Is the contraction of an harmonic form harmonic?

As I'm still a beginner in complex differential geometry, as soon as I tried reading an article I got stuck on what (I think) should be a minor detail. I hope someone can help me a bit.

Let $M$ be a complex manifold of complex dimension $m$, and let $g$ be a Kähler metric on $M$ with Kähler form $\omega$ and Ricci form $\rho$.

Let $h=h_{i\bar{j}}dz^{i}\wedge d\bar{z}^{j}$ be the harmonic part of $\rho$ in the Hodge decomposition for $\bigwedge\nolimits^{1,1}M$. Can I conclude that the function $\varphi=g^{i\bar{j}}h_{i\bar{j}}$ is also harmonic?

I think that the answer should be "yes", but I'm struggling to prove it. So far I've reasoned as follows: since $\varphi$ is a function, it is enough to show that $d\varphi=0$; moreover for functions we know that $d=\nabla$, if $\nabla$ is the Levi-Civita (or Chern) connection on $M$. Since $g$ is $\nabla$-parallel we have, for every $a=1,\dots,m,\bar{1},\dots,\bar{m}$ $$\nabla_a(g^{i\bar{j}}h_{i\bar{j}})=(\nabla_ag^{i\bar{j}})h_{i\bar{j}}+g^{i\bar{j}}(\nabla_ah_{i\bar{j}})=g^{i\bar{j}}(\nabla_ah_{i\bar{j}})$$ so to prove that $\varphi$ is harmonic it is enough to show that $h$ is parallel.

However, I fail to see why this should be true. Could someone please tell me what I'm missing? Any help is greatly appreciated.

edit: Since there are various different notions of "Laplacian" on a Kähler manifold I should have specified that the Laplacian I am talking about here is the Hodge Laplacian, $\Delta:=\bar{\partial}^*\bar{\partial}+\bar{\partial}\bar{\partial}^*$, where $\bar{\partial}^*$ is the adjoint of $\bar{\partial}$.

• A similar calculation should directly show $\Delta \varphi = g^{ij} \nabla_i \nabla_j \varphi = 0$ - just commute both derivatives past the metric. Establishing $d \varphi = 0$ will involve some kind of global argument (e.g. integration). – Anthony Carapetis Jun 22 '16 at 6:35
• @AnthonyCarapetis Thanks for the comment! Doing the calculation you suggest I get $\Delta\varphi=g^{k\bar{l}}g^{i\bar{j}}\nabla_k\nabla_{\bar{l}}h_{i\bar{j}}$, so I guess you are saying that, since $h$ is harmonic, $g^{k\bar{l}}\nabla_k\nabla_{\bar{l}}h_{i\bar{j}}=0$. But these look like the components of the "rough Laplacian" $\nabla^*\nabla h$, rather than those of $\Delta h$. Am I wrong? – Johnny Jun 22 '16 at 8:23
• Sorry, I'm not too familiar with complex geometry - which Laplacian are you talking about? If it's something like the Hodge laplacian then it should be related to the rough one by some Weitzenboeck-y formula that should help you. – Anthony Carapetis Jun 22 '16 at 10:27
• Johnny's formula for the (Hodge) Laplacian is correct. $g^{k\bar l}\nabla_k \nabla_{\bar l} f = \Delta f$ for a function $f$. To prove this, use normal coordinates at a point. – Quinton Westrich Jun 22 '16 at 15:01
• @AnthonyCarapetis I'm sorry, I probably should have specified in the question that I am using the Hodge laplacian. I will edit the question right away. Thanks for suggesting the Weitzenböck formulas. – Johnny Jun 22 '16 at 15:10

One of the various Kähler identities says that $$[\Lambda,\mathrm{d}]=-\delta^{\mathrm{c}}$$ where
1. $\Lambda$ is the formal adjoint of the Lefschetz operator, which is defined by $L(\eta)=\omega\wedge\eta$;
2. $[\Lambda,\mathrm{d}]$ is the usual commutator of the two operators $\Lambda$ and $\mathrm{d}$;
3. $\delta^{\mathrm{c}}$ is the formal adjoint of the twisted differential $\mathrm{d}^{\mathrm{c}}=\mathrm{i}(\bar{\partial}-\partial)$.
We can use this identity to solve my question. Indeed, if $\alpha=\alpha_{i\bar{j}}\mathrm{d}z^i\wedge\mathrm{d}\bar{z}^j$ is a $(1,1)$-form then it is quite easy to compute $$\Lambda(\alpha)=g^{i\bar{j}}\alpha_{i\bar{j}}.$$ In particular if $\alpha$ is closed $$[\Lambda,\mathrm{d}](\alpha)=-\mathrm{d}\left(\Lambda(\alpha)\right)=-\mathrm{d}\left(g^{i\bar{j}}\alpha_{i\bar{j}}\right)$$ so for a closed $(1,1)$-form $\alpha$ the Kähler identity recalled above says that $\alpha$ is harmonic if and only if $g^{i\bar{j}}\alpha_{i\bar{j}}$ is a constant.