# How is the Jacobian or Sandpile group of a graph computed?

From what I understand, given a graph, the Jacobian group and the Sandpile group refer to the same object. Until now, I have been computing this group in the way detailed in Chapter 1 of this undergraduate thesis. The approach is to use a theorem that says for any graph $G$ on $n$ vertices, $$\text{Jac}(G) \cong \mathbb{Z}^{n-1}/\text{im}\left(\tilde{\Delta}\right),$$ where $\tilde{\Delta}$ is the reduced Laplacian of the graph. To find the image of $\tilde{\Delta}$, it is first reduced to Smith normal form. I am looking for more insight into the process and was wondering if anyone could explain or link an alternative method for arriving at the Jacobian of a graph. In particular, I am interested in graphs of hypercubes.

• The SNF seems to me to be the fastest way to compute the isomorphism type of the Jacobian of arbitrary graphs. There are other things you could want though... like representatives (recurrent configurations or superstable configurations). – Sam Hopkins Feb 3 '17 at 14:04
• Hi! Thank you for the response! I would indeed like the isomorphism type of the Jacobian, but I'm not necessarily concerned with efficiency. I'm looking for intuition about the Jacobian and was hoping to find a different method that might shed a different light on the subject. – Aaron Bagheri Feb 13 '17 at 10:56
• I recommend this book in progress as a starting point to learn about Jacobians of graphs: people.reed.edu/~davidp/divisors_and_sandpiles. In particular, Exercise 12.3 might be interesting to you: it shows that deleting an edge from your graph G can increase the minimal number of generators of the Jacobian. – Sam Hopkins Feb 14 '17 at 18:45

In a survey article on chip-firing, the sandpile group is defined on page 7 (definition 2.7) as $$\mathbb{Z}^{n-1} / \mathbb{Z}^{n-1} \tilde{\Delta}$$
Note that the authors use $$\Delta'(G)$$ to denote the reduced Laplacian. In the paper, $$\mathbb{Z}^{n-1} \Delta'(G)$$ refers to the abelian group generated by taking integer-linear combinations of the rows of $$\Delta'(G)$$, which I assume is what you mean by $$im(\tilde{\Delta})$$.
• You say "Here $\mathbb Z^{n - 1}\Delta'(G)$ refers to …", but you haven't used $\mathbb Z^{n - 1}\Delta'(G)$. Did you mean $\mathbb Z^{n - 1}\tilde\Delta$? – LSpice Nov 20 '18 at 14:47
• I just meant to clarify that $\Delta'(G)$ in the linked paper refers to the same thing as $\tilde \Delta$ does in this question (the reduced Laplacian), since that notation is defined in an earlier section of the paper and I was linking to the middle of it. – Akhil Jalan Nov 21 '18 at 4:45