The integral LKB module is free of rank n choose two

The classes $v_{i,j}$ are $K$-linearly independent. Extend the primed pairing of [F2] to $K{\otimes }_{\mathrm{\Lambda }}{H}_2(\tilde{C},\tilde{\nu })\times K{\otimes }_{\mathrm{\Lambda }}{H}_2(\tilde{C},\mathrm{\partial }\tilde{C})$ by $K$-sesquilinearity. Its matrix $(⟨v_{i^{\mathrm{\prime }},j^{\mathrm{\prime }}}^{\mathrm{\prime }},x_{i,j}{⟩}^{\mathrm{\prime }})$ with respect to the dual classes is triangular with unit diagonal by [F2], hence invertible over $\mathrm{\Lambda }$ and over $K$. The pairing matrix for the actual absolute cycles $v_{i,j}$ is this primed matrix with each row multiplied by its displayed nonzero closing factor $(1-q{)}^2$, $(1-q{)}^2(1+qt)$, or $(1-q{)}^2(1+qt)(1-t)$. These factors are not asserted to be Laurent units; they are invertible over $K$, so the actual matrix remains invertible over $K$. If ${\sum }_{i<j}{c}_{i,j}{v}_{i,j}=0$ with $c_{i,j}\in K$, pairing the relation with each $x_{i,j}$ and applying invertibility gives $c_{i,j}=0$ for all $i<j$. Since the $v_{i,j}$ lie in the image of $H_2(\tilde{C};\mathbb{Z})$ and by [F1] that image spans a subspace of dimension $\left(\genfrac{}{}{0px}{}{n}{2}\right)$, which equals the number of pairs $(i,j)$, the classes $v_{i,j}$ form a $K$-basis of $K{\otimes }_{\mathrm{\Lambda }}{H}_2(\tilde{C};\mathbb{Z})$.

The rational comparison and cyclic lattice. By [F4], $K{\otimes }_{\mathrm{\Lambda }}{H}_2$ and the Krammer matrix representation are isomorphic as $B_n$-representations, and the latter has integral realization $L={⨁}_{i<j}\mathrm{\Lambda }{E}_{ij}$ in the cellular kernel. The parameter match is explicit: set $x_{ij}=q^{1-i}{E}_{ij}$ and $t_{\mathrm{K}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{m}\mathrm{e}\mathrm{r}}=-t$. This is a diagonal Laurent-unit basis change. Substitution in the seven displayed cases gives the Krammer table: for $k=i$ the next-row coefficient becomes $q$ and the adjacent coefficient becomes $t_{\mathrm{K}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{m}\mathrm{e}\mathrm{r}}q(q-1)$; for $k=j-1$ the adjacent exponent becomes $q^{j-i}$; in the interior it becomes $q^{k-i}$; for $k=i-1$ the previous-row coefficient becomes1; for $k=j$ the next-column coefficient remains$q$; for an adjacent pair the eigenvalue is $t_{\mathrm{K}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{m}\mathrm{e}\mathrm{r}}{q}^2$; all other basis vectors are fixed. Thus the integral realization is the specified Krammer matrix lattice with exactly its frozen sign translation, not an unspecified rational basis change. The $E$ cycles have only one nonzero $A$ coordinate, namely $S$ in coordinate $(i,j)$, so are independent. They are generated over $\mathrm{\Lambda }[B_n]$ by $E_{12}$: the $k=j$ formula builds $E_{1,j+1}=q^{-1}({\sigma }_j-(1-q))E_{1j}$ along the first row; the $k=i<j-1$ formula then builds row $i+1$ from row $i$ and its already known adjacent element. Induction gives every pair.

The classes $v_{i,j}$ span $H_2(\tilde{C};\mathbb{Z})$ over $\mathrm{\Lambda }$. Let $v\in H_2(\tilde{C};\mathbb{Z})$. By step 1.1 write $v={\sum }_{i<j}{c}_{i,j}{v}_{i,j}$ with $c_{i,j}\in K$. Since $v$ is integral, [F3] gives $c_{i,j}\in \mathrm{\Lambda }$ for all $i<j$. Hence $v$ is a $\mathrm{\Lambda }$-linear combination of the $v_{i,j}$.

A fixed-parameter equivariant map cannot be surjective. For $n\ge 3$, the action of ${\sigma }_1$ on $K\otimes H_2$ has the eigenvalue $-q^{2}t$ on $E_{12}$. Modulo this line, each span of $E_{1j},E_{2j}$, $j\ge 3$, has matrix with characteristic polynomial $(X-1)(X+q)$, and the remaining $E_{ij}$ with $i\ge 3$ are fixed. Since $-q^{2}t$ differs from $1,-q$ as a rational function, its eigenspace is exactly $K{E}_{12}$. Any fixed-parameter $\mathrm{\Lambda }[B_n]$-isomorphism from $L$ onto $H_2$ must therefore send $E_{12}$ to $\lambda E_{12}$ for $\lambda \in K$. In integral cellular coordinates the $A_{12}$ and $B_{13}$ entries of this image are $\lambda (t-1)(qt+1)$ and $\lambda (q-1)$. They belong to $\mathrm{\Lambda }$. Set $A=\lambda (q-1)$ and $B=\lambda (t-1)(qt+1)$ in $\mathrm{\Lambda }$. Then $A(t-1)(qt+1)=B(q-1)$. Evaluate $q=1$ into the Laurent domain $\mathbb{Z}[t^{\pm 1}]$ of [F5]: the factor $(t-1)(t+1)$ is nonzero, so $A(1,t)=0$. The kernel of this evaluation is $(q-1)$: multiply a Laurent polynomial by a sufficiently large power of $q$, then use the finite identities $q^r-1=(q-1)(1+\cdots +q^{r-1})$ to subtract its value at1. Hence $A=(q-1)C$ for $C\in \mathrm{\Lambda }$, and cancellation in $K$ gives $\lambda =C\in \mathrm{\Lambda }$. This proves the needed denominator removal directly, without asserting an undeclared UFD theorem. Cyclic generation from step 1.2 forces the entire image into $L$.

Freeness and rank. A $\mathrm{\Lambda }$-linear relation ${\sum }_{i<j}{\lambda }_{i,j}{v}_{i,j}=0$ with ${\lambda }_{i,j}\in \mathrm{\Lambda }\subseteq K$ is in particular a $K$-linear relation, so step 1.1 gives ${\lambda }_{i,j}=0$. Together with step 2.1 this shows that $\{v_{i,j}\}$ is a $\mathrm{\Lambda }$-basis of $H_2(\tilde{C};\mathbb{Z})$; in particular the module is free of rank $\left(\genfrac{}{}{0px}{}{n}{2}\right)$. The displayed relative images of the basis elements are exactly those recorded in [F2].

The integral kernel is strictly larger. Define $X_{13}=(qt+1)(A_{12}+A_{23}-A_{13})-(q-1)B_{21}+(q^2-1)B_{22}-(q-1)B_{23}\mathrm{.}$ Substituting [F1]'s absolute differential gives zero: the $A$ part contributes $(q-1)(qt+1)(a_2-b_2)$, The $B$ part is $(q-1)(-d{B}_{21}+(q+1)d{B}_{22}-d{B}_{23})$. Its $a_2$ coefficient inside the parentheses is $-(1-t)-(q+1)t=-(1+qt)$ and its $b_2$ coefficient is $(q+1)t-(t-1)=1+qt$; the $c_2,c_3$ coefficients are $1-(q+1)+q=0$ and $-q+(q+1)-1=0$. Thus it is the negative of the $A$ contribution. There are no degree-three boundaries, so $X_{13}\in H_2$. If it lay in $L$, its $A_{13}$ coordinate would force its $E_{13}$ coefficient to equal $-(qt+1)\mathrm{/}S=-1\mathrm{/}(t-1)$, which is not in $\mathrm{\Lambda }$. Hence $X_{13}\mathrm{\notin }L$ for every $n\ge 3$. Step 2.2 rules out an equivariant isomorphism onto $H_2$. This proves the fixed-parameter nonisomorphism, including $n=3$; it does not rely on the source's stronger parameter-twisted maximality statement whose $n=3$ argument was left to the reader. Both underlying modules are free of the same rank, so are abstractly $\mathrm{\Lambda }$-isomorphic by sending one finite basis to the other.