In this document, on page 7 below, it is said that for the travelling wave solutions $$ u_k^t=U_L(t\pm k-m) $$ with $L\geq e+r+1$ and $$ U_L(j)=\begin{cases}j, & 0\leq j\leq e+r\\0, & e+r+1\leq j\leq L,\text{ and }U_L(j+L)=U_L(j)\end{cases} $$
we have that the winding number (which is defined on the same page and the previous page) is $$ W_L(u^t)\equiv 1. $$
I do not see that!
Let for example $e=1=r, L=4, m=2$, $u_k^t=U_4(t+k-2)$.
Then at $t=0$ I have the configuration $$ \ldots 120012001~\dot{2}0012~0012001200\ldots, $$ where the dot marks position 0. For this example I do not get $W_4(u^0)=1$.
But I do not see where my mistake is.
Using (2.14)on p. 6, I have to compute $$ W_4(u^0)=\sum_{k=1}^4\sigma(u_k^0,u_{k-1}^0) $$ and as far as I see, it is $$ \sigma(u_1^0,u_0^0)=\sigma(0,2)=-1,\\\sigma(u_2^0,u_1^0)=\sigma(0,0)=0,\\\sigma(u_3^0,u_2^0)=\sigma(1,0)=-1,\\\sigma(u_4^0,u_3^0)=\sigma(2,1)=-1 $$
and so I get $W_4(u^0)=-3$.
Maybe ìn the linked text, there is the factor $-1/(e+r+1)$ missing before $W_L(u^t)$? Dont know. But going from position 0 to position L, the states are going around the circle exactly one time counterclockwise, so the result should indeed be 1.
I think you're right. A factor of $1/(e+r+1)$ is missing, as becomes clear by comparison with $(2.13)$, and the sign also seems to be wrong. The most natural way to fix things would seem to be to include the factor $1/(e+r+1)$ in $W_L(u^t)$ and to use $\sigma(u_{k-1}^t,u_k^t)$ instead of $\sigma(u_k^t,u_{k-1}^t)$ (or equivalently to change the sign in the definition of $\sigma$).