Rank-two type-A Soergel bimodule decompositions

The four maps: define $R$-bimodule maps ${\mu }_s:B_s\to R$ by ${\mu }_s(p\otimes q)=pq$, ${\mu }_s^a:R\to B_s$ by ${\mu }_s^a(1)={\alpha }_s\otimes 1+1\otimes {\alpha }_s$, ${\kappa }_s:B_s{\otimes }_R{B}_s\to B_s$ by ${\kappa }_s(p\otimes q\otimes h)=\frac{1}{2}p{\mathrm{\partial }}_s^{\beta }(q)\otimes h$ and ${\kappa }_s^a:B_s\to B_s{\otimes }_R{B}_s$ by ${\kappa }_s^a(p\otimes q)=p\otimes 1\otimes q$, and let ${\mu }_t,{\mu }_t^a$ be the same constructions for $t$.

Degrees: ${\mu }_s$ and ${\mu }_s^a$ have degree $+1$ and ${\kappa }_s,{\kappa }_s^a$ have degree $-1$: an element of $B_s$ of degree $d$ is an element of $R{\otimes }_{R^s}R$ of degree $d+1$ by $B_s=R{\otimes }_{R^s}R(1)$, so $pq$ has degree $d+1$ in $R$, ${\alpha }_s\otimes 1$ and $1\otimes {\alpha }_s$ have degree $1$ in $B_s$ because $\mathrm{deg}{\alpha }_s=2$ and $1\otimes 1$ has degree $-1$, and ${\mathrm{\partial }}_s^{\beta }$ lowers degrees by $2$.

Balancedness of the four maps: ${\mu }_s$ and ${\kappa }_s^a$ are visibly balanced in the middle slots; ${\kappa }_s$ is balanced because ${\mathrm{\partial }}_s^{\beta }$ is $R^s$-linear, so ${\mathrm{\partial }}_s^{\beta }(r_{s}q)=r_s{\mathrm{\partial }}_s^{\beta }(q)$ and ${\mathrm{\partial }}_s^{\beta }(q{r}_s)={\mathrm{\partial }}_s^{\beta }(q)r_s$ for $r_s\in R^s$, whence $p{\mathrm{\partial }}_s^{\beta }(r_{s}q)\otimes h=p{r}_s{\mathrm{\partial }}_s^{\beta }(q)\otimes h$ and $p{\mathrm{\partial }}_s^{\beta }(q{r}_s)\otimes h=p{\mathrm{\partial }}_s^{\beta }(q)\otimes r_{s}h$.

The adjoint ${\mu }_s^a$ is well defined: a map out of the regular bimodule $R$ must send $f$ to the same element through $f\cdot {\mu }_s^a(1)$ and ${\mu }_s^a(1)\cdot f$, so it suffices that $Z_s:={\alpha }_s\otimes 1+1\otimes {\alpha }_s$ commutes with $R$. Every $f$ is $f^{+}+{\alpha }_s{f}^{-}$ with $f^{\pm }\in R^s$ by [F3], and elements of $R^s$ slide across the tensor divider and commute with $Z_s$. For the remaining generator ${\alpha }_s$, one has ${\alpha }_s{Z}_s={\alpha }_s^2\otimes 1+{\alpha }_s\otimes {\alpha }_s$ and $Z_s{\alpha }_s={\alpha }_s\otimes {\alpha }_s+1\otimes {\alpha }_s^2$; these are equal because ${\alpha }_s^2\in R^s$. Thus $f{Z}_s=Z_{s}f$ for every $f$, and ${\mu }_s^a(f):=f{Z}_s$ is a well-defined $R$-bimodule map of degree $+1$ because $Z_s$ has degree $1$.

The large summand: by [F5] $\mathrm{im}(1-e)$ is generated as an $R$-bimodule by $1^{\otimes }$ inside $B_{\underset{‾}{i,i+1,i}}=R{\otimes }_{R^s}R{\otimes }_{R^t}R{\otimes }_{R^s}R(3)$; the assignment $p\otimes q\mapsto p\otimes 1\otimes 1\otimes q$ is a well-defined degree-zero $R$-bimodule map from $R{\otimes }_{R^{W_{i,i+1}}}R(3)$ because an element $g\in R^{W_{i,i+1}}$ is both $s$- and $t$-invariant and therefore slides across both dividers of $1^{\otimes }$. It is surjective onto $\mathrm{im}(1-e)$ because the bimodule generated by $1^{\otimes }$ is the set of finite sums of elements $p{1}^{\otimes }q$ with $p,q\in R$, the images of finite sums of $p\otimes q$.

The composite identity: for the adjacent pair, $s({\alpha }_t)={\alpha }_t+{\alpha }_s$ because $s$ exchanges $x_i$ and $x_{i+1}$ and fixes every other coordinate, so ${\mathrm{\partial }}_s^{\beta }({\alpha }_t)=({\alpha }_t-s({\alpha }_t))\mathrm{/}{\alpha }_s=-1$. Applying ${\kappa }_s\circ {\mu }_t\circ {\mu }_t^a\circ {\kappa }_s^a$ to $p\otimes q\in B_s$, the successive images are $p\otimes 1\otimes q$ under ${\kappa }_s^a$, then $p\otimes {\alpha }_t\otimes 1\otimes q+p\otimes 1\otimes {\alpha }_t\otimes q$ under the middle-slot insertion ${\mu }_t^a$, then $2p\otimes {\alpha }_t\otimes q$ under the middle-slot multiplication ${\mu }_t$, and finally $\frac{1}{2}\cdot 2p{\mathrm{\partial }}_s^{\beta }({\alpha }_t)\otimes q=-(p\otimes q)$ under ${\kappa }_s$; hence ${\kappa }_s\circ {\mu }_t\circ {\mu }_t^a\circ {\kappa }_s^a=-{\mathrm{id}}_{B_s}$.

The idempotent: with the identities understood in the sense of tensor slots (as in the sources), let $e:=-{\mu }_t^a\circ {\kappa }_s^a\circ {\kappa }_s\circ {\mu }_t\in {\mathrm{End}}_{R\text{-}R}(B_s{B}_t{B}_s)$; 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 2.1, so $e$ is an idempotent, and it is homogeneous of degree $0$ by step 1.2.

The splitting and its small summand: $1-e$ is an idempotent orthogonal to $e$, so the graded bimodule splits as $B_s{B}_t{B}_s=\mathrm{im}(e)\oplus \mathrm{im}(1-e)$. The composite identity of step 2.1 makes $-{\kappa }_s{\mu }_t$ a left inverse to ${\mu }_t^a{\kappa }_s^a$, so $\mathrm{im}(e)=\mathrm{im}({\mu }_t^a{\kappa }_s^a)\cong B_s$ without a separate injectivity or surjectivity claim; the comparison has degree zero by step 1.2.

Graded dimension count: let $u$ record the degree of a homogeneous free left-$R$ generator. By [F1] and [F2], $B_s$ and $B_t$ each have graded left-$R$ rank $u^{-1}+u$, and their tensor product $B_s{B}_t{B}_s$ has rank $(u^{-1}+u{)}^3$. By step 4.1 the Hilbert series of the complementary summand is the Hilbert series of $R$ times $(u^{-1}+u{)}^3-(u^{-1}+u)=u^{-3}+2u^{-1}+2u+u^3$. This is also the Hilbert series of $B_{i,i+1,i}\cong R(-3)\oplus R(-1{)}^{\oplus 2}\oplus R(1{)}^{\oplus 2}\oplus R(3)$ from [F4]. Thus the surjection of step 1.5 compares $k$-vector spaces of equal finite dimension in every degree.

Conclusion: a graded surjection that is a comparison of finite-dimensional $k$-vector spaces of equal dimension in each degree is an isomorphism, so $\mathrm{im}(1-e)\cong R{\otimes }_{R^{W_{i,i+1}}}R(3)=B_{i,i+1,i}$ with a degree-zero identification; combined with step 4.1 this gives $B_i{B}_{i+1}{B}_i\cong B_{i,i+1,i}\oplus B_i$. The second decomposition is the same argument with $s$ and $t$ interchanged, the identity ${\mathrm{\partial }}_t^{\beta }({\alpha }_s)=-1$ being symmetric, so $B_{i+1}{B}_i{B}_{i+1}\cong B_{i,i+1,i}\oplus B_{i+1}$. ∎