A metric graph satisfying w41=1 that cannot be lifted to a curve satisfying dim⁡ (W41)=1

1 Introduction

Let K be an algebraically closed field complete with respect to some non-trivial non-Archimedean valuation. Let R be the valuation ring of K, let m_R be its maximal ideal and k = R/m_R the residue field. Let X be a smooth complete curve of genus g defined over K. Associated to a semistable formal model 𝔛 over R of X there exists a so-called skeleton Γ = Γ_𝔛 which is a finite metric subgraph of the Berkovich analytification X^an of X together with an augmentation function a: Γ → ℤ⁺ such that a(v) = 0 except for at most finitely many points (see e.g. [1]). In the case all components of the special fiber xk are rational then this augmentation function is identically zero and we can consider Γ as a metric graph. This is the situation we consider in this paper. In this situation, from the point of view of the metric graph Γ, we say the curve Xis a lift of Γ.

There exists a theory of divisors and linear equivalence on Γ very similar to the theory on curves and, in case Xis a lift of Γ, those theories on Xand Γ are related by means of a specialisation map

For a divisor E on Γ one defines a rank rk(E) and for a divisor D on X the specialisation theorem says (see e.g. [2], since we restrict to the case of zero augmentation map this is in principle considered in [3])

In the hyperelliptic case many classical results on linear systems on curves also do hold for linear systems on metric graphs. As an example, if the graph Γ has a very special linear system g2rr then Γ has a g21 and g2rr=rg21 (see [4, 5]). Hence the theory of linear systems gdr of Clifford index 0 (meaning d – 2r = 0) is the same for graphs as for curves (see Section 2.2 for this terminology). This is not true for the theory of linear systems of Clifford index more than 0. As an example: the theory of linear systems of Clifford index 1 concerns the theory of linear systems gdr satisfying d – 2r = 1, r ≥ 1 and d ≤ r + g – 2. In the case of curves, H. Martens Theorem (see [6]) states that if C is a non-hyperelliptic curve and C has a linear system gdr with Clifford index 1 and d ≤ g – 1 then either d = 3 or d = 5. In [7] we obtain for all r ≥ 1 and d = 2r + 1 ≤ g – 1 the existence of a non-hyperelliptic graph Γ of genus g having a linear system gdr.

For a curve X the complete linear systems gdn with n ≥ r are parametrized by a closed subscheme Wdr of the Jacobian J(X) and d−2r−dim⁡(Wdr) gives a kind of generalisation of the Clifford index for moving linear systems on X. In particular in the case r ≤ g – 1 then dim⁡(Wdr)≤d−2r and dim⁡(Wdr)=d−2r for some r < g – 1 if and only if X is hyperelliptic (see [8]). In [9] it is shown that using the dimension of a similar subspace of the Jacobian J(Γ) of a metric graph Γ this statement is not true. Moreover in that paper the authors do introduce a much better invariant wdr as a replacement for dim⁡(Wdr) in the case of graphs which is more close to the definition of the rank of a divisor on a graph (see Section 2.2). In [10] it is proved that wdr≥dim⁡(Wdr) in the case Γ is a metric graph and the curve X is a lift of Γ. At the moment is seems not known whether wdr≥d−2r for some 0 < r < g – 1 implies Γ is hyperelliptic.

Concerning the next case, in [11, Appendix] one finds Mumford’s classification of all curves X such that dim⁡(Wdr)=d−2r−1. In the case C is a non-hyperelliptic curve such that dim⁡(Wdr)=d−2r−1 for some r ≥ 1 and d ≤ g – 1 then either C is a trigonal curve or C is a smooth plane curve of degree 5 or C is a bi-elliptic curve (meaning that there exists a double covering π: C → E with E an elliptic curve). In [11] the author assumes char(k) ≠ 2 although it is not so clear whether this is also necessary for the arguments in the appendix of that paper. In the appendix of this paper I give a very short proof that in the case C is neither hyperelliptic, not trigonal of genus g ≥ 10 and if C has two different linear systems g41 then C is bi-elliptic not using any assumption on char(k). Many more generalisations are proved by different authors (see e.g. [12]). In this paper we show that the theory of curves satisfying dim⁡(W41)=1 is different from the theory of graphs satisfying w41=1. In particular for all genus g> 10 we prove the existence of a metric graph G_n of genus g (n = g – 3) satisfying w41=1 that cannot be lifted to a curve satisfying dim⁡(W41)=1.

Related to this result it should be mentioned that non-liftable linear systems on graphs are also known. We say that a linear system gdr on a graph G is liftable if there exists a lift X of G and a divisor D on X with dim(|D|) = r such that τ*(D)∈gdr. As an example, from [13, Theorem 4.8] it is known that the linear system g21 on a hyperelliptic graph having a vertex v adjacent to at least 3 different bridges (a bridge of a graph G is an edge e such that G \ e is disconnected) is not liftable to a hyperelliptic curve because of the violation of some Hurwitz condition. In [14, Example 5.13] one finds an example due to Luo of a graph with a g31 that cannot be lifted to a curve. Also in [7] one finds lots of types of linear systems gdr that cannot be lifted to curves, e.g. so-called free linear systems g52 of graphs that cannot be lifted to a curve because a curve with a plane model of degree 5 has genus at most 6.

In Section 2 we recall some generalities on graphs and the theory of divisors on graphs. For generalities on the specialisation map and the relation between the metric graphs and skeleta inside Berkovich curves we refer to the references. It is not needed to understand the arguments used in this paper, it is important for the motivation. In Section 3 we give the description of the graph denoted by G_n (n an integer at least equal to 2) and we prove it satisfies w41=1 and it has Clifford index equal to 2. In Section 4 first we explain that in the case G_n could be lifted to a curve satisfying dim⁡(W41)=1 then for some tropical modification Γ˜ of G_n there would exist a finite harmonic morphism (see Section 2.3) π:Γ˜→Γ of degree 2 with Γ a metric graph of genus 1. Finally in Section 4 we prove that such harmonic morphism does not exist.

2 Generalities

3 The example

In the proof of this section we are going to use some lemmas concerning linear systems on graphs.

Fig. 1

The graph G₀

Here υ₁ and υ₂ are two points of valence 3 (all other points have valence 2) and they are connected by three edges e₀, e₁ and e₂ of mutually different lengths. For 0 ≤ / ≤ 2 the point m_i is the midpoint of e_i.

Proof. Clearly 2m_i ∈ |υ₁ + υ₂ | for 0 ≤ i ≤ 2 and in the case υ ∈ e_i \ {υ₁, υ₂, m_i} then taking υ′ on e_i such that the distance on e_i from υ to υ₁ is equal to the distance of υ′ to υ₂, then υ + υ′ ∈ |υ₁ + υ₂| (clearly υ ≠ υ′). This proves rk(υ₁ + υ₂) = 1 (it cannot have rank 2 because g(G₀) ≠ 0). It is well-known that a graph of genus at least 2 has at most one g21 (see [15, Proposition 5.5], the proof given for a finite graph also holds for a metric graph). Indeed, for this graph G₀, if υ ≠ υ₂ then υ₁ + υ is clearly υ₂-reduced, hence |v1+v−v2|=0 and therefore rk(υ₁ + υ) = 0. This proves the uniqueness of g21 on G₀.

Finally for υ + υ′ ∈ |υ₁ + υ₂| as before (including the possibility υ + υ′ = υ₁ + υ₂), since υ′ is a u-reduced divisor one has |v′−v|=0 hence |v1+v2−2v|=0. This proves 2υ ∈ |υ₁ + υ₂| implies A = m_i for some 0 ≤ i ≤ 2. □

As indicated in figure 1 we fix q_i ∈]υ_i, m_i, [⊂ e_i, for i = 1,2.

Proof. For i = 1,2 we take q′i∈ei such that the distance on e_i from υ₁ to q_i is equal to the distance on e_i, from υ₂ to q′i. Assume g31 on G₀ with |g31−2m0|≠0, hence there exists υ ∈ G₀ such that |g31=|g21+v|.

First assume υ ∈ (e₀ ∪ e₂) \ {υ₁}. Then |q1′+v+q1|=|g31-2q1|=∅ and clearly q′1+v is a q₁-reduced divisor. This implies |q'1+v-q1|=|g31-2q1|=∅. In the case υ ∉ (e₀ ∪ e₂) \ {υ₁} then certainly υ ∈ (e₀ ∪ e₁) \ {υ₂} and using similar arguments we find |g31−2q2|  =0. This finishes the proof of the lemma. □

Now for an integer n ≥ 1 we make a graph G_n as follows. In case n ≥ 3 we fix some more different points q₃, … , q_n on G₀ \ {υ₁, υ₂, m₀, m₁, m₂, q₁, q₃}. Then, for all n ≥ 1, the graph G_n is obtained from G₀ by attaching a loop γ₀ at m₀ and loops γ_i, at q_i for each 1 ≤ i ≤ n (we also are going to denote m₀ by q₀). As an example see a possible picture of G₆ in figure 2. Clearly g(G_n) = n + 3. We prove that the Clifford index of G_n is at least 2 in case n ≥ 2.

Fig. 2

The graph G₆

Proof. First we show G_n has no linear system g21 in the case n ≥ 1. Assume there is a g21 on G_n. Take 1 ≤ i ≤ n and let υ ∈ γi \ {q_i} and υ′ ∈ G_n such that v+v′∈g21. Let Gn0 be the closure of G_n \ γ_i, and assume v′∈Gn0. It follows from Lemma 3.3 that rkGn0(v′)≥1, but because g(Gn0) > 0 this is impossible. Hence υ′ ∈ γ_i. On γ_i there exists υ″ such that υ + υ′ is linearly equivalent to q_i + υ″ as divisors on γ_i, hence qi+v″∈g21 because of Lemma 3.1. But we proved this implies υ″ = q_i, hence 2qi∈g21. This implies rk_G0 (2q_i) = 1. Indeed, take p ″ G₀ \ {q_i} and let D_p be the p-reduced divisor on Go linearly equivalent to 2q_i. We need to show D_p − p ≥ 0. The burning algorithm applied to G_n implies D_p is a p-reduced divisor on G_n too. Since rk_Gn(2q_i ) = 1 it follows D_p − p ≥ 0. So we obtain rkG₀ (2q_i) = 1, but from Lemma 3.5 we know this cannot be true, hence G_n has no linear system g21.

From now on assume n ≥ 2. Fix some integer r satisfying 1 ≤ r ≥ n − 1 and assume G_n has a linear system g2r+1r. For 0 ≤ i ≤ r + 1 fix υi ∈ γ_i with υ_i ≠ q_i. For 0 ≤ i ≤ 2 there exists Ei∈g2r+1r satisfying

For j ∈ {3, … , r + 1} ∪ {i} let D_i,j = E_i ∩ (γ_j \ {q_j}), hence D_{i, j} − υ_j > 0. In the case that for some j the q_j-reduced divisor on γ_j linearly equivalent to D_{i, j} contains a point v′j different from q_j (then the point v′j is unique) then there is an effective divisor E′ on Gn\γj¯ of degree 2r such that E′+v′j∈g2r+1r. From Lemma 3.3 it follows rkGn\γj¯(E′) = r, hence Gn\γj¯ has a linear system g2rr. Since (Gn\γj¯)=n+2 and 2 ≤ 2r ≤ 2n − 2 this would imply Gn\γj¯ is a hyperelliptic graph (Lemma 3.4). But we proved Gn\γj¯ is not hyperelliptic (it is a graph G_{n − 1}; the proof of that argument also works on Gn\γ0¯ in the case j = 0). Since υ_j is not linearly equivalent to q_j as a divisor on γ_j it follows D_{i, j} is linearly equivalent to m_i, _jq_j for some m_{i, j} ≥ 2 on γ_i.

So we obtain E′i ∈ g2r+1r on G_n such that

for 0 ≤ i ≤ 2 and E′i is contained in G₀. Because of Lemma 3.2 those divisors E′i are linearly equivalent as divisors on G₀. It follows E″i=E′i−(2q3+⋯+2qr+1) with 0 ≤ i ≤ 2 are effective linearly equivalent divisors on G₀ of degree 3 with E″i−2qi≥0. Since g(G₀) = 2 each divisor of degree 3 on G₀ defines a g31 and we obtain a contradiction to Lemma 3.6.

Finally, if G_n has an g2n+1n then because of the Theorem of Riemann-Roch |KGn−g2n+1n|=g31, we already excluded this case. □

In order to prove this proposition we need the existence of many linear systems g41 on G_n. Those can be obtained from divisors of degree 4 on Go using the following lemma.

Proof. Since D is an effective divisor of degree 4 on G₀ from the Riemann-Roch Theorem it follows rk_G0(D) = 2. This implies for each 0 ≤ i ≤ n there is an effective divisor D′ ≥ 2q_i linearly equivalent to D on G₀. Because of Lemma 3.1 the divisor D′ is linearly equivalent to D on G_n. Moreover, for υ ∈ γ_i, there is an effective divisor on γ_i linearly equivalent to 2q_i containing υ and using the same lemma we obtain the existence of an effective divisor on G_n linearly equivalent to D containing υ. Similarly, for υ ∈ G₀ we obtain an effective divisor on G₀ linearly equivalent to D and containing υ and again this divisor is also linearly equivalent to D as a divisor on G_n. This proves rkG_n(D) ≥ 1. □

Proof ofProposition 3.8. Fix υ₁, υ₂ ∈ G_n. We need to prove that there exists a g41 on G_n such that |g41−v1−v2|≠0.

In the case υ₁, υ₂ ∈ G₀ we can use any effective divisor D on G₀ containing υ₁ + υ₂. Then we have rkG_n (D) ≥ 1 because of Lemma 3.9 and |D−v1−v2|≠0. Next assume υ₁ ∈ G₀ and υ₂ ∈ γ_i \ {q_i} for some 0 ≤ i ≤ n. On G₀ take any effective divisor D of degree 4 containing υ₁ + 2q_i. Since 2q_i is linearly equivalent to v2+v′2 for some v′2∈γi as a divisor on λ_i, again using Lemma 3.1 we find that D−2qi+v2+v′2 is an effective divisor linearly equivalent to D on G_n. Hence rkG_n(D) ≥ 1 because of Lemma 3.9 and |D−v1−v2|≠0. Assume υ₁ ∈ λ_i1\{q_i1} and υ₂ ∈ λ_i2\{q_i2} for some i₁ ≠ i₂. Then one makes a similar argument using the divisor 2q_i1 + 2q_i1 on Go. Finally, if υ₁, υ₂ ∈ λ_i, \ {q_i} for some i then one uses any effective divisor D of degree 4 on G₀ containing 3q_i. Since 3q_i is linearly equivalent on λ_i to an effective divisor containing υ₁ + υ₂ again we find rkG_n(D) ≥ 1 and |D−v1−v2|≠0.

4 The lifting problem

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 dim⁡(W41(X))=1? In that case, taking into account the result from [10] mentioned in the introduction, this would give a geometric explanation for w41(Gn)=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 w41=1 is different from the classification of smooth curves satisfying dim⁡(W41)=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 dim⁡(W41(X))=1 (for arbitrary characteristic and g ≥ 10 see the Appendix): X is trigonal (has a g31), X is a smooth plane curve of degree 5 (hence has genus 6 and has g52) 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 g31 and no g52, 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 Γ˜ (resp. Γ) of X (resp. E) such that π induces a finite harmonic morphism Γ˜→Γ of degree 2 (Section 4 of [16] uses no assumption on the characteristic of k). Since Γ˜ 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 dim⁡(W41(X))=1 then there exist a tropical modification Γ˜ of G_n and a metric graph Γ of genus 1 such that there exists a finite harmonic morphism π˜:Γ˜→Γ 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.

Proof. Assume T is a subtree of T′ not being one point and assume ϕ(T) is contained in a loop Γ of Γ₂. 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(Γ₁), 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(Γ₁) with d_ϕ(q)(v′) = w′. Let f′ be the edge in Γ₁ 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 T′\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 Γ₁. □

Proof. Assume Γ˜ is a tropical modification of G_n and ϕ:Γ˜→Γ is a finite harmonic morphism of degree 2 of metric graphs with g(Γ) = 1. Since g(Γ) = 1 we know g(ϕ(G₀)) ≤ 1. In case g(ϕ(G₀)) = 0 then in step 1 we are going to prove that the restriction of ϕ to G₀ 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(G₀)) = 0. Then ϕ(G₀) looks as in Figure 3 with ϕ|_ei: e_i → [ϕ(v₁), ϕ(m_i)] having degree 2, ♯((ϕ|_ei)⁻¹(q)) = 2 for all q ∈ [ϕ{v₁), ϕ(m_i)[and (ϕ|_ei)⁻ (ϕ(m_i)) = {m_i} for 0 ≤ i < 2.

Fig. 3

In the case g(ϕ(G₀) = 0

Assume g(ϕ(G₀)) = 0. Consider the loop c₁ = e₁ ∪ e₀ (remember Figure 1). Since ϕ(G₀) has genus 0 it follows ϕ(c_i) is a subtree T₁ of ϕ(G₀). Since T₁ is the image of a loop and deg(ϕ) = 2, it follows that ♯((ϕ|_c1)⁻¹(q′)) = 1 for some q′ ∈ ϕ(c₁) if and only if q′ is a point of valence 1 of ϕ(c₁). Since deg(ϕ) = 2 it follows d_q(ϕ) = 2 for q ∈ c₁ such that ϕ(c₁) has valence 1 on ϕ(c₁) and d_q(ϕ) = 1 for q ∈ c₁ if ϕ(q) does not have valence 1 on ϕ(c₁). In the case ϕ(c₁) would have a point q′ of valence 3 then there exist at least 3 different points q on e₁ with ϕ(q) = q′, contradicting deg(ϕ) = 2. Hence ϕ(c₁) can be considered as a finite edge with two vertices. Also for each q ∈ c₁ and v ∈ T_q(c₁) one has d_b(ϕ) = 1.

Assume ϕ(v₁) ≠ ϕ(v₂). Then ϕ(e₂) is a path from ϕ(v₁) to ϕ(v₂) outside of ϕ(c₁). This would imply g(ϕ(G₀)) ≥ 1, contradicting g(ϕ(G₀)) = 0, hence ϕ(v₁) = ϕ(v₂). This also implies that ϕ(m₀) and ϕ(m₁) are the two points of valence 1 on ϕ(c₁). Repeating the previous arguments for the loop c₂ = e₂ ∪ e₀ one obtains the given description for ϕ|G₀: G₀ → ϕ(G₀).

Step 2:g(ϕ(G₀)) = 1.

In case g(ϕ(G₀)) = 0 then we have the description for ϕ|_G0:G₀ → ϕ(G₀) obtained in Step 1. We are going to prove that this description cannot hold. Since g(ϕ(G₀)) ≤ 1, this implies g(ϕ(G₀)) = 1.

Consider ϕ(q₁) ∈ ϕ(e₁) and q′1≠q1 on e₁ with ϕ(q1)=ϕ(q′1) (see Figure 2). Assume ϕ(γ₁) is a tree. If ϕ(γ₁) would have valence 1 at ϕ(q₁) 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(q₁) cannot be a point of valence 1 on ϕ(γ₁). Hence there exists q″1∈γ1\{q1} with ϕ(q1)=ϕ(q″1) and we obtain ♯(ϕ⁻¹(ϕ(q₁))) ≥ 3, contradicting deg(ϕ) = 2. It follows g(ϕ(γ₁)) = 1, hence ϕ(γ₁) contains a loop e′1 in Γ. In case ϕ(q1)∉e′1 then again we obtain ♯(ϕ⁻¹(ϕ(q₁))) ≥ 3, contradicting deg(ϕ) = 2. Therefore ϕ(q₁) ∈ ϕ(q1)∈e′1 and e′1∩ϕ(G0)={ϕ(q1)}.

Repeating the arguments using q₂ and γ₂ we obtain a loop e′2 in Γ such that e′2∩ϕ(G0)={ϕ(q2)}, hence e′1∩e′2=0. Since g(Γ) = 1 this is impossible. As a conclusion we obtain g(ϕ(G₀)) = 1 finishing the proof of step 2.

In the case ϕ(c₁) would have genus 0 (c₁ as in the proof of Step 1), then from the arguments used in Step 1 it follows that for q ∈ e₂ \ {v₁, v₂} one has ϕ(q) ∉ ϕ(c₁). In the case ϕ(v₁) ≡ ϕ{v₂) it implies ϕ(c₂) has genus 1 (again c₂ as in the proof of Step 1). In the case ϕ(v₁) = ϕ(v₂) and g(ϕ(c₂)) = O too, it would imply g(ϕ(G₀)) = 0

so this cannot occur. Therefore without loss of generality, we can assume ϕ(c₁) has genus 1 (but then ϕ(c₂) could have genus 0).

Step 3:ϕ|_c1: c₁ → ϕ(c₁) is an isomorphism (meaning it is finite harmonic of degree 1)

Since g(ϕ(c₁)) = 1 it follows there is a loop e in ϕ(c₁), finitely many points r₁, · · ·, r_t on e and finitely many trees T_i inside ϕ(c₁) such that

T_i ∩ e = {r₁}

Ti∩Tj=0 in the case i ≡ j

ϕ(c₁) = e ∪ T₁ ∪ · · · ∪ T_t

(of course t = 0, hence ϕ(c₁) is a loop, is also possible; we are going to prove that t = 0).

Fig. 4

g(ϕ(c₁)) = 1

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 c₁, contradicting deg(ϕ) = 2. So we obtain a situation like in Figure 4. Let r ∈ {r₁, · · · , r_t} and let T be the associated subtree of ϕ(c₁). Then ϕ⁻¹(r) = {r′, r″} ⊂ c₁ with r′ ≡ r″. The tangent space T_r(ϕ(c₁)) consists of 3 elements (see Figure 5). Hence there exists w ∈ T_r′(G₀) \ T_r′(c₁) 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'n defining w hence ϕ(f) ⊂ e. Because of Lemma 4.1 this implies r′ is one of the points q_i on G₀ 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(c₁) = e is a loop.

Fig. 5

In the case t ≡ 0

Because deg(ϕ) = 2 we cannot go back and forth on e moving along c₁ and taking the image under ϕ. In principle it could be the case that there exist different points q′, q″ on c₁ such that the image of the closure of both components of c₁ \ {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 c₁ 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 c₁ 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∈Tq'(G'n) with d_ϕ(q′)(υ) ∈ T_ϕ(q′)(e) \ d_ϕ(q′)(T_q′ (c₁)). Hence the edge f of Gn' 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 c₁ such that ϕ(q′) = ϕ(q″) then ϕ|_c1 : c₁ → e is harmonic of degree 2 and d_q (ϕ|_c1) = 1 for all q ∈ c₁. In this case ϕ(e₂) ∩ e = {ϕ{υ₁), ϕ{υ₂)} since deg(ϕ) = 2. In the case ϕ(υ₁) ≠ ϕ(υ₂) then this contradicts g(Γ) = 1. In the case ϕ(υ₁) = ϕ(υ₂) then because of the description of ϕ|_c1 one has l(e₁) = l(e₀). We assume this is not the case, so we can assume ϕ:_c1 → e is bijective.

In the case for each edge f on Γ˜ with f ⊂ c₁ one has d_f(ϕ) = 2 then again, since ϕ(υ₁) ≠ ϕ(υ₂) we have ϕ(e₂) ∩ e = {ϕ(υ₁), ϕ(υ₂)}, contradicting g(Γ) = 1. Assume there exists q ∈ c₁ being a vertex of Γ˜ and two edges e′, e″ of Γ˜ contained in c₁ 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'∈Tq(Γ˜) correspond to e′ then there exists v∈Tq(Γ˜) with υ ∉ T_q (c₁) such that d_ϕ(q)(υ) = d_ϕ(q)(υ′). Let f be the edge of Γ˜ 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 : c₁ → e is an isomorphism of metric graphs.

Step 4: Finishing the proof of the theorem.

It follows that ϕ(υ₁) and ϕ(υ₂) do split e into two parts e′ and e″ of lengths l(e₁) and l(e₀). Since g(Γ) = 1 it follows ϕ(e₂) contains e′ or e″, we assume it contains e′. In the case ϕ(e₂) would contain q˜∈e"\{ϕ(v1), ϕ(v2)} then because of g(Γ) = 1 it follows ϕ(e₂) contains one of the connected components of e"\{q˜}. 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 c₁ and two different points of e₂.

So we obtain different points r₁, … , r_t on e′ and trees T₁, … , T_t with T_i ∩ e′ = r_i for 1 ≤ i ≤ t and Ti∩Tj=0 for i ≠ j such that ϕ(e₂) = e′ ∪ T₁ ∪ … ∪ T_t. It is possible (and we are going to prove) that t = 0, hence e′ = ϕ(e₂). In the case r_i ∉ {ϕ(υ₁), ϕ(υ₂)} then there exist two different points r′, r″ on e₂ such that ϕ(r′) = ϕ(r″) = r_i. Since r_i is also the image of a point on c₁ we get a contradiction to deg(ϕ) = 2. Hence t ≤ 2 and r_i ∈ {ϕ(υ₁), ϕ(υ₂)}.

Assume r_i = ϕ(υ₁). We obtain q ∈ e₂ with q ∉ {υ₁, υ₂} and ϕ(q) = ϕ(υ₁). 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 Γ˜ 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(e₂) = e′. Since deg(ϕ) = 2 it also implies d_f(ϕ) = 1 for each edge f of Γ˜ contained in e₂, hence l(e′) = l(e₂). Since l(e₂) ∉ {l(e₀), l(e₁)} we obtain a contradiction, finishing the proof of the theorem. □

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