Soergel Intersection Forms and Hodge Theory — Examples

The character of the two-letter product. By [F1] and multiplicativity, $\mathrm{ch}(B_s{B}_t)=(H_s+v)(H_t+v)=H_s{H}_t+v{H}_s+v{H}_t+v^2$; since $st$ is reduced, [F2] gives $H_s{H}_t=H_{st}$, so the standard-basis expansion is $H_{st}+v{H}_s+v{H}_t+v^2$. Its coefficients give the multiplicities $1,1,1,1$ displayed, and reading the layers through [F1] the reduced degrees are $2,0,0,-2$, so the graded dimension is $v^2+2+v^{-2}$ and the ordinary real dimension is $4$.

The Kazhdan-Lusztig element of $st$. Put $A:=v-v^{-1}$ and $J:=H_{st}+v(H_s+H_t)+v^2$; then $J\in H_{st}+v\mathbb{Z}[v]H_s+v\mathbb{Z}[v]H_t+v^2\mathbb{Z}[v]$. Since $H_s^{-1}=H_s+A$, $H_t^{-1}=H_t+A$ and $\overline{H_{st}}=H_{ts}^{-1}=H_s^{-1}{H}_t^{-1}=(H_s+A)(H_t+A)=H_{st}+A(H_s+H_t)+A^2$ (the last identity uses $H_s{H}_t=H_{st}$), applying the semilinear involution gives $\bar{J}=\overline{H_{st}}+v^{-1}(\overline{H_s}+\overline{H_t})+v^{-2}=H_{st}+A(H_s+H_t)+A^2+v^{-1}(H_s+H_t)+2v^{-1}A+v^{-2}$, whose coefficients are $A+v^{-1}=v$ at $H_s+H_t$ and $A^2+2v^{-1}A+v^{-2}=(v^2-2+v^{-2})+(2-2v^{-2})+v^{-2}=v^2$ at $H_e$; hence $\bar{J}=J$, and by uniqueness [F2], $J={\underset{‾}{H}}_{st}$.

Indecomposability of $B_s{B}_t$. The unshifted tensor $M=S{\otimes }_{S^s}S{\otimes }_{S^t}S$ is generated as a bimodule by $1\otimes 1\otimes 1$. To see this, write a middle polynomial as $h_0+x_1{h}_1$ with $h_0,h_1\in S^s$, which slide to the first slot. The remaining $x_1$ slides to the last slot because $t=s_2$ fixes $x_1$. Therefore every tensor is a sum of $p(1\otimes 1\otimes 1)q$. After the shift $(2)$ its lowest component is one-dimensional in degree $-2$, by its finite left tensor basis and $S_0=\mathbb{R}$. A degree-zero bimodule endomorphism must send the generator to a scalar multiple of itself, and generation forces that same scalar on all of $M$. Hence its degree-zero endomorphism algebra is $\mathbb{R}$, with no nontrivial idempotent, so $B_s{B}_t$ is indecomposable. The balanced map $p\otimes q\otimes r\mapsto ps(q)st(r)$ is surjective onto the $st$ graph bimodule, showing that this indecomposable has the top graph of its reduced word; it is the normalized summand $B_{st}$. Thus $B_s{B}_t\cong B_{st}$ with no lower summands, independently of the Hom formula.

The three-letter braid relation. By multiplicativity and the products computed in steps 1.1 and 2.1, $\mathrm{ch}(B_s{B}_t{B}_s)={\underset{‾}{H}}_s{\underset{‾}{H}}_t{\underset{‾}{H}}_s={\underset{‾}{H}}_{st}{\underset{‾}{H}}_s$; expanding with the Hecke relations gives ${\underset{‾}{H}}_{st}{\underset{‾}{H}}_s={\underset{‾}{H}}_{sts}+{\underset{‾}{H}}_s$ Indeed the expansion is $J_0+H_s+v$ where $J_0=H_{sts}+v(H_{st}+H_{ts})+v^2(H_s+H_t)+v^3$. The element $J_0={\underset{‾}{H}}_{st}{\underset{‾}{H}}_s-{\underset{‾}{H}}_s$ is bar fixed, with triangular lower coefficients in $v\mathbb{Z}[v]$, so uniqueness in [F2] identifies it with ${\underset{‾}{H}}_{sts}$. The ordinary real dimensions of the reductions add to $6+2=8$. Exchanging $s$ and $t$ in [F3] gives $B_t{B}_s{B}_t\cong B_{sts}\oplus B_t$, whose character differs by ${\underset{‾}{H}}_t-{\underset{‾}{H}}_s=H_t-H_s\mathrm{\ne }0$, so these Bott-Samelson products are not isomorphic; the rank-two Kazhdan-Lusztig data $P_{e,st}=P_{s,st}=P_{t,st}=1$ are the coefficients displayed in step 2.1. This verifies the example.

The bimodule and its reduction. Symmetric polynomials in three variables are generated by $e_1,e_2,e_3$: for a homogeneous symmetric polynomial, its lexicographically leading monomial has exponents $a\ge b\ge c$; subtracting its coefficient times $e_1^{a-b}{e}_2^{b-c}{e}_3^c$ lowers that leading monomial. The finite set of monomials of the fixed total degree makes this reduction terminate. Thus $S^{S_3}=\mathbb{R}[e_1,e_2,e_3]$ and [F1] gives the reduced module $C(3)$ with $C=S\mathrm{/}(e_1,e_2,e_3)$. Eliminating $x_3=-x-y$ gives $C=\mathbb{R}[x,y]\mathrm{/}(x^3,y^2+xy+x^2)$: the $e_3$ relation is $x^{2}y+x{y}^2=0$, and subtracting this from $x(x^2+xy+y^2)=0$ gives $x^3=0$, with the same identities proving the reverse inclusion of ideals. Over $\mathbb{R}[x]\mathrm{/}(x^3)$, division by the monic quadratic in $y$ gives the free basis $1,y$; hence $1,x,y,x^2,xy,x^{2}y$ is a real basis and the dimensions are $1,2,2,1$ in degrees $0,2,4,6$, centered at $-3,-1,1,3$ after shifting by $(3)$. The tensor bimodule is generated as a bimodule by $1\otimes 1$ and its bottom component is one dimensional, so every degree-zero endomorphism is scalar; it is therefore indecomposable. The balanced surjection $f\otimes g\mapsto f{w}_0(g)$ onto $S_{w_0}$ proves that it has the $w_0$ graph in its support. In the rank-two decomposition it is the summand carrying the $w_0$ graph (the other summand $B_{s_1}$ has only the $e,s_1$ graphs), so its label is $w_0$.

The Kazhdan-Lusztig element. Put $A:=v-v^{-1}$ and $J:=H_{w_0}+v(H_{s_1{s}_2}+H_{s_2{s}_1})+v^2(H_{s_1}+H_{s_2})+v^3$. The element $J$ lies in $H_{w_0}+{\sum }_{y<w_0}v\mathbb{Z}[v]H_y$, since all coefficients belong to $v\mathbb{Z}[v]$. For bar-invariance we expand: from $H_s^{-1}=H_s+A$ and the reduced products $H_{s_1}{H}_{s_2}=H_{s_1{s}_2}$ and $H_{s_1{s}_2}{H}_{s_1}=H_{w_0}$ one computes $\overline{H_{s_1{s}_2}}=H_{s_1{s}_2}+A(H_{s_1}+H_{s_2})+A^2$ and $\overline{H_{w_0}}=H_{w_0}+A(H_{s_1{s}_2}+H_{s_2{s}_1})+A^2(H_{s_1}+H_{s_2})+A+A^3$, while $\overline{H_{s_1}}=H_{s_1}+A$ and $\overline{H_{s_2}}=H_{s_2}+A$. Applying the semilinear involution to $J$ gives $\bar{J}=\overline{H_{w_0}}+v^{-1}(\overline{H_{s_1{s}_2}}+\overline{H_{s_2{s}_1}})+v^{-2}(\overline{H_{s_1}}+\overline{H_{s_2}})+v^{-3}$; collecting coefficients in the standard basis, the coefficient of $H_{s_1{s}_2}+H_{s_2{s}_1}$ becomes $A+v^{-1}=v$, the coefficient of $H_{s_1}+H_{s_2}$ becomes $A^2+2v^{-1}A+v^{-2}=v^2$, and the coefficient of $H_e$ becomes $A+A^3+2v^{-1}{A}^2+2v^{-2}A+v^{-3}=v^3$. Hence $\bar{J}=J$, and by the uniqueness clause of [F2] $J={\underset{‾}{H}}_{w_0}$.

Character and multiplicities. By [F4], $\mathrm{ch}(B_{w_0})=(H_{s_1}+v)(H_{s_2}+v)(H_{s_1}+v)-(H_{s_1}+v)=J$, the last equality following by the quadratic relation $H_s^2=1+(v^{-1}-v)H_s$ and reduced multiplication. Step 1.2 gives $J={\underset{‾}{H}}_{w_0}$, so the $\mathrm{\Delta }$-multiplicities of $B_{w_0}$ are exactly the six coefficients of $J$, one for each of the six elements of $S_3$. Reading them through [F3], the layer ${\mathrm{\Delta }}_{w_0}(0)$ contributes in degree $\mathrm{\ell }(w_0)=3$, the two layers ${\mathrm{\Delta }}_{s_1{s}_2}(1),{\mathrm{\Delta }}_{s_2{s}_1}(1)$ contribute in degree $1$ with multiplicity two, the two layers ${\mathrm{\Delta }}_{s_1}(2),{\mathrm{\Delta }}_{s_2}(2)$ contribute in degree $-1$ with multiplicity two and the layer ${\mathrm{\Delta }}_e(3)$ contributes in degree $-3$; this is the graded dimension $1,2,2,1$ of the centered coinvariant algebra computed in step 1.1, and the value at $v=1$ is $6=\dim C$. The coefficients $P_{e,w_0}=P_{s_i,w_0}=P_{s_1{s}_2,w_0}=P_{s_2{s}_1,w_0}=1$ are the classical normalization of these coefficients. This verifies the example.

Non-dominance. The reflection $s$ exchanges $x_1$ and $x_2$, so ${\rho }^{\mathrm{\prime }}=x_1+x_2$ is $s$-invariant and ${\rho }^{\mathrm{\prime }}({\alpha }^{\vee })=(x_1+x_2,x_1-x_2)=1-1=0$ in the standard form of the coordinate normalization; in particular ${\rho }^{\mathrm{\prime }}$ is not strictly dominant.

The left action of ${\rho }^{\mathrm{\prime }}$ equals its right action. Since ${\rho }^{\mathrm{\prime }}\in S^s$, the balanced tensor relation gives ${\rho }^{\mathrm{\prime }}\otimes 1=1\otimes {\rho }^{\mathrm{\prime }}$ in $B_s$; hence for every $b\in B_s$ the left multiplication ${\rho }^{\mathrm{\prime }}\cdot b$ equals the right multiplication $b\cdot {\rho }^{\mathrm{\prime }}$: indeed both actions are $S$-linear and the identity holds on the two basis elements $c_{\mathrm{i}\mathrm{d}}$ (where ${\rho }^{\mathrm{\prime }}\cdot c_{\mathrm{i}\mathrm{d}}={\rho }^{\mathrm{\prime }}\otimes 1=1\otimes {\rho }^{\mathrm{\prime }}=c_{\mathrm{i}\mathrm{d}}\cdot {\rho }^{\mathrm{\prime }}$) and $c_s$ (where both equal $c_s{\rho }^{\mathrm{\prime }}$ by [F1]).

The operator vanishes on the reduced module. By step 2.1, ${\rho }^{\mathrm{\prime }}\cdot c_{\mathrm{i}\mathrm{d}}=c_{\mathrm{i}\mathrm{d}}\cdot {\rho }^{\mathrm{\prime }}=1\otimes {\rho }^{\mathrm{\prime }}$ lies in $B_s{S}_{+}$, since it is the right product of $c_{\mathrm{i}\mathrm{d}}=1\otimes 1$ with ${\rho }^{\mathrm{\prime }}\in S_{+}$; by [F2] its class in $\overline{B_s}$ is zero, so ${\rho }^{\mathrm{\prime }}\cdot \overline{c_{\mathrm{i}\mathrm{d}}}=0$. Similarly ${\rho }^{\mathrm{\prime }}\cdot c_s=c_s\cdot {\rho }^{\mathrm{\prime }}=c_s{\rho }^{\mathrm{\prime }}$ lies in $B_s{S}_{+}$ and reduces to zero. Since $\overline{c_{\mathrm{i}\mathrm{d}}}$ and $\overline{c_s}$ span $\overline{B_s}$, the induced operator $L={\rho }^{\mathrm{\prime }}\cdot (-)$ is the zero endomorphism of $\overline{B_s}$.

Failure of hard Lefschetz. The reduced module has $\dim (\overline{B_s}{)}^{-1}=\dim (\overline{B_s}{)}^1=1$ with bases $\overline{c_{\mathrm{i}\mathrm{d}}}$ and $\overline{c_s}$. The degree-one Lefschetz map $L:(\overline{B_s}{)}^{-1}\to (\overline{B_s}{)}^1$ is zero by step 3.1, hence not an isomorphism, and hard Lefschetz fails for $L$ in the sense of [F3]. The bottom Lefschetz form $⟨-,L(-)⟩$ is likewise identically zero, so the Hodge-Riemann relations with any nonzero sign also fail.

The general degree-two case. Let ${\rho }^{\mathrm{\prime }\mathrm{\prime }}=a{x}_1+b{x}_2$ satisfy ${\rho }^{\mathrm{\prime }\mathrm{\prime }}({\alpha }^{\vee })=a-b=0$; then $a=b$ and ${\rho }^{\mathrm{\prime }\mathrm{\prime }}=a{\rho }^{\mathrm{\prime }}$ is a scalar multiple of ${\rho }^{\mathrm{\prime }}$. If $a=0$ the operator is trivially zero; if $a\mathrm{\ne }0$ the same computation as in steps 2.1-4.1 applies because invariance and positivity of degree are unaffected by the scalar. Hence every degree-two element with vanishing evaluation on ${\alpha }_s^{\vee }$ yields the zero operator on the reduced rank-one module and fails hard Lefschetz. This proves the claimed failure and shows that allowing arbitrary degree-two elements invalidates the conclusion; no choice principle is used.

Products and the trace. The multiplication rules are those of [F1]. By definition the trace extracts the coefficient of $c_s$ in the right basis, so $\mathrm{Tr}(c_{\mathrm{i}\mathrm{d}}a+c_{s}b)=b$ for $a,b\in S$.

The Gram matrix and non-degeneracy. Using $⟨b,b^{\mathrm{\prime }}⟩=\mathrm{Tr}(b{b}^{\mathrm{\prime }})$ and the products of step 1.1, $⟨c_{\mathrm{i}\mathrm{d}},c_{\mathrm{i}\mathrm{d}}⟩=\mathrm{Tr}(c_{\mathrm{i}\mathrm{d}})=0$, $⟨c_{\mathrm{i}\mathrm{d}},c_s⟩=\mathrm{Tr}(c_s)=1$ and $⟨c_s,c_s⟩=\mathrm{Tr}(c_s{\alpha }_s)={\alpha }_s$, so the Gram matrix in the basis $(c_{\mathrm{i}\mathrm{d}},c_s)$ is $\left(\begin{array}{cc}0& 1\\ 1& {\alpha }_s\end{array}\right)$; its determinant is $-1$, hence the form is non-degenerate, and by [F2] so is the induced form on the reduced module.

The reduced action of $\rho$. Let $\rho ={\alpha }_s\mathrm{/}2$. Since $\rho$ is $s$-anti-invariant, $s(\rho )=-\rho$ and ${\mathrm{\partial }}_s(\rho )=1$, so by [F1] we get $\rho \cdot c_{\mathrm{i}\mathrm{d}}=c_{\mathrm{i}\mathrm{d}}\cdot (-\rho )+c_s=c_s-c_{\mathrm{i}\mathrm{d}}\rho$; reducing kills the right multiple $c_{\mathrm{i}\mathrm{d}}\rho$ by [F2], so $\rho \cdot \overline{c_{\mathrm{i}\mathrm{d}}}=\overline{c_s}$. Similarly $\rho \cdot c_s=c_s\rho$ by [F1] and $c_s\rho$ is a right multiple by a positive-degree element, so $\rho \cdot \overline{c_s}=0$. Also $\rho ({\alpha }_s^{\vee })=1>0$, so $\rho$ is strictly dominant.

Hard Lefschetz, primitivity and the sign. The reduced module has real basis $\overline{c_{\mathrm{i}\mathrm{d}}}$ in degree $-1$ and $\overline{c_s}$ in degree $+1$, hence both graded pieces are one-dimensional; by step 2.2 the degree-one Lefschetz map $\rho :(\overline{B_s}{)}^{-1}\to (\overline{B_s}{)}^1$ sends a basis vector to a basis vector and is an isomorphism, which is hard Lefschetz by [F3]. The primitive subspace is $P^{-1}=\ker ({\rho }^2{\mathrm{\mid }}_{(\overline{B_s}{)}^{-1}})=(\overline{B_s}{)}^{-1}$, since ${\rho }^2$ maps the one-dimensional space to the higher space in two steps and ${\rho }^2$ lands in degree $+3$, which is zero; the Lefschetz form on it is $⟨\overline{c_{\mathrm{i}\mathrm{d}}},\rho \overline{c_{\mathrm{i}\mathrm{d}}}⟩=⟨\overline{c_{\mathrm{i}\mathrm{d}}},\overline{c_s}⟩=1>0$, the reduction of the Gram entry $1$, and the standard sign in degree $-1$ is $(-1{)}^{(1-1)\mathrm{/}2}=+1$, so HR holds with the standard sign and the bottom form is positive.

The character. Multiplication $B_s\to S(1)$ is surjective with kernel generated on the right by $k=c_s-c_{\mathrm{i}\mathrm{d}}{\alpha }_s$, of degree $1$. The left-action rule gives $rk=ks(r)$, so the kernel is $S_s(-1)={\mathrm{\Delta }}_s(0)$ and the quotient is $S(1)={\mathrm{\Delta }}_e(1)$. These two layers give $\mathrm{ch}(B_s)=H_s+v={\underset{‾}{H}}_s$. The top graph layer has generator degree $+1$ and the unit quotient has generator degree $-1$. The rational model has the same formulas after real scalar extension.