The Casimir and place-permutation operators satisfy the degenerate affine Hecke relations

The infinitesimal braid identities. For distinct labels $a,b,c$ one has $[{\mathrm{\Omega }}_{ab},{\mathrm{\Omega }}_{ac}+{\mathrm{\Omega }}_{bc}]=0$: expanding $\mathrm{\Omega }=\sum E_{uv}\otimes E_{vu}$ and using [F3], the double commutator sum splits into terms $[E_{uv},E_{xy}]\otimes E_{vu}\otimes E_{yx}$ and $E_{uv}\otimes [E_{vu},E_{xy}]\otimes E_{yx}$, and the two resulting sums cancel term by term after relabelling; equivalently, this is the identity $[{\mathrm{\Omega }}_{12},{\mathrm{\Omega }}_{13}+{\mathrm{\Omega }}_{23}]=0$ in $U(\mathfrak{g}{)}^{\otimes 3}$, which is the derivative of the invariance of $\mathrm{\Omega }$ under the diagonal adjoint action. Consequently $[{\mathrm{\Omega }}_{ab},{\mathrm{\Omega }}_{ac}]=-[{\mathrm{\Omega }}_{ab},{\mathrm{\Omega }}_{bc}]=[{\mathrm{\Omega }}_{ac},{\mathrm{\Omega }}_{bc}]$. In particular, for three natural factors this is the identity $[P_{ab},P_{ac}]+[P_{ab},P_{bc}]=0$ for the flips.

The Coxeter relations for $s_i$. $s_i=P_{i,i+1}$ is an involution; $s_i{s}_j=s_j{s}_i$ for $\mathrm{\mid }i-j\mathrm{\mid }>1$ because the pairs of interchanged labels are disjoint; and $s_i{s}_{i+1}{s}_i=s_{i+1}{s}_i{s}_{i+1}$ because both sides are the permutation operator inducing the same permutation of the three labelled factors.

Cross relations. Conjugation by $s_i=P_{i,i+1}$ interchanges the labels $i$ and $i+1$ in any operator supported on those factors, so $s_i{\mathrm{\Omega }}_{0,i+1}={\mathrm{\Omega }}_{0i}{s}_i$ and $s_i{P}_{k,i+1}=P_{ki}{s}_i$ for $k<i$. Therefore $s_i{x}_{i+1}=s_i{\mathrm{\Omega }}_{0,i+1}+{\sum }_{k<i}{s}_i{P}_{k,i+1}+s_i{P}_{i,i+1}={\mathrm{\Omega }}_{0i}{s}_i+{\sum }_{k<i}{P}_{ki}{s}_i+1=x_i{s}_i+1$, and symmetrically $x_{i+1}{s}_i=s_i{x}_i+1$. For $j<i$, all labels in $x_j$ are disjoint from $\{i,i+1\}$, so $s_i$ commutes with every summand. For $j>i+1$, ${\mathrm{\Omega }}_{0j}$ and the terms $P_{kj}$ with $k\mathrm{\notin }\{i,i+1\}$ commute with $s_i$, while conjugation by $s_i$ interchanges $P_{ij}$ and $P_{i+1,j}$. Their sum is invariant, so again $s_i{x}_j=x_j{s}_i$; the two overlapping summands need not commute individually.

Polynomial commutation. Fix $i<j$. Since $x_i={\mathrm{\Omega }}_{0i}+{\sum }_{k<i}{P}_{ki}$ and $x_j={\mathrm{\Omega }}_{0j}+{\sum }_{l<j}{P}_{lj}$, the commutator is $[{\mathrm{\Omega }}_{0i},{\mathrm{\Omega }}_{0j}]+{\sum }_{l<j}[{\mathrm{\Omega }}_{0i},P_{lj}]+{\sum }_{k<i}[P_{ki},{\mathrm{\Omega }}_{0j}]+{\sum }_{k<i}{\sum }_{l<j}[P_{ki},P_{lj}]$. The second sum reduces to the single term $[{\mathrm{\Omega }}_{0i},P_{ij}]$ (all other $P_{lj}$ are disjoint from the labels $0,i$), and the third sum vanishes (all pairs $\{k,i\}$, $\{0,j\}$ are disjoint since $k<i<j$). In the fourth sum only the terms with $k=l$ or $l=i$ can fail to vanish, so it equals ${\sum }_{k<i}([P_{ki},P_{kj}]+[P_{ki},P_{ij}])$. Hence $[x_i,x_j]=([{\mathrm{\Omega }}_{0i},{\mathrm{\Omega }}_{0j}]+[{\mathrm{\Omega }}_{0i},P_{ij}])+{\sum }_{k<i}([P_{ki},P_{kj}]+[P_{ki},P_{ij}])$. The first bracket vanishes by step 1.1 with $a=0$, $b=i$, $c=j$ together with ${\mathrm{\Omega }}_{ij}=P_{ij}$; each summand of the second bracket vanishes by step 1.1 for the three distinct labels $k,i,j$. Therefore $[x_i,x_j]=0$.

Naturality, commutation with $\mathfrak{g}$, and the right action. The place permutations are natural in $M$ and commute with $\mathfrak{g}$; the Casimir operators are intertwiners by [F5]; hence every $x_j$ and $s_i$ is natural and $\mathfrak{g}$-linear. The relations established in steps 1.1, 1.2, 2.1 and 2.2 are exactly the defining relations of $H_d$ [F4], so the assignment $x_j\mapsto x_j$, $s_i\mapsto s_i$ defines a left $H_d$-action. The generator-fixing anti-automorphism $\omega$ of $H_d$ (which preserves each defining relation because the reversed cross relation $x_{i+1}{s}_i=s_i{x}_i+1$ is also a defining relation) converts it into a right action by $e\cdot u:=\rho (\omega (u))e$; the anti-automorphism property gives $(e\cdot u)\cdot v=\rho (\omega (v))\rho (\omega (u))e=\rho (\omega (uv))e=e\cdot (uv)$. This proves both assertions.