# What can we say about tropical maps $\mathbb{P}^1 \to A$ for an Abelian variety $A$?

It's well known that maps $\mathbb{P}^1_\mathbb{C} \to A$ are constant for any Abelian variety $A$ (in fact, for any complex torus).

Is there any similar statement in the tropical case? Naively, the answer seems to be no. We can see this in the following way: a tropical $\mathbb{P}^1$, $\mathbb{TP}^1$ is just a metric tree, and (for example) a circle (with a view vertices for good measure) produces an example of a one-dimensional tropical abelian variety. We can include $\mathbb{TP}^1 \hookrightarrow C$ as one of the edges quite easily which is a tropical map (unless I'm making a silly mistake).

So naively, the answer seems to be that there are non-constant tropical maps. Is there some way to eliminate these in a meaningful way? That is, it would be nice to have a statement of the form

Theorem: All (possibly restricted) tropical maps $\mathbb{TP}^1 \to A$ are constant (up to some appropriate notion of equivalence).

Edit: As pointed out in the comments, there is a balancing condition to be considered. For the case of maps to a tropical elliptic curve, this is given as the following.

For each point in $C$, the target, and each pre-image $p$ of $C$, the sum of all weights of the edges incident to $p$ must sum to zero (paying attention to outward orientation of edges).

One sees immediately that this implies that no $\mathbb{TP}^1$s can map to $C$, since the image of a leaf of a tree can never satisfy the balancing condition (unless the weight of every edge is zero).

However, this doesn't address maps to higher dimensional abelian varieties, but one can hope that a similar balancing condition would be true.

• Is there really any $\mathbb{TP}^1$ at all in $\mathbb{R}^g / \Lambda$? There is a balancing condition that needs to be fulfilled. – Vesselin Dimitrov Oct 5 '15 at 15:34
• @VesselinDimitrov - Maybe there isn't! I'll have to look at that a little more closely. – Simon Rose Oct 6 '15 at 8:31

Here is why. A tropical curve of genus zero is a metric tree $\Gamma$. In particular, it is simply connected. So any continuous map $\phi: \Gamma \to \mathbb{R}^g/\Lambda$ lifts to a continuous map $\Gamma \to \mathbb{R}^g$, which can't obey the balancing condition at a leaf.
One caveat: I could imagine someone making a definition (though I haven't see it) which would allow $\mathbb{R}$, with its standard metric, as a tropical genus zero curve with two punctures; $\mathbb{R}/\mathbb{Z}$ with the quotient metric as a tropical genus $1$ curve, and the covering map as a tropical map. This is precisely analogous to the analytic map $\mathbb{C}^{\ast} \to \mathbb{C}^{\ast}/q^{\mathbb{Z}} \cong E$ for some $q \in \mathbb{C}$ with $0 < |q|<1$.
• This was exactly the example I wanted to give showing that there are morphisms from a tropical $\mathbf P^1$ to an abelian variety. Actually, some papers define $\mathbf T\mathbf P^1$ as $\mathbf R^2/(1,1)\R$. Moreover, the analogy with non-archimedean geometry makes this example very reasonable. When you say that the map $\Gamma\to\mathbf R^g$ can't obey the balancing condition at a leaf, this presumes the existence of a leaf... – ACL Oct 7 '15 at 20:25