Estimate on currents in Cayley graphs Take a Cayley graph $\Gamma$ (thought of as an electrical network with all edges having equal resistance) and break one edge $e$ and put a battery there. (Assume the graph has only one end* so that this operation does not disconnect the graph.) Are there any estimates on the resulting current?
In more mathematical terms, write $e=(s,t)$, then define the voltage at a vertex $x$ to be [Edit: Formula incorrect see Ori's comment after Victor's answer] 
$$
V_x = 
\mathbb{P}(\textrm{a simple random walk starting at } x \textrm{ hits } s \textrm{ before it hits } t) = \mathbb{P}^x(X_{\tau_{ \lbrace s,t \rbrace }}=s)
$$
$V_x$ is (up to a constant**) a harmonic function, and the current on the edge $(x,y)$ is just $V_y - V_x$. It turns out $V_x$ may be chosen so that $(x,y) \mapsto V_y-V_x$ is in $\ell^2(E)$.
Are there any finer estimates on the decay of these functions (on the edges)? For example, might it even belong to $\ell^1(E)$?
$\textrm{ }$
*[Edit] Having one end implies being infinite, not X where X stands for cyclic, free, amalgamated product or HNN extension over a finite subgroup (a theorem of Stallings). One could add virtually before these adjectives but it is redundant (see HW's comment)
**[Edit] it could happen that the function is even less uniquely defined. For example, if there is a harmonic function with $\ell^2$ gradient on the graph $\Gamma \setminus e$. However, the element in $\overline{\nabla \ell^2(X)}^{\ell^2(E)}$ (see PPS below) is unique up to a constant. As far as my little knowledge of the topic is concerned, it might happen that this element is also not the best in terms of decay.
PS: it does not matter whether the edge $e$ is removed or not when defining the probability $V_x$.
PPS: this is classical, but a way to see that the current belongs in $\ell^2(E)$, is that $\ell^2(E)$ decomposes as $\overline{\nabla \ell^2(X)} + \ker \textrm{Div}$, projecting the Dirac at $e$ on the first factor gives the desired current (except on the edge $e$). [Edit: another way is given in Victor's answer below]
 A: Do you ask the random walk on your group to be recurrent? As otherwise, you can have strange effects (take $\mathbb{Z}^3$, then the random walk on it hits a given point with probability less than one, so from one point charge you can have a current "flowing to infinity")...
Two remarks from the point of view of physical intuition: 
0) [most probably, useless]  The fact that the current function is in $l^2$ means that the total power dispersed by the flow is finite (on an edge $e$, you disperse the power $I_e^2 R$), and as the it should be equal to $U^2/R'$, where $R'$ is the equivalent resistance of your network, it means that $R'$ is positive (and this is rather natural: no matter what happens afterwards, you have only $2n$ edges attached to your starting point).
1) If you're asking for your function to be in $l^1$: this is a total number of electrons that are on their way in a given moment. It seems that its finiteness is equivalent for a random walk to be not only recurrent, but even having a finite expectation time of hitting $y$... Am I right here? If yes, it seems to me quite improbable to have such a possibility in an infinite group: I would expect that the distance to a given point can have a zero or positive drift, but not a negative one...
A: The current from a vertex to infinity in any graph is never in $\ell^1(E)$. Any cutset (that is, a set of edges separating the origin vertex from infinity) will contribute at least a constant to the $\ell^1$ norm, and there are infinitely many disjoint cutsets.
A: [Edit: add a heuristic argument for it not being $\ell^1$]
Trying to recycle the answer below and understand what was behind Ori and Victor' ideas:
The current from $1$ to $\infty$ is not in $\ell^1(E)$: take $S_n$ to be the sphere around $1$, and $T_n$ the edges between $S_n$ and $S_{n+1}$. Then the quantity of current in $T_n$ is constant for any $n$.
Assume very naively that $|T_n|$ is roughly $g'(n)$ where $g(n)$ is the growth function (i.e. $g(n) = | \cup_{0\leq i \leq n} S_n|$). Assume very optimistically that the current is uniformly distributed on $T_n$, i.e. $f(e') = 1/|T_n|$ if $e' \in T_n$.
Take the difference between this current $f$ and the one translated to get a current $\tilde{f}$ between $1$ and a neighbour. Assume very crudely that $\tilde{f}$ has values approximately the derivative of $f$. Then $\tilde{f}$ is roughly $g''(n)/g'^2(n)$ on edges at distance $n$ from $1$. So it would never be in $\ell^1$.
There is an obvious flaw in this argument: even if $\tilde{f}$ is a satisfactory current, it might not be the most appropriate (as in the [otherwise excluded] case where $e$ disconnects the graph, see Misha's comment above)
(This is also unhelpful for a decay intermediate between $\ell^1$ and $\ell^2$.)
[Edit: As Ori points out below, this answer is flawed (flow described does not satisfy the two Kirchhoff laws).]
This is just a partial answer, for the case $G=\mathbb{Z}^d$ ($d>1$). [Edit: For the usual generating set, answer may depend on generating set.]
If I'm not mistaken, there's an explicit formula for the current between $0$ and $\infty$ given as follows. Define the current on the first $2^d$-ant (points $x=(x_1,\ldots,x_d)$ with $x_i \geq 0$) as in Bollobás' "Modern Graph theory" proof of Theorem 14 p.309, namely the current between $x$ and $x+ e_i$ (where $e_i$ is a standard basis vector) is:
$$
\frac{(x_i+1)}{(n+1)(n+2)\cdots(n+d)} \qquad \textrm{where } n = \sum_{i=1}^d x_i
$$
Then sum this with similar currents defined on each $2^d$-ant to get the required harmonic current (except at $0$).
This gives a current between $0$ and $\infty$, obtain one between, say $(1,0,\ldots,0)$ and $\infty$ by simply translating. Take the difference of these two currents to get a (relatively explicit) formula for the current between $0$ and one of its neighbours. A rough estimate (inequalities are up to constants) for the $p$-summability is 
$$
\sum_{n\geq 0} \sum_{\sum x_i = n} \sum_{i=1}^d  \frac{(x_i+1)^p}{n^p (n+1)^p \cdots (n+d)^p} \leq \sum_{n\geq 0} \binom{n+d}{d}\frac{1}{(n+1)^p \cdots (n+d)^p} \leq \sum_{n\geq 0} \frac{1}{n^{d(p-1)}} 
$$
So it is not in $\ell^1(E)$ but it belongs to $\ell^p(E)$ for $p> 1+\tfrac{1}{d}$.
So allow me to refine the initial question in:


*

*Are there similar estimates in more general groups $G$ on the current between the identity and $\infty$?

*Suppose $G$ is of exponential (or maybe superpolynomial) growth, does the current belongs to $\ell^1(E)$? (I guess that some monotonicity may be used to conclude directly it belongs to $\ell^p(E)$ for $p \in ]1,2]$). 
