Yangians, Quantum Affine Algebras, and Transfer Matrices

In the coefficient of the auxiliary row $(i,k)$ and column $(j,l)$, the $xI$ part of the left side is $x{t}_{ij}(u)t_{kl}(v)$ and the $iP$ part is $i{t}_{kj}(u)t_{il}(v)$, because the flip sends the row $(i,k)$ to $(k,i)$. On the right side the corresponding terms are $x{t}_{kl}(v)t_{ij}(u)$ and $i{t}_{kj}(v)t_{il}(u)$, since the flip acts on the column. Thus equality of those matrix entries gives $x{t}_{ij}(u)t_{kl}(v)+i{t}_{kj}(u)t_{il}(v)=x{t}_{kl}(v)t_{ij}(u)+i{t}_{kj}(v)t_{il}(u)$.

Subtract the two $x$-terms and move the two $i$-terms to the other side. The result is exactly the displayed commutator relation. Conversely, if that relation holds for every four indices, the coefficient computation in step 1.1 makes every auxiliary matrix entry of the RTT difference zero, hence the full matrix relation holds.

Each coefficient of $T^{[1]}(u)T^{[2]}(u)$ is a finite sum: the coefficient of $u^{-n}$ uses only the finitely many pairs of exponents adding to $n$ and the finite index sum $1\le k\le N$. Thus the proposed coproduct lands in the ordinary algebraic tensor product. For two independent Yangian factors, write $T_s^{[r]}$ for the matrix acting on auxiliary leg $s\in \{1,2\}$ and coefficient factor $r\in \{1,2\}$. Then $T_1^{[2]}$ commutes with $T_2^{[1]}$, and $T_1^{[1]}$ commutes with $T_2^{[2]}$, because each pair acts on disjoint tensor factors. Reordering these pairs and applying the RTT relation first in coefficient factor $1$ and then in factor $2$ gives $R_{12}{T}_1^{[1]}{T}_1^{[2]}{T}_2^{[1]}{T}_2^{[2]}=T_2^{[1]}{T}_2^{[2]}{T}_1^{[1]}{T}_1^{[2]}{R}_{12}$. The middle matrix product is precisely the RTT relation for $\mathrm{\Delta }(T)$, so $\mathrm{\Delta }$ preserves the defining relation and extends to a unital algebra homomorphism.

Iterating the matrix product yields $(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(T)=T^{[1]}{T}^{[2]}{T}^{[3]}=(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }(T)$, so the coproduct is coassociative coefficientwise. Substitution $T\mapsto I$ sends the RTT relation to $R=R$, so it defines a unital counit homomorphism; it also gives $(ϵ\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(T)=T=(\mathrm{i}\mathrm{d}\otimes ϵ)\mathrm{\Delta }(T)$ and $ϵ(1)=1$.

Since $T(u)=I+O(u^{-1})$, recursively solving $T(u)T(u{)}^{-1}=I$ gives a unique right inverse; solving in the reverse order gives a left inverse, and associativity makes them equal. Each inverse coefficient is a finite polynomial in the coefficients of $T-I$, because only products of at most $n$ positive-degree terms contribute to $u^{-n}$. Write $A=T_1(u)$ and $B=T_2(v)$. In $RAB=BAR$, multiply on the left by $A^{-1}{B}^{-1}=(BA{)}^{-1}$ and on the right by $B^{-1}{A}^{-1}=(AB{)}^{-1}$ to obtain $A^{-1}{B}^{-1}R=R{B}^{-1}{A}^{-1}$. Define an anti-homomorphism on the free generator algebra by sending the coefficient matrix $T(u)$ entrywise to this inverse matrix. Applying it to the defining RTT relator reverses coefficient multiplication, so its image is $R{B}^{-1}{A}^{-1}=A^{-1}{B}^{-1}R$, exactly the same identity. Hence it preserves the RTT ideal and descends to the quotient, with $S(T(u))=T(u{)}^{-1}$.

The two matrix inverse identities give ${\sum }_{k}S(t_{ik}(u))t_{kj}(u)={\delta }_{ij}={\sum }_k{t}_{ik}(u)S(t_{kj}(u))$, the left and right antipode identities on generators. For the left identity on a product $xy$, use $\mathrm{\Delta }(xy)=\sum x_{(1)}{y}_{(1)}\otimes x_{(2)}{y}_{(2)}$ and anti-multiplicativity of $S$ to obtain $\sum S(y_{(1)})S(x_{(1)})x_{(2)}{y}_{(2)}=ϵ(x)\sum S(y_{(1)})y_{(2)}=ϵ(x)ϵ(y)1$; the right identity follows from $\sum x_{(1)}{y}_{(1)}S(y_{(2)})S(x_{(2)})$. Induction on word length and linearity extend both identities to the generated algebra, giving $m(S\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }=\eta ϵ=m(\mathrm{i}\mathrm{d}\otimes S)\mathrm{\Delta }$.

Let $e_1,\dots ,e_N$ be the standard column basis. By [F3], $E_{ij}{e}_s={\delta }_{js}{e}_i$, so $E_{ij}{E}_{kl}{e}_s={\delta }_{jk}{\delta }_{ls}{e}_i=({\delta }_{jk}{E}_{il})e_s$ for every $s$. Hence $E_{ij}{E}_{kl}={\delta }_{jk}{E}_{il}$ and $[E_{ij},E_{kl}]={\delta }_{jk}{E}_{il}-{\delta }_{li}{E}_{kj}$. On a dual basis vector $f^s$, the explicit commutator is $[{\rho }^{\vee }(E_{ij}),{\rho }^{\vee }(E_{kl})]f^s={\delta }_{ks}{\delta }_{il}{f}^j-{\delta }_{is}{\delta }_{jk}{f}^l$. The right side is also ${\rho }^{\vee }({\delta }_{jk}{E}_{il}-{\delta }_{li}{E}_{kj})f^s$, by the same formula. Thus the dual-natural operators satisfy the matrix-unit bracket and form a ${\mathfrak{g}\mathfrak{l}}_N$-representation.

Put $x=u-v$, $U=u-a$, and $V=v-a$. By [F4], the required component identity is the RTT relation to check. Under [F2], its left side is $-x({\delta }_{jk}\rho (E_{il})-{\delta }_{li}\rho (E_{kj}))\mathrm{/}(UV)$ by the bracket relation. In the parenthesized difference on the right, the quadratic terms cancel because both are $(-i{)}^2\rho (E_{kj})\rho (E_{il})\mathrm{/}(UV)$. Its linear terms are $-i{\delta }_{kj}\rho (E_{il})(U^{-1}-V^{-1})-i{\delta }_{il}\rho (E_{kj})(V^{-1}-U^{-1})=ix({\delta }_{jk}\rho (E_{il})-{\delta }_{li}\rho (E_{kj}))\mathrm{/}(UV)$, since $V-U=-x$. Multiplication by the outer factor $i$ therefore gives the same left side. Thus every component RTT relation holds.

The component relations are exactly the coefficients of the matrix relation [F1]. Thus the images of every defining relation vanish, so the assignment on the free generators factors through the RTT quotient and defines the representation $W(a)$. For $W=V_N^{\vee }$, its auxiliary matrix is $I+i{P}_N\mathrm{/}(u-a)$ by the explicit sum ${\sum }_{i,j}{E}_{ij}\otimes E_{ji}=P_N$; multiplying by $u-a$ gives $R_N(u-a)$.

Direct substitution gives $R_i(-i(z-w))=-i((z-w)I-P_N)=-i{R}_0(z-w)$.

Substitute $u=-iz$ and $v=-iw$ in the RTT relation [F1]. By step 1.1, each side has the same nonzero scalar factor $-i$, which cancels; the resulting relation is $R_0(z-w){\tilde{T}}_1(z){\tilde{T}}_2(w)={\tilde{T}}_2(w){\tilde{T}}_1(z)R_0(z-w)$.

In the evaluation module, substitute $u=-iz$ into [F2]. Since $-iz-a=-i(z-ia)$, one obtains $t_{ij}(-iz)={\delta }_{ij}I+\rho (E_{ij})\mathrm{/}(z-ia)$. Thus the shifted standard evaluation parameter is exactly $ia$.

For $N=2$, the ordered basis and flip in [F1] are exactly those in [F2], so $R_2(x)=R(x)$ on the same ${\mathbb{C}}^2\otimes {\mathbb{C}}^2$. The un-checked Yang–Baxter identity for this case is [F3].

For $N\ge 3$, put $s=P_{12}$, $t=P_{13}$, and $r=P_{23}$. Acting on any basis tensor $e_p\otimes e_q\otimes e_{\mathrm{\ell }}$, the products $st,tr,rs$ all send $(p,q,\mathrm{\ell })$ to $(q,\mathrm{\ell },p)$, while $sr,rt,ts$ all send it to $(\mathrm{\ell },p,q)$; also $str=rts$ as both send $(p,q,\mathrm{\ell })$ to $(\mathrm{\ell },q,p)$. Hence, as operators, with $A=st=tr=rs$ and $B=sr=rt=ts$, those permutation identities hold on the whole basis.

The flip satisfies $P_N^2=I$, so direct multiplication gives $(xI+i{P}_N)(xI-i{P}_N)=(x^2+1)I=(xI-i{P}_N)(xI+i{P}_N)$. Over $\mathbb{C}$, $x^2+1\mathrm{\ne }0$ exactly when $x\mathrm{\notin }\{i,-i\}$; division by this scalar gives the stated inverse.

Expanding the left side gives $xyzI+i(yzs+xyt+xzr)-(yst+zsr+xtr)-istr$; the right side gives $xyzI+i(yzs+xyt+xzr)-(xrt+zrs+yts)-irts$. By step 1.2, the quadratic parts are respectively $-((x+y)A+zB)$ and $-(zA+(x+y)B)$, which agree since $x+y=z$; the scalar and linear parts match termwise and the cubic terms match by $str=rts$. This proves the Yang–Baxter identity for all $N\ge 3$, and step 1.1 covers $N=2$.

The operators $P_{+}=(I+P_N)\mathrm{/}2$ and $P_{-}=(I-P_N)\mathrm{/}2$ are complementary projectors because $P_N^2=I$. Their images have bases $e_j\otimes e_j$, $e_j\otimes e_k+e_k\otimes e_j$ and $e_j\otimes e_k-e_k\otimes e_j$ for $j<k$, respectively, so their dimensions are $N(N+1)\mathrm{/}2$ and $N(N-1)\mathrm{/}2$. On these subspaces $R_N(x)$ acts by $x+i$ and $x-i$; therefore at $x=i$ it has kernel $\mathrm{im}{P}_{-}={\bigwedge }^2{V}_N$, and at $x=-i$ it has kernel $\mathrm{im}{P}_{+}={\mathrm{Sym}}^2{V}_N$, with the ranks stated.

On the same ordered space $W_a\otimes W_b$, the coproduct action has auxiliary matrix $T_0(u)=T_{0a}(u)T_{0b}(u)$ by [F1]. The opposite coproduct reverses those two matrix factors, giving $T_{0b}(u)T_{0a}(u)$. Multiplying both by the scalar $(u-a)(u-b)$ gives the polynomial products $L_{0a}(u-a)L_{0b}(u-b)$ and $L_{0b}(u-b)L_{0a}(u-a)$, since [F2] and [F4] give $L_{0c}(u-c)=R_{0c}(u-c)$ for $c=a,b$.

Apply the spectral Yang–Baxter identity [F3] on auxiliary space $0$ and quantum spaces $a,b$, using their indexed-basis identifications and spectral points $u,a,b$. It gives $L_{0a}(u-a)L_{0b}(u-b)R_{ab}(a-b)=R_{ab}(a-b)L_{0b}(u-b)L_{0a}(u-a)$. Since $(u-a)(u-b)$ is nonzero and invertible in the scalar Laurent-series field at infinity, cancel it from the two polynomial Lax products in step 1.1. The resulting identity is the claimed intertwining relation for $T_0(u)$, and equality of its matrix entries gives it for each $t_{ij}(u)$ coefficient.

If $a-b\mathrm{\notin }\{i,-i\}$, [F3] makes $R_{ab}$ invertible, so the intertwiner is an isomorphism. At $a-b=i$ its kernel is antisymmetric, and at $a-b=-i$ its kernel is symmetric, by [F3]; the identity in step 2.1 still holds because the Yang–Baxter identity is polynomial and no inverse was used.

On a fixed site $j$, apply [F2] to auxiliary spectral points $u,v$ and the site parameter $a_j$. By the indexed-basis convention of [F1], the local relation is $R_{{00}^{\mathrm{\prime }}}(u-v)L_{0j}(u)L_{0^{\mathrm{\prime }}j}(v)=L_{0^{\mathrm{\prime }}j}(v)L_{0j}(u)R_{{00}^{\mathrm{\prime }}}(u-v)$. If $j\mathrm{\ne }k$, then $L_{0j}(u)$ and $L_{0^{\mathrm{\prime }}k}(v)$ act on disjoint auxiliary and quantum tensor factors, so their coefficients commute and their operators commute.

Assume it holds for $m-1$, and write $M_0^{(m)}=M_0^{(m-1)}{L}_{0m}$ and $M_{0^{\mathrm{\prime }}}^{(m)}=M_{0^{\mathrm{\prime }}}^{(m-1)}{L}_{0^{\mathrm{\prime }}m}$.

For $m=1$, the required equation is exactly step 1.1.

Moving $L_{0m}$ past $M_{0^{\mathrm{\prime }}}^{(m-1)}$ is valid by step 1.1. The induction hypothesis gives $R{M}_0^{(m)}{M}_{0^{\mathrm{\prime }}}^{(m)}=M_{0^{\mathrm{\prime }}}^{(m-1)}{M}_0^{(m-1)}R{L}_{0m}{L}_{0^{\mathrm{\prime }}m}$. Apply the local relation at site $m$, then commute $L_{0^{\mathrm{\prime }}m}$ past $M_0^{(m-1)}$ because they act on distinct sites. The result is $M_{0^{\mathrm{\prime }}}^{(m)}{M}_0^{(m)}R$, proving the induction step and the claim for every finite $m$.

Suppose $u-v\mathrm{\notin }\{i,-i\}$. Multiplying the RTT identity [F2] on the right by $R_{{00}^{\mathrm{\prime }}}(u-v{)}^{-1}$, which exists by [F3], gives $R_{{00}^{\mathrm{\prime }}}{M}_0(u)M_{0^{\mathrm{\prime }}}(v)R_{{00}^{\mathrm{\prime }}}^{-1}=M_{0^{\mathrm{\prime }}}(v)M_0(u)$. Take the ordinary trace over the two auxiliary spaces. For a finite-dimensional auxiliary matrix $X$ with coefficients in $\mathrm{End}(\mathcal{H})$ and an invertible scalar auxiliary matrix $R$, $\mathrm{tr}(RX{R}^{-1})=\mathrm{tr}(X)$: writing entries, ${\sum }_{i,j,k}{R}_{ij}{X}_{jk}(R^{-1}{)}_{ki}={\sum }_{j,k}{X}_{jk}{\sum }_i(R^{-1}{)}_{ki}{R}_{ij}={\sum }_j{X}_{jj}$, since scalar entries commute with the coefficients and $R^{-1}R=I$. Thus the trace of the left side is ${\mathrm{tr}}_{{00}^{\mathrm{\prime }}}(M_0(u)M_{0^{\mathrm{\prime }}}(v))$.