The full boundary twist acts on LKB by the scalar q to two n t squared

For $n=1$ the representation definition supplies the zero absolute module; all stated action identities hold there, without detecting any exponent. For the remaining proof assume $n\ge 2$. The scalar identity in the fraction-field model. Fix $1\le i<j\le n$ and write $i^{\mathrm{\prime }}=n+1-j$, $j^{\mathrm{\prime }}=n+1-i$, so that $1\le i^{\mathrm{\prime }}<j^{\mathrm{\prime }}\le n$ and the assignment $(i,j)\mapsto (i^{\mathrm{\prime }},j^{\mathrm{\prime }})$ is an involution of the set of pairs. Applying [F1] to the pair $(i,j)$ gives $\mathrm{\Delta }{x}_{i^{\mathrm{\prime }}{j}^{\mathrm{\prime }}}=t{q}^{i+j-1}{x}_{ij}$; applying [F1] to the pair $(i^{\mathrm{\prime }},j^{\mathrm{\prime }})$ gives $\mathrm{\Delta }{x}_{ij}=t{q}^{i^{\mathrm{\prime }}+j^{\mathrm{\prime }}-1}{x}_{i^{\mathrm{\prime }}{j}^{\mathrm{\prime }}}$, because the pair associated with $(i,j)$ is again $(i^{\mathrm{\prime }},j^{\mathrm{\prime }})$. Combining the two identities, ${\mathrm{\Delta }}^2{x}_{i^{\mathrm{\prime }}{j}^{\mathrm{\prime }}}=\mathrm{\Delta }\left(t{q}^{i+j-1}{x}_{ij}\right)=t{q}^{i+j-1}\mathrm{\Delta }{x}_{ij}=t{q}^{i+j-1}\cdot t{q}^{i^{\mathrm{\prime }}+j^{\mathrm{\prime }}-1}{x}_{i^{\mathrm{\prime }}{j}^{\mathrm{\prime }}}=t^2{q}^{2n}{x}_{i^{\mathrm{\prime }}{j}^{\mathrm{\prime }}},$ since $i^{\mathrm{\prime }}+j^{\mathrm{\prime }}=2n+2-i-j$, so $i+j-1+i^{\mathrm{\prime }}+j^{\mathrm{\prime }}-1=(i+j)+(i^{\mathrm{\prime }}+j^{\mathrm{\prime }})-2=2n$. As $(i^{\mathrm{\prime }},j^{\mathrm{\prime }})$ runs over all pairs, every basis element of the fraction-field model is an eigenvector of ${\mathrm{\Delta }}^2$ with eigenvalue $t^2{q}^{2n}$, so $({\rho }_K(\mathrm{\Delta }){)}^2=t^2{q}^{2n}\mathrm{id}$, a matrix identity whose entries lie in $\mathrm{\Lambda }$.

Transfer to the absolute integral module. By [F2] the scalar extension $K{\otimes }_{\mathrm{\Lambda }}{H}_2(\tilde{C};\mathbb{Z})$ is isomorphic to $K{\otimes }_{\mathrm{\Lambda }}V$ as a $B_n$-module, and the isomorphism intertwines the two actions of ${\mathrm{\Delta }}^2$. Hence step 1.1 shows that ${\mathrm{\Delta }}^2$ acts on $K{\otimes }_{\mathrm{\Lambda }}{H}_2(\tilde{C};\mathbb{Z})$ by the scalar $q^{2n}{t}^2$. Let $x\in H_2(\tilde{C};\mathbb{Z})$. The normalized lift of ${\mathrm{\Delta }}^2$ gives ${\rho }_{\mathrm{L}\mathrm{K}\mathrm{B}}({\mathrm{\Delta }}^2)x\in H_2(\tilde{C};\mathbb{Z})$ because the action preserves the integral lattice, and by definition of the scalar extension its image in $K{\otimes }_{\mathrm{\Lambda }}{H}_2(\tilde{C};\mathbb{Z})$ equals $q^{2n}{t}^2$ times the image of $x$. Both classes lie in the image of the integral lattice, and the natural map is injective by [F2]; therefore ${\rho }_{\mathrm{L}\mathrm{K}\mathrm{B}}({\mathrm{\Delta }}^2)x=q^{2n}{t}^{2}x$ already in $H_2(\tilde{C};\mathbb{Z})$. As $x$ was arbitrary, ${\rho }_{\mathrm{L}\mathrm{K}\mathrm{B}}({\mathrm{\Delta }}^2)=q^{2n}{t}^2\mathrm{id}$.

Powers and nontriviality. Multiplying the scalar identity, for every $k\ge 0$ one has ${\rho }_{\mathrm{L}\mathrm{K}\mathrm{B}}({\mathrm{\Delta }}^{2k})=(q^{2n}{t}^2{)}^k\mathrm{id}=q^{2nk}{t}^{2k}\mathrm{id}$; the same identity with $k=-1$ follows by inverting the scalar $q^{2n}{t}^2$, and inverting again gives the stated formula for every $k\in \mathbb{Z}$. Since $n\ge 2$, the free module has a nonzero basis vector. If $q^{2nk}{t}^{2k}$ were the identity matrix, equality on that vector would imply the two Laurent monomials $q^{2nk}{t}^{2k}$ and $q^0{t}^0=1$ would be equal in $\mathrm{\Lambda }$; distinct monomials with distinct exponent vectors are distinct elements of $\mathrm{\Lambda }$, so $(2nk,2k)=(0,0)$ and $k=0$.

The Bigelow sign convention. The source's geometric computation [F3] exhibits the eigenvalue of $({\mathrm{\Delta }}^2{)}^k$ on the class of the standard $N_1$-fork as a unit multiple of $(q^{2n}{t}^2{)}^k$; the two computations agree on the scalar $q^{2n}{t}^2$ and differ only in the fixed unit contributed by the normalization of the pairing, which is immaterial for the matrix identity above.