The hard Lefschetz theorem is a theorem in complex differential geometry, more specifically Kahler geometry, which determines an isomorphisms between certain cohomology spaces. Together with the Poincare duality, Serre duality it shows just how rigid the cohomology groups of a compact Kahler manifold are. In this post the theorem and its proof will be stated, with a view towards the more rudimentary setting of symplectic geometry.

The basic prerequisites

This obviously does not constitute an introduction to complex geometry, though I have gone a little further than just fixing notation. All manifolds will be assumed to have real dimension equal to $2n$ .

A Kahler manifold is a complex Riemann manifold $(M, J, g)$ with an integrable isometric almost complex structure, for which the $2$ -form

$\omega(X, Y) := g(X, JY)$

is a closed (equivalently $\nabla_XJ = 0$ for every vector field $X$ ).

Let $L: \Omega^p(M) \to \Omega^{p+2}(M)$ be the Lefschetz operator given by

$L(\alpha) = \omega \wedge \alpha$

and $\Lambda$ the dual operator with respect to the $L^2$ -inner product on $\Omega^\star(M) = \bigoplus \Omega^i(M)$ given on the direct summand $\Omega^p(M)$ by

$\langle \alpha, \beta \rangle_2 := \int_M \alpha \wedge \star \beta = \int_M (\alpha, \beta)_g Vol_g$

and extended by orthogonality. Here $\star$ is the Hodge-star operator and $(-, -)_g$ denotes the inner product on $\Omega^p(M)$ induced by the metric, which is

$(\alpha , \beta)_p = det(g^{-1}(\alpha_i, \beta_j))$

The $L^2$ -inner product can be extended to a sesquilinear form on the complexification $\Omega^k$ , by

$\langle \lambda\alpha, \lambda'\beta\rangle = \lambda \overline{\lambda'}\langle \alpha, \beta\rangle$

for $\lambda, \lambda' \in \mathbb{C}$ and $\alpha, \beta$ real $p$ -forms.

Another important adjoint operator which we will need is $d^\star$ (the adjoint of the exterior differential $d$ ). Together with the usual differential $d$ , they form the Laplace-Beltrami (or generalized Laplacian) operator given by

$\Delta = d^\star d + dd^\star.$

This is an extremely important operator at the epicenter of Hodge theory. A p-form $\alpha$ for which $\Delta \alpha = 0$ is called harmonic.

It is customary to substitute the the tangent space $TM$ with $TM_\mathbb{C} = TM\otimes_\mathbb{R} \mathbb{C}$ and give this local coordinates by the Wirtingen derivatives

(1) $\begin{align*} \frac{\partial }{\partial z_i} &= \frac{1}{2}\left( \frac{\partial }{\partial x_i} - i \frac{\partial }{\partial y_i}\right)\\ \frac{\partial }{\partial \overline{z}_i} &= \frac{1}{2}\left( \frac{\partial }{\partial x_i} + i \frac{\partial }{\partial y_i}\right). \end{align*}$

with associated covectors

(2) $\begin{align*} dz_i &= dx_i + i dy_i\\ d\overline{z}_i &= dx_i - i dy_i \end{align*}$

The tangent space can then be decomposed into $T^kM\otimes_\mathbb{R}\mathbb{C} = \bigoplus_{p+q = k} T^{(p,q)}M$ where $T^{(p,q)}M$ is the vector bundle of p partial derivatives of the form $\frac{\partial }{\partial z_i}$ and $q$ of the form $\frac{\partial }{\partial \overline{z}_i}$ . This is actually the eigenspace decomposition of $TM_\mathbb{C}$ given by the (lift of the) the almost complex structure $J$ , which acts on $T^{(0,1)}M$ by multiplication by $-i$ and on $T^{(1,0)}$ by multiplication by $i$ . The decompositions are also inherited (in the obvious way) by arbitrary tensor fields. Explicitly on forms we have

$\Omega^{(p,q)}(M) =\underbrace{\Omega^{(1,0)}\wedge...\wedge\Omega^{(1,0)}}_{p}\wedge \underbrace{\Omega^{(0,1)}\wedge ... \wedge \Omega^{(0, 1)}}_{q}$

It is also customary to decompose the exterior differential $d$ as

$d = \partial + \overline{\partial}$

where $\partial$ and $\overline{\partial}$ are defined on $k$ -forms by

$\partial\left(\sum_{|I \cup J|= k} \alpha_{I,J}dz_I\wedge d\overline{z}_J\right)= \sum_{|I \cup J|= k }\sum_{r\not\in I}\frac{\partial \alpha_{I, J}}{\partial z_r} dz_r\wedge dz_I\wedge d\overline{z}_J$

$\overline{\partial}\left(\sum_{|I \cup J|= k} \alpha_{I,J}dz_I\wedge d\overline{z}_J\right)= \sum_{|I \cup J|= k }\sum_{r\not\in J}\frac{\partial \alpha_{I, J}}{\partial \overline{z_r}} d\overline{z}_r\wedge dz_I\wedge d\overline{z}_J$

where we sum over all multiindices $I$ and $J$ of $\{1, ...., 2n\}$ . We denote again by $\partial^\star$ $\overline{\partial}^\star$ their (formal, since unbounded) adjoints with respect to the $L^2$ inner product.

There is an explicit formula for the adjoint operators, using the Hodge- $\star$ operator, namely,

$\Lambda = \star^{-1} \circ L \circ \star$

similar expressions exist for all adjoints. This is how one defines adjoints for non-compact manifolds, and most of the results from Hilbert space theory extends word for word to the the non-compact setting by working locally on the (sheaf) $\Omega^k(M)$ .

A manifold $M$ together with a closed 2-form $\omega$ for which $\omega^n$ is a nowhere vanishing volume form, is called a symplectic manifold. Equivalently, $\omega$ is a closed 2-form which gives a non-degenerate pairing of tangent vectors. Note that all Kahler manifolds are symplectic with respect to the Kahler form, but with added structure of a Riemmannian and complex manifold. We also call the operator

$L : \Omega^{r} \to \Omega^{r+2} \qquad \alpha \mapsto \omega \wedge \alpha$

the Lefschetz operator. We will mostly be interested in the k-fold powers of this operator from the $(n-k)$ -forms, that is

$L^k : \Omega^{n-k} \to \Omega^{n+k} \qquad \alpha \mapsto \omega^k \wedge \alpha$

for reasons that will become clear shortly.

The symplectic case

On arbitrary symplectic manifolds we have the following important theorem –

Theorem 1 – With $\omega$ the symplectic form on $M^{2n}$ , the Lefschetz operator

$L^k : \Omega^{n-k} \to \Omega^{n+k} \qquad \alpha \mapsto \omega^k \wedge \alpha$

is an isomorphism of $C^\infty$ -modules.

Proof

Step 1) Let $L ^k$ be as above and let $(U, \phi)$ be a trivializing neighborhood of $TM$ . Write $TM|_U \simeq U \times V$ for a vector space $V$ of dimension $2n$ . On $U$ we have

(3) $\begin{align*} \Omega^{n-k} & \simeq \Gamma(U, \bigwedge^{n-k} V^\star) \simeq C^\infty(U) \otimes \bigwedge^{n-k} V^\star \\ \Omega^{n+k} & \simeq \Gamma(U, \bigwedge^{n+k} V^\star) \simeq C^\infty(U) \otimes \bigwedge^{n+k} V^\star \end{align*}$

The Lefschetz operator acts on this space by

$L^k(f\otimes v) = f\otimes \omega^k\wedge v.$

To show that $L^k$ is bijective it suffices to show it is bijective on a trivializing cover (employ the sheaf axioms here), and on a trivializing cover, we have now reduced the problem to showing that the operator $\omega^k\wedge - : \bigwedge^{n-k} V^\star \to \bigwedge^{n+k} V^\star$ is an isomorphism of vector spaces. Since these are all of equal dimension, we will only need to check that the map is injective.

Step 2) (borrowed from Prop.1.1 of [1]) Let us prove this by induction on $k$ . Assume first that $k = n$ , then

$\omega^n\wedge -: \mathbb{R} \to \bigwedge^nV^\star$

is a map between 1-dimensional vector spaces which is bijective by the assumtion that $\omega^n$ is symplectic. Next assume this holds for some $k$ , we will show it also holds for $k-1$ . If $\omega^{k-1}\wedge \xi = 0$ then clearly $\omega^k \wedge \xi = 0$ , hence for any vector $X \in V$ we have

(4) $\begin{align*} 0 &= \iota_X (\omega^k\wedge \xi) = \iota_X(\omega)\wedge \omega^{k-1} \wedge \xi + \omega \wedge \iota_X(\omega)\wedge \omega^{k-2} \wedge \xi + ... + \omega^{k} \wedge \iota_X(\xi) \\ &= k \iota_X(\omega)\wedge \omega^{k-1}\wedge \xi + \omega^{k} \wedge \iota_X(\xi) \end{align*}$

where $\iota_X$ is the the contraction operator. Now by assumption $\omega^{k+1}\wedge \xi =0$ so we are left with the second term

(by the induction hypothesis) $\begin{align*} &\omega^{k} \wedge \iota_X(\xi) = 0\\ \Rightarrow &\iota_X(\xi) = 0 \\ \Rightarrow &\xi = 0 \tag{since $X$ was arbitrary.}\end{align*}$

This concludes the proof.

Tracing the proof of the above theorem, we can make the following observations

We did not needed $\omega$ was closed, just that $\omega^n$ is a nowhere vanishing volume form on $M$ .
If we are given a 2-form $\omega$ for which the map $\omega^r \wedge -$ is injective for some $0<r<n$ , the the wedge powers of $\omega$ determines an isomorphism
$\Omega^{n-k} \to \Omega^{n+k}$
for all $r\leq k \leq n$ .
Noting that $dL = Ld$ since if $\omega$ is any closed form $d(\omega \wedge \alpha) = \omega \wedge d\alpha$ , so $L^k$ induces a map of cohomology groups
$L^k: H^{n-k} \to H^{n+k}.$
Not much can be said about this map though, it may or may not be an isomorphism. Symplectic manifolds for which these maps are isomorphisms for all $0\leq k \leq n$ are said to have the strong (or hard) Lefschetz property.

The Kahler case

In the comfort of Kahler geometry we can say significantly more about the the induced Leftshcetz operators $L^k$ on the level of cohomology, namely –

Theorem (Hard Lefschetz) Let $(M, g, J)$ be a compact Kahler manifold of real dimension $2n$ . Then the Lefschetz operator induces an isomorphism of chomomology spaces for all $k$

$L^k: H^{n-k}(M) \to H^{n+k}(M)$

The proof of this theorem relies on the two following fundamental theorems of Hodge theory –

Theorem 2(Hodge) Let $(M, g)$ be a compact oriented Riemannian manifold, then every cohomology class has a unique harmonic representative.

Theorem 3 – For any compact Riemannian manifold $(M, g)$ the cohomology spaces are all finite dimensional.

and an equally fundamental theorem in Kahler geometry –

Theorem 4 (Kahler identities) [Prop. 3.1.12 of 2] Let $(M, g, J)$ be a(ny) Kahler manifold, $L$ the Lefschetz operator, $\partial$ , $\overline{\partial}$ the differential defined above, and $\partial^\star$ , $\overline{\partial}^\star$ their (formal) adjoints, then

(5) $\begin{align*} [L, \partial] &= [L, \overline{\partial}] = 0 \\ [L, \partial^\star] &= i \overline{\partial} \\[L, \overline{\partial}^\star] &= - i \partial \end{align*}$

We will also need the following two significantly less fundamental lemmas, the first of which is a consequence of the fact that $0 = d^2 = (\partial + \overline{\partial})^2$ , the second is valid for any Kahler manifold

Lemma 1 –

$\partial^2 = 0 \qquad \overline{\partial}^2 = 0 \qquad \overline{\partial}\partial = -\partial \overline{\partial}$

Lemma 2 – $\Delta = d^\star d + dd^\star = 2(\partial^\star \partial + \partial \partial^\star)$ .

Claim – To prove the hard Lefschetz theorem it is sufficient to show that

$[L, \Delta] = 0.$

To see this, assume $[L, \Delta] = 0$ , then $[L, \Delta]^\star =[\Delta^\star, \Lambda] =[\Delta, \Lambda] = 0$ (since $\Delta$ is self adjoint). Recall that the adjoint of a bijective map is bijective, hence in our case since $L^k$ of Theorem 1 is bijective, $\Lambda^k$ is also bijective. Let $\mathcal{H}^r$ denote the space of harmonic $r$ -forms on $M$ , then in particular both

$L^k: \mathcal{H}^{n-k} \to \mathcal{H}^{n+k}$

and

$\Lambda^k:\mathcal{H}^{n+k} \to \mathcal{H}^{n-k}$

are injective. Their lifts to cohomology induces two injective maps

$H^{n-k} \leftrightarrow H^{n+k}.$

Since these are finite dimensional $\mathbb{R}$ -vector spaces (Theorem 3 above) we conclude that they have the same dimension and $L^k$ is a bijection.

Luckily it turns out that $[L, \Delta]= 0$ , (or else the preceding paragraph would quite the waist of space) and here is why –

(lemma 2) $\begin{align*}L\Delta &= L(d^\star d + dd^\star) = 2L(\partial^\star \partial + \partial \partial^\star) \\&= 2([L, \partial^\star] - \partial^\star)([L, \partial] - \partial L) + ([L, \partial] - \partial)\partial^\star([L, \partial^\star] - \partial^\star) \\&= 2((\partial^\star \partial + \partial \partial^\star)L - 2i (\partial \overline{\partial} + \overline{\partial} \partial)) \tag{Theorem 4}\\&= \Delta L \tag{Lemma 1}\end{align*}$

This concludes the proof of the hard Lefschetz theorem.

Remarks

The isomorphism $L^k: \mathcal{H}^{n-k} \to \mathcal{H}^{n+k}$ holds for any Kahler manifold, even non-compact. For compact manifolds, we defined the adjoint in the beginning by either $L^\star = \star^{-1}L\star$ or as the actual adjoint of $L$ with respect to the $L^2$ -inner product. On non-compact complex manifolds the former definition makes sense and inherits many properties of the Hilbert space adjoint. In our case, the map $L$ is injective if and only if its adjoint $L^\star$ is surjective, and vice verca (just like normal linear operators on Hilbert spaces). To see why this is so, note that it suffices to check the claim for germs since $L$ is a sheaf map of $\Omega^p$ , and morphisms of sheaves are injective/surjective if and only the induced map on the germs is injective/surjective respectively. Given this, we may assume the representative of the germ is compactly supported (why?) so it follows that it suffices to check the claim on compactly supported forms, but the compactly supported forms have a well defined $L^2$ -inner product (as defined above) and the claim follows by noting that the adjoint with respect to this inner product agrees with $\star^{-1}L\star$ (which is straightforward to verify).
The isomorphism of Theorem 2 works well with the usual decomposition the cohomology groups of compact Kahler manifolds, meaning
$L^{n-k}: H^{(p, q)} \to H^{(n-q, n-p)}$
where $k = p + q$ is an isomorphism. This follows since $\omega$ is a $(1, 1)$ -form.

Consequences and counterexamples

For non-compact Kahler manifolds, the theorem fails miserably, take for instance

$\mathbb{R} = H^0(\mathbb{C}\backslash \{0\}) \neq H^2(\mathbb{C}\backslash \{0\}) = 0.$

We do however still have an isomorphism of harmonic forms, which tells us that if the hard Lefschetz theorem fails, some cohomology class must fail to have a unique harmonic representative.

Another fun property of the theorem is that if $b_i$ denotes the i’th Betti number which are the real dimensions of the cohomology spaces, then

$\{b_{2i}\}_{i=0,..., n}$

and

$\{b_{2i-1}\}_{i=1,.., n}$

are non-decreasing sequences. For the first one since $L^n = \underbrace{L\circ ... \circ L}_{n} : H^0(M) \to H^{2n}(M)$ is an isomorphism we must have that

$L: H^{2i}(M)\to H^{2(i+1)}(M)$

are all injective. Similarly for the odd Betti numbers.

Lefschetz decomposition

Another important theorem which puts further restrains on the cohomology structure of Kahler manifolds is the Lefschetz decomposition theorem. Let’s define

$P^k= \{ \alpha \in \Omega^k(M) ~|~ \Lambda(\alpha) = 0 \}$

Elements of $P^k$ are called primitive forms. We have an equivalent description of primitive forms given by

$P^k = ker(L^{n-k+1}: H^k \to H^{2n-k+2})$

By abuse of notation, we will also denote by $P^k$ the equivalence class of these elements in $H^k(M)$ . The statement of the theorem is as follows

Theorem 5 (Lefschetz decomposition) Let $(M, g, J)$ be a Kahler manifold (not necessarily compact) , then we have the following decomposition

$H^k(M) \simeq \bigoplus_{r\in \mathbb{N}} L^{r}(P^{k-2r}).$

Proof

Define $H=[L,\Lambda]$ . See [2] Proposition 1.2.26 for the proof that $H = \sum_{0\leq k\leq n} (k-n) \pi_k$ , where $\pi_k: \Omega^\star(M) \to \Omega^k(M)$ is the projection. Most of the complexity of the proof is hidden in this assertion, which is where we need the requirement that $M$ is Kahler. If one believes this, then it is easy to verify that we have the commutator relations

$[H, L] = 2L \qquad [H, \Lambda] = -2\Lambda \qquad [L, \Lambda] = H$

Recall that the Lie algebra of the special linear group of $2\times 2$ -matrices with coefficients in $\mathbb{C}$ (i.e. $sl_2(\mathbb{C})$ ) is spanned by

\begin{array}
& X = \begin{pmatrix} 0&0\\ 1&0
\end{pmatrix} &
Y = \begin{pmatrix} 0&1\\ 0&0
\end{pmatrix} &
Z = \begin{pmatrix} 1&0\\ 0&-1
\end{pmatrix}
\end{array}

with relations

$[Z, X] = 2X \qquad [Z, Y] = -2Y \qquad [X,Y] = Z$

hence we get a representation of $sl_2(\mathbb{C})$ by the assignemnt

$X\mapsto L\qquad Y\mapsto \Lambda \qquad Z\mapsto H$

Since $sl_2(\mathbb{C})$ is semisimple, and $H^*_\mathbb{C}(M)$ is a finite dimensional complex vector space, Weil’s theorem asserts that the representation is totally decomposable, that is, it can be written as a linear combination of irreducible representations.

Since $H$ is diagonal (and invertible), we may decompose $H^\star_\mathbb{C}(M)$ into eigenspaces of $H$ . If $Hv = \lambda v$ , then $H\Lambda v = (2 + \lambda )\Lambda v$ , so $\Lambda v$ is an eigenvector of $H$ with distinct eigenvalue hence must be linearly independent from $v$ . Continuing inductively, and using the fact that the vector space is finite dimensional, there must be a non-zero $v$ such that $\Lambda v = 0$ . Now let

$W_v = span\{v, Lv, L^2v, ... \}.$

It is straightforward to verify that $L(W)\subset W$ and $H(W)\subset W$ . It turns out that $\Lambda(W)\subset W$ hence $W$ is a subrepresentation which is clearly irreducible, being cyclic. Continuing the process we can decompose $H^\star_\mathbb{C}(M)$ into irreducible components spanned by $L^n(P^k)$ for $k, n \in \mathbb{N}$ . This concludes the proof.

The above decomposition is also induces a decomposition of

$\Omega^k_\mathbb{C} = \bigoplus_{r\in \mathbb{N}} L^{r}(P^{k-2r})$

which is orthogonal with respect to the $L^2$ -inner product.

[1] Bryant, Griffiths, Grossman – Exterior Differential Systems andEuler-Lagrange Partial Differential Equations.
[2] Huybrechts – Complex geometry
[3] Voisin – Hodge theory and complex algebraic geometry.

The hard Lefschetz Theorem and Lefschetz Decomposition.

The basic prerequisites

The symplectic case

The Kahler case

Lefschetz decomposition

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