Relating the toric rank of a semistable curve and the first Betti number of its reduction graph

Question

Introduction

Let $k$ be a local field. Let $C$ be the spectrum of $\mathcal{O}_{k}$. Let $X/k$ be a smooth projective curve with a semistable model $\mathcal{X}/C$.

Let $J$ be the Jacobian of $X$. The identity component of the reduction $\tilde{J}$ fits into an exact sequence \[ 1 \to L \to \tilde{J}^{0} \to A \to 0, \] where $L$ is a linear group, and $A$ is an abelian variety. Furthermore $L$ can be decomposed in a torus $T$ and unipotent group $U$. (Since we have a semistable model, $U = 0$.) Let $t$ denote the rank of this torus $T$.

We can also associate a reduction graph $G$ to $X$, see e.g., [Liu, §10.4], but as Liu points out there, there are different ways of defining a reduction graph. To this graph we can associate its first Betti number, which is the rank of the first homology group. (Observe that $\beta(G)$, equals $\#E - \#V + 1$. I.e., the number of edges left after removing a spanning tree.)

Claim

In [Zha, lem 5.2.2] it is suggested that $t$ equals the $\beta(G)$.

I am looking for a proof of this claim. The first result that comes to mind is [Liu, prp 10.1.51(c)] but this is not exactly the result I am looking for. If we let $u$ denote the rank of $U$ (see above), then this proposition says $\beta(G) \le t + u$. Since $u = 0$ in our case, this does give an inequality in one direction.

What makes the question more difficult is that (as far as I see) Zhang does not specify which definition of reduction graph he uses. I guess he uses the same definition as [Yam, 1.10], since that article is very much related to the results of [Zha]. This definition is pretty much the same as the one in [Liu], although I think [Yam] allows loops in the reduction graph, for self-intersections of irreducible components. The problem, of course, is that allowing or disallowing loops has quite an impact on the rank of the first homology group. Allowing loops raises $\beta(G)$, and therefore the above inequality ($\beta(G) \le t$) is no longer necessarily true.

Question

Since I am only taking my first steps into the theory of reduction graphs I do not know whether the ramblings above or the question that follows makes any sense. M question essentially boils down to:

Under which conditions are $t$ and $\beta(G)$ equal, and why?

---

References

[Liu] Liu, Qing. Algebraic geometry and arithmetic curves. Translated from the French by Reinie Erné. Oxford Graduate Texts in Mathematics, 6. Oxford Science Publications. Oxford University Press, Oxford, 2002. xvi+576 pp.

[Yam] Kazuhiko Yamaki. Graph invariants and the height of the Gross-Schoen cycle. 2009. url: http://jairo.nii.ac.jp/0019/00096332/en.

[Zha] Shou-Wu Zhang. “Gross–Schoen Cycles and Dualising Sheaves”. arXiv:math/0812.0371.

Answer 1

Your "reduction" $\widetilde{J}$ is really the identity component of the reduction. Also, I think you should assume $\mathcal{X}$ is regular (as may be arranged). The big theorem that is relevant here is due to Raynaud (see 9.5/4 in "Neron Models"): the relative identity component of the Neron model is the separated open subgroup scheme ${\rm{Pic}}^0_{\mathcal{X}/C}$ inside the algebraic space ${\rm{Pic}}_{\mathcal{X}/C}$, so passing to special fibers gives that $\widetilde{J}$ is isomorphic to ${\rm{Pic}}^0_{\mathcal{X}_{\kappa}/\kappa}$, where $\kappa$ is the residue field of $k$.

So now the problem has nothing to do with Neron models and is entirely about how to compute the toric part of ${\rm{Pic}}^0_{X/K}$ for a connected semistable curve $X$ over an algebraically closed field $K$. We claim that the character group of the torus is canonically isomorphic to the integral homology of the (connected) graph associated to this semistable curve (so the rank $t$ of the torus is always the degree-1 Betti number of the graph). This is nicely explained (modulo a few details) in 9.2/8 of "Neron Models" using some exact sequences for the etale topology; the definition of the graph that is used there is given in the paragraph preceding 9.2/8 (using loops for "self-crossing" irreducible components).

Answer 2

I never studied this, but here is I guess the natural approach:

Suppost that the reduction is just a cycle of smooth curves $C_1, \ldots, C_k$ with $C_i$ intersecting $C_{(i+1)\mod k}$ in one point $p_i = q_{i+1\mod k}$ with $p_i\neq q_i$.

How do we construct a line bundle on this cycle? Such a bundle $L$ gives us by pullback a collection of line bundles $L_i$ on $C_i$, hence a (surjective) map $\tilde J\to A:= \prod Pic^0(C_i)$. What is the kernel? Suppose $L$ is trivial on every $C_i$ and choose a trivialization. This gives us isomorphisms $L_{p_i} \to L_{q_i}$. The composition $L_{p_1} \to L_{q_2} \to L_{p_2} \to \ldots \to L_{q_2} \to L_{p_1}$ is an automorphism of a $1$-dimensional vector space, i.e., an element $t$ of a $1$-dimensional torus $\mathbb{G}_m$. This is all the data we need.

In case there are more cycles, we will get such a $t$ for every cycle, hence the dimension of the toric part of $\tilde J$ will be equal to the number of cycles in the graph.