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.

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

The answer should be that there are no non-constant maps from tropical curves of genus zero to tropical abelian varieties. By "should" I mean that there are a lot of definitions of these things in the literature and it would be frustrating to write an answer which is correct for every definition, but any definition which doesn't have this consequence is a "bad" definition.

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 $0 < |q|<1$.

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    $\begingroup$ 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... $\endgroup$ – ACL Oct 7 '15 at 20:25

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