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Hard Lefschetz fails for a non-dominant degree-two element: the rank-one case

Statement refuted

In the rank-one case n=2 the element ρ′:=x1+x2 is a degree-two element of S with ρ′(αs∨)=0, so it is not strictly dominant. Its action on the reduced rank-one Soergel module is zero: ρ′=e1=x1+x2 is s-invariant and of positive degree, and x1+x2=1⋅e1 acts on the right basis element cid‾ as the right action by an element of S+, hence ρ′⋅cid‾=0; a fortiori it cannot map (Bs‾)−1=R cid‾ isomorphically to (Bs‾)1. Therefore for ρ′ the hard Lefschetz property fails (the map on the −1 degree is the zero map), even though ρ′ is a symmetric degree-two endomorphism of the reduced module; the same failure occurs for every degree-two element with ρ′(αs∨)=0. Consequently the hard Lefschetz and Hodge-Riemann conclusions do not hold for arbitrary degree-two elements when strict dominance is omitted. This does not assert that every Lefschetz element is strictly dominant.

Facts & Assumptions

Given: The rank-one setting n=2, S=R[x1,x2] with deg⁡xi=2, the simple reflection s=s1 with coordinate root α=αs=x1−x2, the generator Bs=S⊗SsS(1) with right basis cid=1⊗1 of degree −1 and cs=12(α⊗1+1⊗α) of degree +1, its reduced module Bs‾=Bs/BsS+ with the induced basis cid‾,cs‾, and the degree-two element ρ′=x1+x2 (The Soergel bimodule Bi of a simple reflection, The real type-A Soergel category by scalar extension, The reduced Soergel module and the dominant Lefschetz operator).

[F1]

The left and right actions on Bs satisfy r⋅cid=cid⋅sr+∂s(r)cs and r⋅cs=cs⋅r, and cscs=csα, cid is the unit of the commutative ring Bs(−1) (The trace, the invariant intersection form on Bott-Samelson bimodules and its reduction, The Soergel bimodule Bi of a simple reflection).

[F2]

The reduction functor kills right multiplication by S+: M‾=M/MS+, and mr+MS+=MS+ for r∈S+ (The reduced Soergel module and the dominant Lefschetz operator).

[F3]

Hard Lefschetz for a degree-two symmetric operator L on a graded space with lowest degree −d requires L:( )−1→( )1 to be an isomorphism in the case d=1; definiteness and symmetry conventions as in Primitive subspaces, hard Lefschetz and the Hodge-Riemann signs: the dictionary of hypotheses S, hL and HR and Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms.

Counterexample

technique · direct computation in the rank-one reduced module
1.1givenF1

Non-dominance. The reflection s exchanges x1 and x2, so ρ′=x1+x2 is s-invariant and ρ′(α∨)=(x1+x2,x1−x2)=1−1=0 in the standard form of the coordinate normalization; in particular ρ′ is not strictly dominant.

2.1step 1.1givenF1

The left action of ρ′ equals its right action. Since ρ′∈Ss, the balanced tensor relation gives ρ′⊗1=1⊗ρ′ in Bs; hence for every b∈Bs the left multiplication ρ′⋅b equals the right multiplication b⋅ρ′: indeed both actions are S-linear and the identity holds on the two basis elements cid (where ρ′⋅cid=ρ′⊗1=1⊗ρ′=cid⋅ρ′) and cs (where both equal csρ′ by [F1]).

3.1step 2.1F1F2

The operator vanishes on the reduced module. By step 2.1, ρ′⋅cid=cid⋅ρ′=1⊗ρ′ lies in BsS+, since it is the right product of cid=1⊗1 with ρ′∈S+; by [F2] its class in Bs‾ is zero, so ρ′⋅cid‾=0. Similarly ρ′⋅cs=cs⋅ρ′=csρ′ lies in BsS+ and reduces to zero. Since cid‾ and cs‾ span Bs‾, the induced operator L=ρ′⋅(−) is the zero endomorphism of Bs‾.

4.1step 3.1F3

Failure of hard Lefschetz. The reduced module has dim⁡(Bs‾)−1=dim⁡(Bs‾)1=1 with bases cid‾ and cs‾. The degree-one Lefschetz map L:(Bs‾)−1→(Bs‾)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.

5.1step 4.1∎

The general degree-two case. Let ρ′′=ax1+bx2 satisfy ρ′′(α∨)=a−b=0; then a=b and ρ′′=aρ′ is a scalar multiple of ρ′. If a=0 the operator is trivially zero; if a≠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 αs∨ 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.

Remark

The counterexample isolates the hypothesis at the very bottom of the induction: the strict dominance of ρ is used on this page exactly through the positivity of the scaling factors (wρ)(αs∨) for reduced prefixes, and at the simple reflection itself the vanishing of the coroot evaluation makes the operator vanish on the whole reduced module. Since ρ′ is still symmetric of degree two, the failure is not an artifact of a badly normalized form.

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