6
$\begingroup$

Let $G$ be a vertex-transitive locally finite graph and $c_n$ the number of self-avoiding walks in $G$ starting from some fixed vertex $v_0$. One can easily see that $c_{m+n} \leq c_m c_n$ and hence Fekete's subadditive lemma gives that there exists the limit $\mu := \lim_{n\to\infty} c_n^{1/n}$. This quantity is known in the literature as the connective constant of $G$.

My question is basically what is the source of this term? I.e. it seems to be a strange choice of words for this quantity, so is there some application where it plays a role in some kind of "connectedness" which would excuse the name?

$\endgroup$
1
  • 4
    $\begingroup$ I think it sort of makes sense as terminology interpreted in a straightforward way. A regular tree of degree r is the degree r graph that will have the highest connective constant, and a tree is the least “connected” such graph in the sense of joining up with itself. The more “connected” the graph is, the more the walks will collide with themselves, and the less SAWs there will be. $\endgroup$ Commented Nov 14, 2022 at 12:55

1 Answer 1

4
$\begingroup$

Q: Is there some application where $\mu$ plays a role in some kind of "connectedness" which would excuse the name?

A: The application is to crystalline structure. The name originates from Hammersley, who introduced [1,2] the "connective constant" to characterize the bonds in a crystal.

In a certain sense [the connective constant] measures the richness of the connexions in a crystal.

For example, if each atom in a branching process has $M$ direct descendant atoms to each of which there is a one-way bond, then $\mu=\log M$.

[1] J. M. Hammersley, Percolation processes II. The connective constant, Proc. Camb. Phil. Soc. 53 (1957), 642–645.
[2] S. R. Broadbent and J. M. Hammersley, Percolation processes (1957).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.