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 natural map $j:R{\otimes }_{R^{W_{i,i+1}}}R(3)\to T:=B_s{B}_t{B}_s$, $j(p\otimes q)=p\otimes 1\otimes 1\otimes q$, is balanced and has degree zero: a parabolic invariant slides across all three dividers. Its injectivity will follow from the explicit perfect trace pairing below, rather than by an assumed description of the complement.

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 ${\kappa }_s\circ {\mu }_t\circ {\mu }_t^a\circ {\kappa }_s^a=-{\mathrm{id}}_{B_s}$.

Injectivity of the large-summand map. On $k[x,y,z]$ put $\tau ={\mathrm{\partial }}_s^{\beta }{\mathrm{\partial }}_t^{\beta }{\mathrm{\partial }}_s^{\beta }$, a map of degree $-6$ into the symmetric invariant ring. Explicitly, its value is the alternating sum of the six permuted inputs divided by $(x-y)(x-z)(y-z)$: expanding the three operators gives this identity, and the alternating numerator is divisible by each of these distinct linear factors. Therefore it is symmetric and invariant-ring-linear. It sends $x^{2}y$ to $1$ (successively the three operators give $xy,x,1$) and kills polynomials of degree less than $6$. The Gram determinant of $\tau (fg)$ in the six-element basis of [F5] is homogeneous of degree zero: the sum of the six basis degrees is $18$, so every determinant term has degree $2\cdot 18-6\cdot 6=0$. Evaluate at zero positive-degree invariants. Complementary degrees pair by $1$ in degrees $0,6$ and by the matrix $\left(\begin{array}{cc}0& 1\\ 1& -1\end{array}\right)$ in degrees $2,4$, using $x^3=0$ and $x{y}^2=-x^{2}y$; all noncomplementary pairings vanish. The determinant is nonzero. Thus the original determinant is a nonzero constant, and $\tau (fg)$ is a perfect pairing over the invariant ring. On the unshifted commutative tensor algebra underlying $T$, successive rank-one traces give a right-$R$-linear map ${\mathrm{Tr}}_T$ of degree $-6$. Indeed a rank-one trace sends $a\otimes b$ to ${\mathrm{\partial }}_s^{\beta }(a)b$ in the right-coefficient convention, and iterating gives ${\mathrm{Tr}}_T(j(p\otimes q))=\tau (p)q$. Termwise multiplication consequently gives ${\mathrm{Tr}}_T(j(u)j(v))$ equal to the scalar extension of the perfect $\tau$ pairing, for $u,v\in R{\otimes }_{R^{W_{i,i+1}}}R$. If $j(u)=0$, all these pairings vanish, and perfection forces $u=0$. Hence $j$ is injective.

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.

Its image lies in the complementary summand. The split projection $d=-{\kappa }_s{\mu }_t:T\to B_s$ from step 4.1 kills $1^{\otimes }$, because the middle polynomial entering ${\mathrm{\partial }}_s^{\beta }$ is $1$. By bimodule linearity it kills every $j(p\otimes q)$. With the section $a={\mu }_t^a{\kappa }_s^a$, the identity $da=\mathrm{id}$ shows $e=ad$, so $\ker d=\mathrm{im}(1-e)$. Thus $j$ injects into that complement.

Write $u$ for the degree of a homogeneous left-$R$ basis generator. The triple tensor has rank $(u^{-1}+u{)}^3$, while the split small summand has rank $u^{-1}+u$. Thus the complement has Hilbert series equal to the Hilbert series of $R$ times $u^{-3}+2u^{-1}+2u+u^3$, precisely the series of the source of $j$ by the proved six-element basis. In each degree these vector spaces have equal finite dimension. The injection of step 5.1 is therefore surjective degree by degree, and $j$ is an isomorphism onto $\mathrm{im}(1-e)$.

Conclusion: the injection and Hilbert-series comparison in the preceding injectivity and Hilbert-series calculations prove that $\mathrm{im}(1-e)\cong R{\otimes }_{R^{W_{i,i+1}}}R(3)=B_{i,i+1,i}$ 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 $B_{i+1}{B}_i{B}_{i+1}\cong B_{i,i+1,i}\oplus B_{i+1}$.