The rank-one Soergel category

The invariant ring: by [F2] the substitution $T_1\mapsto e_1$, $T_2\mapsto e_2$ identifies $\mathbb{Q}[T_1,T_2]$ with $R^s$, so $R^s=\mathbb{Q}[e_1,e_2]$ with $e_1,e_2$ algebraically independent of degrees $2$ and $4$; and ${\alpha }^2=x_1^2-2x_1{x}_2+x_2^2=e_1^2-4e_2$ lies in $R^s$.

The decomposition and the basis: by [F1] every $f\in R$ has the unique expression $f=g+\alpha h$ with $g,h\in R^s$, $g=\frac{1}{2}(f+s(f))$ and $h=\frac{1}{2}{\mathrm{\partial }}_s(f)$; equivalently, in the normalization $f=g+\delta g^{-}$ of [F5] with $\delta =\alpha \mathrm{/}2$, one has the unique $g^{-}=2h={\mathrm{\partial }}_s(f)=(f-s(f))\mathrm{/}\alpha$ in $R^s$. Applying this to the second tensor factor of $B_s=R{\otimes }_{R^s}R(1)$, the elements $u=1\otimes 1$ of degree $-1$ and $1\otimes \alpha =1\otimes 2\delta =2w_0$ of degree $+1$ form a homogeneous left $R$-basis of $B_s$ by [F4], hence so does $(u,w_0)$, and $B_s\cong R(-1)\oplus R(1)$ by [F3].

The right action: by [F3] the right action is $(r\otimes r^{\mathrm{\prime }\mathrm{\prime }})r^{\mathrm{\prime }}=r\otimes r^{\mathrm{\prime }\mathrm{\prime }}{r}^{\mathrm{\prime }}$, so $u\cdot g=1\otimes g=g^{+}u+g^{-}{w}_0$ and $w_0\cdot g=1\otimes \delta g$; since $\delta g=\delta g^{+}+{\delta }^2{g}^{-}$ with ${\delta }^2{g}^{-}$ invariant and $\delta g^{+}$ anti-invariant, the unique decomposition $\delta g=(\delta g{)}^{+}+\delta (\delta g{)}^{-}$ of [F5] has $(\delta g{)}^{+}={\delta }^2{g}^{-}$ and $(\delta g{)}^{-}=g^{+}$, hence $w_0\cdot g={\delta }^2{g}^{-}u+g^{+}{w}_0$.

The two sub-bimodules: expanding with step 1.3, $(\delta u+w_0)\cdot g=\delta (g^{+}u+g^{-}{w}_0)+({\delta }^2{g}^{-}u+g^{+}{w}_0)=(\delta g^{+}+{\delta }^2{g}^{-})u+(g^{+}+\delta g^{-})w_0=\delta gu+g{w}_0=g(\delta u+w_0)$, while $(\delta u-w_0)\cdot g=(\delta g^{+}-{\delta }^2{g}^{-})u+(\delta g^{-}-g^{+})w_0=s(g)(\delta u-w_0)$, because $s(g)=s(g^{+}+\delta g^{-})=g^{+}-\delta g^{-}$ by [F1] and $\delta g=\delta g^{+}+{\delta }^2{g}^{-}$, ${\delta }^2{g}^{-}\in R^s$; both elements have degree $1$ by step 1.2, so $R{\rho }_{+}\cong R(-1)$ and $R{\rho }_{-}\cong R_s(-1)$ as graded bimodules generated in degree $1$, as [F5] records.

The square and its idempotents: $B_s{\otimes }_R{B}_s=(R{\otimes }_{R^s}R{\otimes }_{R^s}R)\{-2\}$ by [F3], and by [F4] the four elements $u\otimes u=1\otimes 1\otimes 1$, $u\otimes w_0=1\otimes 1\otimes \delta$, $w_0\otimes u=1\otimes \delta \otimes 1$, $w_0\otimes w_0=1\otimes \delta \otimes \delta$ form a homogeneous left $R$-basis of degrees $(-2,0,0,2)$; the middle-slot projections ${\mathrm{pr}}_{+}:g\mapsto g^{+}$ and ${\mathrm{pr}}_{-}:g\mapsto \delta g^{-}$ are $(R^s,R^s)$-bilinear, hence induce well-defined $R$-bimodule endomorphisms $e_{+},e_{-}$ of the balanced tensor, and $g^{+}$ and $\delta g^{-}$ are the two components of $g$, so $e_{+}+e_{-}=\mathrm{id}$, $e_{+}{e}_{-}=e_{-}{e}_{+}=0$ and $e_{\pm }^2=e_{\pm }$.

The kernels: by step 1.2 every element of $B_s$ is uniquely $Au+B{w}_0$ with $A,B\in R$, and the maps displayed in claim 3 are $R$-balanced and degree zero with ${\psi }_{+}(u)=1$, ${\psi }_{+}(w_0)=s(\delta )=-\delta$, ${\psi }_{-}(u)=1$ and ${\psi }_{-}(w_0)=\delta$; hence ${\psi }_{+}(Au+B{w}_0)=A-\delta B$ and ${\psi }_{-}(Au+B{w}_0)=A+\delta B$, so $\ker {\psi }_{+}=R{\rho }_{+}$ and $\ker {\psi }_{-}=R{\rho }_{-}$. The inclusions $1\mapsto {\rho }_{\pm }$ are injective because ${\rho }_{\pm }\mathrm{\ne }0$ in the free left $R$-module $B_s$ by step 1.2, and both maps are surjective because ${\psi }_{\pm }(u)=1$ generates the rank-one target; this verifies the two exact sequences of claim 3 and their degree-zero maps.

The matrix and the split: on the basis of step 2.2 the map $e_{+}$ fixes $u\otimes u$ and $u\otimes w_0$, whose middle slot is $1$, and kills $w_0\otimes u$ and $w_0\otimes w_0$, whose middle slot is $\delta$ with ${\delta }^{+}=0$; so $e_{+}=\mathrm{diag}(1,1,0,0)$ and $e_{-}=\mathrm{diag}(0,0,1,1)$ in that basis, and $B_s{\otimes }_R{B}_s=\mathrm{im}(e_{+})\oplus \mathrm{im}(e_{-})$ with $\mathrm{im}(e_{+})$ free on $u\otimes u,u\otimes w_0$ and $\mathrm{im}(e_{-})$ free on $w_0\otimes u,w_0\otimes w_0$, the two blocks having different degree sets $\{-2,0\}$ and $\{0,2\}$ and hence different graded ranks.

The identification of the summands: the multiplication $p\otimes r\otimes q\mapsto pr\otimes q$ is a degree-zero isomorphism $R{\otimes }_{R^s}{R}^s{\otimes }_{R^s}R\{-2\}\to (R{\otimes }_{R^s}R)\{-2\}=B_s\{-1\}$ by [F3], and $p\otimes \delta r\otimes q\mapsto p\otimes r\otimes q$ is a degree-zero isomorphism $R{\otimes }_{R^s}\delta R^s{\otimes }_{R^s}R\{-2\}\to (R{\otimes }_{R^s}R)\{0\}=B_s\{1\}$; since $\delta R^s=\alpha R^s$ (as $2$ is invertible in $\mathbb{Q}$), this realizes the two idempotent images of [F6], and the non-isomorphism of the two summands is their differing graded rank from step 3.2.

Conclusion: for $n=2$ the invariant ring is $R^s=\mathbb{Q}[e_1,e_2]$ with the homogeneous $R^s$-basis $\{1,\alpha \}$ of $R$ and ${\alpha }^2=e_1^2-4e_2$ (claim 1, step 1.1); the two rank-one exact sequences of claim 3 hold with the explicit maps and kernels of steps 2.1 and 3.1; and the square splits through the explicit middle-slot idempotents $e_{\pm }$ of steps 2.2 and 3.2 into the summands $B_s\{-1\}\oplus B_s\{1\}$ identified in step 4.1, that is into $B_s(1)\oplus B_s(-1)$ in the external shift, with the two summands of different graded rank. Every object and map used is an explicit finite free module with a displayed homogeneous basis, so no choice principle is used. ∎