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Existence and uniqueness of the Kazhdan–Lusztig basis

Facts & Assumptions

Given: n≥1, the normalized Hecke algebra Hv(n), its bar involution, and the Laurent coefficients rx,y of the bar images in the standard basis.

[F1]

The standard elements Hy form an A=Z[v±1]-basis of Hv(n) (The normalized type-A Hecke algebra and its bar involution, The standard basis of the generic type-A Hecke algebra).

[F2]

The bar is a semilinear algebra involution with v‾=v−1 (The Hecke bar involution is well defined).

[F3]

The coefficients satisfy rx,y=0 unless x≤y, ry,y=1, and both matrix identities RRˉ=RˉR=I; for x≤y, their degree bounds, parity, and top coefficient are as stated in the R-coefficient theorem (The R-coefficient recursion, support, degree bounds and inversion).

[F4]

The A-linear anti-automorphism ♭ commutes with bar and sends Hy to Hy−1 (Reversal anti-involution commutes with the Hecke bar).

[F5]

Bruhat order on Sn is a finite graded order, strict inequalities raise length, and inversion preserves the order (Basic properties of the Bruhat order on Sn).

[F6]

Elias–Williamson, Corollary 1.2(1), states that the coefficients hy,w of their triangular bar-fixed basis belong to Z≥0[v]. Their §3.2 uses Hs2=1+(v−1−v)Hs, and Remark 3.2 fixes q=v−2 and hy,w=vℓ(w)−ℓ(y)Py,w(v−2). This is the single original-source positivity fact authorized for this item; its Soergel–Hodge proof is not a local prerequisite.

Statement

For each w∈Sn there is a unique element H‾w∈Hv(n) with (i) H‾w‾=H‾w and (ii) H‾w∈Hw+∑y<wvZ[v] Hy (the sum over the lower Bruhat ideal of w). The elements {H‾w}w∈Sn form an A-basis of Hv(n), and writing H‾w=∑y≤wpy,wHy one has pw,w=1, py,w=0 unless y≤w, py,w∈vZ[v] for y<w, the bar-duality px,w=∑yrx,y py,w‾ (matrix form P=RPˉ), and the symmetry py−1,w−1=py,w. Moreover for y<w, with d=ℓ(w)−ℓ(y): py,w=vd+terms of strictly smaller degree and py,w∈vd Z[v−2]; in particular every H‾w has integer nonnegative coefficients in the standard basis with py,w of fixed parity d mod 2.

Proof

technique · construct the triangular bar-fixed element by descending induction in the finite Bruhat order, then use its uniqueness and coefficient comparison
1.1F2F3F5algebra

Construct the coefficients. Fix w and descend on d=ℓ(w)−ℓ(x) over the finite lower Bruhat ideal {x:x≤w}, starting with pw,w=1. Suppose x<w and py,w has been constructed for every x<y≤w, satisfying py,w=∑y≤z≤wry,zpz,w‾. Put ax:=∑x<y≤wrx,ypy,w‾. Then ax‾=∑x<y≤wrx,y‾py,w=∑x<y≤z≤wrx,y‾ry,zpz,w‾=−∑x<z≤wrx,zpz,w‾=−ax. The second equality uses the induction equations; for x<z, the identity RˉR=I makes the sum over x≤y≤z zero, and its omitted diagonal term is rx,x‾rx,z=rx,z. Write ax=∑m∈Zγmvm. Anti-invariance gives γ−m=−γm and γ0=0. Define px,w:=∑m>0γmvm. Then px,w∈vZ[v] and px,w−px,w‾=ax. The induction is finite and uses no choice principle.

2.1F1F2F3F5step 1.1algebra

Bar invariance and the coefficient equations. Set H‾w:=∑y≤wpy,wHy. The coefficient of Hx in H‾w‾ is ∑x≤y≤wrx,ypy,w‾. For x=w this is 1; for x<w it is px,w‾+ax=px,w by step 1.1. If x≰w, no y≤w can satisfy x≤y, so support from [F3] gives coefficient zero. Thus H‾w‾=H‾w and px,w=∑x≤y≤wrx,ypy,w‾, which is the bar-duality formula.

3.1F1F2F3step 2.1algebra

Uniqueness. If two bar-invariant elements satisfy the triangular condition, their difference is a bar-fixed sum h=∑y<wcyHy with each cy∈vZ[v]. If h≠0, choose a Bruhat-maximal x in its finite support. The coefficient of Hx in h‾ is cx‾: no supported y>x contributes, and support of rx,y requires x≤y. Since h=h‾, cx=cx‾; but vZ[v] and v−1Z[v−1] intersect only in 0, a contradiction. Thus the element is unique. The construction also gives pw,w=1 and py,w=0 unless y≤w.

3.2F1F5step 2.1algebra

Basis. The transition from (Hw) to (H‾w) is unitriangular on the finite Bruhat poset: each H‾w=Hw+∑y<wpy,wHy. A finite unitriangular matrix over A is invertible, so (H‾w)w∈Sn is an A-basis.

3.3F3step 1.1step 2.1algebra

Degree, leading term, and parity. Induct on d=ℓ(w)−ℓ(x), with pw,w=1. For x<w, the equation in step 1.1 has ax=∑x<y≤wrx,ypy,w‾. By [F3] and induction, every term has exponents congruent to d modulo 2. The term y=w is rx,w, whose highest term is vd with coefficient 1. For each y<w, put d1=ℓ(y)−ℓ(x) and d2=ℓ(w)−ℓ(y)≥1, so d1+d2=d; the highest degree of rx,ypy,w‾ is at most d1−1≤d−2, since py,w∈vZ[v]. Therefore ax has highest term vd with coefficient 1 and only exponents of parity d. As px,w is its positive-degree part by step 1.1, it follows that px,w=vd+ terms of strictly smaller degree and px,w∈vdZ[v−2]. This proves the degree and parity clauses.

4.1F4F5step 3.1algebra

Inverse-index symmetry. By [F4], ♭(H‾w) is bar-fixed. By [F5], inversion preserves Bruhat order, so this element has the form Hw−1+∑y<wpy,wHy−1 with lower terms in vZ[v]. Uniqueness from step 3.1 gives ♭(H‾w)=H‾w−1. Comparing coefficients yields py−1,w−1=py,w.

5.1F1F2F6step 2.1step 3.1step 3.3algebra∎

Normalize and apply the authorized positivity result. The identity on v and on each Hs identifies our presented algebra with the type-A algebra in Elias–Williamson’s Hecke section: the quadratic and braid relations agree by [F1], and its bar agrees on v and all generators by [F2]. Its standard element Hw is the same reduced-word product as ours. Their basis has precisely the bar-invariance and triangularity established in step 2.1, so uniqueness in step 3.1 identifies it with our H‾w. Comparing standard-basis coefficients gives py,w=hy,w. The authorized positivity conclusion in [F6] therefore gives py,w∈Z≥0[v]. For y<w, step 3.3 and py,w∈vZ[v] give the polynomial Py,w(q)=v−dpy,w, q=v−2, exactly as in the normalization remark of [F6]; this substitution preserves individual integer coefficients. The diagonal and unsupported coefficients are respectively 1 and 0. Thus coefficientwise nonnegativity and every asserted boundary case hold.

Remarks

Existence, uniqueness, basis, support, bar-duality, degrees, leading terms, parity, inverse symmetry and the normalization comparison are proved locally. Only coefficientwise positivity invokes the owner's exact original-source fallback, recorded in research/frontier-43-complex-representation-15-kl-positivity-citation-authorization.json. The triangular construction alone does not imply positivity, and no local proof of the Soergel–Hodge theorem is asserted. All local inductions are finite and use no Choice.

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