The type-A quantum presentation descends to its coproduct tensor action

On $V_N$, each $L_j$ is diagonal and invertible, and the $L_j$ commute. Conjugating the matrix unit $E_i:e_{i+1}\mapsto e_i$ by $L_j$ multiplies its sole nonzero entry by $q^{{\delta }_{j,i}-{\delta }_{j,i+1}}$; conjugating $F_i:e_i\mapsto e_{i+1}$ gives the inverse exponent. Also $K_i$ has eigenvalue $q$ on $e_i$, $q^{-1}$ on $e_{i+1}$ and $1$ on every other basis vector, so $(K_i-K_i^{-1})\mathrm{/}(q-q^{-1})$ is the diagonal matrix with entries $1,-1,0$ in those positions. Directly, $[E_i,F_j]=0$ for $i\mathrm{\ne }j$ and $[E_i,F_i]$ is that diagonal matrix. Products $E_i{E}_j$ and $E_j{E}_i$ (and the corresponding $F$ products) both vanish when $\mathrm{\mid }i-j\mathrm{\mid }>1$; $E_i^2=F_i^2=0$, and each middle three-letter Serre product also vanishes because its matrix-unit indices do not compose. Thus every defining relation holds, and the quotient universal property [F1] gives the vector representation.

Define $\mathrm{\Delta }$ on the free generators by the displayed formulas in The type-A quantum algebra and its tensor-action formulas. The $\mathrm{\Delta }(L_j)$ are mutually commuting units with the displayed inverses. Conjugation by $\mathrm{\Delta }(L_j)$ gives the required weights on each of the two terms of $\mathrm{\Delta }(E_i)$ and $\mathrm{\Delta }(F_i)$, since each $K_i$ is a Laurent monomial in the commuting $L$'s. For any $i,j$, expansion gives $[\mathrm{\Delta }(E_i),\mathrm{\Delta }(F_j)]=[E_i,F_j]\otimes K_i+K_j^{-1}\otimes [E_i,F_j]+[E_i\otimes K_i,K_j^{-1}\otimes F_j]$. The last cross term is zero: its first product can be reordered as $q^{a_{ji}-a_{ij}}{K}_j^{-1}{E}_i\otimes F_j{K}_i=K_j^{-1}{E}_i\otimes F_j{K}_i$, which is its reverse product by [F3]. If $i\mathrm{\ne }j$, the remaining terms vanish; if $i=j$, they sum to $(K_i\otimes K_i-K_i^{-1}\otimes K_i^{-1})\mathrm{/}(q-q^{-1})=\mathrm{\Delta }((K_i-K_i^{-1})\mathrm{/}(q-q^{-1}))$. For $\mathrm{\mid }i-j\mathrm{\mid }>1$, $a_{ij}=0$ and the $K_i$ commute with $E_j,F_j$; expanding the two commutators therefore proves that the distant $E$ and $F$ relations are preserved.

Let $\gamma =q+q^{-1}$ and $S_{ij}(X,Y)=X^{2}Y-\gamma XYX+Y{X}^2$ for adjacent $i,j$. For the $E$ relation write $\mathrm{\Delta }(E_s)=A_s+B_s$ with $A_s=E_s\otimes K_s$, $B_s=1\otimes E_s$. In each of the three words $iij,iji,jii$, choose an $A$ at each selected position and a $B$ at each other position, then move the selected $K$ factors to the right using $K_r{E}_s=q^{a_{rs}}{E}_s{K}_r$. Grouping the resulting 24 terms by the number of selected positions gives: degree $0$ in the first factor is $1\otimes S_{ij}(E_i,E_j)$; degree $3$ is $S_{ij}(E_i,E_j)\otimes K_i^2{K}_j$; degree $1$ has respective coefficients $q+q^{-1}-\gamma$, $q^2+1-\gamma q$, $1+q^{-2}-\gamma q^{-1}$ on $E_i\otimes E_i{E}_j{K}_i$, $E_i\otimes E_j{E}_i{K}_i$, $E_j\otimes E_i^2{K}_j$; degree $2$ has respective coefficients $q^{-2}+1-\gamma q^{-1}$, $q^2+1-\gamma q$, $q+q^{-1}-\gamma$ on $E_i^2\otimes E_j{K}_i^2$, $E_i{E}_j\otimes E_i{K}_i{K}_j$, $E_j{E}_i\otimes E_i{K}_j{K}_i$. All six mixed coefficients are zero by the definition of $\gamma$, and the two pure terms vanish by the defining Serre relation. Thus $\mathrm{\Delta }$ preserves every positive Serre relation.

The assignment $\varphi (L_j)=L_j^{-1}$, $\varphi (E_i)=F_i$, $\varphi (F_i)=E_i$ preserves the presentation: it exchanges the two weight relations, sends the mixed commutator to its negative with $K_i$ replaced by $K_i^{-1}$, and exchanges the distant and Serre relations. It is an involution. On generators $\mathrm{\Delta }(F_i)=\tau (\varphi \otimes \varphi )\mathrm{\Delta }(E_i)$, where $\tau (x\otimes y)=y\otimes x$ is an algebra isomorphism. Since step 1.3 proves that the positive Serre polynomial maps to zero, applying the algebra maps $\tau$ and $\varphi \otimes \varphi$ proves that the negative Serre polynomial maps to zero as well. Together steps 1.2–2.1 show every relation is killed, so [F1] descends $\mathrm{\Delta }$ to the quotient. The counit assignment kills the relations directly: each relation becomes either $1=1$ or $0=0$, so it too descends.

Compare $\alpha \circ (\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\circ \mathrm{\Delta }$ and $(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\circ \mathrm{\Delta }$ as maps to $U_N\otimes (U_N\otimes U_N)$, and compare $\lambda \circ (ϵ\otimes \mathrm{i}\mathrm{d})\circ \mathrm{\Delta }$ and $\rho \circ (\mathrm{i}\mathrm{d}\otimes ϵ)\circ \mathrm{\Delta }$ with ${\mathrm{i}\mathrm{d}}_{U_N}$. All these composites are algebra homomorphisms by [F2] and [F4], so it suffices to check generators. The two counit composites return $L_j,E_i,F_i$. After applying the associator, both iterated coproducts on $L_j$ give $L_j\otimes L_j\otimes L_j$; on $E_i$, both give $E_i\otimes K_i\otimes K_i+1\otimes E_i\otimes K_i+1\otimes 1\otimes E_i$; on $F_i$, both give $F_i\otimes 1\otimes 1+K_i^{-1}\otimes F_i\otimes 1+K_i^{-1}\otimes K_i^{-1}\otimes F_i$. Hence $\mathrm{\Delta }$ is coassociative with counit $ϵ$. The vector representation of step 1.1 and the iterated coproduct therefore define actions on every $V_N^{\otimes n}$. They are finite sums and products of matrices with entries $0,1,q^{\pm 1}$, so all entries lie in $\mathbb{Z}[q,q^{-1}]$.