property that $\mathbb{G}_m(A)=A^*$ [how do I have to understand this property? how can this group be applied to some algebra?] for some $K$-algebra $A$ as well as that a $K$-split torus is exactly an algebraic group isomorphic to $\mathbb{G}_m^n$ for some $n$.
  2. I think that a character lattice has to be understood as a lattice generated by these characters above, i.e. group homomorphisms $\chi_i$ from the group $\mathbb{G}_m^n$ to the multiplicative group of a field, which I assume here is ${\mathbb{C}^*}^n$ [not sure about those sets though...I also thought about the $\chi_i$ starting from $(\mathbb{G}_m^n)^{an}$]
  3. I further know that for $X=Spec(B)$ an affine $K$-scheme of finite type, the analytification in the sense of Berkovich is the set of seminorms on a certain ring $B$ extending the absolute value on $K$.[Now, I'm not sure whether I have this case here or if I need a more general definition of analytification?]

Based on these thoughts, I'm not sure how to understand those $\gamma$ in the defintion of the map? I get that $\chi_i$ each map to $\mathbb{R}$ or $\mathbb{C}$ and I think that the value applied to $\chi_i(\gamma)$ is just the well-known absolute value in order to apply $\log$.

However, I'm not sure about these thoughts and I would be very grateful if anyone could help me out with my questions or correct me if any of my previous thoughts were already wrong!


Let me answer 1 and 2 by recollecting some properties of split Tori. First, $\mathbb{G}_{m, K}$ is just $Spec(K[T, T^{-1}])$ confirming your formula $\mathbb{G}_{m, K}(X) = \mathcal{O}_X(X)^{\times}$ for any $K$-scheme $X$ as well as showing that this functor is actually represented by an (affine!) scheme.

A split torus $G$ of rank $r$ is then an algebraic group $G$ which is isomorphic to $\mathbb{G}_m^r$, without fixing such an isomorphism. Also recall that a (split) lattice over $K$ of rank $r$ is an algebraic group $\Lambda$ over $K$ which is isomorphic to the constant group $\mathbb{Z}_K^r$, without fixing such an isomorphism.

Those two notions are dual to each other in the following sense: Given a split torus $G$, the group scheme $\Lambda(G):=\mathcal{Hom}(G, \mathbb{G}_{m,k})$ is a split lattice, the character lattice of $G$. On the other hand, starting with a split lattice $\Lambda$ we obtain a split torus $G(\Lambda) = \mathcal{Hom}(\Lambda, \mathbb{G}_{m,k})$ over $K$. Finally, choosing an isomorphism $\Lambda \cong \mathbb{Z}_K^r$ corresponds precisely to choosing an isomorphism $\mathbb{G}_{m,K}^r \cong G(\Lambda)$ (you should quickly check that this works).

To then (hopefully) answer your first two questions: A basis $\{\chi_1, \ldots, \chi_r\}$ of $\Lambda$ consists of morphisms $\chi_i: G \to \mathbb{G}_{m,K}$ such that there is an isomorphism $$\phi: G \to \mathbb{G}_{m,K}^r$$ with $\chi_i = pr_i \circ \phi$. So after a convenient choice of isomorphism, you lose nothing in thinking about $\chi_i$ as being coordinate projections. In particular, and very importantly, on $K$-rational points they map $(K^{\times})^r$ to $K^{\times}$, not to $\mathbb{C}$!

Coming to your question 3 we observe that all schemes above are affine over $K$, so your definition of analytification is sufficient for our purpose. Now $\chi_i$ corresponds to an injective ring homomorphism $$\xi_i: K[T, T^{-1}] \to K[S_1, S_1^{-1}, \ldots, S_r, S_r^{-1}] =: B$$ and any mult. seminorm $\gamma$ on the $B$ restricts to such a mult. seminorm $\gamma \circ \xi_i$ on $K[T,T^{-1}]$ which is uniquely determined by the value $$|\chi_i(\gamma)| := \gamma(\xi_i(T)) \in \mathbb{R}_{>0}$$

At this point you should rightfully object that the notation on the left hand side is very misleading. The motivation comes from considering only "algebraic points of $\mathbb{G}_{m,K}$". Indeed recall that any $K$-rational point $\Gamma = (\Gamma_1, \ldots, \Gamma_r) \in \mathbb{G}_{m,K}^r(K) = (K^{\times})^r$ gives rise to a point $\gamma \in (\mathbb{G}_{m,K}^r)^{an}$: The point $\Gamma$ corresponds precisely to a $K$-algebra homomorphism $$\psi: B \to K$$ and $\gamma: s \mapsto |\psi(s)|$ gives a multiplicative seminorm on $B$ extending the one on $K$, that is, a point in the analytification.

Now we can play the above game and observe that $\gamma \circ \xi_i (T) = \Gamma_i$, so in particular, $$|\chi_i(\gamma)| = |\Gamma_i|$$ where the LHS is the formal expression defined above, but the RHS is to be taken litterally, with $|\bullet|$ being the absolute value on $K$. Evidently, this is where the notation on the left hand side comes from.

Let me mention in this context that Bieri-Groves tells you that you can calculate the image of the tropicalization by taking the closure of the values on algebraic points (provided your valuation on $K$ is non-trivial). Hence, if one only cares about the image of $trop$, it is enough an often convenient to never think about the analytification; in this context the above notation changes from being suggestive and technically wrong to being suggestive and correct.

  • 1
    $\begingroup$ Thank you very much for your quick and elaborate answer. I still need some time to work through this thoroughly, so I'll hope it'll be okay if might come up with some questions later on... :) $\endgroup$ – SallyOwens Mar 15 '18 at 10:17

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