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I am having a little difficulty with a calculation in "The connective constant of the honeycomb lattice equals $\sqrt{2+\sqrt2}$", Hugo Duminil-Copin, Stanislav Smirnov, (arXiv:1007.0575).

On page 4 of the paper, in the second equation block, it is shown that $c( \gamma_1) + c(\gamma_ 2) = 0$.

Would someone be able to fill out the details of this calculation?

In particular where does $j$ come from and how does the $(p-v)$ factor get there. I can sort of see how it works and how it uses the equation block above but I am left with extra factors. I can also 'reverse engineer' j to determine its value. Is it perhaps just a placeholder and the authors haven't made this clear? I know that sometimes $j$ is used instead of $i$ but I don't think it is being used like this.

Many thanks

Mark

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$j$ is introduced on p. $2$, right above the heading of section $2$: $j=\mathrm e^{\mathrm i2\pi/3}$, a third root of unity.

$W_{\gamma_1}(a,q)$ and $W_{\gamma_2}(a,r)$ are substituted from the previous equation block. Then the terms contain a common factor $\mathrm e^{-\mathrm i\sigma W_{\gamma_1}(a,p)}x_c^{l(\gamma_1)}$, and factoring that out leaves

$ (q-v)\mathrm e^{\mathrm i\sigma4\pi/3}+(r-v)e^{-\mathrm i\sigma4\pi/3}=(q-v)\bar\lambda^4+(r-v)\lambda^4 $

(with $\lambda$ introduced on p. $3$). Now $q-v$, $r-v$ and $p-v$ are just rotated versions of each other, with $q-v=j(p-v)$ and $r-v=\bar j(p-v)$, so this yields

$ (p-v)\left(j\bar\lambda^4+\bar j\lambda^4\right)\;. $

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    @Mark: You're welcome!2012-12-14