In principle, this lemma can be verified by a direct calculation, by writing out the six terms in the sum explicitly and using trigonometric identities. Below, we give a slightly more conceptual proof, in which we use symmetry arguments to reduce the number of expressions significantly.
For the matrix involving
B, we obtain
$$\begin{aligned}&\big (B:[\sigma _k]^2\big |B:[\sigma _{k+1}]^2\big ) \\&= \begin{pmatrix} b_{111}\sigma _{k,1}^2+b_{122}\sigma _{k,2}^2+2b_{112}\sigma _{k,1}\sigma _{k,2} &{} b_{111}\sigma _{k+1,1}^2+b_{122}\sigma _{k+1,2}^2+2b_{112}\sigma _{k+1,1}\sigma _{k+1,2} \\ b_{211}\sigma _{k,1}^2+b_{222}\sigma _{k,2}^2+2b_{212}\sigma _{k,1}\sigma _{k,2} &{} b_{211}\sigma _{k+1,1}^2+b_{222}\sigma _{k+1,2}^2+2b_{212}\sigma _{k+1,1}\sigma _{k+1,2} \end{pmatrix}, \end{aligned}$$
while clearly
$$\begin{aligned} \big (\sigma _k\big |\sigma _{k+1}\big )^{-1} = \frac{2}{\sqrt{3}} \begin{pmatrix} \sigma _{k+1,2} &{} -\sigma _{k+1,1} \\ -\sigma _{k,2} &{} \sigma _{k,1} \end{pmatrix}. \end{aligned}$$
The sum of the diagonal entries of the matrix product are easily calculated,
$$\begin{aligned} T_k:={{\mathrm{tr}}}\big [\big (\sigma _k\big |\sigma _{k+1}\big )^{-1}\big (B:[\sigma _k]^2\big |B:[\sigma _{k+1}]^2\big )\big ] = \frac{2}{\sqrt{3}}\sum _{p,q,r=1}^2 b_{pqr}\gamma _{pqr,k}, \end{aligned}$$
with the trigonometric expressions
$$\begin{aligned}&\gamma _{111,k}=\sigma _{k,1}^2\sigma _{k+1,2}-\sigma _{k+1,1}^2\sigma _{k,2},\quad \gamma _{122,k}=\sigma _{k,2}^2\sigma _{k+1,2}-\sigma _{k+1,2}^2\sigma _{k,2}, \\&\gamma _{112,k}=\gamma _{121,k}=\sigma _{k,1}\sigma _{k,2}\sigma _{k+1,2}-\sigma _{k+1,1}\sigma _{k+1,2}\sigma _{k,2}, \\&\gamma _{211,k}=\sigma _{k+1,1}^2\sigma _{k,1}-\sigma _{k,1}^2\sigma _{k+1,1},\quad \gamma _{222,k}=\sigma _{k+1,2}^2\sigma _{k,1}-\sigma _{k,2}^2\sigma _{k+1,1},\\&\gamma _{212,k}=\gamma _{221,k}=\sigma _{k+1,1}\sigma _{k+1,2}\sigma _{k,1}-\sigma _{k,1}\sigma _{k,2}\sigma _{k+1,1}. \end{aligned}$$
To key step is to calculate the sum over
\(k=0,1,\ldots ,5\) of the products of
\(T_k\) with the respective vector
$$\begin{aligned} \eta _k ={\mathbb {J}}(\sigma _{k}-\sigma _{k+1})= \begin{pmatrix} \sigma _{k+1,2}-\sigma _{k,2}\\ \sigma _{k,1}-\sigma _{k+1,1} \end{pmatrix}. \end{aligned}$$
Several simplifications of this sum can be performed, thanks to the particular form of the
\(\gamma _{pqr,k}\) and elementary trigonometric identities. First, observe that
\(\sigma _{k+3}=-\sigma _k\), and hence that
\(\gamma _{pqr,k+3}=-\gamma _{pqr,k}\). Since further
\(\eta _{k+3}=-\eta _k\), it follows that
$$\begin{aligned} \gamma _{pqr,k+3}\eta _{k+3}=\gamma _{pqr,k}\eta _k. \end{aligned}$$
(B.4)
Second,
\(\eta \) can be evaluated explicitly for
\(k=1,2,3\):
$$\begin{aligned} \eta _0 =\frac{1}{2} \begin{pmatrix} \sqrt{3} \\ 1 \end{pmatrix}, \quad \eta _1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \quad \eta _2 =\frac{1}{2} \begin{pmatrix} -\sqrt{3} \\ 1 \end{pmatrix}. \end{aligned}$$
(B.5)
Third, since
\(\sigma _{0,1}=-\sigma _{3,1}\) and
\(\sigma _{1,1}=-\sigma _{2,1}\), as well as
\(\sigma _{0,2}=\sigma _{3,2}\) and
\(\sigma _{1,2}=\sigma _{2,2}\), we obtain that
$$\begin{aligned} \gamma _{pqr,1} = 0 \quad \text {if }p+q+r\text { is odd}, \quad \text {and}\quad \gamma _{pqr,2} = (-1)^{p+q+r}\gamma _{pqr,0}. \end{aligned}$$
(B.6)
By putting this together, we arrive at
$$\begin{aligned} \sum _{k=0}^5\gamma _{pqr,k}\eta _k&{\mathop {=}\limits ^{\mathrm{B.4}}}2\sum _{k=0}^2\gamma _{pqr,k}\eta _k \\&{\mathop {=}\limits ^{\mathrm{B.5}}} \begin{pmatrix} \sqrt{3}\big (\gamma _{pqr,0}-\gamma _{pqr,2}\big ) \\ \gamma _{pqr,0}+2\gamma _{pqr,1}+\gamma _{pqr,2} \end{pmatrix} \\&{\mathop {=}\limits ^{\mathrm{B.6}}} \begin{pmatrix} \sqrt{3}\big (1-(-1)^{p+q+r}\big )\gamma _{pqr,0}\\ \big (1+(-1)^{p+q+r}\big )\big (\gamma _{pqr,0}+\gamma _{pqr,1}\big ) \end{pmatrix} = \begin{pmatrix} 2\sqrt{3}\,\gamma _{pqr,0}\,(1-\mathfrak {e}_{pqr})\\ 2\big (\gamma _{pqr,0}+\gamma _{pqr,1}\big )\,\mathfrak {e}_{pqr} \end{pmatrix}, \end{aligned}$$
where
\(\mathfrak {e}_{pqr}=1\) if
\(p+q+r\) is even, and
\(\mathfrak {e}_{pqr}=0\) if
\(p+q+r\) is odd. By elementary computations,
$$\begin{aligned} \begin{array}{llll} p+q+r\text { odd, }k=0: &{}\gamma _{111,0}=\frac{\sqrt{3}}{2}, &{}\gamma _{122,0}=0, &{}\gamma _{212,0}=\gamma _{221,0}=\frac{\sqrt{3}}{4};\\ p+q+r\text { even, }k=0: &{}\gamma _{211,0}=-\frac{1}{4}, &{}\gamma _{222,0}=\frac{3}{4}, &{}\gamma _{112,0}=\gamma _{121,0}=0;\\ p+q+r\text { even, }k=1: &{}\gamma _{211,1}=\frac{1}{4}, &{}\gamma _{222,1}=\frac{3}{4}, &{}\gamma _{112,1}=\gamma _{121,1}=\frac{3}{4}, \end{array} \end{aligned}$$
and so the final result is:
$$\begin{aligned}&\sum _{k=0}^5 {{\mathrm{tr}}}\big [\big (\sigma _k\big |\sigma _{k+1}\big )^{-1}\big (B:[\sigma _k]^2\big |B:[\sigma _{k+1}]^2\big )\big ]\,{\mathbb {J}}(\sigma _{k}-\sigma _{k+1}) \\&\quad =\sum _{k=0}^5T_k\eta _k =\frac{2}{\sqrt{3}}\sum _{p,q,r=1}^2\left( b_{pqr}\sum _{k=0}^5\gamma _{pqr,k}\eta _k\right) =2\sqrt{3} \begin{pmatrix} b_{111}+b_{212} \\ b_{222}+b_{112} \end{pmatrix}, \end{aligned}$$
which is (
B.3).
\(\square \)