**2. Preliminaries**

Let (A, *d*) denote the complex valued differential forms of an almost complex manifold (*<sup>M</sup>*, *J*). For any Hermitian metric, define the associated Hodge-star operator

$$\star : \mathcal{A}\_{\mathbf{x}}^{p,q} \to \mathcal{A}\_{\mathbf{x}}^{n-q,n-p} \text{ by } \omega \wedge \star \eta = \langle \omega, \eta \rangle \text{vol}\_{\star}$$

where *ω* is the fundamental (1, 1)-form, and vol = 1*n*!*ω<sup>n</sup>* ∈ A*<sup>n</sup>*,*<sup>n</sup>* is the volume form determined by the Hermitian metric. Note <sup>2</sup> = (−<sup>1</sup>)*<sup>k</sup>* on A*k*.

Define *d*∗ = − *d*, so that *d*<sup>∗</sup> = (−<sup>1</sup>)*<sup>k</sup>*+<sup>1</sup> *d* on A*k*. Similarly, consider the bidegree decomposition of the exterior differential

$$d = \mu + \bar{\theta} + \theta + \mu\_r$$

where the bidegree of each component is given by

$$|\bar{\mu}| = (-1, 2), |\bar{\partial}| = (0, 1), |\partial| = (1, 0) \text{ and } |\mu| = (2, -1).$$

We then let ¯ *δ*∗ = − *δ* for *δ* = *μ*¯, ¯ *∂*, *∂*, *μ* and we have the bidegree decomposition

$$d^\* = \mathfrak{h}^\* + \mathfrak{d}^\* + \mathfrak{d}^\* + \mu^\*.$$

where

$$|\vec{\mu}^\*| = (1, -2), |\overleftarrow{\theta}^\*| = (0, -1), |\overleftarrow{\theta}^\*| = (-1, 0) \text{ and } |\mu^\*| = (-2, 1).$$

Let *L* : A*p*,*q* → A*p*+1,*q*+<sup>1</sup> be the real (1, <sup>1</sup>)-operator given by *<sup>L</sup>*(*η*) = *ω* ∧ *η*. Let Λ = *L*∗ = <sup>−</sup>1*L*. Then Λ = *L* and *L* = Λ. Let *P<sup>k</sup>* = Ker Λ ∩ A*<sup>k</sup>* denote the primitive forms of total degree *k*.

It is well known that {*<sup>L</sup>*, Λ, [*L*, Λ]} defines a representation of *sl*(2, C) and induces the Lefschetz decomposition on forms:

**Lemma 1.** *We have*

$$\mathcal{A}^k = \bigoplus\_{r=0}^{k/2} L^r P^{k-2r} \nu$$

*and this direct sum decomposition respects the* (*p*, *q*) *bigrading.*

Let [*<sup>A</sup>*, *B*] = *AB* − (−<sup>1</sup>)|*A*||*B*|*BA* be the graded commutator, where |*A*| denotes the total degree of *A*. This defines a graded Poisson algebra

$$[A,BC] = [A,B]C + (-1)^{|A||B|}B[A,C]$$

The following is well known (e.g., [10] Corollary 1.2.28):

**Lemma 2.** *For all j* ≥ 0 *and α* ∈ A*<sup>k</sup>*

$$[L^j, \Lambda] \mathfrak{a} = j(k - n + j - 1)L^{j - 1}\mathfrak{a}.$$

By induction, and the fact that [*d*, *L*] and *L* commute, we have:

**Lemma 3.** *For all n* ≥ 1

$$[d, L^n] = n[d, L]L^{n-1}{}\_{\prime}$$

*and*

$$\star[d,L]\mathfrak{a} = (-1)^{k+1}[d^\*,\Lambda]\star\mathfrak{a} \quad \text{for } \mathfrak{a} \in \mathcal{A}^k.$$

Let I be the extension of *J* to all forms as an algebra map with respect to wedge product, so that <sup>I</sup>*p*,*<sup>q</sup>* acts on A*p*,*q* by multiplication by *ip*−*q*. Then <sup>I</sup>2*p*,*<sup>q</sup>* = (−<sup>1</sup>)*p*+*<sup>q</sup>* so that I−<sup>1</sup> *p*,*q* = (−<sup>1</sup>)*p*+*q*I*p*,*q*. Note that I and commute, and I and *Ln* commute for all *n* ≥ 0. The following is a direct calculation.

**Lemma 4.** *If an operator Tr*,*<sup>s</sup>* : A*p*,*q* → A*p*+*r*,*q*+*<sup>s</sup> has bidegree* (*<sup>r</sup>*,*<sup>s</sup>*)*, then*

$$
\mathbb{T}^{-1}\_{r+p,s+q} \circ T\_{r,s} \circ \mathbb{I}\_{p,q} = (-i)^{r-s} T\_{r,s}.
$$

The above result readily implies that

$$
\mathbb{T}^{-1} \circ d \circ \mathbb{T} = -i(\mu - \overleftarrow{\partial} + \overleftarrow{\partial} - \mu).
$$

Finally, the following is well known (e.g., [10] Proposition 1.2.31):

**Lemma 5.** *If M is an almost Hermitian manifold of dimension* 2*n, then for all j* ≥ 0 *and all α* ∈ *Pk,*

$$\star L^j \mathfrak{a} = (-1)^{\frac{k(k+1)}{2}} \frac{j!}{(n-k-j)!} L^{n-k-j} \, \mathbb{I} \mathfrak{a} \, \mathfrak{a}$$

## **3. Almost Hermitian Identities**

By the previous section, any differential form *η* can be written as *η* = *Ljα* for unique *j*, *k* ≥ 0 and *α* ∈ *Pk*. We now state the main result:

**Theorem 1.** *For any almost Hermitian manifold of dimension* 2*n, let α* ∈ *Pk, with dα written as*

$$d\mathfrak{a} = \mathfrak{a}\_0 + L\mathfrak{a}\_1 + L^2\mathfrak{a}\_2 + \cdots \mathfrak{,} \tag{2}$$

*for unique αr* ∈ *Pk*+1−2*r. Then, for all j* ≥ 0*,*

$$\begin{aligned} [\Lambda, d]L^j \mathfrak{a} - \star \mathbb{T}^{-1} d \, \mathbb{T} \star L^j \mathfrak{a} &= \frac{1}{j+1} \operatorname{\mathbb{T}}^{-1} [d^\*, \Lambda] \, \mathbb{T} L^{j+1} \mathfrak{a} \\ &+ j \Lambda [d, L] L^{j-1} \mathfrak{a} + j(j-1)(k - n + j - 1) [d, L] L^{j-2} \mathfrak{a} \\ &+ \sum\_{r=2}^{\infty} f\_{n, j, k}(r) L^{j+r-1} \mathfrak{a}\_{r, r} \end{aligned}$$

*where*

$$f\_{n,k,j}(r) = (r(n-k+r)-j) + (-1)^r \frac{j!(n-k-j+r)!}{(j+r-1)!(n-k-j)!}.$$

**Remark 1.** *In the almost Kähler case we have* [*d*<sup>∗</sup>, <sup>Λ</sup>]=[*d*, *L*] = 0*, and dα* = *α*0 + *Lα*1*, so we recover the identity*

$$[\Lambda, d] = \star \mathbb{I}^{-1} d \, \mathbb{I} \star\_{\star}$$

*as expected.*

**Proof.** The proof consists of several calculations using the lemmas in the previous section. Using [I, *L*] = 0, and I2 = (−<sup>1</sup>)*<sup>k</sup>* on A*k*, we have

$$\begin{aligned} \star \mathbb{I}^{-1} \, d\mathbb{I} \, \star \eta &= \star \mathbb{I}^{-1} \, d\mathbb{I} \left( (-1)^{\frac{k(k+1)}{2}} \frac{j!}{(n-k-j)!} L^{n-k-j} \, \mathbb{I} \alpha \right) \\ &= (-1)^{\frac{k(k+1)}{2} + k} \frac{j!}{(n-k-j)!} \, \star \mathbb{I}^{-1} \, dL^{n-k-j} \alpha. \end{aligned}$$

By Lemma 3 this is equal to

$$(-1)^{\frac{k(k+1)}{2} + k} \frac{j!}{(n-k-j)!} \star \mathbb{I}^{-1} L^{n-k-j} da + (-1)^{\frac{k(k+1)}{2} + k} \frac{j!}{(n-k-j-1)!} \star \mathbb{I}^{-1} [d, L] L^{n-k-j-1} a. \tag{3}$$

We first simplify each of these last two summands. By Equation (2), the fact that commutes with I, and Lemma 5 applied to *αr* ∈ *<sup>P</sup>k*+1−2*r*, the first summand of Equation (3) is equal to:

$$\begin{aligned} (-1)^{\frac{k(k+1)}{2}+k} \frac{j!}{(n-k-j)!} \, \star \mathbb{I}^{-1} \left( \sum\_{r=0}^{\infty} L^{n-k-j+r} a\_r \right) \\ &= (-1)^{\frac{k(k+1)}{2}+k} \frac{j!}{(n-k-j)!} \, \mathbb{I}^{-1} \left( \sum\_{r=0}^{\infty} (-1)^{\frac{(k+1-2r)(k-2r+2)}{2}} \frac{(n-k-j+r)!}{(j+r-1)!} L^{j+r-1} \, \|a\_r \right) \\ &= \sum\_{r=0}^{\infty} (-1)^{r+1} \frac{j! (n-k-j+r)!}{(j+r-1)! (n-k-j)!} L^{j+r-1} \, a\_r. \end{aligned}$$

For the second summand, we use the fact that for all *m* ≥ 0 and *β* ∈ A*k*,

$$\star L^m[d,L]\beta = \star [d,L]L^m\beta = (-1)^{k+1}[d^\*,\Lambda] \star L^m\beta.$$

So, the second summand in Equation (3) is equal to

$$\begin{aligned} (-1)^{\frac{k(k+1)}{2}+k} \frac{j!}{(n-k-j-1)!} \star \mathbb{I}^{-1} \left[ d, L \right] L^{n-k-j-1} a \\ &= (-1)^{\frac{k(k+1)}{2}+1} \frac{j!}{(n-k-j-1)!} \mathbb{I}^{-1} \left[ d^\*, \Lambda \right] \star L^{n-k-j-1} a \\ &= (-1)^{\frac{k(k+1)}{2}+1} \frac{j!}{(n-k-j-1)!} \mathbb{I}^{-1} \left[ d^\*, \Lambda \right] (-1)^{\frac{k(k+1)}{2}} \frac{(n-k-j-1)!}{(j+1)!} L^{j+1} \mathbb{I} a \\ &= \frac{-1}{j+1} \mathbb{I}^{-1} \left[ d^\*, \Lambda \right] \mathbb{I} L^{j+1} a, \end{aligned}$$

where in the second to last step we used Lemma 5.

In summary, we have

$$\star \star \mathbb{I}^{-1} d \, \mathbb{I} \star \eta = \sum\_{r=0}^{\infty} (-1)^{r+1} \frac{j! (n-k-j+r)!}{(j+r-1)! (n-k-j)!} L^{j+r-1} a\_r - \frac{1}{j+1} \mathbb{I}^{-1} [d^\*, \Lambda] \, \mathbb{I} L^{j+1} a. \tag{4}$$

We now compute [<sup>Λ</sup>, *d*]*η*, by first computing <sup>Λ</sup>*dLj<sup>α</sup>*, using that all *αr* are primitive. By Equation (2), Lemma 2, and Lemma 3, we have:

$$\begin{split} \Lambda \boldsymbol{L} \boldsymbol{L}^{j} \boldsymbol{a} &= \Lambda \boldsymbol{L}^{j} \boldsymbol{a} \boldsymbol{a} + \Lambda [\boldsymbol{d}, \boldsymbol{L}^{j}] \boldsymbol{a} \\ &= \Lambda \boldsymbol{L}^{j} \left( \sum\_{r=0}^{\infty} \boldsymbol{L}^{r} \boldsymbol{a}\_{r} \right) + j \Lambda [\boldsymbol{d}, \boldsymbol{L}] \boldsymbol{L}^{j-1} \boldsymbol{a} \\ &= \sum\_{r=0}^{\infty} \Lambda \boldsymbol{L}^{j+r} \boldsymbol{a}\_{r} + j \Lambda [\boldsymbol{d}, \boldsymbol{L}] \boldsymbol{L}^{j-1} \boldsymbol{a} \\ &= -\sum\_{r=0}^{\infty} (j+r)(k+1-2r-n+j+r-1) \boldsymbol{L}^{j+r-1} \boldsymbol{a}\_{r} + j \Lambda [\boldsymbol{d}, \boldsymbol{L}] \boldsymbol{L}^{j-1} \boldsymbol{a}. \end{split}$$

Next using, *α* is primitive, and Lemma 2 again, we have

$$\begin{aligned} d\Lambda \dot{L}^j \dot{a} &= -j(k - n + j - 1) dL^{j-1} a \\ &= -j(k - n + j - 1)L^{j-1} da - j(k - n + j - 1)(j - 1)[d, L]L^{j-2} a \\ &= -j(k - n + j - 1)\left(\sum\_{r=0}^{\infty} L^{j+r-1} a\_r\right) - j(k - n + j - 1)(j - 1)[d, L]L^{j-2} a. \end{aligned}$$

So,

$$[\Lambda, d]\eta = \sum\_{r=0}^{\infty} (r(n-k+r)-j)L^{j+r-1}a\_r + j\Lambda [d, L]L^{j-1}a + j(j-1)(k-n+j-1)[d, L]L^{j-2}a.$$

Using this last equation and combining with Equation (4) we obtain the desired result:

$$\begin{aligned} [\Lambda, d] \eta - \star \mathbb{T}^{-1} d \, \mathbb{T} \star \eta &= \frac{1}{j+1} \mathbb{T}^{-1} [d^\*, \Lambda] \, \mathbb{T} L^{j+1} \mathfrak{a} \\ &+ j \Lambda [d, L] L^{j-1} \mathfrak{a} + j(j-1)(k - n + j - 1) [d, L] L^{j-2} \mathfrak{a} \\ &+ \sum\_{r=0}^{\infty} f\_{n, k, j}(r) L^{j+r-1} \mathfrak{a}\_{r, r} \end{aligned}$$

where

$$f\_{n,k,j}(r) = (r(n-k+r)-j) + (-1)^r \frac{j!(n-k-j+r)!}{(j+r-1)!(n-k-j)!}.$$

It is a curious fact that *f*(0) = *f*(1) = 0, whereas for *r* ≥ 2, *f*(*r*) is in general non-zero.
