Soergel Intersection Forms and Hodge Theory

Well-definedness and bilinearity (1), (2). Since $\{H_w\}$ is an $A$-basis by [F1], prescribing $\epsilon (H_w)={\delta }_{e,w}$ and extending $A$-linearly is well defined and unique; the reversal map $a$ is a well-defined $A$-linear anti-automorphism by [F2], and multiplication is $A$-bilinear, so $(h,h^{\mathrm{\prime }})=\epsilon (a(h)h^{\mathrm{\prime }})$ is a well-defined $A$-bilinear form, for $(ph,q{h}^{\mathrm{\prime }})=\epsilon (pa(h)q{h}^{\mathrm{\prime }})=pq\epsilon (a(h)h^{\mathrm{\prime }})$ with $p,q\in A$. This proves (1) and (2).

The key identity: base case and recursion. We prove by induction on $\mathrm{\ell }(u)$ that $\epsilon (H_u{H}_y)={\delta }_{u,y^{-1}}$ for all $u,y\in S_n$; for $u=e$ this reads $H_e{H}_y=H_y$ and $\epsilon (H_y)={\delta }_{e,y}={\delta }_{u,y^{-1}}$, and for $u\mathrm{\ne }e$ we choose a simple reflection $s$ with $\mathrm{\ell }(us)=\mathrm{\ell }(u)-1$ and note that [F1] applied to $w=us$ gives $H_{us}{H}_s=H_u$, since $(us)s=u$ has length $\mathrm{\ell }(u)=\mathrm{\ell }(us)+1$.

The left multiplication rule. Applying the anti-automorphism $a$ of [F2] to the two cases of [F1] and using $a(H_w{H}_s)=a(H_s)a(H_w)=H_s{H}_{w^{-1}}$ and $(ws{)}^{-1}=s{w}^{-1}$ gives $H_s{H}_w=H_{sw}$ if $\mathrm{\ell }(sw)=\mathrm{\ell }(w)+1$, and $H_s{H}_w=H_{sw}+(v^{-1}-v)H_w$ if $\mathrm{\ell }(sw)<\mathrm{\ell }(w)$; this is the rule used in the two cases below.

The ascent case. Keep $u\mathrm{\ne }e$ and its right descent $s$ from step 2.1, and let $y\in S_n$ satisfy $\mathrm{\ell }(sy)>\mathrm{\ell }(y)$. Then $H_s{H}_y=H_{sy}$ by step 3.1, so $\epsilon (H_u{H}_y)=\epsilon (H_{us}{H}_s{H}_y)=\epsilon (H_{us}{H}_{sy})$; the induction hypothesis applies to the pair $(us,sy)$, whose first length is $\mathrm{\ell }(u)-1$, giving $\epsilon (H_{us}{H}_{sy})={\delta }_{us,(sy{)}^{-1}}={\delta }_{us,y^{-1}s}$, and $us=y^{-1}s$ holds if and only if $u=y^{-1}$ (multiply on the right by $s$ and use $s^2=e$), so $\epsilon (H_u{H}_y)={\delta }_{u,y^{-1}}$.

The descent case. Keep $u,s$ as above and let $\mathrm{\ell }(sy)<\mathrm{\ell }(y)$. Then $H_s{H}_y=H_{sy}+(v^{-1}-v)H_y$ by step 3.1, so $\epsilon (H_u{H}_y)=\epsilon (H_{us}{H}_{sy})+(v^{-1}-v)\epsilon (H_{us}{H}_y)$; the induction hypothesis gives $\epsilon (H_{us}{H}_{sy})={\delta }_{us,y^{-1}s}$ and $\epsilon (H_{us}{H}_y)={\delta }_{us,y^{-1}}$. If $us=y^{-1}$, then $y=(us{)}^{-1}=s{u}^{-1}$ and, since $us$ is a right descent of $u$, taking inverses gives $\mathrm{\ell }(y)=\mathrm{\ell }(s{u}^{-1})=\mathrm{\ell }(us)=\mathrm{\ell }(u)-1$, while $\mathrm{\ell }(sy)=\mathrm{\ell }(u^{-1})=\mathrm{\ell }(u)>\mathrm{\ell }(y)$, contradicting $\mathrm{\ell }(sy)<\mathrm{\ell }(y)$; hence ${\delta }_{us,y^{-1}}=0$ and $\epsilon (H_u{H}_y)={\delta }_{us,y^{-1}s}={\delta }_{u,y^{-1}}$.

The induction concludes. For every $u\mathrm{\ne }e$ the cases $\mathrm{\ell }(sy)>\mathrm{\ell }(y)$ and $\mathrm{\ell }(sy)<\mathrm{\ell }(y)$ exhaust all $y$ for the chosen right descent $s$, and each case reduced the first argument's length; with the base case this proves $\epsilon (H_u{H}_y)={\delta }_{u,y^{-1}}$ for all $u,y\in S_n$.

Standard orthonormality (4). For all $x,y$, using $a(H_x)=H_{x^{-1}}$ from [F2] and step 6.1 with $u=x^{-1}$, we get $(H_x,H_y)=\epsilon (a(H_x)H_y)=\epsilon (H_{x^{-1}}{H}_y)={\delta }_{x^{-1},y^{-1}}={\delta }_{x,y}$, the last equality being injectivity of $w\mapsto w^{-1}$; this proves (4).

The generator trace identity. By step 6.1, for every $w$ and simple reflection $s$ one has $\epsilon (H_w{H}_s)={\delta }_{w,s^{-1}}={\delta }_{w,s}$ and $\epsilon (H_s{H}_w)={\delta }_{s,w^{-1}}={\delta }_{w,s}$. Extending over the standard basis gives $\epsilon (z{H}_s)=\epsilon (H_{s}z)$ for every $z\in H$. In particular the common value for $w=s$ is $1$.

$\epsilon$ is a trace and yields the adjunctions (3). Every element of $H$ is a finite $A$-linear combination of words in the generators, and moving each letter of a word for the second factor to the front one at a time with step 7.2 and bilinearity shows $\epsilon (xy)=\epsilon (yx)$ for all $x,y\in H$; consequently, for simple $s$ and $h,h^{\mathrm{\prime }}\in H$, using $a(H_s)=H_s$ from [F2], $(h{H}_s,h^{\mathrm{\prime }})=\epsilon (a(h{H}_s)h^{\mathrm{\prime }})=\epsilon (H_{s}a(h)h^{\mathrm{\prime }})=\epsilon (a(h)h^{\mathrm{\prime }}{H}_s)=(h,h^{\mathrm{\prime }}{H}_s)$ and $(H_{s}h,h^{\mathrm{\prime }})=\epsilon (a(h)a(H_s)h^{\mathrm{\prime }})=\epsilon (a(h)H_s{h}^{\mathrm{\prime }})=(h,H_s{h}^{\mathrm{\prime }})$, which proves (3) and shows both left and right multiplication by $H_s$ are self-adjoint for $(-,-)$.

Values on the Kazhdan-Lusztig basis (5). Write ${\underset{‾}{H}}_x={\sum }_{u\le x}{h}_{u,x}{H}_u$ with $h_{u,x}=p_{u,x}\in \mathbb{Z}[v]$, $h_{x,x}=1$ and $h_{u,x}\in v\mathbb{Z}[v]$ for $u<x$, all other coefficients zero, as in [F3]; then bilinearity (2) and orthonormality from step 7.1 give $({\underset{‾}{H}}_x,{\underset{‾}{H}}_y)={\sum }_u{h}_{u,x}{h}_{u,y}$. A nonzero term requires $u\le x$ and $u\le y$; if $x\mathrm{\ne }y$, every surviving term has a factor in $v\mathbb{Z}[v]$, because $u<x$ or $u<y$ is [F3], and $u=x\mathrm{\ne }y$ gives $h_{x,y}\in v\mathbb{Z}[v]$ when $x<y$ and $h_{x,y}=0$ otherwise, symmetrically for $u=y$; hence $({\underset{‾}{H}}_x,{\underset{‾}{H}}_y)\in v\mathbb{Z}[v]$ for $x\mathrm{\ne }y$. For $x=y$ the term $u=x$ contributes $h_{x,x}^2=1$ and every other term has $u<x$ and contributes to $v^2\mathbb{Z}[v]$, so $({\underset{‾}{H}}_x,{\underset{‾}{H}}_x)\in 1+v\mathbb{Z}[v]$.

The standard pairing (6) and conclusion. By (4) the form $(-,-)$ is the $A$-bilinear form whose matrix in the normalized standard basis $\{H_x\}$ is the identity; the published type-A Soergel Hom formula uses the standard pairing with $⟨H_x,H_y⟩={\delta }_{xy}$ on that normalized basis, and the dictionary of [F4] fixes the normalization $H_x=v^{\mathrm{\ell }(x)}{T}_x$ and the resulting Kazhdan-Lusztig coefficients, so the transported pairing agrees with $(-,-)$ on the basis and hence on $H$ by bilinearity. All six clauses are proved over $A=\mathbb{Z}[v^{\pm 1}]$, using only existence and triangularity of the Kazhdan-Lusztig basis and no positivity and no choice.

A finite-dimensional endomorphism algebra and the absence of nontrivial idempotents imply locality. Indeed for an endomorphism $f$ its polynomial minimal relation splits into a power of $T$ and a factor coprime to $T$. The polynomial Chinese remainder identity produces an idempotent separating the nilpotent and invertible parts; indecomposability forces one part to vanish. Thus every $f$ is invertible or nilpotent, and a nonunit $f$ has $1-f$ invertible. Nonunits form a two-sided ideal: multiplication cannot make a nonunit invertible in a finite-dimensional algebra; if two nonunits summed to a unit, dividing their sum would express $1$ as two nonunits, contradicting the invertibility of $1-f$. This is the unique maximal ideal. Its powers eventually vanish: if a power stabilizes, take a minimal finite generating list for that stable ideal and express its last generator using coefficients in the maximal ideal; the invertibility of $1-a$ for such a coefficient eliminates that generator, a contradiction.

Split idempotents until indecomposable summands remain. Each split reduces the positive right rank of its summand, so there are only finitely many splits. The resulting summands have local endomorphism rings by step 1.1. Decompositions are unique: in two finite decompositions, express the identity of one indecomposable as the sum of its composites through the summands of the other. The local maximal ideal cannot contain every composite; an invertible composite gives a split inclusion into one indecomposable summand, hence an isomorphism to it. Remove the common summand and repeat. This proves finite Krull-Schmidt existence and uniqueness.

Decompose a reduced word for $x$. Its top layer is a single ${\mathrm{\Delta }}_x$ by [F4], and the top-layer functor is additive by [F2]. Exactly one indecomposable summand therefore has this graph, with precisely this normalized layer; call it $B_x$. All other summands have support strictly below $x$. This constructs a normalized candidate without any character conjecture.

Let $M$ be any nonzero Soergel object and choose a maximal support label $x$. The two graph layers ${\mathrm{\Gamma }}_{x}M$ and ${\mathrm{\Gamma }}^{x}M$ are identified by multiplication by $p_x$ as in [F2]. Choose one homogeneous basis element $m$ of the latter, and let $d$ make its degree equal to the generator degree of ${\mathrm{\Gamma }}^x{B}_x(d)$. Choose graph maps $f_0:{\mathrm{\Gamma }}^x{B}_x(d)\to {\mathrm{\Gamma }}^{x}M$ taking that generator to $m$, and $g_0:{\mathrm{\Gamma }}_{x}M\to {\mathrm{\Gamma }}_x{B}_x(d)$ which takes the corresponding element $p_{x}m$ to the generator $p_{x}b$ and kills the other basis elements. These maps are homogeneous of degree zero and satisfy the identity on the rank-one $x$ layer.

Lift these maps to bimodule maps $f:B_x(d)\to M$ and $g:M\to B_x(d)$. The quotient map $M\to {\mathrm{\Gamma }}^{x}M$ is the last step of a $\mathrm{\nabla }$ flag with $x$ last; its kernel is flagged, so [F3] lifts the map $B_x(d)\to {\mathrm{\Gamma }}^{x}M$. The inclusion ${\mathrm{\Gamma }}_{x}M\to M$ is the first step of a $\mathrm{\Delta }$ flag with $x$ first; its quotient is flagged, so [F3] lifts $g_0$ followed by ${\mathrm{\Gamma }}_x{B}_x(d)\to B_x(d)$. The first lifted map on the quotient and the second on the submodule agree with the prescribed graph maps. The image identity ${\mathrm{\Gamma }}_{x}M={\mathrm{\Gamma }}^{x}M{p}_x$ makes their composite $gf$ act as the identity on ${\mathrm{\Gamma }}_x{B}_x(d)$. Thus $gf$ is not nilpotent, and is invertible by step 1.1. Rescale $g$ by its inverse to exhibit $B_x(d)$ as a summand of $M$.

If $M$ is indecomposable, step 5.1 gives $M\cong B_x(d)$. If two normalized candidates were obtained from different reduced expressions for $x$, applying that same splitting argument to either candidate makes them isomorphic, because their normalized rank-one top degrees force $d=0$. Distinct labels cannot be isomorphic, by their maximal support; distinct shifts at a fixed label cannot be isomorphic, by their top generator degrees. This proves classification and reduced-word independence.

Restriction of degree-zero endomorphisms of $B_x$ to its rank-one top layer is a unital homomorphism onto $k$, since homogeneous scalars act there and its endomorphism ring in degree zero is $k$. Its kernel is a maximal ideal. Locality in step 1.1 makes this the unique maximal ideal, so the residue field is $k$. All selections were from finite supports and finite homogeneous bases, and no choice axiom beyond finite choice is used. The statement follows from steps 2.1, 3.1, 6.1 and this residue calculation.

The unit has the perfect multiplication form. If a right-free bimodule $M$ has an invariant form, put $a(b)=b{c}_{\mathrm{i}\mathrm{d}}$ and $b(b^{\mathrm{\prime }})=b^{\mathrm{\prime }}{c}_s$ in $M{B}_s$ and define the induced form by $⟨a(u),a(v)⟩={\mathrm{\partial }}_s⟨u,v⟩$, $⟨a(u),b(v)⟩=⟨b(u),a(v)⟩=⟨u,v⟩$, and $⟨b(u),b(v)⟩=⟨u,v⟩{\alpha }_s$. Extend on a right basis by right bilinearity. The relation of [F2] and the identity ${\mathrm{\partial }}_s(fr)={\mathrm{\partial }}_s(f)r+s(f){\mathrm{\partial }}_s(r)$ verify that these formulas agree for arbitrary $u,v$, and that left multiplication is self-adjoint. Symmetry follows from symmetry of the preceding form. If $e_i,e_i^{\ast }$ are dual right bases for that form, the pairing between bases $(a(e_i),b(e_i))$ and $(b(e_i^{\ast }),a(e_i^{\ast }))$ has matrix $\left(\begin{array}{cc}I& 0\\ {\alpha }_{s}I& I\end{array}\right)$, using ${\mathrm{\partial }}_s(1)=0$, and has determinant one. Thus perfection is preserved. Iteration yields a perfect symmetric invariant form on every word, exactly its successive rank-one trace multiplication form.

For finite right-free modules the graded double-dual evaluation is an isomorphism, so $D$ is an involutive contravariant equivalence and takes indecomposables to indecomposables. The invariant perfect word form of step 1.1 identifies a word with its right dual. Generic graph support is unchanged by this duality: in a right basis left multiplication matrices transpose under $D$, so their simultaneous eigencharacters, and hence the graph labels after passing to the polynomial fraction field, are unchanged. A right-free summand has precisely the support closure of those graph labels. Therefore $D{B}_x$ is an indecomposable summand with the $x$ graph of the same reduced word. Uniqueness in [F1] identifies it with $B_x$ with no shift.

A word of length $m$ has a homogeneous right basis given by tensor products of the two basis elements in [F2]; its generator degrees lie between $-m$ and $m$ and have parity $m$. Right reduction preserves a split direct-sum decomposition. Since $B_x$ is an unshifted summand of its reduced word of length $m=\mathrm{\ell }(x)$, its reduction has these same bounds and parity. The word's bottom degree is the single $1$-tensor. The balanced evaluation of that word on its top graph, obtained by successively twisting the tensor-slot polynomials by the successive prefix permutations, sends this $1$-tensor to $1$. It factors through the rank-one top quotient, and is an isomorphism onto $R_x(m)$ because both generators have degree $-m$. All other summands have no $x$ support; consequently the projection of the bottom tensor into $B_x$ is nonzero. Its bottom dimension is therefore one.

Right duality reverses the degrees of the finite homogeneous basis after reduction. The unshifted duality $D{B}_x\cong B_x$ from step 2.1 therefore makes the reduced dimensions symmetric and gives top dimension one from the bottom dimension in step 3.1. The perfect trace forms, duality, degree bounds, parity and extremal dimensions have now all been established locally.

For finite right-free rational $M,N$, a homogeneous bimodule map is determined by the images of a finite right basis. In a fixed degree the finitely many left-generator commutation relations are finite linear equations with rational coefficients. Their kernel commutes with the flat field extension to $\mathbb{R}$. Thus every homogeneous Hom space, and its composition, commutes with extension. The support cut on a specified graph union is the common kernel of its finitely generated graph ideal acting on the carrier: the carrier is right-free and generically a sum of reduced graph modules, so the relations hold before localization by torsion-freeness. These kernels commute with flat extension as well. Therefore the canonical support submodules, quotients, graph generator degrees and multiplicities commute with extension. Right reduction commutes with it by tensor associativity and the augmentation map.

Let $A={\mathrm{End}}^0(B_x)$ over $\mathbb{Q}$. By [F1] it is finite-dimensional local with residue $\mathbb{Q}$, and its maximal ideal $J$ is nilpotent by the finite-algebra proof there. Step 1.1 gives ${\mathrm{End}}^0(B_x^R)=A{\otimes }_{\mathbb{Q}}\mathbb{R}$. The ideal $J\otimes \mathbb{R}$ is nilpotent and its quotient is $\mathbb{R}$; an element with nonzero residue is invertible by a finite geometric series after subtracting its scalar part. Hence this algebra is local with residue $\mathbb{R}$, and $B_x^R$ is indecomposable. Its top support layer is unchanged by step 1.1, so the real classification [F1] identifies it with the same normalized real label. Every real word is the extension of its rational word and its rational finite decomposition; the real classification and uniqueness imply that all real indecomposables and all real decompositions are the corresponding labels and shifts.

Duality exchanges the $\mathrm{\Delta }$ and $\mathrm{\nabla }$ flags: dualize their right-split canonical layer sequences, which reverses order and the homogeneous degrees of their graph generators. A ${\mathrm{\Delta }}_y(d)$ layer dualizes to ${\mathrm{\nabla }}_y(-d)$. Therefore normalized self-duality gives $h_{\mathrm{\Delta }}(B_y)=h_{\mathrm{\nabla }}(B_y)$ by the weight conventions. The same is true for $B_x{B}_s$: an invariant isomorphism $B_x\to D{B}_x$ can be extended by the induced rank-one form construction in [F2], whose mixed-basis matrix has determinant one and yields an invariant isomorphism $B_x{B}_s\to D(B_x{B}_s)$. If $S(y)$ holds, both characters of $B_y$ are consequently ${\underset{‾}{H}}_y$. By [F3],[F4], its endomorphism Hom rank has no negative powers and constant term one. Since the polynomial ring has degree-zero part $\mathbb{R}$ and no negative degrees, its degree-zero endomorphism algebra is then exactly $\mathbb{R}$.