# Monday July 13th ## Ring structure on $K_0^\oplus(\id)$. :::{.definition title="Monoidal Categories"} A *monoidal category* is a tuple $(\mcc, \wait \tensor \wait, 1, \alpha, \ell, r)$ such that - $\mcc$ is a category - $\wait \tensor \wait: \mcc \cross \mcc \to \mcc$ is a bifunctor. - $1\in \mcc$ - Natural isomorphisms \[ \alpha_{X,Y,Z}: (X\tensor Y)\tensor Z \mapsvia{\cong} X\tensor (Y\tensor Z) \] for all $X,Y,Z\in \mcc$ (associators). - Natural isomorphisms \[ \ell_X: 1\tensor X &\mapsvia{\cong} X\\ r_X:X\tensor 1 &\mapsvia{\cong}X \] and for all $X\in \mcc$. Along with coherence axioms: for all $W,X,Y,Z\in \mcc$, ![](figures/image_2020-07-13-11-09-53.png) \begin{tikzcd} X \tensor(1\tensor Y)\ar[rd, "\id_x \tensor \phi_Y"] \ar[rr, "\alpha_{X, 1, Y}"] & & X\tensor 1 \tensor Y \ar[ld, "r_X \tensor \id_Y"] \\ & X\tensor Y & \end{tikzcd} ::: :::{.remark} If $\mcc$ is additive, we require $\wait \tensor \wait$ to be biadditive, i.e. $X\tensor \wait$ and $\wait \tensor Y$ are additive functors. In particular, \[ X\tensor (V\oplus W) \cong (X\tensor V) \oplus (X\tensor W) \] and similarly \[ (V\oplus W) \tensor Y \cong (V\tensor Y) \oplus (W\tensor Y) .\] ::: :::{.example} $\rmod$ with $R$ a commutative unital ring, take $\tensor \definedas \tensor_R$ with $1$ the "regular left $R\dash$module" ${}_R R$ with $R$ acting on the left by multiplication. Similarly, $R\dash$bimodules, take $1 = {}_R R_R$. ::: :::{.proposition title="?"} If $\mca$ is additive and $(\mca, \tensor, 1, \alpha, \ell, r)$ is monoidal, then setting $[X] \cdot [Y] \definedas [X\tensor Y]$ defines a ring structure on $K_0^\oplus(\mca) = F(\mca) / N(\mca)$. ::: :::{.proof} \hfill - This is well-defined on $F(\mca)$. - Unital: Check $[X][1] = [X\tensor 1] = [X] = [1\tensor X] = [1][X]$ - Associativity: \[ ([X][Y])[Z] &= [X\tensor Y][Z] \\ &= [(X\tensor Y) \tensor Z] \\ &= [X\tensor (Y\tensor Z)] \\ &= [X][Y\tensor Z] = X([Y][Z]) .\] - Distributive: Check. Therefore $F(\mca)$ is a unital ring. - Check $N(\mca) \subseteq F(\mca)$ is a two-sided ideal (use the isomorphism from the earlier remark.) ::: :::{.example} The group morphism $\bar \phi: K_0^\oplus(\kmod) \mapsvia{\cong} \ZZ$ is in fact a ring morphism. - Check \[ \bar \phi([V][W]) &= \bar\phi([V\tensor_k W]) \\ &= \dim(V\tensor_k W) \\ &= \dim(V) \dim(W) \\ &= \bar\phi([V]) \bar\phi([W]) .\] - Check $\bar\phi([k]) = \dim k = 1$. For $\mca$ an additive category, for all $i\in \ZZ$ there exist additive functors \[ (i): \mca &\to \mca \\ X &\mapsto (i)(X) = X(i) .\] ::: :::{.remark} These satisfy $(j) \circ (i) = (i+j)$ and $(0) = \id_\mca$, so they will correspond to degree shifts. ::: :::{.proposition title="?"} Setting $v^i[X] \definedas [X(i)]$ defines a $\ZZ[v, v\inv]\dash$module structure on $K_0^\oplus(\mca)$. ::: :::{.proof} \hfill - Check that this is well-defined on $F(\mca)$; the module axioms will follow from the above remark. - Check that is descends to the quotient, i.e \[ v^i([X\oplus Y] -[X] - [Y]) &= v^i]X\oplus Y - v^i[X] - v^i [Y] \\ &= [(X\oplus Y)(i)] - [X(i)] - [Y(i)] \\ &= [X(i)\oplus Y(i)] - [X(i)] - [Y(i)] .\] ::: :::{.exercise title="?"} Show that $K_0^\oplus(k\dash\text{grmod}) \cong \ZZ[v, v\inv]$ where $[v] \mapsto \sum_{k\in \ZZ} \dim(V_n)v^n$ is an isomorphism of $\ZZ[v,v\inv]\dash$modules (and in fact an isomorphism of $\ZZ[v,v\inv]\dash$algebras). ::: :::{.remark} For $(\mca, \tensor, 1, \alpha, \ell, r)$ a monoidal category with additive functors $(i)$ as above, if \[ (i) \circ (X\tensor \wait) \cong (X\tensor \wait) \circ (i) \\ (i) \circ (\wait \tensor Y) \cong (\wait \tensor Y) \circ (i) \] using the fact that \[ (X\tensor Y)(i) \cong X \tensor (Y(i)) \cong (X(i)) \tensor Y .\] Thus $K_0^\oplus(\mca)$ is a $\ZZ[v, v\inv]\dash$algebra. ::: Recall that $H_n^R(q, q-1)$ taking $R = \ZZ[v, v\inv]$ with $q=v^{-2}$ and $q-1 = z$ was the Iwahari-Hecke algebra, generated by $\theset{T_i}_{i\leq n-1}$ and the braid/skein relations. Substitute $Hs_i = vT_i$ (Soergel's correction) to obtain a new presentation of $H_n^{\ZZ[v, v\inv]}(v^{-2}, v^{-2}-1)$. The generators are now $\ts{ H_{s_i} \st i\leq n-1}$ and \[ H_{s_i} H_{s_{i+1}} H_{s_i} &= H_{s_{i+1}} H_{s_i} H_{s_{i+1}} \\ H_{s_i} H_{s_j} &= H_{s_j} H_{s_i} && \abs{i-j} \geq 2 \\ H_{s_i}^2 &= v^2 T_i^2 \\ &= v^2 \qty{ (v^{-2} - 1)T_i + v^{-2}1} \\ &= (1-v^{-2}) T_i + 1 \\ &= (v\inv - v) H_{s_i} +1 .\] Notation: we'll abbreviate $\mch(S_n) = H_n^{\ZZ[v, v\inv]} (v^{-2}, v^{-2} - 1)$. There is a standard basis \[ H_w \definedas H_{s_{i_1}} \cdots H_{s_{i_r}} = v^{\ell(w)} T_w && w\in S_n,\,\, w = s_{i_1} \cdots s_{i_r}, \,\, \ell(w) = r .\] where $w$ is written as a minimal length reduced expression. ## Some technical tools (1) The Bruhat order. This is a partial order on the symmetric group $S_n$ where $w'\leq w$ iff there exists a word for $w'$ obtained by deleting some $s_i$ from the reduced expression for $w$. :::{.example} For $S_3$: ![](figures/image_2020-07-13-11-54-42.png) ::: (2) The Bar involution. There is a ring morphism \[ \mch(S_n) &\to \mch(S_n) \\ h &\mapsto \bar h .\] uniquely determined by $\bar{H_{s_i}} = H_{s_i\inv}$ (which incidentally equals $H_{s_i} + (v-v\inv)1$) and $\bar v = v\inv$. :::{.theorem title="KL-Soergel"} For all $w\in S^n$ there exists a unique $C_w \in \mch(S_n)$ such that 1. $\bar{C_w} = C_w$, self-duality 2. $C_w = H_w + \sum_{x < w} h_{x, w} H_x \in v\ZZ[v]$, upper triangularity. ::: :::{.definition title="?"} $\theset{C_w \suchthat w\in S_n}$ is the *KL-basis* of $\mch(S_n)$. ::: This is a basis because we can refine $\leq$ to a total order, then write a change-of-basis matrix from the standard basis to this: ![](figures/image_2020-07-13-11-59-25.png) The elements $h_{x, w} \in \ZZ[v, v\inv]$ are called the *KL-polynomials* where we set $h_{w, w} = 1$ and $h_{x, w} = 0$ when $x\not\leq w$. :::{.example} Note $C_e = H_e$ and \[ C_{s_1} &= H_{s_1} + v1 \\ C_{s_2} &= H_{s_2} + v1 .\] Thus (2) is satisfied, and (1) follows from \[ \bar{C_{s_i}} &= \bar{H_{s_i} + v_1} \\ &= \bar{H_{s_1}} + \bar{v} 1 \\ &= H_{s_i\inv} + v\inv 1 \\ &= H_{s_i} + (v-v\inv)1 + v\inv 1 \\ &= H_{s_i} + v1 .\] Can also check that \[ C_{s_1 s_2} &= C_{s_1 s_2} &&\text{automatically self-dual} \\ &= (H_{s_1} + v) (H_{s_2} + v) \\ &= H_{s_1} H_{s_2} + v H_{s_2} + vH_{s_1} + v^2 .\] Similarly expand $C_{s_2 s_1} = H_{s_2 s_1} + vH_{s_1} + vH_{s_2} + v^2$. Finally compute \[ C_{s_2} C_{s_1} C_{s_2} &= (H_{s_2 s_1} +v H_{s_1} + vH_{s_2} + v^2 )(H_{s_2} + v) \\ &= H_{s_2 s_1 s_2} + v H_{s_1 s_2} + v H_{s_2}^2 + v^2 H_{s_2} = v H_{s_2 s_1} + v^2 H_{s_1} + v^2 H_{s_2} + v^4 .\] Note that coefficients need to be contained in $v\ZZ[v]$ but we still need self-duality. :::