# Circuit Reduction on Dual Graph of an Algebraic curve

I want to compute the resistance function r(p,q) between any two vertices of a fairly complicated graph. This resistance function is the one in Admissible pairing on a curve by Shouwu Zhang, section 3 page 179.

The graph is too complicated and merely applying series and parallel reductions I am getting nowhere. Can one direct me to a good reference for circuit reductions on metrized graphs?

## 1 Answer

In general, you shouldn't expect to be able to compute the effective resistance of an electrical circuit using only series and parallel reductions; the class of graphs (called series-parallel graphs) for which this can be done is quite special. There are other transformations for computing resistances such as star-triangle / delta-Y transformations but similar comments hold.

However, resistance computations for metrized graphs are really not so bad. They always boil down to solving a linear system of equations involving the Laplacian matrix of a weighted graph which models the metrized graph. This expository paper of Baker and Faber seems to be a good reference; see Theorem 4 for the relation between the Laplacian operator of a metrized graph and the Laplacian matrix of a weighted graph model.

Since the two points you care about are vertices of the metrized graph, you can proceed as follows. First, write down the Laplacian matrix for a weighted graph model of your metrized graph with the same vertex set $V$ (see Definition 7 of Baker and Faber). This is a symmetric $|V|\times|V|$ matrix $Q$. Let $\delta_j$ be the $|V|$-dimensional vector whose components are all zero except for the $j$th component, which is 1. Then it is possible to find a vector $v$ such that:

$$Qv=\delta_p-\delta_q.$$

(Note that $Q$ is not invertible, so $v$ is only defined up to addition of scalar multiples of the vector $(1,1,1,\dots,1)$).

The effective resistance you want is $r(p,q)=v_p-v_q$; note that this difference is well-defined even if $v$ is not.

You can derive this formally from the properties of $r(p,q)$ (see e.g. section 6 of Baker and Faber), but I wrote it down using the following physical intuition: suppose one unit of current is injected at $p$ ($+\delta_p$) and removed at $q$ ($-\delta_q$), then the effective resistance is equal to the potential difference between $p$ and $q$.

One way to compute $r(p,q)$ without worrying about the kernel of $Q$ is to delete any row and column of $Q$ to get a matrix $\tilde Q$ (now with full rank $|V|-1$) and delete the corresponding rows of $\delta_p,\delta_q$ to get vectors $\tilde\delta_p,\tilde\delta_q$. Then there is a unique solution to

$$\tilde Q\tilde v=\tilde\delta_p-\tilde\delta_q,$$

(throw this into your favorite computer algebra / numerical linear algebra program) and $\tilde v_p-\tilde v_q$ is the effective resistance you want.

• Is this method of deleting a row and column of $Q$ to produce an invertible matrix $\tilde{Q}$ also useful for finding the Moore-Penrose inverse of $Q$? This may be a naive question so I apologise in advance. – Chitrabhanu Jan 16 '18 at 10:56
• @Chitrabhanu I'm not sure exactly what you have in mind, but I suppose the Moore-Penrose inverse can be used to compute an orthogonal projection operator onto the orthogonal complement of the kernel of $Q$. Deleting a row and column is applying a different projection operator so I don't think it can be used in the way you're asking. – j.c. Jan 16 '18 at 13:16
• Thanks @j.c. for helpful answer. Can you give some reference for the row and column deletion method that you outlined above? – debargha Jan 17 '18 at 14:52
• @debargha It's not hard to see that extending $\tilde v$ to $v$ by adding a zero in the deleted component results in a solution to $Qv=\delta_p-\delta_q$, but you can also see for instance the paragraph after Kirchhoff's Voltage Law on page 15 of these notes by David G Wagner math.uwaterloo.ca/~dgwagner/Networks.pdf . – j.c. Jan 17 '18 at 19:37
• @j.c. thanks a lot for a prompt reply. This is really helpful. – debargha Jan 18 '18 at 4:37