Let $G=(E,V)$ be a connected Graph with $V \subset \mathbb{Z}^n$ Am I right with my assumption that if $(x,y) \in E$ meaning x and y are connected with an edge $e$ with the conductance $c(e)$. Then the effective resistance between those two points is at most $R(x \leftrightarrow y) \leq \frac{1}{c(e)}$ ?
Thanks in advance
On
This is one of the things that's quick to see from the random-walk definition of effective resistance. (Knowing the monotonicity law is also quick, of course, but less generally useful.)
In that interpretation, we define $R(x \leftrightarrow y)$ in terms of the probability that the conductance-weighted random walk starting at $x$ reaches $y$ before it returns to $x$. If that probability is $p$, and if $c_x$ is the sum of the conductances of all the edges incident to $x$, then $$R(x \leftrightarrow y) := \frac{1}{p \cdot c_x}.$$
Well, if there is an edge $e$ from $x$ to $y$, there is a probability of $\frac{c(e)}{c_x}$ that the conductance-weighted random walk goes directly from $x$ to $y$. If that happens, then it definitely reaches $y$ before returning to $x$. So we have $$p \ge \frac{c(e)}{c_x}$$ which tells us that $p \cdot c_x \ge c(e)$, and therefore $R(x \leftrightarrow y) \le \frac1{c(e)}.$
I believe that is correct, as adding any edges to any graph cannot increase the effective resistance between any two vertices.