LetF be a complex of singular Soergel bimodules. Following the notation of [EW14] we indicate the homological degree on the left, that is:
F := [. . .→i−1F →iF →i+1F →. . .].
We denote by{−}the homological shift, so thati(F{1}) =i+1F.
LetKb(SBimI)be the bounded homotopy category of complexes ofI-singular Soergel bimodules.
We definepK≥0 :=pKb(SBimI)≥0to be the full subcategory ofKb(SBimI)with objects complexes inKb(SBimI)which are isomorphic to a complexF which satisfiesτ≤−i−1iF = 0 for alli∈Z.
Similarly, we define pK≤0 := pKb(SBimI)≤0 to be the full subcategory with objects complexes inKb(SBimI)which are isomorphic to a complexF which satisfiesiF =τ≤−iiF for alli∈Z. LetpK0 =pK≥0∩pK≤0.
Fors∈S let Fs denote the complex
Fs= [0→B0s−→ds R[1]→0]
where ds(f ⊗g) = f g.2 Then tensoring with Fs on the left induces an equivalence on the categoryKb(SBimI). The inverse is given by tensoring on the left with the complex Es= [0→R[−1] d
0s
−→B0s→0]. Hered0s(1) =cs= 12(αs⊗1 + 1⊗αs).
For anyx∈WIwe consider the complexFs1. . . Fsk for any reduced expressions1. . . sk of x. As an object in Kb(SBim) it does not depend on the chosen reduced expression [Rou06, Proposition 9.2]. Hence, also (Fs1. . . Fsk)I does not depend on the reduced ex-pression as an object inKb(SBimI).
We choose FxI
⊕
⊆(Fs1. . . Fsk)I to be the corresponding minimal complex (see [EW14,
§6.1]), so FxI ∼= (Fs1. . . Fsk)I in Kb(SBimI) and the complex FxI does not contain any contractible direct summand. We callFxI asingular Rouquier complex.
Observe that if Fx is the Rouquier complex for x ∈ Kb(SBim), i.e. is the minimal complex forFs1. . . Fsk, then FxI can also be obtained as the minimal complex of Fx,I :=
(Fx)I inKb(SBimI).
2We use−0 to indicate where the object in homological degree0is placed.
Lemma 4.4.1. Let x∈WI and s∈S.
i) If sxwI> xwI then FsBxI ∈pK≥0. ii) If sxwI< xwI then FsBxI ∼=BIx[−1].
Proof. i) From Theorem 4.1.5 we have ch(BsBxI) = HsHIx =HIsx+P
z∈W I
z<xs
mzHIz with mz ∈Z≥0. Hence
BsBIx∼=BxsI ⊕ M
z∈W I
z<xs
(BIz)⊕mz.
Then the complex
FsBxI= [0→
0
BsBxI →BxI[1]→0]
is manifestly inpK≥0.
ii) We have ch(BsBxI) =HsHIx =HsHxwI = (v+v−1)HIx. Therefore BsBxI ∼=BxI[1]⊕ BxI[−1] and
FsBxI = [0→
0
BxI[1]⊕BxI[−1]→BxI[1]→0].
Tensoring withFs induces an equivalence on the categoryKb(SBimI), and sinceBxIis indecomposable also the complexFsBxI must be indecomposable. Therefore the map BxI[1] → BxI[1] cannot be 0, otherwise
0
BIx[1] would be a non-trivial direct summand ofFsBxI. Since BxI[1]→BxI[1] is non zero, it is an isomorphism and BxI[1]→BIx[1]is a contractible direct summand that we can remove from the complex. In this way we obtainFsBxI∼=BxI[−1]∈pK≥0.
Lemma 4.4.2. Let F ∈pK≥0. ThenFsF ∈pK≥0.
Proof. We denote by ω≥k the truncation of complexes, that is ω≥kF = [0→kF →k+1F →. . .].
We have distinguished triangles
ω≥k+1F →ω≥kF →kF{−k}−→[1]
Fs(ω≥k+1F)→Fs(ω≥kF)→Fs(kF{−k})−→[1]
SinceFs(kF{−k})∈pK≥0 by Lemma 4.4.1, the statement follows by induction onkusing the analogue of [EW14, Lemma 6.1].
Corollary 4.4.3. For anyx∈WI we have FxI∈pK≥0.
Proof. Since FxI ∼= Fs1Fs2. . . Fsk(RI) ∈ Kb(SBimI) this follows by induction on `(x) directly from Lemma 4.4.1 and Lemma 4.4.2.
4.4.1 Singular Rouquier complexes are ∆-split
We can identifyR⊗RRI with the ring of regular functions onh×(h/WI). The inclusion R⊗RRI ,→R⊗RR corresponds to the projection map π:h×h→h×(h/WI).
For a cosetp∈W/WI letGrI(p)be the image Gr(p) underπ, whereGr(p)⊆h×his the twisted graph defined in §3.1. Similarly, ifC ⊆W/WI let GrI(C) =S
p∈CGrI(p).
Let BI ∈ SBimI. For C ⊆ W/WI let ΓCB = {b ∈ B | suppb ∈ GrI(C)}. The functorΓC extends to a functorΓC from Kb(SBimI) to the homotopy category of graded (R, RI)-bimodules, which we denote by Kb(R-Mod-RI).
Let q : W → W/WI denote the projection. For y ∈ W/WI let us denote by y− the minimal element in the coset y. The bijection WI ∼= W/WI induces a Bruhat order on W/WI, i.e. we say y≤z if and only if y−≤z−. The projectionq is a strict morphism of posets:
Lemma 4.4.4. Let w≥v in W. Then q(w)≥q(v).
Proof. This follows from [Dou90, Lemma 2.2].
For any B ∈SBim and anyC ⊆W/WI we have by [Wil11, Prop 6.1.6]
(Γq−1(C)B)I = ΓC(BI). (4.3) Note that q−1(≥ y) = {x ∈ W | x ≥ y−}. If x ∈ WI we write ΓI≥x for the functor Γ{y∈WI|y≥x} on SBimI, to differentiate it from the functor Γ≥x on SBim.
We choose an enumerationy1, y2, y3, . . .of W/WI refining the Bruhat order onW/WI. For any cosetyi ∈W/WI we choose an enumeration yi,1, yi,2, yi,3. . . of the elements in yi refining the Bruhat order. Let
z1 =y1,1, z2 =y1,2, . . . , z|WI|=y1,|WI|, z|WI|+1 =y2,1, z|WI|+2=y2,2. . . .
By virtue of Lemma 4.4.4, z1, z2, z3. . . is also an enumeration of W which refines the Bruhat order.
We denote byΓI≥m the functor Γ{yi:i≥m} onSBimI and by Γ≥m the functor Γ{zi:i≥m}
onSBim. For l≥k, let
ΓI≥k/≥lB := (ΓI≥kB)/(ΓI≥lB).
The functorΓI≥k/≥lextends to a functorΓI≥k/≥l:Kb(SBimI)→ Kb(R-Mod-RI). Similarly, we define the functorsΓ≥k/≥l,ΓI≥x/≥y,Γ≥x/≥y. They also extend to functors between the respective homotopy categories.
Fix y = ym ∈ W/WI and x ∈WI. We have (ym)− = zk for some k and (ym+1)− = zk+|WI|. Then by the hin-und-her Lemma for singular Soergel bimodules [Wil11, Lemma 6.3.2] we have
ΓI≥y/>y(FxI)∼= ΓI≥y/>y(Fx,I)∼= ΓI≥m/≥m+1(Fx,I)∼= (Γ≥k/≥k+|WI|Fx)I ∈ Kb(R-Mod-RI) Assume x 6∈ ym. Then x = zj with j < k or j ≥ k+|WI|. For any i such that 1≤i≤ |WI|we have a distinguished triangle in Kb(R-Mod-R)
Γ≥k/≥k+i−1Fx →Γ≥k/≥k+iFx →Γ≥k+i−1/≥k+iFx
−→[1]
and the last term is0by [LW14, Prop 3.7]. It follows by induction thatΓ≥k/≥k+|WI|Fx∼= 0, hence
ΓI≥y/>y(FxI)∼= ΓI≥y/>y(Fx,I)∼= (Γ≥k/≥k+|WI|Fx)I ∼= 0∈ Kb(R-Mod-RI).
Assume nowx=y−, so thatx=zk. Let Rx,I := (Rx)I. Since Γ≥k/≥k+1Fx=Rx[−`(x)]
the same argument as above showsΓ≥k/≥k+|WI|Fx∼=Rx[−`(x)], hence ΓI≥x/>x(FxI)∼= ΓI≥x/>x(Fx,I)∼= (Γ≥k/≥k+|WI|Fx)I ∼=Rx,I[−`(x)].
We obtain the singular version of [LW14, Prop 3.7]:
Lemma 4.4.5. Let x, y∈WI. Then ΓI≥y,>y(FxI) =
(0 if y6=x, Rx,I[−`(x)] if y=x.
Remark 4.4.6. If we view Fs as a complex of graded left R-modules it splits, i.e. we have Fs ∼= R[−1] in Kb(R-mod). Similarly (Fs1Fs2. . . Fsk)I ∼= R[−k]in Kb(R-mod). For x∈ WI, letFxI =R⊗RFxI. It is a complex of graded real vector spaces. It follows that we have:
Hi(FxI) = (
R[−`(x)] if i= 0, 0 if i6= 0.
4.4.2 Singular Rouquier complexes are linear
For us it is important to understand how singular Rouquier complexes look. The idea is to use the first differential in a singular Rouquier complex as a replacement for “weak Lefschetz” in the inductive proof of hard Lefschetz. More precisely, the first differential will have the role of the mapφin [EW14, Lemma 2.3]. For this, we first have to show that the first differential is a map of degree1between perverse singular Soergel bimodules.
Lemma 4.4.7. Let x ∈WI. For i >0 if iFxI contains a direct summand isomorphic to BzI[j], theni−1FxI contains a direct summand isomorphic toBzI0[j0]with z0 > z andj0< j.
Proof. The proof is the same of [EW14, Lemma 6.11]. From Theorem 4.1.5 (and the definition of the map ch, cf. [Wil11, §6.3]) we have that for anyy, z ∈WI the bimodule ΓI≥z/>z(ByI) is generated in degree< `(z) if y > z andΓI≥z/>z(BzI)∼=Rz,I[−`(z)].
The image ofBIz[j]ini+1FxI is contained inτ<−j(i+1FxI) because of (4.1): in fact any non-zero homomorphism in degree0is an isomorphism and thus yields a contractible direct summand.
Applying ΓI≥z/>z to FxI the direct summand BzI[j] induces a summand Rz,I[j−`(z)].
This cannot be a direct summand inΓI≥z/>z(τ<−ji+1FxI) and cannot survive in the coho-mology of the complex because of Lemma 4.4.5. ThusRz,I[j−`(z)]must be the image of a direct summandRz,I[j−`(z)]inΓ≥z/>z(τ>−j(i−1Fx)).
This implies that there is a direct summandByI[k]ini−1Fx withy > z andk < j.
Theorem 4.4.8. Let x∈WI and FxI be a singular Rouquier complex. Then:
i) 0FxI=BxI.
ii) For i≥1, iFxI =L
(BzI[i])⊕mz,i with z < x, z∈WI andmz,i∈Z≥0. In particular,FxI ∈pK0.
Proof. We apply the previous lemma. The same proof shows that since −1(FxI) = 0, the only direct summands occurring in0(FxI) isBxI. By induction we have iFxI =τ≤−iFxI for anyi >0. Now ii) follows since we already know FxI ∈pK≥0 from Corollary 4.4.3.
Remark 4.4.9. We can define the character of a complexF ∈ Kb(SBimI) as ch(F) =X
i∈Z
(−1)ich(iF).
Ifx∈WI andx=s1s2. . . sk is a reduced expression we have ch(FxI) = ch((Fs1Fs2. . . Fsk)I) =HxHw
I =:HIx.
An immediate consequence of Theorem 4.1.5 is that there is a non trivial morphism of degreei between BxI and ByI for x, y ∈WI if and only ifi and `(x)−`(y) have the same parity. Because of Theorem 4.4.8 we can write
HIx =X
i≥0
(−1)ich(iFx) =X
y≤x
gy,xHIy
with gx,x(v) = 1 and gy,x(v) = P
i>0my,i(−v)i. The polynomials gy,x are the parabolic inverse Kazhdan-Lusztig polynomials. We obtain that for any y ≤ x the polynomial (−1)`(y)−`(x)gy,x has non-negative coefficients.
4.4.3 Singular Rouquier complexes are Hodge-Riemann
The complex FxI is a direct summand of (Fs1. . . Fsm)I for a reduced expression x = s1. . . sm. Hence, for anyj,jFxI is a direct summand of j(Fs1. . . Fsm)I, that is
jFxI ⊆⊕ M
x0∈π(x,j)
BS(x0)I[j]
whereπ(x, j)is the set of all subexpressions of xobtained by omittingjsimple reflections.
Fixλ= (λx0)x0∈π(x,j) a tuple of strictly positive real numbers. We define a symmetric formh−,−iλ on L
BS(x0)I by hb, b0iλ = X
x0∈π(x,j)
λx0hbx0, b0x0iBS(x0) for allb= (bx0), b0 = (b0x0)∈ M
x0∈π(x,j)
BS(x0)I (4.4) whereh−,−iBS(x0) is the intersection form on BS(x0)I =BS(x0) defined in §3.1.1.
We say that FxI satisfies the Hodge-Riemann bilinear relations if we can choose an embedding FxI
⊕
⊆ (Fs1. . . Fsm)I such that for all tuples λ as above, multiplication by ρ on the right onjFx[−j]satisfies the Hodge-Riemann bilinear relations with respect to the formh−,−iλ.
Proposition 4.4.10. Assume HR(s, y) for all s∈ S and y ∈ WI with y < x such that sywI> ywI. Then FxI satisfies the Hodge-Riemann bilinear relations.
Proof. Fix a reduced expression x =s1. . . sm and let s=s1, y=s2. . . sm. By induction assumeFyI satisfies Hodge-Riemann so that we can find an appropriate embeddingFyI ⊆⊕ (Fs2. . . Fsm)I.
NowFxI is a direct summand of FsFyI, so we have an embedding
jFxI[−j]⊆⊕ j(BsFyI)[−j]⊕j−1FyI[−j+ 1]⊆⊕ M
y0∈π(y,j)
BsBS(y0)I⊕ M
y00∈π(y,j−1)
BS(y00)I =
= M
x0∈π(x,j)
BS(x0).
We fix a tuple(µx0)x0∈π(x,j), or equivalently two tuples(λy0)y0∈π(y,j) and (σy00)y00∈π(y,j) of positive real numbers.
From Theorem 4.4.8, we know that jFyI[−j] = L
(BIz)⊕mz with mz ∈ Z≥0. Let
jFyI[−j] =B↑⊕B↓ with B↑ = M
szwI>zwI
(BzI)⊕mz and B↓ = M
szwI<zwI
(BzI)⊕mz.
The decomposition is orthogonal with respect of the restriction of the form h−,−iλ since Hom0(B↑,DB↓) = 0. Then also BsB↑ and BsB↓ are orthogonal with respect of the restriction of the form h−,−iµ on Bs(jFyI[−j]). In fact, for any b ∈ B↑ and b0 ∈ B↓ we have:
hcidb, cidb0iµ=∂s(hb, b0iλ) = 0 (4.5) hcsb, cidb0iµ=hcidb, csb0iµ=hb, b0iλ = 0 (4.6) hcsb, csb0iµ=αshb, b0iλ= 0. (4.7) The bimoduleBsB↑ is perverse whileBsB↓=B↓[−1]⊕B↓[1]. So we have a decompo-sition
jFxI[−j]⊆⊕ BsB↓⊕BsB↑⊕j−1FyI[−j+ 1]. (4.8) The inclusion is, by definition, an isometry. This decomposition is orthogonal with respect to the form h−,−iµ. Moreover by (4.1) the image of jFxI[−j] → BsB↓ is contained in B↓[1].
Claim 4.4.11. The restriction of the Lefschetz form (−,−)−kρ = h−,− ·ρkiµ to (B↓[1])−k is zero.
Proof of the claim. Let BIz ⊆⊕ B↓ and let z = t1. . . tl be a reduced expression. Then z0 := t2. . . tl ∈ WI and t1z0wI > z0wI. Since BzI ⊆⊕ Bt1BzI0, the hypothesis HR(t1, z0) implies that multiplication byρ satisfies hard Lefschetz onBzI, hence onB↓, i.e.
ρi : (B↓)−i −∼→(B↓)i for all i≥0.
By shifting we getρi :B↓[1]−i−1 −→∼ B↓[1]i−1. Let Pρ−1−i = Kerρi+1 ⊆B↓[1]−i−1 so that for anym≤0 we have
B↓[1]m= M
j≥max{m+12 ,0}
Pρm−2j·ρj.
Letx∈Pρm−2j,y∈Pρm−2k for some j≥k≥0, then (xρj, yρk)mρ =hx, yρj+k−miµ= 0 becausej+k−m≥2k−m andy∈Ker(ρ2k−m).
So the image of jFxI[−j] in BsB↓ does not contribute to the Lefschetz form. We can consider the projection onto the other two factors
i:jFxI[−j]→BsB↑⊕j−1FyI[−j+ 1]
which is an isometry.
Furthermore, the mapiis injective, in factjFxI
⊕
⊆j(FsFyI)is a split inclusion and since when we project toSBimI/rad (SBimI), the image is contained inBsB↑⊕j−1FyI[−j+ 1], then alsoimust be a split injective morphism.
Using the fact thatjFxIis stable under the Lefschetz operator and that Hodge-Riemann holds by induction for bothBsB↑ and j−1FyI[−j+ 1], the thesis follows.