Frobenius biadjunction for the type-A Soergel generators

Since $\alpha$ is $s$-anti-invariant, ${\mathrm{\partial }}_s$ is $R^s$-linear and ${\mathrm{\partial }}_s(h)=0$ for $h\in R^s$, while ${\mathrm{\partial }}_s(\delta )=\frac{1}{2}{\mathrm{\partial }}_s(\alpha )=1$ by [F1]; and ${\delta }^2={\alpha }^2\mathrm{/}4\in R^s$ lies in the kernel, so in the basis $(1,\delta )$ the form $⟨f,g{⟩}_s={\mathrm{\partial }}_s(fg)$ has $⟨1,1{⟩}_s=0$, $⟨1,\delta {⟩}_s=1$, $⟨\delta ,\delta {⟩}_s=0$, that is Gram matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, whence $(1,\delta )$ and $(\delta ,1)$ are mutually dual bases.

The assignment ${\mathrm{\Phi }}_M(f\otimes m)(g):={\mathrm{\partial }}_s(fg)m$ is well defined on the balanced tensor product, since ${\mathrm{\partial }}_s(hfg)m={\mathrm{\partial }}_s(fg)hm$ for $h\in R^s$, and its values are $R^s$-linear in $g$. For the left $R$-action $(r\cdot \psi )(g)=\psi (rg)$ on ${\mathrm{Hom}}_{R^s}(R,M)$, one has ${\mathrm{\Phi }}_M(rf\otimes m)(g)={\mathrm{\Phi }}_M(f\otimes m)(rg)$, so ${\mathrm{\Phi }}_M$ is $R$-linear. On the unshifted tensor product its formula lowers degree by two because ${\mathrm{\partial }}_s$ has degree $-2$; hence the domain shift $R\{-2\}$ makes ${\mathrm{\Phi }}_M$ homogeneous of degree zero.

Hom-tensor adjunction over $R^s$ as in [F3] turns $B_s{\otimes }_{R}M=R{\otimes }_{R^s}(R\{-1\}{\otimes }_{R}M)$ into the hom space ${\mathrm{Hom}}_{R^s\text{-}R}(R\{-1\}{\otimes }_{R}M,N)$, and a second adjunction identifies the latter with ${\mathrm{Hom}}_{R\text{-}R}(M,{\mathrm{Hom}}_{R^s}(R\{-1\},N))$.

Put $b_1=1$, $b_2=\delta$ and $b^1=\delta$, $b^2=1$. The Gram matrix of step 1.1 gives the two dual-basis identities $r={\sum }_i{b}_i{\mathrm{\partial }}_s(b^{i}r)={\sum }_i{\mathrm{\partial }}_s(r{b}_i)b^i$ for every $r\in R$. For any graded $R^s$-module $M$ define ${\mathrm{\Psi }}_M(\psi ):=1\otimes \psi (\delta )+\delta \otimes \psi (1)$ in $R\{-2\}{\otimes }_{R^s}M$ for $\psi \in {\mathrm{Hom}}_{R^s}(R,M)$. The first identity and $R^s$-linearity of $\psi$ give ${\mathrm{\Phi }}_M{\mathrm{\Psi }}_M(\psi )(r)=\psi (r)$. Conversely, writing $f=h_0+\delta h_1$ with $h_0,h_1\in R^s$, we have ${\mathrm{\partial }}_s(f\delta )=h_0$ and ${\mathrm{\partial }}_s(f)=h_1$, so ${\mathrm{\Psi }}_M{\mathrm{\Phi }}_M(f\otimes m)=1\otimes h_{0}m+\delta \otimes h_{1}m=f\otimes m$. If $\psi$ has degree $d$, then $1\otimes \psi (\delta )$ has degree $-2+(d+2)=d$ and $\delta \otimes \psi (1)$ has degree $0+d=d$; thus ${\mathrm{\Psi }}_M$ is degree zero. Both formulas commute with every graded $R^s$-linear map $M\to M^{\mathrm{\prime }}$, so ${\mathrm{\Phi }}_M$ is a natural degree-zero $R$-linear isomorphism for every graded $M$, without a freeness assumption.

Applying step 2.1 to the graded $R^s$-module $N\{1\}$ gives ${\mathrm{Hom}}_{R^s}(R\{-1\},N)\cong {\mathrm{Hom}}_{R^s}(R,N\{1\})\cong R\{-2\}{\otimes }_{R^s}N\{1\}\cong R\{-1\}{\otimes }_{R^s}N$. The last object is $B_s{\otimes }_{R}N$: by the balanced tensor relation, $(R{\otimes }_{R^s}R\{-1\}){\otimes }_{R}N\cong R{\otimes }_{R^s}N\{-1\}\cong R\{-1\}{\otimes }_{R^s}N$, with no extra $R$ factor. These are degree-zero $(R,R)$-bimodule identifications by the shift convention; composing them with the two Hom-tensor adjunctions of step 1.3 gives the first displayed natural bijection.

The flip $f\otimes g\mapsto g\otimes f$ on $R{\otimes }_{R^s}R\{-1\}$ is well defined because $R$ is commutative and $R^s$ is central, is homogeneous because the factor degrees add, and is its own inverse. It intertwines the left and right $R$-actions, so it is a graded bimodule isomorphism $B_s\to B_s^{\mathrm{o}\mathrm{p}}$. In particular it fixes $u=1\otimes 1$ and sends $w_0=1\otimes \delta$ to $\delta \otimes 1$, both with their original degrees; it does not swap $u$ and $w_0$. Applying the first adjunction of step 3.1 to opposite bimodules yields the second displayed natural degree-zero bijection ${\mathrm{Hom}}_{R\text{-}R}(M{\otimes }_R{B}_s,N)\cong {\mathrm{Hom}}_{R\text{-}R}(M,N{\otimes }_R{B}_s)$.

For completeness, the two triangle maps can be checked directly. Under $B_s{\otimes }_R{B}_s\cong R{\otimes }_{R^s}R{\otimes }_{R^s}R\{-2\}$, evaluation is the degree-zero bimodule map $\epsilon ((a\otimes b)\otimes (c\otimes d))=a{\mathrm{\partial }}_s(bc)d$, and coevaluation is the degree-zero bimodule map sending $1\in R$ to $E:={\sum }_i(b_i\otimes 1)\otimes (1\otimes b^i)$, with the dual bases of step 2.1. The underlying degree of $E$ is two and the total shift is $-2$; $E$ is central for the outer $R$-actions, as is checked on the generators $R^s$ and $\delta$ using ${\delta }^2\in R^s$. For $z=a\otimes b\in B_s$, the composite $({\mathrm{id}}_{B_s}\otimes \epsilon )(E\otimes z)$ equals ${\sum }_i{b}_i\otimes {\mathrm{\partial }}_s(b^{i}a)b=a\otimes b$, while $(\epsilon \otimes {\mathrm{id}}_{B_s})(z\otimes E)$ equals ${\sum }_{i}a{\mathrm{\partial }}_s(b{b}_i)\otimes b^i=a\otimes b$, by the two dual-basis identities of step 2.1. Tensoring these maps with any graded bimodule gives the unit and counit triangle identities for both adjunctions; all maps are natural and degree zero. ∎