We now consider the following lifting problem associated to *G*_{n}. Let *K* be an algebraically closed complete non-archimedean valued field and let *X* be a smooth algebraic curve of genus *g*. Let *X*^{an} be the analytification of *X*(as a Berkovich curve). Let *R* be the valuation ring of *K*, assume 𝔛 is a strongly semistable model of *X* over *R* (meaning the special fiber is nodal with smooth irreducible components) such that the special fiber has only rational components and let Γ be the associated skeleton. Is it possible to obtain this situation such that Γ = *G*_{n} and $\mathrm{dim}({W}_{4}^{1}(X))=1?$ In that case, taking into account the result from [10] mentioned in the introduction, this would give a geometric explanation for ${w}_{4}^{1}({G}_{n})=1$. This lifting problem will be the motivation for considering the existence of a certain harmonic morphism associated to *G*_{n}. Making this motivation we are going to refer to some suited papers for terminology and some definitions. The definitions necessary to understand the question on the existence of the harmonic morphism are given in Section 2.3. Finally we are going to prove that the harmonic morphism does not exist, proving that the lifting problem has no solution. In particular we obtain that the classification of metric graphs satisfying ${w}_{4}^{1}=1$ is different from the classification of smooth curves satisfying $\mathrm{dim}({W}_{4}^{1})=1$.

Assume the lifting problem has a solution. The curve *X* of that solution cannot be hyperelliptic since *G*_{n} is not hyperelliptic. This follows from the specialisation Theorem from [2] or [3] already mentioned in the introduction. From [11] one obtains the following classification in case char(*k*) ≠ 2 of non-hyperelliptic curves *X* of genus at least 6 satisfying $\mathrm{dim}({W}_{4}^{1}(X))=1$ (for arbitrary characteristic and *g* ≥ 10 see the Appendix): *X* is trigonal (has a ${g}_{3}^{1}$), *X* is a smooth plane curve of degree 5 (hence has genus 6 and has ${g}_{5}^{2}$) or *X* is bi-elliptic (there exists a double covering *π*: *X* → *E* with *g*(*E*) = 1). From Proposition 3.7 we know *G*_{n} has no ${g}_{3}^{1}$ and no ${g}_{5}^{2}$, hence the curve *X* has to be bi-elliptic.

So assume there exists a morphism *π*: *X* → *E* with *g*(*E*) = 1 of degree 2. This induces a map *π*^{an}: *X*^{an} → *E*^{an} between the Berkovich analytifications. In the case *E* is not a Tate curve then each strong semistable reduction of *E* contains a component of genus 1 in its special fiber, in particular the augmentation map of the associated skeleton has a unique point with value 1. Otherwise such skeleton can be considered as a metric graph of genus 1. Each skeleton associated to a semistable reduction of *X* is tropically equivalent to the graph *G*_{n}, in particular it can be considered as a metric graph. From the results in [16, Section 4, especially Corollaries 4.26 and 4.28] it follows that there exist skeletons $\tilde{\Gamma}$ (resp. Γ) of *X* (resp. *E*) such that *π* induces a finite harmonic morphism $\tilde{\Gamma}\to \Gamma $ of degree 2 (Section 4 of [16] uses no assumption on the characteristic of *k*). Since $\tilde{\Gamma}$ is a metric graph (augmentation map identically zero) this is also the case for Γ hence Γ is a metric graph. So in the case *G*_{n} is liftable to smooth curve *X* satisfying $\mathrm{dim}({W}_{4}^{1}(X))=1$ then there exist a tropical modification $\tilde{\Gamma}$ of *G*_{n} and a metric graph Γ of genus 1 such that there exists a finite harmonic morphism $\tilde{\pi}:\tilde{\Gamma}\to \Gamma $ of degree 2. We are going to prove that such finite harmonic morphism does not exist. In the proof the following lemma will be useful.

*Let ϕ*: (Γ_{1}, *V*_{1}) → (Γ_{1}, *V*_{2}) *be a finite harmonic morphism between metric graphs with vertex sets. Let* (*T*′, *V*′) ⊂ (Γ_{1}, *V*_{1}) *be a subgraph such that T*′ *is a tree*, $\overline{{\Gamma}_{1}\backslash {T}^{\prime}}\subset {\Gamma}_{1}$ *is connected and* $\overline{{\Gamma}_{1}\backslash {T}^{\prime}}\cap {T}^{\prime}$ *consists of a unique point t* (*in particular t* ∈ *V*_{1}). *There is no subtree* (*T, V*) *of* (*T*′, *V*′) *different from a point such that ϕ*(*T*) *is contained in a loop* Γ ⊂ Γ_{2}.

*Proof*. Assume *T* is a subtree of *T*′ not being one point and assume *ϕ*(*T*) is contained in a loop Γ of Γ_{2}. Let *l*(*T*) (resp. *l*(Γ)) be the sum of the lengths of all the edges of *T* (resp. Γ). By definition one has *l*(*T*) ≤ deg(*ϕ*)*l*(Γ), in particular *l*(*T*) is finite. We are going to prove that we have to be able to enlarge *T* such that *l*(*T*) grows with a fixed lower bound. Repeating this a few times gives a contradiction to the upper bound deg(*ϕ*)*l*(*Γ*).

Let *q* ∈ *V* be a point of valence 1 on *T* such that *q* ≠ *t* and let *f* be the edge of *T* having *q* as a vertex point. This edge *f* defines *v* ∈ *T*_{q}(Γ_{1}), let *w* = *d*_{ϕ}(*q*)(*v*), hence *ϕ*(*q*) ∈ Γ and *w* ∈ *T*_{ϕ(q>)}(Γ). Since Γ is a loop there is a unique *w*′ ∈ *T*_{ϕ(q)}(Γ) with *w*′ ≠ *w* and since *ϕ* is harmonic there exists *v*′ ∈ *T*_{q}(Γ_{1}) with *d*_{ϕ}(*q*)(*v*′) = *w*′. Let *f*′ be the edge in Γ_{1} having *q* as a vertex point and defining *v*′. From *d*_{ϕ}(*q*)(*v*′) ∈ *T*_{ϕ(q)}(Γ) it follows *ϕ*(*f*′) ⊂ Γ, hence *l*(*f*′) is finite. Since *f*′ ≠ *f* and *q* ≠ *t* one has *f*′ is an edge of $\overline{{T}^{\prime}\backslash T}$. Since *T*′ is a tree, also *T* ∪ *f*′ is a tree and one has *ϕ*(*T* ∪ *f*′) ⊂ Γ. Moreover *l*(*T* ∪ *f*′) = *l*(*T*) + *l*(*f*′) and *l*(*f*′) has as a fixed lower bound the minimal length of an edge contained in Γ_{1}. □

*There does not exist a tropical modification* $\tilde{\Gamma}$ *of G*_{n} such that there exists a graph Γ *with g*(T) = 1 *and a finite harmonic morphism* $\varphi :\tilde{\Gamma}\to \Gamma $ *of degree 2.*

*Proof*. Assume $\tilde{\Gamma}$ is a tropical modification of *G*_{n} and $\varphi :\tilde{\Gamma}\to \Gamma $ is a finite harmonic morphism of degree 2 of metric graphs with *g*(Γ) = 1. Since *g*(Γ) = 1 we know *g*(*ϕ*(*G*_{0})) ≤ 1. In case *g*(*ϕ*(*G*_{0})) = 0 then in step 1 we are going to prove that the restriction of *ϕ* to *G*_{0} has a very particular description and next in step 2 we are going to prove that this does not occur.

Step 1: Assume *g*(*ϕ*(*G*_{0})) = 0. Then *ϕ*(*G*_{0}) looks as in Figure 3 with *ϕ*|_{ei}: *e*_{i} → [*ϕ*(*v*_{1}), *ϕ*(*m*_{i})] having degree 2, ♯((*ϕ*|_{ei})^{−1}(*q*)) = 2 for all *q* ∈ [*ϕ*{*v*_{1}), *ϕ*(*m*_{i})[and (*ϕ*|_{ei})^{−} (*ϕ*(*m*_{i})) = {*m*_{i}} for 0 ≤ *i* < 2.

Fig. 3 In the case *g*(*ϕ*(*G*_{0}) = 0

Assume *g*(*ϕ*(*G*_{0})) = 0. Consider the loop *c*_{1} = *e*_{1} ∪ *e*_{0} (remember Figure 1). Since *ϕ*(*G*_{0}) has genus 0 it follows *ϕ*(*c*_{i}) is a subtree *T*_{1} of *ϕ*(*G*_{0}). Since *T*_{1} is the image of a loop and deg(*ϕ*) = 2, it follows that ♯((*ϕ*|_{c1})^{−1}(*q*′)) = 1 for some *q*′ ∈ *ϕ*(*c*_{1}) if and only if *q*′ is a point of valence 1 of *ϕ*(*c*_{1}). Since deg(*ϕ*) = 2 it follows *d*_{q}(*ϕ*) = 2 for *q* ∈ *c*_{1} such that *ϕ*(*c*_{1}) has valence 1 on *ϕ*(*c*_{1}) and *d*_{q}(*ϕ*) = 1 for *q* ∈ *c*_{1} if *ϕ*(*q*) does not have valence 1 on *ϕ*(*c*_{1}). In the case *ϕ*(*c*_{1}) would have a point *q*′ of valence 3 then there exist at least 3 different points *q* on *e*_{1} with *ϕ*(*q*) = *q*′, contradicting deg(*ϕ*) = 2. Hence *ϕ*(*c*_{1}) can be considered as a finite edge with two vertices. Also for each *q* ∈ *c*_{1} and *v* ∈ *T*_{q}(*c*_{1}) one has *d*_{b}(*ϕ*) = 1.

Assume *ϕ*(*v*_{1}) ≠ *ϕ*(*v*_{2}). Then *ϕ*(*e*_{2}) is a path from *ϕ*(*v*_{1}) to *ϕ*(*v*_{2}) outside of *ϕ*(*c*_{1}). This would imply *g*(*ϕ*(*G*_{0})) ≥ 1, contradicting *g*(*ϕ*(*G*_{0})) = 0, hence *ϕ*(*v*_{1}) = *ϕ*(*v*_{2}). This also implies that *ϕ*(*m*_{0}) and *ϕ*(*m*_{1}) are the two points of valence 1 on *ϕ*(*c*_{1}). Repeating the previous arguments for the loop *c*_{2} = *e*_{2} ∪ *e*_{0} one obtains the given description for *ϕ*|*G*_{0}: *G*_{0} → *ϕ*(*G*_{0}).

Step 2: *g*(*ϕ*(*G*_{0})) = 1.

In case *g*(*ϕ*(*G*_{0})) = 0 then we have the description for *ϕ*|_{G0}:*G*_{0} → *ϕ*(*G*_{0}) obtained in Step 1. We are going to prove that this description cannot hold. Since *g*(*ϕ*(*G*_{0})) ≤ 1, this implies *g*(*ϕ*(*G*_{0})) = 1.

Consider *ϕ*(*q*_{1}) ∈ *ϕ*(*e*_{1}) and ${{q}^{\prime}}_{1}\ne {q}_{1}$ on *e*_{1} with $\varphi ({q}_{1})=\varphi ({{q}^{\prime}}_{1})$ (see Figure 2). Assume *ϕ*(*γ*_{1}) is a tree. If *ϕ*(*γ*_{1}) would have valence 1 at *ϕ*(*q*_{1}) then from the arguments used in Step 1 it follows *d*_{q1}(*ϕ*) = 2. But from Step 1 we know *d*_{q1}(*ϕ)* = 1, therefore *ϕ*(*q*_{1}) cannot be a point of valence 1 on *ϕ*(*γ*_{1}). Hence there exists ${{q}^{\u2033}}_{1}\in {\gamma}_{1}\backslash \left\{{q}_{1}\right\}$ with $\varphi ({q}_{1})=\varphi ({{q}^{\u2033}}_{1})$ and we obtain ♯(*ϕ*^{−1}(*ϕ*(*q*_{1}))) ≥ 3, contradicting deg(*ϕ*) = 2. It follows *g*(*ϕ*(*γ*_{1})) = 1, hence *ϕ*(*γ*_{1}) contains a loop ${{e}^{\prime}}_{1}$ in Γ. In case $\varphi ({q}_{1})\notin {{e}^{\prime}}_{1}$ then again we obtain ♯(*ϕ*^{−1}(*ϕ*(*q*_{1}))) ≥ 3, contradicting deg(*ϕ*) = 2. Therefore *ϕ*(*q*_{1}) ∈ $\varphi ({q}_{1})\in {{e}^{\prime}}_{1}$ and ${{e}^{\prime}}_{1}\cap \varphi ({G}_{0})=\{\varphi ({q}_{1})\}$.

Repeating the arguments using *q*_{2} and *γ*_{2} we obtain a loop ${{e}^{\prime}}_{2}$ in Γ such that ${{e}^{\prime}}_{2}\cap \varphi ({G}_{0})=\{\varphi ({q}_{2})\}$, hence ${{e}^{\prime}}_{1}\cap {{e}^{\prime}}_{2}=\overline{)0}$. Since *g*(Γ) = 1 this is impossible. As a conclusion we obtain *g*(*ϕ*(*G*_{0})) = 1 finishing the proof of step 2.

In the case *ϕ*(*c*_{1}) would have genus 0 (*c*_{1} as in the proof of Step 1), then from the arguments used in Step 1 it follows that for *q* ∈ *e*_{2} \ {*v*_{1}, *v*_{2}} one has *ϕ*(*q*) ∉ *ϕ*(*c*_{1}). In the case *ϕ*(*v*_{1}) ≡ *ϕ*{*v*_{2}) it implies *ϕ*(*c*_{2}) has genus 1 (again *c*_{2} as in the proof of Step 1). In the case *ϕ*(*v*_{1}) = *ϕ*(*v*_{2}) and *g*(*ϕ*(*c*_{2})) = O too, it would imply *g*(*ϕ*(*G*_{0})) = 0

so this cannot occur. Therefore without loss of generality, we can assume *ϕ*(*c*_{1}) has genus 1 (but then *ϕ*(*c*_{2}) could have genus 0).

Step 3: *ϕ*|_{c1}: *c*_{1} → *ϕ*(*c*_{1}) is an isomorphism (meaning it is finite harmonic of degree 1)

Since *g*(*ϕ*(*c*_{1})) = 1 it follows there is a loop *e* in *ϕ*(*c*_{1}), finitely many points *r*_{1}, · · ·, *r*_{t} on *e* and finitely many trees *T*_{i} inside *ϕ*(*c*_{1}) such that

*T*_{i} ∩ *e* = {*r*_{1}}

${T}_{i}\cap {T}_{j}=\overline{)0}$ in the case *i* ≡ *j*

*ϕ*(*c*_{1}) = *e* ∪ *T*_{1} ∪ · · · ∪ *T*_{t}

(of course *t* = 0, hence *ϕ*(*c*_{1}) is a loop, is also possible; we are going to prove that *t* = 0).

In the case val_{ϕ(c1)} (*r*_{i}) > 3 for some 1 ≤ *i* ≤ *t* then *ϕ*^{−1}(*v*_{i}) contain at least 3 different points on *c*_{1}, contradicting deg(*ϕ*) = 2. So we obtain a situation like in Figure 4. Let *r* ∈ {*r*_{1}, · · · , *r*_{t}} and let *T* be the associated subtree of *ϕ*(*c*_{1}). Then *ϕ*^{−1}(*r*) = {*r*′, *r*″} ⊂ *c*_{1} with *r*′ ≡ *r*″. The tangent space *T*_{r}(*ϕ*(*c*_{1})) consists of 3 elements (see Figure 5). Hence there exists *w* ∈ *T*_{r′}(*G*_{0}) \ *T*_{r′}(*c*_{1}) such that *d*_{ϕ}(*r*′)(*w*) ∈ *T*_{r}(*e*). (In Figure 5 it is the tangent vector corresponding to the direction on *e* indicated by the number 2.) Let *f* be the edge of $G{\text{'}}_{n}$ defining *w* hence *ϕ*(*f*) ⊂ *e*. Because of Lemma 4.1 this implies *r*′ is one of the points *q*_{i} on *G*_{0} and *f* ⊂ *γ*_{i}. Since deg(*ϕ*) = 2 and g(Γ) = 1 it follows *e* ⊂ *ϕ*(*γ*_{i}). Repeating the same argument using *r*′ instead of *r*′ one obtains a contradiction to deg(*ϕ*) = 2. This proves *t* = 0, hence *ϕ*(*c*_{1}) = *e* is a loop.

Because deg(*ϕ*) = 2 we cannot go back and forth on *e* moving along *c*_{1} and taking the image under *ϕ*. In principle it could be the case that there exist different points *q*′, *q*″ on *c*_{1} such that the image of the closure of both components of *c*_{1} \ {*q*′, *q*″} is equal to *e* with *ϕ*(*q*′) = *ϕ*(*q*″) and *d*_{q′}(*ϕ*) = *d*_{q″}(*ϕ*) = 2. This would correspond to something like shown in Figure 6. This figure has to be understood as follows. Moving along *c*_{1} from *q*′ to *q*″ in the direction indicated by 1 (left hand side of the figure) the image under *ϕ* is equal to *e* while moving in the direction indicated by 1 (right hand side of the figure). Moving on *c*_{1} from *q*″ to *q*′ in the direction indicated by 2 (left hand side of the figure) the image under *ϕ* is equal to *e* while moving in the direction indicated by 2 (right hand side of the figure).

Fig. 6 A case that cannot occur

In that case there should exist $v\in {T}_{q\text{'}}\left(G{\text{'}}_{n}\right)$ with *d*_{ϕ}(*q*′)(*υ*) ∈ *T*_{ϕ(q′)}(*e*) \ *d*_{ϕ}(*q*′)(*T*_{q′} (*c*_{1})). Hence the edge *f* of ${G}_{n}^{\text{'}}$ defining *υ* satisfies *ϕ*(*f*) ⊂ *e*, contradicting deg(*ϕ*) = 2. Hence the situation from Figure 6 cannot occur.

It follows that in the case there exist *q*′ ≠ *q*″ on *c*_{1} such that *ϕ*(*q*′) = *ϕ*(*q*″) then *ϕ*|_{c1} : *c*_{1} → *e* is harmonic of degree 2 and *d*_{q} (*ϕ*|_{c1}) = 1 for all *q* ∈ *c*_{1}. In this case *ϕ*(*e*_{2}) ∩ *e* = {*ϕ*{*υ*_{1}), *ϕ*{*υ*_{2})} since deg(*ϕ*) = 2. In the case *ϕ*(*υ*_{1}) ≠ *ϕ*(*υ*_{2}) then this contradicts *g*(Γ) = 1. In the case *ϕ*(*υ*_{1}) = *ϕ*(*υ*_{2}) then because of the description of *ϕ*|_{c1} one has *l*(*e*_{1}) = *l*(*e*_{0}). We assume this is not the case, so we can assume *ϕ*:_{c1} → *e* is bijective.

In the case for each edge *f* on $\tilde{\Gamma}$ with *f* ⊂ *c*_{1} one has *d*_{f}(*ϕ*) = 2 then again, since *ϕ*(*υ*_{1}) ≠ *ϕ*(*υ*_{2}) we have *ϕ*(*e*_{2}) ∩ *e* = {*ϕ*(*υ*_{1}), *ϕ*(*υ*_{2})}, contradicting *g*(Γ) = 1. Assume there exists *q* ∈ *c*_{1} being a vertex of $\tilde{\Gamma}$ and two edges *e*′, *e*″ of $\tilde{\Gamma}$ contained in *c*_{1} with vertex end point *q* such that *d*_{e′} (*ϕ*) = 1 and *d*_{e″} = 2. In particular it follows that *d*_{q} (*ϕ*) = 2. Let $v\text{'}\in {T}_{q}(\tilde{\Gamma})$ correspond to *e*′ then there exists $v\in {T}_{q}(\tilde{\Gamma})$ with *υ* ∉ *T*_{q} (*c*_{1}) such that *d*_{ϕ}(*q*)(*υ*) = *d*_{ϕ}(*q*)(*υ*′). Let *f* be the edge of $\tilde{\Gamma}$ defining *υ*, then *ϕ*(*f*) ⊂ *e*. From Lemma 4.1 it follows that *q* is one of the points *q*_{i} and *f* ⊂ *γ*_{i}. Since *g*(Γ) = 1 and deg(*ϕ*) = 2 it follows that *e* ⊂ *ϕ*(*γ*_{i}), but this is impossible because *d*_{e″} (*ϕ*) = 2 and deg(*ϕ*) = 2. This proves *ϕ*|_{c1} : *c*_{1} → *e* is an isomorphism of metric graphs.

Step 4: Finishing the proof of the theorem.

It follows that *ϕ*(*υ*_{1}) and *ϕ*(*υ*_{2}) do split *e* into two parts *e*′ and *e*″ of lengths *l*(*e*_{1}) and *l*(*e*_{0}). Since *g*(Γ) = 1 it follows *ϕ*(*e*_{2}) contains *e*′ or *e*″, we assume it contains *e*′. In the case *ϕ*(*e*_{2}) would contain $\tilde{q}\in e"\backslash \{\varphi ({v}_{1}),\text{\hspace{0.17em}}\varphi ({v}_{2})\}$ then because of *g*(Γ) = 1 it follows *ϕ*(*e*_{2}) contains one of the connected components of $e"\backslash \left\{\tilde{q}\right\}$. On that connected component we get a contradiction to deg(*ϕ*) = 2. Indeed, for a point on that connected component the inverse image under *ϕ* contains one point of *c*_{1} and two different points of *e*_{2}.

So we obtain different points *r*_{1}, … , *r*_{t} on *e*′ and trees *T*_{1}, … , *T*_{t} with *T*_{i} ∩ *e*′ = *r*_{i} for 1 ≤ *i* ≤ *t* and ${T}_{i}\cap {T}_{j}=\overline{)0}$ for *i* ≠ *j* such that *ϕ*(*e*_{2}) = *e*′ ∪ *T*_{1} ∪ … ∪ *T*_{t}. It is possible (and we are going to prove) that *t* = 0, hence *e*′ = *ϕ*(*e*_{2}). In the case *r*_{i} ∉ {*ϕ*(*υ*_{1}), *ϕ*(*υ*_{2})} then there exist two different points *r*′, *r*″ on *e*_{2} such that *ϕ*(*r*′) = *ϕ*(*r*″) = *r*_{i}. Since *r*_{i} is also the image of a point on *c*_{1} we get a contradiction to deg(*ϕ*) = 2. Hence *t* ≤ 2 and *r*_{i} ∈ {*ϕ*(*υ*_{1}), *ϕ*(*υ*_{2})}.

Assume *r*_{i} = *ϕ*(*υ*_{1}). We obtain *q* ∈ *e*_{2} with *q* ∉ {*υ*_{1}, *υ*_{2}} and *ϕ*(*q*) = *ϕ*(*υ*_{1}). There exists *υ* ∈ *T*_{q}(Γ) such that *d*_{ϕ}(*q*)(*υ*) is the element of *T*_{ϕ(q)} (Γ) defined by *e*″. Let *f* be the edge of $\tilde{\Gamma}$ defining *υ*. From Lemma 4.1 it follows *q* is one of the points *q*_{i} and *f* ⊂ *γ*_{i}. Since *g*(Γ) = 1 and deg(*ϕ*) = 2 we obtain *ϕ*(*γ*_{i}) contains *e*. This implies that for *P* ∈ *e*′ there are at least 3 points contained in *ϕ*^{−1}(*P*), a contradiction. This proves *t* = 0, hence *ϕ*(*e*_{2}) = *e*′. Since deg(*ϕ*) = 2 it also implies *d*_{f}(*ϕ*) = 1 for each edge *f* of $\tilde{\Gamma}$ contained in *e*_{2}, hence *l*(*e*′) = *l*(*e*_{2}). Since *l*(*e*_{2}) ∉ {*l*(*e*_{0}), *l*(*e*_{1})} we obtain a contradiction, finishing the proof of the theorem. □

As a corollary of the theorem we obtain the goal of this paper.

*For each genus g* ≥ 5 *there is metric graph G of genus g satisfying* ${w}_{4}^{1}\ge 1$ *that has no divisor of Clifford index at most 1 and is not tropically equivalent to a metric graph* $\tilde{\Gamma}$ *such that there exists a finite harmonic morphism $\pi :\tilde{\Gamma}\to \Gamma $ **of degree 2 with g*(Γ) = 1. *In particular in the case g* ≥ 10 the graph G cannot be lifted to a curve X of genus g satisfying dim $({W}_{4}^{1})=1$.

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