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\def\ppp{{\mathbb{P}}}
\def\hhh{{\mathbb{H}}}
\def\aaa{{\mathbb{A}}}
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\def\Pic{{\rm Pic}}
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\begin{document}


\author{A. Ksir, C. Melles, D. Joyner}
\title{A question about tropical $\Pic(X)$ as a $G$-module}
\date{7-7-2008}
\maketitle

\begin{abstract}
This paper addresses the following question:
Let $X$ be a tropical curve and let $G$ be a 
finite subgroup of the automorphism group of $X$. 
Let $D$ be a divisor and assume that its 
equivalence class $[D]$ is $G$-invariant.

\noindent
{\bf Question}: Is there always a $D'\in [D]$ which is
$G$-equivariant?

This was answered by Goldstein, Guralnick, and Joyner \cite{GGJ} in
the case of an irreducible algebraic curve over an algebraically
closed field. We try to extend the arguments of \cite{GGJ} to the 
tropical case. For instance, in the tropical case, we prove a
tropical analog of Hilbert's Theorem 90 and of 
Tsen's Theorem (in the abelian case). As our 
main result, we show that, the answer to the above question is ``yes''
for the class of tropical ``$\qqq$-divisors''.
\end{abstract}

%\tableofcontents

\section{Introduction}


Let $X$ be a connected tropical curve 
and let $G$ denote a (finite) subgroup of the
automorphism group of $X$. 

\begin{lemma}
A tropical curve can be identified with a metric
graph with leaves (possibly with infinite terminal edges,
possibly with loops). The automorphism group of a 
tropical curve can be identified with the isometry group of
that graph.
\end{lemma}

\pf
This is an immediate consequence of Proposition 5.3 in \cite{M1}.
\qed


Let $\ttt=\rrr\cup \{\pm \infty\}$.

Let $M(X)$ denote the (tropical) ring of {\it rational functions} on $X$,
i.e., continuous piecewise-linear $\ttt$-valued functions 
with integral slopes. Let $S\subset \qqq$ denote any (non-trivial,
additive) subgroup and let $M_S(X)$ denote the ring of continuous 
piecewise-linear functions on $X$ with $S$-valued slopes.
In this case we call $S$ the {\it slope group}. When
$S=\zzz$ we omit the subscript.
Let $M_S(X)^\sharp$ denote those functions
which are not equal to $\pm \infty$ on any part.

 Let $\Div(X)$ denote the group of divisors (the free abelian group
generated by the points of $X$) and call
$\Div_S(X)=\Div(X)\otimes S$ the {\it $S$-divisor classes}.
Define the map ${\rm div}:M_S(X)^\sharp \to \Div_S(X)$ by
${\rm div}(f)=\sum_{P\in X} {\rm ord}_P(f)\cdot P$, where
${\rm ord}_P(f)$ is the sum of all ($S$-valued) slopes of $f$ for all
edges emanating from $P$. Let

\[
\div: M_S(X)^\sharp\to \Div_S(X),
\]
and let
$\Prin_S(X)=\div (M_S(X)^\sharp)$ denote the subgroup of {\it principal divisors}
Let $\Pic_S(X)=\Div_S(X)/\Prin_S(X)$ denote
the {\it Picard group} of $S$-divisor classes.

For background on tropical rings, see for example Mikhalkin \cite{M1}.

\section{Cohomology}

The basic idea is to use group cohomology to attack this
question. For background on cohomology,
we reference Serre \cite{S}, ch. VII, or the survey
\cite{J}.


Let $S$ be a slope group and let $M=M_S(X)^\sharp$. The group
$G$ acts on the (tropical-multiplicative) 
abelian group $M$ via its action on $X$.

Define the {\it $1$-cocycles} on $G$ with coefficients in $M$ 
by

\[
Z^1(G,M) = \{f:G\to M\ |\ \forall g_1,g_2\in G,\ f(g_1)g_1f(g_2)=f(g_1g_2)\},
\]
the {\it $1$-coboundaries} by

\[
B^1(G,M) = \{f:G\to M\ |\ \exists m\in M \suchthat 
\forall g\in G,\ f(g)=gm\cdot m^{-1}\},
\]
and the {\it $1$-cohomology} by

\[
H^1(G,M) =Z^1(G,M)/B^1(G,M).
\]

The following result is roughly analogous to Hilbert's Theorem 90:

\begin{proposition}
\label{prop:thrm90}
$H^1(G,M_\qqq(X)^\sharp)=0$.
\end{proposition}

\pf
We prove something slightly more general (while at the same time 
illustrating some difficulties with computing
$H^1(G,M(X)^\sharp)$). Let $M_{|G|\zzz}(X)$ denote the ring of continuous 
piecewise-linear functions on $X$ with integer slopes, all of which are
multiples of $|G|$. We shall show that $H^1(G,M_{|G|\zzz}(X)^\sharp)$
is trivial.

Let $M=M_{|G|\zzz}(X)^\sharp$, $f\in Z^1(G,M)$ represent a class in
$H^1(G,M)$, so $f(g)=f(gx)gf(x)^{-1}$. Taking the product over all $x\in G$,
this implies $f(g)^{|G|}=(\prod_{x\in G}f(x))\cdot g(\prod_{x\in G}f(x))^{-1}$.
Taking (tropical) $|G|$-th roots, $f(g)=m\cdot gm^{-1}$, 
where $m=\prod_{x\in G}f(x)^{1/|G|}$. 
This implies $f(g)^{-1}\in B^1(G,M)$. Since $B^1(G,M)$ is an abelian group,
$f(g)\in B^1(G,M)$, so the class which $f$ represents is trivial.

The same proof applies if you replace $M=M_{|G|\zzz}(X)^\sharp$ by
$M=M_\qqq(X)^\sharp$ (since taking $|G|$-th roots at the last step is
still allowed).
\qed

\begin{remark}
What is $\div(M_{|G|\zzz}(X)^\sharp)$?
The map $f(x)\to f(x)^{|G|}$ defines a surjection
$M(X)\to M_{|G|\zzz}(X)$. Therefore, $\div(M_{|G|\zzz}(X)^\sharp)
=|G|\div(M(X)^\sharp)$.
\end{remark}

\begin{lemma}
For any non-trivial subgroup $S\subset \qqq$,
we have a short exact sequence of $\zzz[G]$-modules,

\[
0\rightarrow \rrr\oplus S^d \rightarrow M_S(X)^\sharp 
\rightarrow \Prin(X) \rightarrow 0,
\]
where $d$ is 
the number of leaves minus the number of non-terminal vertices
adjacent to a leaf.
\end{lemma}

\begin{remark} 
\begin{itemize}
\item
If $X$ is compact then $d=0$.

\item
This is the tropical analog of the well-known
short exact sequences

\[
1\rightarrow F^\times \rightarrow F(X)^\times 
\rightarrow \Prin(X) \rightarrow 0,
\]
for an irreducible non-singular 
algebraic curve $X$ over an algebraically closed field $F$.
\end{itemize}
\end{remark}

\pf
The order of a function $f$ at a point is the sum of the outgoing 
slopes. So for $f$ to be in the kernel, it must have
the the sum of the slopes at each point equal to $0$.

First, look at the metric graph of $X$ without its leaves. This 
is a compact 
set, so $f$ must take a maximum or minimum somewhere. But at 
the point where the minimum is attained, all outgoing slopes are 
greater than or equal to $0$. Therefore, $f$ must be constant 
on this part of the graph.

Second, if there multiple leaves at one vertex then the slop of $f$ 
must sum to $0$ at that vertex. Otherwise, the slopes can be 
arbitrary. 
\qed

\vskip .15in

Define the {\it $2$-cocycles} on $G$ with coefficients in $M$ 
by

\[
\begin{array}{l}
Z^2(G,M) = \{f:G\times G\to M\ |\ 
\forall g_1,g_2,g_3\in G,\\ 
\qquad \qquad \qquad \qquad \qquad %\qquad \qquad \qquad \qquad 
g_1f(g_2,g_3)f(g_1g_2,g_3)^{-1}f(g_1,g_2g_3)f(g_1,g_2)^{-1}=1\},
\end{array}
\]
the {\it $2$-coboundaries}\footnote{It is straightforward to check that 
$B^2(G,M)\subset Z^2(G,M)$.} by

\[
\begin{array}{l}
B^2(G,M) = \{f:G\times G\to M\ |\ \exists h:G\to M \suchthat 
\\ 
\qquad \qquad \qquad \qquad \qquad
\forall g_1,g_2\in G,\ f(g_1,g_2)=h(g_1)g_1h(g_2)h(g_1g_2)^{-1}\},
\end{array}
\]
and the {\it $2$-cohomology} by

\[
H^2(G,M) =Z^2(G,M)/B^2(G,M).
\]

By definition, we have a short exact sequence

\[
0\rightarrow \Prin(X) \rightarrow \Div(X) 
\rightarrow \Pic(X) \rightarrow 0,
\]
as $\zzz[G]$-modules.
The covariant functor of $G$-invariants,
$M\longmapsto H^0(G,M)=M^G$ is left exact.
Therefore, for $M=M_S(X)^\sharp$ we have

\begin{equation}
\label{eqn:longer}
\begin{split}
1\rightarrow H^0(G,\rrr\oplus S^d) \rightarrow H^0(G,M )
\rightarrow H^0(G,\Prin(X))\\ \rightarrow 
H^1(G,\rrr\oplus S^d) \rightarrow 
H^1(G,M)
\rightarrow  
H^1(G,\Prin(X)) \rightarrow \\
H^2(G,\rrr\oplus S^d)\rightarrow 
H^2(G,M)\rightarrow H^2(G,\Prin(X)) \rightarrow ...\ ,
\end{split}
\end{equation}
and

\begin{equation}
\label{eqn:long}
\begin{split}
0\rightarrow H^0(G,\Prin(X)) \rightarrow H^0(G,\Div(X)) 
\rightarrow H^0(G,\Pic(X)) \\
\rightarrow 
H^1(G,\Prin(X)) 
\rightarrow H^1(G,\Div(X)) 
\rightarrow H^1(G,\Pic(X)) \rightarrow  ...\ .
\end{split}
\end{equation}

The following result is a corollary of Proposition \ref{prop:thrm90}.

\begin{corollary}
\label{cor:4}
For $M=M_\qqq(X)^\sharp$ we have exact sequences

\[
0\rightarrow (\rrr\oplus \qqq^d)^G \to M^G
\rightarrow \Prin(X)^G \rightarrow 
H^1(G,\rrr\oplus S^d) \rightarrow 
0
\]
and

\[
0 \rightarrow  H^1(G,\Prin(X)) \rightarrow H^2(G,\rrr\oplus S^d)\rightarrow 
...\ .
\]

\end{corollary}



\section{Tropical curves}

Let $X$ be a connected tropical curve. The following computation
will not be needed for our main result.

%We claim that the answer to the above question 
%is ``no'' in general. 

\begin{proposition}
\label{prop:h1div=0} 
$H^1(G,\Div(X))=0$.
\end{proposition}

%This result is similar to a result of Tate on the Galois cohomology 
%of the divisor group of a variety over a $p$-adic field - see
%Tate's appendix in Lang \cite{L}. However, this result does 
%not follow from Tate's result since the $G$-action is different.

\begin{remark}
\begin{itemize}
\item
The analogous result for algebraic curves is proven
in \cite{GGJ} and the proof below follows the same idea.
\item
Much of the proof below (but not all) goes through for 
$H^2(G,\Div(X))$ as well.
\end{itemize}
\end{remark}

\pf
Let $GX$ denote the set of all orbits of points of $X$ 
under the action of $G$. Let $GX/G$ denote a complete set of 
representatives in $X$ of these orbits. For each
$P\in X$, let $G_P=\{g\in G\ |\ gP=P\}$ denote its 
stabilizer (or inertia group). 
We have a direct sum decomposition

\[
\Div(X)=\oplus_{P\in GX/G} \oplus_{g\in G/G_P} \zzz [gP].
\]
Let 
\[
X_{ram}=\{P\in X\ |\ G_P\not= 1\},
\]
\[
X_{unram}=\{P\in X\ |\ G_P= 1\}.
\]
Let $\Div(X)_{ram}$ and $\Div(X)_{unram}$ denote the corresponding
$\zzz[G]$-submodules of $\Div(X)$. Unlike for the 
algebraic curve analog, $\Div(X)_{ram}$ can have infinite rank
(except in the compact case).

We have 

\[
\begin{split}
\Div(X)_{unram}
&=\oplus_{P\in (GX/G)_{unram}} 
\oplus_{g\in G} 
\zzz [gP]\\
&= \oplus_{g\in G} \oplus_{P\in (GX/G)_{unram}} 
\zzz [gP],
\end{split}
\]
so $\Div(X)_{unram}$ is induced in the sense
of Serre \cite{S}, page 110. This implies
$H^1(G,\Div(X)_{unram})=0$, by Prop. 1 in \S VII.2
of \cite{S}.

Recall $H^1(G,L_1\oplus L_2)= H^1(G,L_1)\oplus H^1(G,L_2)$,
for any $G$-modules $L_1,L_2$. 
Suppose that $H^1(G,\Div(X)_{ram})\not= 0$ and let
$f\in Z^1(G,\Div(X)_{ram})$ denote a non-trivial cocycle.
Since $G$ is finite, $f$ is supported on finitely many points.
Let $\mathcal{P}$ denote the $G$-orbit of these points
and let $L$ denote the
sublattice of $\Div(X)_{ram}$ generated by $\mathcal{P}$.
There is no harm in assuming that $G$ is transitive
on the basis of $L$ ($L$ is a direct 
sum of such lattices). So, let
$L=\oplus_{g\in G/G_P} \zzz [gP]$, for some $P\in GX/G$.
So $L \cong ind_H^G(\zzz)$, where $H\cong G_P$ is the stabilizer of one
of the basis elements. Here $G$ acts on the induced module

\[
ind_H^G(\zzz)=\{f:G\rightarrow \zzz \ |\ f(hg)=hf(g),\ \ 
\forall h\in H,\ g\in G\},
\]
which are just the $\zzz$-valued functions on $H\backslash G$,
by right multiplication.
By Shapiro's Lemma, $H^1(G,ind_H^G(\zzz)) \cong H^1(H,\zzz)$.
Now $H^1(H,\zzz)=Hom(H,\zzz)=0$, since 
$H$ is finite.

\qed

\begin{corollary}
The map

\[
H^0(G,\Pic(X)) \rightarrow H^1(G,\Prin(X)) 
\]
is surjective.
\end{corollary}

\begin{proposition}
\label{prop:tsen}
If $G$ is abelian then $H^2(G,M_\qqq(X)^\sharp)$ is trivial.
\end{proposition}

\pf
First, assume $G$ is cyclic and let $M=M_\qqq(X)^\sharp$.
Then $H^2(G,M)=M^G/NM$, where $N:M\to M$ is the norm map.
Since (tropical) $|G|$-th roots in $M$ always exist,
$N$ is surjective, so $H^2G,M)$ is trivial in this case.

In general, $G$ is a product of cyclic groups. In this case, the
result follows by induction and the inflation-restriction 
sequence $0\to H^2(G/H,A^H)\to H^2(G,A)\to H^2(H,A)$.
\qed

%\begin{corollary}
%If $G$ is abelian then $H^2(G,\Prin(X))\cong 
%H^2(G,\rrr\oplus \qqq^d)$.
%\end{corollary}

\begin{remark}
\begin{itemize}
%\item
%We want to know if $H^1(G,\Prin(X))$ is trivial or not, as 
%then (thanks to the above proposition) the
%answer to the question will follow from (\ref{eqn:long}).
%
\item
In the case of an algebraic curve,
$H^2(G,F^\times(X)))=1$ by Tsen's theorem 
(a function field 
over an algebraically closed field is a $C^1$ field;
see the Corollaries on pages 96 and 
109 of Shatz \cite{Sh}, or \S 4 and \S 7 of chapter X
in \cite{S}). An analog of Tsen's theorem for tropical curves is,
for $G$ abelian, given in Proposition \ref{prop:tsen}.
An analog for general finite groups,
if it exists, would be very interesting and useful.

\item
In the case of an algebraic curve, $H^1(G,F(X)^\times )=1$,
by Hilbert's Theorem 90.
A tropical curve analog of Hilbert's Theorem 90 is 
Proposition \ref{prop:thrm90}.
\end{itemize}
\end{remark}

%\begin{corollary}
%If $X$ is compact and $G$ has trivial Schur multiplier
%then $H^1(G,\Prin(X))$ is trivial and the map
%
%\[
%Div(X)^G\to Pic(X)^G
%\]
%is surjective.
%\end{corollary}
%
%\begin{remark}
%This implies that the answer to the question raised in the 
%introduction is ``yes'' if
%(a) $X$ is compact, (b) $G$ has trivial Schur multiplier.
%\end{remark}
%
%\pf
%The hypothesis implies $d=0$ and $H^2(G,\rrr )=0$, so
%by (\ref{eqn:longer}), we have 
%
%\[
%H^1(G,M_\qqq(X)^\sharp )\to H^1(G,\Prin(X) )\to 0.
%\]
%By by Proposition \ref{prop:thrm90}, $H^1(G,M_\qqq(X)^\sharp )=0$,
%so $H^1(G,\Prin(X))$ is trivial. The second statement now follows from
%Corollary \ref{cor:4}.
%\qed

\begin{proposition}
\label{prop:H2constants}
$H^2(G,\rrr\oplus \qqq^d)$ is trivial.
\end{proposition}

\pf
Let $S\subset \qqq$ be any $|G|$-divisible subgroup.
By Rotman \cite{R} Proposition 10.119, we have
$|G|\cdot H^2(G,\rrr\oplus S^d)=0$. Since $\rrr\oplus S^d$ is
torsion-free,
this means that if $f\in Z^2(G,\rrr\oplus S^d)$ then
$|G|\cdot f\in B^2(G,\rrr\oplus S^d)$. Since
$S\subset \qqq$ is $|G|$-divisible, so is
$B^2(G,\rrr\oplus S^d)$. This forces, $f\in B^2(G,\rrr\oplus S^d)$.
\qed

\begin{remark}
If $S\subset \qqq$ is finitely generated (which means
$S=a\zzz$, for some $a\in \qqq$) then $H^2(G,\rrr\oplus S^d)$ is 
finite. Indeed, 
$H^2(G,\rrr\oplus \qqq^d)= H^2(G,\rrr)\oplus H^2(G,S^d)$ and
$H^2(G,\rrr)=0$ (this is a corollary of the proof above).
Since $S$ is finitely generated, so is $S^d$.
Now apply Rotman \cite{R} Corollary 10.120. \qed
\end{remark}

The following is our main result and implies
that the answer to the question raised in the 
introduction is ``yes'' for the $\qqq$-divisors.

\begin{theorem}
If $S=\qqq$ then 
\[
\Div(X)^G\to \Pic(X)^G
\]
is surjective 
\end{theorem}

\pf
By Corollary \ref{cor:4}, the map 
$H^1(G,\Prin(X))\to H^2(G,\rrr\oplus \qqq^d)$ is injective.
Therefore, $H^1(G,\Prin(X))=0$ by the previous result.
The result now follows from (\ref{eqn:long}).
\qed

We return to the case where $S\subset \qqq$ can be arbitrary.
Consider $\Pic^0(X)=\Pic_S^0(X)$, the subgroup
of $\Pic(X)$ of degree $0$ divisors (i.e. the Jacobian).
The degree map $deg_S$ defined on $\Div_S(X)$ factors through
$\Pic_S(X)$. Let $B$ be $deg(\Div_S(X)^G)$.  We identify 
this with a subgroup of $S\cong \Pic_S(X)/\Pic_S^0(X)$.  
Clearly, $B$ contains $|G|\cdot S$ and may
be bigger.  

The analog of Proposition \ref{prop:h1div=0} is the following result.

\begin{lemma}  
\label{lemma:guralnik}
$H^1(G,\Div_S^0(X)) \cong S/B$
and $B=bS$,
where $b={\rm min}\, \{|G|/|I|\ |\ I\subset G\ {\rm an\ inertia\ subgroup}\}$.
\end{lemma}

\begin{remark}
The analogous result for algebraic curves is also true
and the proof below, is basically the same argument as
that given in \cite{GGJ}.
\end{remark}

\pf  
First, we prove the isomorphism.
Consider the sequence
$ 0 \rightarrow \Div_S^0(X) \rightarrow \Div_S(X) \rightarrow S \rightarrow 0$.
The map from $\Div_S(X)$ to $S$ is $deg$.  Using Proposition
\ref{prop:h1div=0} yields:

\[
(0\rightarrow \Div_S^0(X)^G\rightarrow )\ \ 
\Div_S(X)^G \rightarrow S \rightarrow H^1(G,\Div_S^0(X)) \rightarrow 0,
\]
as asserted.

Now we prove the claim about $B$.
Observe that the smallest possible degree of a divisor 
fixed by $G$ is the size of the smallest orbit of $G$ acting
on $X$. Since each inertia group $I$ is the stabilizer
of a point $P$, the orbit of $P$ under $G$ is 
in one-to-one correspondence with $G/I$.
Therefore, the size of the smallest possible degree of a divisor 
fixed by $G$ is equal to the minimum of the $|G|/|I|$,
as $I$ ranges over all the inertia subgroups. 
Since $d(\Div_S(X)^G)$ is an abelian group, this 
proves the claim.
\qed

Now consider the short exact sequence
\[
0 \rightarrow \Prin_S(X) \rightarrow 
\Div_S^0(X) \rightarrow \Pic_S^0(X) 
    \rightarrow 0.
\]
Taking fixed points leads to the long exact sequence for cohomology:

\[
\begin{split}
0 \rightarrow \Prin_S(X)^G\rightarrow 
\Div_S^0(X)^G \stackrel{\phi}{\rightarrow} 
\Pic_S^0(X)^G  \rightarrow \\
H^1(G,\Prin_S(X)) \rightarrow H^1(G,\Div_S^0(X)) \rightarrow H^1(G,\Pic_S^0(X)).
\end{split}
\]

\section{Remarks on the general case}

This section, to be written, will explain how to
weaken the condition $S=\qqq$ to $S=\zzz$.

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\end{document}