Type-A Soergel Bimodules and Hecke Categorification — Examples

Take $n=2$, so that $R=\mathbb{Q}[x_1,x_2]$, that $S_2=\{e,s\}$ has the simple reflection $s=s_1$ exchanging $x_1$ and $x_2$, and put $\alpha ={\alpha }_1=x_1-x_2$ and $\delta =\alpha \mathrm{/}2$. Write $B_s=B_1=R{\otimes }_{R^s}R(1)$ for the Soergel bimodule of $s$, $u:=1\otimes 1$, $w_0:=1\otimes \delta$, and let $R_s$ be the standard bimodule $R$ with the right action $x\cdot g=xs(g)$ of Standard graph bimodules, support filtrations and characters, so that $(k)$ denotes the external shift $M(k)=M\{-k\}$ of that item. Then:

1. The invariant ring and a homogeneous basis of $R$. The invariant ring is $R^s=\mathbb{Q}[e_1,e_2]$ for the elementary symmetric polynomials $e_1=x_1+x_2$ and $e_2=x_1{x}_2$, of degrees $2$ and $4$, so it is a graded polynomial ring in two algebraically independent generators; and $\{1,\alpha \}$ is a homogeneous $R^s$-basis of $R$, of degrees $0$ and $2$, with ${\alpha }^2=(x_1-x_2{)}^2=(x_1+x_2{)}^2-4x_1{x}_2=e_1^2-4e_2\in R^s\mathrm{.}$
2. The right action and the two sub-bimodules. $\{u,w_0\}$ is a homogeneous left $R$-basis of $B_s$ of degrees $-1$ and $+1$, so $B_s\cong R(1)\oplus R(-1)$ as a graded left $R$-module and $B_s$ is free of rank two on each side. Writing $g=g^{+}+\delta g^{-}$ with $g^{\pm }\in R^s$, that is $g^{+}=\frac{1}{2}(g+s(g))$ and $g^{-}={\mathrm{\partial }}_s(g)={\displaystyle \frac{g-s(g)}{\alpha }}$, the right action on this basis is $u\cdot g=g^{+}u+g^{-}{w}_0,w_0\cdot g={\delta }^2{g}^{-}u+g^{+}{w}_0,$ and the two degree-one elements ${\rho }_{+}:=\delta u+w_0$ and ${\rho }_{-}:=\delta u-w_0$ satisfy ${\rho }_{+}\cdot g=g{\rho }_{+},{\rho }_{-}\cdot g=s(g){\rho }_{-},$ so that $R{\rho }_{+}\cong R(-1)$ and $R{\rho }_{-}\cong R_s(-1)$ are graded $(R,R)$-sub-bimodules of $B_s$ generated in degree $1$.
3. The two rank-one exact sequences. The two degree-zero surjections ${\psi }_{+}:B_s\to R_s(1),f\otimes g\mapsto fs(g),{\psi }_{-}:B_s\to R(1),f\otimes g\mapsto fg$ take the values ${\psi }_{+}(u)=1$, ${\psi }_{+}(w_0)=-\delta$ and ${\psi }_{-}(u)=1$, ${\psi }_{-}(w_0)=\delta$ on the basis, so that $\ker {\psi }_{+}=\{Au+B{w}_0:A=\delta B\}=R{\rho }_{+}$ and $\ker {\psi }_{-}=\{Au+B{w}_0:A=-\delta B\}=R{\rho }_{-}$; consequently $0\to R(-1)\stackrel{1\mapsto {\rho }_{+}}{\to }{B}_s\stackrel{{\psi }_{+}}{\to }{R}_s(1)\to 0,0\to R_s(-1)\stackrel{1\mapsto {\rho }_{-}}{\to }{B}_s\stackrel{{\psi }_{-}}{\to }R(1)\to 0$ are exact sequences of graded $(R,R)$-bimodules with degree-zero maps.
4. The square and its explicit splitting. $B_s{\otimes }_R{B}_s$ is canonically $R{\otimes }_{R^s}R{\otimes }_{R^s}R\{-2\}$, free of rank four on each side, with the homogeneous left $R$-basis $(u\otimes u,u\otimes w_0,w_0\otimes u,w_0\otimes w_0)$ of degrees $(-2,0,0,2)$; the middle-slot maps $e_{+}:p\otimes g\otimes q\mapsto p\otimes g^{+}\otimes q,e_{-}:p\otimes g\otimes q\mapsto p\otimes \delta g^{-}\otimes q$ are degree-zero $R$-bimodule endomorphisms with $e_{+}^2=e_{+}$, $e_{-}^2=e_{-}$, $e_{+}{e}_{-}=e_{-}{e}_{+}=0$ and $e_{+}+e_{-}=\mathrm{id}$, given in the displayed basis by the diagonal matrices $\mathrm{diag}(1,1,0,0)$ and $\mathrm{diag}(0,0,1,1)$. Hence $B_s{\otimes }_R{B}_s=\mathrm{im}(e_{+})\oplus \mathrm{im}(e_{-})\cong \left(R{\otimes }_{R^s}{R}^s{\otimes }_{R^s}R\right)\{-2\}\oplus \left(R{\otimes }_{R^s}\delta R^s{\otimes }_{R^s}R\right)\{-2\}\cong B_s\{-1\}\oplus B_s\{1\},$ with $\mathrm{im}(e_{+})$ free on the basis $(u\otimes u,u\otimes w_0)$ in degrees $\{-2,0\}$ and $\mathrm{im}(e_{-})$ free on the basis $(w_0\otimes u,w_0\otimes w_0)$ in degrees $\{0,2\}$; the two summands have different graded ranks as left $R$-modules and are therefore not isomorphic as graded bimodules. This realizes the rank-one square $B_s{\otimes }_R{B}_s\cong B_s(1)\oplus B_s(-1)$ of The rank-one Soergel bimodule square splits by explicit idempotents.

Take $n=3$ and $i=1$, so that $R=\mathbb{Q}[x_1,x_2,x_3]$, that $s=s_1$ and $t=s_2$ are the two adjacent simple reflections, and that $W_{1,2}=⟨s,t⟩=S_3$ acts by permuting $x_1,x_2,x_3$. Put ${\alpha }_s=x_1-x_2$, ${\alpha }_t=x_2-x_3$, ${\delta }_s={\alpha }_s\mathrm{/}2$, ${\delta }_t={\alpha }_t\mathrm{/}2$, and let $u_s=1\otimes 1$, $w_s=1\otimes {\delta }_s$ be the rank-one basis of $B_s$, with $u_t,w_t$ the corresponding basis of $B_t$. Then:

1. The two Demazure values that carry the rank-two calculus. For the adjacent pair, $s({\alpha }_t)=x_1-x_3={\alpha }_t+{\alpha }_s$ and $t({\alpha }_s)=x_1-x_3={\alpha }_s+{\alpha }_t$, so ${\mathrm{\partial }}_s^{\beta }({\alpha }_t)=\frac{{\alpha }_t-s({\alpha }_t)}{{\alpha }_s}=-1,{\mathrm{\partial }}_t^{\beta }({\alpha }_s)=\frac{{\alpha }_s-t({\alpha }_s)}{{\alpha }_t}=-1,$ while ${\mathrm{\partial }}_s^{\beta }({\alpha }_s)=2$, ${\mathrm{\partial }}_t^{\beta }({\alpha }_t)=2$ and ${\mathrm{\partial }}_s^{\beta }({\delta }_s)=1$, ${\mathrm{\partial }}_t^{\beta }({\delta }_t)=1$. All roots and Demazure operators in this example are in the coordinate normalization ${\alpha }_s={\beta }_s=x_1-x_2$, ${\alpha }_t={\beta }_t=x_2-x_3$ of The standard type-A reflection realization and its polynomial ring, so the off-diagonal values just computed are ${\mathrm{\partial }}_s^{\beta }({\beta }_t)={\mathrm{\partial }}_t^{\beta }({\beta }_s)=-1$; in the balanced normalization ${\mathrm{\partial }}_r^{\mathrm{b}\mathrm{a}\mathrm{l}}={\epsilon }_r{\mathrm{\partial }}_r^{\beta }$ and each root insertion and Demazure contraction of color $r$ gains the factor ${\epsilon }_r$, while multiplication and unit insertion are unchanged. As ${\epsilon }_s{\epsilon }_t=-1$, the zig-zag changes from $-\mathrm{id}$ to $+\mathrm{id}$; the balanced idempotent uses the corresponding positive composite and equals the coordinate idempotent. The two graphs $Gr(s),Gr(t)$ of the adjacent reflections meet in codimension two: $s^{-1}t=st$ is a 3-cycle whose fixed space is the line $x_1=x_2=x_3$, of dimension one in the ambient space of dimension three.
2. The four maps on the rank-one bases. With ${\mu }_s:B_s\to R$, ${\mu }_s^a:R\to B_s$, ${\kappa }_s:B_s{\otimes }_R{B}_s\to B_s$ and ${\kappa }_s^a:B_s\to B_s{\otimes }_R{B}_s$ the maps ${\mu }_s(p\otimes q)=pq,{\mu }_s^a(1)={\alpha }_s\otimes 1+1\otimes {\alpha }_s,{\kappa }_s\text{ contracts the middle slot by }\frac{1}{2}{\mathrm{\partial }}_s^{\beta },{\kappa }_s^a(p\otimes q)=p\otimes 1\otimes q,$ and ${\mu }_t,{\mu }_t^a$ the same constructions for $t$; that is, ${\kappa }_s$ sends $p\otimes q\otimes h$ to $\frac{1}{2}p{\mathrm{\partial }}_s^{\beta }(q)\otimes h$ and ${\kappa }_s^a$ inserts the unit $1\in R^s$ in the middle slot $({\kappa }_s^a(1\otimes 1)=1\otimes 1\otimes 1)$. Under $(a\otimes b)\otimes (c\otimes d)\mapsto a\otimes bc\otimes d$, the evaluations on the four left-$R$ basis tensors of $B_s{\otimes }_R{B}_s$ are ${\kappa }_s(u_s\otimes u_s)=0,{\kappa }_s(u_s\otimes w_s)=0,{\kappa }_s(w_s\otimes u_s)=\frac{1}{2}{u}_s,{\kappa }_s(w_s\otimes w_s)=\frac{1}{2}{w}_s\mathrm{.}$ Moreover ${\kappa }_s^a(u_s)=u_s\otimes u_s$ and ${\kappa }_s^a(w_s)=u_s\otimes w_s$. The dot evaluations are ${\mu }_t(u_t)=1$, ${\mu }_t(w_t)={\delta }_t$ and ${\mu }_t^a(1)={\alpha }_t{u}_t+2w_t$, so ${\mu }_t{\mu }_t^a(1)=2{\alpha }_t$.
3. The zig-zag identity. Evaluating the composite ${\kappa }_s\circ {\mu }_t\circ {\mu }_t^a\circ {\kappa }_s^a$ on a general element $p\otimes q\in B_s$ gives, step by step, $p\otimes 1\otimes q⟼p\otimes {\alpha }_t\otimes 1\otimes q+p\otimes 1\otimes {\alpha }_t\otimes q⟼2p\otimes {\alpha }_t\otimes q⟼\frac{1}{2}\cdot 2p{\mathrm{\partial }}_s^{\beta }({\alpha }_t)\otimes q=-(p\otimes q),$ so ${\kappa }_s\circ {\mu }_t\circ {\mu }_t^a\circ {\kappa }_s^a=-{\mathrm{id}}_{B_s}$; this is the first of the two matrix identities, verified here on the basis $(u_s,w_s)$ by claim 1 and claim 2.
4. The idempotent and the second identity. With $e:=-{\mu }_t^a\circ {\kappa }_s^a\circ {\kappa }_s\circ {\mu }_t\in {\mathrm{End}}_{R\text{-}R}(B_s{\otimes }_R{B}_t{\otimes }_R{B}_s)$ the identity of claim 3 gives $e^2={\mu }_t^a{\kappa }_s^a\left({\kappa }_s{\mu }_t{\mu }_t^a{\kappa }_s^a\right){\kappa }_s{\mu }_t=e$, so $e$ is a degree-zero idempotent endomorphism; consequently $1-e$ is an idempotent orthogonal to $e$ and $B_1{\otimes }_R{B}_2{\otimes }_R{B}_1=\mathrm{im}(e)\oplus \mathrm{im}(1-e),$ which is the second matrix identity together with the splitting it produces.
5. The two decompositions and their rank count. For $n=3$ the rank-two decomposition theorem applies to the 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 $B_1$, so $B_1{B}_2{B}_1\cong B_{1,2,1}\oplus B_1,B_2{B}_1{B}_2\cong B_{1,2,1}\oplus B_2,$ the second decomposition being the $s\leftrightarrow t$ instance. The ranks match: $R$ is a free $R^{S_3}$-module on six generators of degrees $0,2,2,4,4,6$, so $B_{1,2,1}$ is free of graded dimension $d_{-3}+2d_{-1}+2d_1+d_3$ and $B_1$ of graded dimension $d_{-1}+d_1$ in the notation $R(a{)}_e=R_{e+a}$, while $B_1{B}_2{B}_1$ is free of graded dimension $(d_{-1}+d_1{)}^3$; and indeed $(d_{-1}+d_1{)}^3-(d_{-1}+d_1)=d_{-3}+3d_{-1}+3d_1+d_3-(d_{-1}+d_1)=d_{-3}+2d_{-1}+2d_1+d_3\mathrm{.}$