The type $A_2$ rank-two Soergel decomposition

The adjacent Demazure values: $s$ exchanges $x_1$ and $x_2$ and fixes $x_3$, so $s({\alpha }_t)=s(x_2-x_3)=x_1-x_3={\alpha }_t+{\alpha }_s$ and ${\mathrm{\partial }}_s^{\beta }({\alpha }_t)=({\alpha }_t-{\alpha }_t-{\alpha }_s)\mathrm{/}{\alpha }_s=-1$ by [F1]; symmetrically $t$ exchanges $x_2$ and $x_3$ and fixes $x_1$, so $t({\alpha }_s)=t(x_1-x_2)=x_1-x_3={\alpha }_s+{\alpha }_t$ and ${\mathrm{\partial }}_t^{\beta }({\alpha }_s)=({\alpha }_s-{\alpha }_s-{\alpha }_t)\mathrm{/}{\alpha }_t=-1$, while ${\mathrm{\partial }}_s^{\beta }({\alpha }_s)=2$ and ${\mathrm{\partial }}_t^{\beta }({\alpha }_t)=2$ by [F1] with $h=1$, so ${\mathrm{\partial }}_s^{\beta }({\delta }_s)={\mathrm{\partial }}_t^{\beta }({\delta }_t)=1$.

The evaluations of the four maps: the four tensors $u_s\otimes u_s$, $u_s\otimes w_s$, $w_s\otimes u_s$, $w_s\otimes w_s$ correspond respectively to $1\otimes 1\otimes 1$, $1\otimes 1\otimes {\delta }_s$, $1\otimes {\delta }_s\otimes 1$, $1\otimes {\delta }_s\otimes {\delta }_s$. Contracting their middle polynomial by $\frac{1}{2}{\mathrm{\partial }}_s^{\beta }$ gives $0,0,\frac{1}{2}{u}_s,\frac{1}{2}{w}_s$, using step 1.1. Unit insertion sends $u_s,w_s$ to $u_s\otimes u_s,u_s\otimes w_s$. Multiplication gives ${\mu }_t(u_t)=1$, ${\mu }_t(w_t)={\delta }_t$, so ${\mu }_t({\alpha }_t{u}_t+2w_t)=2{\alpha }_t$. These are exactly the well-typed evaluations of claim 2.

The zig-zag: applying the four maps in the order ${\kappa }_s^a,{\mu }_t^a,{\mu }_t,{\kappa }_s$ to $p\otimes q\in B_s$ produces $p\otimes 1\otimes q$, then $p\otimes {\alpha }_t\otimes 1\otimes q+p\otimes 1\otimes {\alpha }_t\otimes q$, then $2p\otimes {\alpha }_t\otimes q$ by the middle-slot multiplication ${\mu }_t$, and finally, applying ${\kappa }_s$ with its factor $\frac{1}{2}$, the element $2\cdot \frac{1}{2}p{\mathrm{\partial }}_s^{\beta }({\alpha }_t)\otimes q=p{\mathrm{\partial }}_s^{\beta }({\alpha }_t)\otimes q=-(p\otimes q)$ by step 1.1; since $p\otimes q$ was arbitrary in the free left $R$-module $B_s$, ${\kappa }_s\circ {\mu }_t\circ {\mu }_t^a\circ {\kappa }_s^a=-{\mathrm{id}}_{B_s}$. This is the first matrix identity, checked on the basis $(u_s,w_s)$ through the evaluations of step 2.1.

The idempotent: following the construction of [F7], put $e:=-{\mu }_t^a\circ {\kappa }_s^a\circ {\kappa }_s\circ {\mu }_t$ on $B_s{\otimes }_R{B}_t{\otimes }_R{B}_s$; then $e^2={\mu }_t^a{\kappa }_s^a({\kappa }_s{\mu }_t{\mu }_t^a{\kappa }_s^a){\kappa }_s{\mu }_t={\mu }_t^a{\kappa }_s^a(-\mathrm{id}){\kappa }_s{\mu }_t=e$ by step 3.1, so $e$ is an idempotent, and it has degree zero because the four maps, ordered as ${\mu }_t,{\mu }_t^a,{\kappa }_s,{\kappa }_s^a$, are homogeneous of degrees $+1,+1,-1,-1$: the $\mu$ maps have degree $+1$ and the $\kappa$ maps have degree $-1$ of The type-A diagrammatic Soergel category and its candidate bimodule functor, so their degrees sum to zero by step 2.1; hence $1-e$ is an idempotent orthogonal to $e$ and the graded bimodule of endomorphisms splits $B_s{B}_t{B}_s=\mathrm{im}(e)\oplus \mathrm{im}(1-e)$. This is the second matrix identity and the splitting it produces.

The summands: by [F7] applied to the adjacent pair $s,t$ the summand $\mathrm{im}(1-e)$ is isomorphic to $B_{1,2,1}=R{\otimes }_{R^{S_3}}R(3)$ and $\mathrm{im}(e)$ to $B_s$, with degree-zero identifications and no extra shift, because the comparison maps of that theorem are the composites of the four maps used here; hence $B_1{B}_2{B}_1\cong B_{1,2,1}\oplus B_1$, and applying the same statement with $s$ and $t$ interchanged gives $B_2{B}_1{B}_2\cong B_{1,2,1}\oplus B_2$, the parabolic $W_{1,2}$ being symmetric in $s$ and $t$.

The rank count: by [F5] $B_1{B}_2{B}_1$ is free as a left $R$-module with the tensor basis of the left bases $(u_s,w_s)$, $(u_t,w_t)$, $(u_s,w_s)$, so its graded dimension is $(d_{-1}+d_1{)}^3$ in the notation $R(a{)}_e=R_{e+a}$ for the graded dimension of a shift of $R$; by [F2] the invariant ring is the polynomial ring $\mathbb{Q}[e_1,e_2,e_3]$ on generators of degrees $2,4,6$, and by [F6] $B_{1,2,1}$ is free of graded dimension $d_{-3}+2d_{-1}+2d_1+d_3$, using the six generators of $R$ over $R^{S_3}$ in degrees $0,2,2,4,4,6$ shifted by $-3$; and $B_1$ is free of graded dimension $d_{-1}+d_1$; the identity $(d_{-1}+d_1{)}^3-(d_{-1}+d_1)=d_{-3}+2d_{-1}+2d_1+d_3$ holds by expanding $(d_{-1}+d_1{)}^3=d_{-3}+3d_{-1}+3d_1+d_3$, so the ranks of the two sides of step 5.1 agree in every degree, as in the dimension count of [F7].

Conclusion: for $n=3$ the adjacent pair $s=s_1$, $t=s_2$ has ${\mathrm{\partial }}_s^{\beta }({\alpha }_t)={\mathrm{\partial }}_t^{\beta }({\alpha }_s)=-1$, the four rank-two maps take the explicit values of step 2.1 on the tensor basis, the two matrix identities hold by steps 3.1 and 4.1, and the rank-two decomposition theorem gives $B_1{B}_2{B}_1\cong B_{1,2,1}\oplus B_1$ and $B_2{B}_1{B}_2\cong B_{1,2,1}\oplus B_2$ with matching graded ranks as computed in step 6.1. All objects involved are finite free graded $R$-modules with displayed homogeneous bases, so no choice principle is used. ∎