Hodge Theory and Lefschetz Linear Algebra

Abstract

The cohomology rings of smooth complex projective algebraic varieties satisfy a “package” of properties: Poincaré duality, weak Lefschetz, hard Lefschetz, and the Hodge–Riemann bilinear relations. These properties impose conditions on the structure of the cohomology ring, and give rise to interesting linear operators and bilinear forms on the ring. In several contexts this “package” yields important results about geometry, representation theory, or combinatorics. We discuss the above properties purely in terms of linear algebra, so that some of the techniques and intuition from geometry can be brought to bear on more general problems. Our focus is on the techniques which appear in the proof of Soergel’s conjecture, but there have been successes in other fields as well (e.g. in the combinatorics of matroids and polytopes), and it seems there will be more to come.

This chapter is based on expanded notes of a lecture given by the authors and taken by

Elijah Bodish and Jay Hathaway

E. Bodish⋅ J. Hathaway

Department of Mathematics, University of Oregon, Fenton Hall, Eugene, OR, USA

1 Introduction

Let $\Bbbk$ be a field and let V  be a finite-dimensional vector space over $\Bbbk$. A bilinear form on V  is a function

$$\displaystyle \begin{aligned} (-,-): V\times V \longrightarrow \Bbbk \end{aligned}$$

which is $\Bbbk$-linear in each variable. We say that the form is symmetric if (v, v′) = (v′, v) for all v, v′∈ V , and nondegenerate if the mapping v↦(−, v) is an isomorphism between V  and V ^∗. A fundamental question in mathematics is to understand the symmetric nondegenerate forms on a fixed vector space V .

A bilinear form is defined independent of any fixed basis, but to see some of its properties we are inclined to choose an ordered basis of V , say (v ₁, …, v _n). Then we may define the Gram matrix of the form with respect to this basis as the matrix $\left ((v_i, v_j)\right )_{1 \le i,j \le n}$. This defines a bijection between bilinear forms on V  and $(\dim V) \times (\dim V)$ matrices with entries in $\Bbbk$. The form is symmetric (resp. nondegenerate) if and only if the Gram matrix is a symmetric (resp. invertible) matrix.

There is an action of $\operatorname {\mathrm {GL}}(V)$ on the set of all symmetric nondegenerate bilinear forms on V  defined by

$$\displaystyle \begin{aligned} T\cdot (-, -) = (T(-), T(-)). \end{aligned}$$

With respect to a fixed basis this action looks like $\operatorname {\mathrm {GL}}_n(\Bbbk )$ acting on symmetric invertible n × n matrices by

$$\displaystyle \begin{aligned} M\mapsto A^{t}MA \end{aligned}$$

The natural notion of equivalence for two forms then translates to their Gram matrices being in the same $\operatorname {\mathrm {GL}}_n(\Bbbk )$ orbit. The nature of these orbits obviously depends on the dimension of V , but the most interesting behavior is observed when the field $\Bbbk$ varies.

Example 17.1

When $\Bbbk =\mathbb {C}$, all symmetric nondegenerate forms are equivalent.

Example 17.2

When $\Bbbk = \mathbb {F}_p$ with p≠2, symmetric nondegenerate forms are determined by the image of the determinant of the Gram matrix in $(\mathbb {F}_p^{\times })/(\mathbb {F}_p^{\times })^2$ (first think about this for the case of 1 × 1 matrices).

Example 17.3

When $\Bbbk = \mathbb {R}$ the Gram matrix is symmetric and invertible, hence it is conjugate, by an orthogonal change of basis, to a diagonal matrix with diagonal entries in $\mathbb {R}^{\times }$. Taking p to be the number of positive eigenvalues and q to be the number of negative eigenvalues, one says then that p − q is the signature of the form. Sylvester’s law of inertia is the statement that nondegenerate symmetric bilinear forms on a real vector space are classified up to equivalence by their signature.

The difference between the real and complex case should be thought of as a consequence of the difference between the topologies of $\mathbb {R}^{\times }$ and $\mathbb {C}^{\times }$. We will take advantage of the disconnectedness of $\mathbb {R}^{\times }$ in the following lemma as well.

Lemma 17.4

Let V  be a finite-dimensional $\mathbb {R}$ -vector space. Suppose that B _λ is a continuous family of nondegenerate symmetric bilinear forms on V , parameterized by an interval $\lambda \in (a,b)\subset \mathbb {R}$ . Then all B _λ have the same signature.

Proof

Since the forms are all symmetric, the eigenvalues of B _λ must remain real as λ ∈ (a, b) varies. The signature of B _λ will only change if the sign of an eigenvalue changes. Since the forms vary continuously, the intermediate value theorem would then imply that there is some μ so that B _μ has 0 as an eigenvalue. Then B _μ is degenerate, a contradiction. □

Remark 17.5

In contrast to the proof above, when $\Bbbk =\mathbb {C}$ the eigenvalues are free to move around 0 in a continuous family. (Again thinking about the case of a form on a one-dimensional vector space is useful!) This gives some intuition as to why all symmetric nondegenerate forms over $\mathbb {C}$ are equivalent.

We will now restrict our attention to forms on $\mathbb {R}$-vector spaces. Any field of characteristic zero will suffice for the discussion of the hard Lefschetz property below, but once we reach the Hodge–Riemann bilinear relations the base field $\mathbb {R}$ will be essential, in order to have a notion of signature.

2 Hard Lefschetz

Fix a finite-dimensional graded $\mathbb {R}$-vector space

$$\displaystyle \begin{aligned} H= \bigoplus_{i\in{\mathbb{Z}}}H^i \end{aligned} $$

and a symmetric nondegenerate graded bilinear form

$$\displaystyle \begin{aligned} \langle -, -\rangle:H\times H\longrightarrow \mathbb{R}. \end{aligned} $$

By “graded” we mean that 〈H ⁱ, H ^j〉 = 0 if i≠ − j.

It is immediate that 〈−, −〉 induces an isomorphism between H ⁻ⁱ and (H ⁱ)^∗. We say that the graded vector space H satisfies Poincaré duality . Thus, if $b_i= \dim (H^i)$ (the i-th Betti number of H ), then b _i = b _−i for all $i\in \mathbb {Z}$. Our convention is such that the mirror of Poincaré duality is in degree zero. We will use the notation $H^{\min }$ to denote the nonzero graded component of H of minimal degree, which is well defined as long as H≠0.

Definition 17.6

Let $L\in \operatorname {\mathrm {End}}_{\mathbb {R}}(H)$ be a degree two operator

$$\displaystyle \begin{aligned} L:H^i\longrightarrow H^{i+2}. \end{aligned} $$

• 1. We say L is a Lefschetz operator if 〈La, b〉 = 〈a, Lb〉 for all a, b ∈ H.

• 2. If L is a Lefschetz operator, then it is said to satisfy hard Lefschetz (hL) if for all i ≥ 0,

  $$\displaystyle \begin{aligned} L^i:H^{-i}\longrightarrow H^i \end{aligned}$$

  is an isomorphism.

The following exercise requires some work, but is extremely worthwhile:

Exercise 17.7

Show that a degree two operator L on H satisfies hard Lefschetz if and only if there is an action of $\mathfrak {sl}_2(\mathbb {R})$ on H with e acting as L and h ⋅ v = kv for all v ∈ H ^k. Moreover, this $\mathfrak {sl}_2(\mathbb {R})$-action is unique. In particular, deduce in this case that

$$\displaystyle \begin{aligned} b_0 \ge b_{-2} \ge \cdots \ge 0 \end{aligned} $$

(17.1a)

and that

$$\displaystyle \begin{aligned} b_{-1} \ge b_{-3} \ge \cdots \ge 0. \end{aligned} $$

(17.1b)

Example 17.8

The cohomology ring $H^*(\mathbb {C} \mathbb {P}^n, \mathbb {R})$ of complex projective space is isomorphic as a graded ring to $\mathbb {R}[x]/(x^{n+1})$, where deg(x) = 2. In order for the mirror of Poincaré duality to be in degree zero, we shift the grading on $H^*(\mathbb {C} \mathbb {P}^n, \mathbb {R})$ down by n. (More generally, when we consider the cohomology ring of a projective variety over $\mathbb {C}$, we will shift the grading down by $\dim _{\mathbb {C}}(X)$.) Consider the graded bilinear pairing on this graded ring where (f, g) is the coefficient of x ⁿ in the product fg. Observe that multiplication by x is a Lefschetz operator satisfying hard Lefschetz. (Geometrically, this bilinear pairing is the Poincaré pairing, and x is the first Chern class of a certain line bundle on $\mathbb {C} \mathbb {P}^n$.)

When thinking about Lefschetz operators, it is helpful to have in mind the picture in Fig. 17.1.

Fig. 17.1

This picture represents a Lefschetz operator satisfying hard Lefschetz. The ⋆ ’s represent basis vectors in each degree, and the arrows represent the action of L. If there is no arrow leaving a ⋆ , then that basis vector is in the kernel of L

The graded form on H restricts to the zero form on each graded component H ⁱ when i≠0. Instead it pairs graded components in complementary degree. We are interested in obtaining a form on each graded component H ⁱ, and to do this we will relate H ⁻ⁱ and H ⁱ by applying L ⁱ. This yields an identification $H^{-i} \overset {\sim }{\to } H^i$ (depending on L) exactly when L satisfies hard Lefschetz.

Definition 17.9

Let L be a Lefschetz operator. For each i ≥ 0, define the Lefschetz form on H ⁻ⁱ with respect to L as

$$\displaystyle \begin{aligned} (a,b)_L^{-i} = \langle a,L^ib\rangle \end{aligned} $$

(17.2)

for a, b ∈ H ⁻ⁱ.

The Lefschetz form is nondegenerate if and only if it induces an isomorphism between H ⁻ⁱ and (H ⁻ⁱ)^∗. Since this isomorphism factors as

(17.3)

the nondegeneracy of the Lefschetz form is equivalent to hard Lefschetz for L. Note that hard Lefschetz is satisfied trivially in degree zero for any Lefschetz operator L, and the Lefschetz form $(-,-)_L^0$ is independent of the choice of L.

Remark 17.10

If the Betti numbers of H make hard Lefschetz possible, i.e. (17.1) holds, then satisfying hard Lefschetz is a generic condition. Namely, the collection of degree two Lefschetz operators forms an affine space, and L ⁱ : H ⁻ⁱ → H ⁱ being an isomorphism is equivalent to the non-vanishing of a polynomial. Thus the operators satisfying hard Lefschetz form an open set in the Zariski topology of an affine space, and therefore are dense. In what follows, we will sometimes consider families of Lefschetz operators on a fixed graded vector space.

Suppose that L is a Lefschetz operator satisfying hard Lefschetz. The $\mathfrak {sl}_2(\mathbb {R})$ action specified in Exercise 17.7 is unique, and the weight spaces of H are the nonzero graded components. To decompose H into irreducible $\mathfrak {sl}_2(\mathbb {R})$ representations it suffices to determine the lowest weight spaces. In our context we introduce some new terminology to discuss the lowest weight spaces (without having to define a lowering operator). For all i ≥ 0 set

$$\displaystyle \begin{aligned} P^{-i}_L = \ker(L^{i+1})\cap H^{-i}. \end{aligned} $$

(17.4)

We will refer to elements of this subspace as primitive vectors of degree − i, and call the subspace a primitive subspace .

Exercise 17.11

Check that $P^{-i}_L$ is indeed the space of lowest weight vectors in H ⁻ⁱ (where we view H as an $\mathfrak {sl}_2(\mathbb {R})$-module via Exercise 17.7).

It is a consequence of the finite-dimensional representation theory of $\mathfrak {sl}_2(\mathbb {R})$ that

$$\displaystyle \begin{aligned} H= \bigoplus_{i\ge 0}\bigoplus_{i\ge j\ge 0}L^j\left(P_L^{-i}\right), \end{aligned} $$

(17.5)

and that for i ≥ 0,

$$\displaystyle \begin{aligned} H^{-i} = P_L^{-i}\oplus L\left(P_L^{-i-2}\right)\oplus L^2\left(P_L^{-i-4}\right)\oplus \cdots. \end{aligned} $$

(17.6)

Observe that $L^j(P_L^{-i})$ is the intersection of the isotypic component of H of highest weight i and the − i + 2j weight space of H. Can you see this decomposition in Fig. 17.1?

Exercise 17.12

Show that if L satisfies hard Lefschetz, then the primitive decomposition (17.6) is orthogonal with respect to the Lefschetz form.

Remark 17.13

A trivial but important formula which holds for any Lefschetz operator L is that

$$\displaystyle \begin{aligned} (La, Lb)_L^{-(i-2)} = (a,b)_L^{-i}. \end{aligned} $$

(17.7)

Consequently, if L satisfies hard Lefschetz then the Lefschetz form is determined by its restriction to the primitive subspaces.

3 Hodge–Riemann Bilinear Relations

Recall that a symmetric $\mathbb {R}$-bilinear form is determined by its signature. In topology one is interested in what is sometimes called the Poincaré form on the cohomology ring of an even-dimensional manifold, defined by cupping complementary cocycles then integrating against the fundamental class. An interesting invariant arises when studying this form: the signature of its restriction to degree zero (again, we shift our grading down by half the dimension of the space), called the signature of the manifold.

When the manifold is a smooth complex projective variety, there are Lefschetz operators given by cupping with the first Chern class of ample line bundles (cf. [41]). These Lefschetz operators allow one to consider not just the signature of the Poincaré form in degree zero, but of the Lefschetz forms arising in all lower degrees as well. The signatures of these Lefschetz forms are also of interest, and the study of their properties is part of “Hodge theory.”

Finite-dimensional representations of $\mathfrak {sl}_2(\mathbb {R})$ are partitioned into even and odd representations, so the same is true for Lefschetz operators satisfying hard Lefschetz. This means we can split H into its even and odd degree parts and study them separately. Recall that $\min$ is the most negative integer such that $H^{\min } \ne 0$.

Definition 17.14

Assume H ^odd = 0 or H ^even = 0 and that L is a Lefschetz operator satisfying hard Lefschetz. We say that (H, 〈−, −〉, L) satisfies the Hodge–Riemann bilinear relations (HR) if the restriction of the Lefschetz form to the primitive subspace

$$\displaystyle \begin{aligned} (-, -)_L^{\min+2i}|{}_{P_L^{\min+2i}} \end{aligned} $$

(17.8)

is (−1)ⁱ-definite.

Example 17.15

Consider the example pictured in Fig. 17.1, and suppose that it satisfies the Hodge–Riemann bilinear relations. The Lefschetz form must be positive definite on H ⁻⁴, since $-4 = \min$ and the whole space is primitive. By (17.7), the Lefschetz form is also positive definite on L(H ⁻⁴) and L ²(H ⁻⁴), i.e. on the two stars directly above the degree − 4 star. Meanwhile, $\dim P^{-2}_L = 2$ and the Lefschetz form restricted to this space is negative definite. Thus the Gram matrix of the Lefschetz form on H ⁻² has one positive and two negative eigenvalues. Again by (17.7), these signs are copied to the image of L inside H ⁰, and the remaining part $P^0_L \subset H^0$ is positive definite. Thus the Gram matrix of the Lefschetz form on H ⁰ has two positive and two negative eigenvalues.

If the Hodge–Riemann bilinear relations are satisfied, then each Lefschetz form is determined as follows. Let $i= \min +2j$. We first observe that the Lefschetz form restricted to the primitive subspace in H ⁻ⁱ is (−1)^j-definite. Then from the primitive decomposition

$$\displaystyle \begin{aligned} H^{-i} = \bigoplus_{j\ge k\ge 0}L^k\left(P_L^{-i-2k}\right) \end{aligned} $$

(17.9)

we see that the Lefschetz form is (−1)^j+k-definite on $L^k\left (P_L^{-i- 2k}\right )$, for all k.

Remark 17.16

The Hodge–Riemann bilinear relations tell us that the mixed signature of the Lefschetz forms is coming from some definiteness on each primitive subspace.

Exercise 17.17

Assume H ^odd = 0. Find a formula for the signature of the Lefschetz form in degree zero in terms of the Betti numbers of H.

For context, let us state a classical result of Hodge theory (with a simplifying assumption).

Example 17.18

Let X be a smooth complex projective variety of complex dimension d, let H ^∗(X) denote its cohomology ring with coefficients in $\mathbb {R}$, and let α ∈ H ²(X) be the first Chern class of an ample line bundle. Assume that in the Hodge decomposition of $H^*(X, \mathbb {C})$ only type (p, p) occurs, so in particular H is concentrated in even degree. (This is the case for any variety whose cohomology is generated by the fundamental classes of algebraic subvarieties. Examples include projective spaces, Grassmannians, flag varieties, and Schubert varieties.) Equip the cohomology ring of X with its Poincaré pairing 〈−, −〉. Then (H ^∗+d(X), 〈−, −〉, α ∪−) satisfies hard Lefschetz and the Hodge–Riemann bilinear relations.

When one has a nondegenerate form on a vector space, and a subspace of the vector space, it need not be the case that the restriction of the form to the subspace is nondegenerate. However, a (positive or negative) definite form will restrict to a definite form, and hence will remain nondegenerate! Here is an analogue of this crucial idea in Hodge theory.

Exercise 17.19

Suppose that (H, 〈−, −〉, L) satisfies Hodge–Riemann and K is an L-invariant graded subspace of H satisfying Poincaré duality. Show that

$$\displaystyle \begin{aligned} (K, \langle-, -\rangle_K, L|{}_K)\end{aligned}$$

satisfies Hodge–Riemann (up to a sign). In particular, the restriction of the Lefschetz form from H ⁻ⁱ to K ⁻ⁱ is nondegenerate, for each i ≥ 0.

This elementary observation is a key ingredient in de Cataldo and Migliorini’s proof of the decomposition theorem [41], and in Elias and Williamson’s proof of the Soergel conjecture [57]. In both cases the main theorem will follow from the nondegeneracy of some form, but proving the nondegeneracy (say, by induction) is seemingly intractable. However, by proving a stronger “package” of results, through a complicated inductive argument, one can deduce nondegeneracy of a form by embedding the form in a larger one which has the Hodge–Riemann bilinear relations. Two important lemmas in this inductive proof will be presented in the next section.

Exercise 17.20

This exercise will explore $H^*( \operatorname {\mathrm {Gr}}(3,6), \mathbb {R})$, the cohomology of the Grassmannian of 3-planes in $\mathbb {C}^6$, using a combinatorial model. Let P(3, 6) denote the set of Young diagrams which fit inside a 3 × 3 rectangle. The degree of a partition will be − 9 plus twice the number of boxes; for example the partition (3,1,1) has degree + 1. Two Young diagrams are complementary if one can be rotated 180 degrees in order to fill the full 3 × 3 rectangle with the other: for example (3,1,0) and (3,2,0) are complementary.

Let H be the graded vector space with basis {v _λ}_{λ ∈ P(3,6)}. Place a symmetric bilinear pairing on H, where 〈v _λ, v _μ〉 = 1 if λ and μ are complementary, and is equal to zero otherwise. Place an operator L : H → H(2) on this space, where Lv _λ =∑_μv _μ is the sum over the Young diagrams μ ∈ P(3, 6) obtained from λ by adding one box.

• 1. Prove that L is a Lefschetz operator.

• 2. Prove that L satisfies hard Lefschetz. Compute a basis for each primitive subspace.

• 3. Prove that L satisfies the Hodge–Riemann bilinear relations.

This exercise may be made easier or harder by replacing P(3, 6) with P(k, n) for different values of k and n, and considering Young diagrams which fit inside an k × (n − k) box. The enterprising reader might try to find a formula for the number of different summands in the Lefschetz decomposition (i.e. the number of i ≥ 0 such that $P^{-i}_L \ne 0$) as a function of k and n.

4 Lefschetz Lemmas

Classically, the “Kähler package” consists of Poincaré duality, hard Lefschetz, the Hodge–Riemann bilinear relations, as well as the Lefschetz hyperplane theorem (weak Lefschetz). We have not yet discussed the Lefschetz hyperplane theorem in this abstract setting. One reason for this omission is that there is not an obvious working candidate for a hyperplane section of a Soergel bimodule (see Sect. 19.1 for a further discussion). The analogue of the Lefschetz hyperplane theorem used in the proof of Soergel’s Conjecture is the following lemma.

Lemma 17.21 (The Weak Lefschetz Argument)

Let V, W be graded vector spaces with graded nondegenerate symmetric bilinear forms 〈−, −〉_V, 〈−, −〉_W and Lefschetz operators L _V, L _W. Let

$$\displaystyle \begin{aligned} \sigma: V\longrightarrow W \end{aligned}$$

be a degree one linear map such that

• 1. σ is injective from negative degrees,

• 2. 〈v, L _Vv′〉_V = 〈σv, σv′〉_W for all v, v′∈ V ,

• 3. σL _V = L _Wσ.

If L _W satisfies the Hodge–Riemann bilinear relations, then L _V satisfies hard Lefschetz.

Proof

Let i > 0 (recall that hard Lefschetz always holds trivially in degree zero). We claim that

$$\displaystyle \begin{aligned} (\sigma(v), \sigma(v'))^{-(i-1)}_{L_W} = (v, v')_{L_V}^{-i} \end{aligned} $$

(17.10)

for all v, v′∈ V ⁻ⁱ. Indeed,

$$\displaystyle \begin{aligned} & (\sigma(v), \sigma(v'))^{-(i-1)}_{L_W} = \langle \sigma(v), L_W^{i-1}\sigma(v')\rangle_W \\ &\qquad \qquad \qquad = \langle \sigma(v), \sigma L_V^{i-1}(v')\rangle_W = \langle v, L_V^i(v') \rangle_V = (v, v')_{L_V}^{-i}. \end{aligned} $$

(17.11)

Since the pairing 〈−, −〉_V gives an isomorphism between V ⁻ⁱ and (V ⁱ)^∗, the Betti numbers of V  satisfy b _i = b _−i. To prove that L _V satisfies hard Lefschetz, it therefore suffices to show that $L_V^{i}$ is injective on V ⁻ⁱ. So let v ∈ V ⁻ⁱ and suppose that $L_V^{i}(v) =0$. Since σL _V = L _Wσ it follows that $L_W^{i}(\sigma (v)) = \sigma (L_V^{i}(v))=0$, so

$$\displaystyle \begin{aligned} \sigma(v)\in \ker(L_W^{i})\cap W^{-(i-1)}= P_{L_w}^{-(i-1)}. \end{aligned} $$

(17.12)

By the Hodge–Riemann bilinear relations, the Lefschetz form $(-, -)_{L_W}^{-(i-1)}$ is positive or negative definite when restricted to the primitive subspace $P_{L_w}^{-(i-1)}$, so from

$$\displaystyle \begin{aligned} (\sigma(v), \sigma(v))_{L_W}^{-(i-1)} \overset{\text{(17.10)}}{=} (v, v)_{L_V}^{-i} =\langle v, L_V^i(v)\rangle_V =\langle v, 0\rangle_V = 0 \end{aligned} $$

(17.13)

we may conclude that σ(v) = 0. Since σ is injective from negative degrees, v = 0. Thus $L_V^{i}$ is injective on V ⁻ⁱ. □

Exercise 17.22

In the lemma above, what conditions would imply that σ sends primitive vectors to primitive vectors?

Lemma 17.23 (Limit Lemma )

Suppose that L _γ is a continuous family of Lefschetz operators on (H, 〈−, −〉) parametrized by $\gamma \in [a,b]\subset \mathbb {R}$. Assume that L _γ satisfies hard Lefschetz for all γ (i.e. that all Lefschetz forms are nondegenerate). Then if one L _γ satisfies the Hodge–Riemann bilinear relations, they all do.

Proof

In each fixed degree the Hodge–Riemann bilinear relations are a statement about signatures of the Lefschetz forms. As the Lefschetz operators vary, the associated Lefschetz forms vary as well, but remain nondegenerate. Thus, if at least one form has eigenvalues distributed so that it satisfies the Hodge–Riemann bilinear relations, then by Lemma 17.4 every form will. □

Remark 17.24

Although the continuous family of Lefschetz forms all restrict on the primitive subspaces to definite forms, the primitive subspaces may change as L _γ varies. However, the primitive subspaces will remain on either the positive or the negative side of the isotropic cone as L _γ varies (the isotropic cone of a form (−, −) on a vector space V  is the union of the lines spanned by vectors v ∈ V  so that (v, v) = 0). The following example illustrates some of the subtlety present in this phenomenon.

Example 17.25

Consider $\overline {B_sB_s} = B_sB_s\otimes _{R}\mathbb {R}$ equipped with the global intersection form 〈−, −〉 (defined in Sect. 12.3), and consider the two-parameter family of Lefschetz operators

$$\displaystyle \begin{aligned} L_{a,b}:=(a\rho\cdot -){\mathrm{id}}_{B_s} + {\mathrm{id}}_{B_s}(b\rho\cdot -), \end{aligned} $$

(17.14)

where $(a,b)\in \mathbb {R}^2$ and $\rho = \frac {1}{2}\alpha _s$. This is left multiplication by aρ plus “middle multiplication” by bρ.

The grading on the vector space $\overline {B_sB_s}$ can be seen explicitly in the decomposition

$$\displaystyle \begin{aligned} \overline{B_sB_s} = (\mathbb{R} \cdot c_{\mathrm{bot}}) \oplus (\mathbb{R} \cdot c_{01}) \oplus (\mathbb{R} \cdot c_{10}) \oplus (\mathbb{R} \cdot c_{\mathrm{top}}), \end{aligned}$$

where we have written $c_{ \underline {e}}$ (where ${ \underline {e}} \subset (s,s)$, and c _bot = c ₀₀, c _top = c ₁₁) for the images of the 01-basis elements (see Sect. 12.1). We have deg c _bot = −2, deg c ₀₁ =deg c ₁₀ = 0, and deg c _top = 2.

With respect to this basis we have

$$\displaystyle \begin{aligned} L_{a,b} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ (b-a) & 0 & 0 & 0 \\ a & 0 & 0 & 0 \\ 0 & a & (b+a) & 0 \end{bmatrix} \quad \text{and} \quad L_{a,b}^2 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 2ab & 0 & 0 & 0 \end{bmatrix}, \end{aligned}$$

so L _a,b satisfies hard Lefschetz if and only if a and b are nonzero. In this case the Lefschetz decomposition is

$$\displaystyle \begin{aligned} (\overline{B_sB_s})^{-2} &= \mathbb{R} \cdot c_{\mathrm{bot}} = P_{L_{a,b}}^{-2}, \\ (\overline{B_sB_s})^0 & = \mathbb{R} \cdot ((b-a) \cdot c_{01} + a \cdot c_{10})\oplus \mathbb{R} \cdot ((b+a)\cdot c_{01} + (-a) \cdot c_{10}) \\ & = L_{a,b}\left(P_{L_{a,b}}^{-2}\right)\oplus P_{L_{a,b}}^0, \\ (\overline{B_sB_s})^2 &= \mathbb{R} \cdot (2ab) \cdot c_{\mathrm{top}} = L_{a,b}^2 \left(P_{L_{a,b}}^{-2}\right). \end{aligned} $$

The Gram matrix of the global intersection form with respect to the basis (c _bot, c ₀₁, c ₁₀, c _top) is

$$\displaystyle \begin{aligned} \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}. \end{aligned}$$

Using this we compute the Lefschetz form restricted to the primitive vectors:

$$\displaystyle \begin{aligned} (c_{\mathrm{bot}}, c_{\mathrm{bot}})_{L_{a.b}}^{-2} = 2ab,\end{aligned}$$

$$\displaystyle \begin{aligned} ((b+a)\cdot c_{01} +(-a)\cdot c_{10}, (b+a)\cdot c_{01} + (-a) \cdot c_{10})_{L_{a,b}}^0 = -2ab. \end{aligned}$$

We see that L _a,b satisfies the Hodge–Riemann bilinear relations if 2ab > 0 and − 2ab < 0, i.e. if a and b are both strictly positive or both strictly negative.

Exercise 17.26

Verify all the computations in this example using diagrammatics.

Remark 17.27

The decomposition $B_sB_s\simeq B_s(-1)\bigoplus B_s(1)$ is a decomposition of R-bimodules, so clearly each summand is preserved¹ by left (resp. right) multiplication by any element of R. So if L is left multiplication by a linear element of R on $\overline {B_sB_s}= B_sB_s\otimes _{R}\mathbb {R}$ (a left R-module), then since

$$\displaystyle \begin{aligned} (\overline{B_sB_s})^{-2}\subset \overline{B_s(-1)} \qquad \text{and}\qquad (\overline{B_sB_s})^{2}\subset \overline{B_s(1)}, \end{aligned}$$

the operator L ² cannot restrict to an isomorphism between $(\overline {B_sB_s})^{-2}$ and $(\overline {B_sB_s})^2$. Thus no operator L induced by left multiplication can satisfy hard Lefschetz. This difficulty explains why we used “middle multiplication” when defining L _a,b in Example 17.25.