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Let $(M,J)$ be a complex manifold and $d^c := - J d J$, where $J$ acts on forms by acting on their arguments and does nothing to functions. I have seen the claim that $d^c$ is proportional to $\partial - \bar \partial$. I can't find this again anywhere and I'm wondering why it holds.

I thought of using that if $Q$ denotes complex conjugation, then $Q \bar\partial Q = \partial$. If this would hold for $J$ instead of $Q$, using that $\partial^2 = \bar\partial^2 = \bar\partial \partial + \partial \bar\partial = 0 $ it would follow immediately that $ d^c = \bar\partial - \partial$. But I don't see why $J \bar\partial J = \partial$ - I think this might be false.

rosecabbage
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1 Answers1

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As you said, $J$ acts on a form by acting on the inputs. If $\alpha$ is a $(p, q)$-form, then it takes $p + q$ vector fields as inputs, and by linearity we can reduce to considering the case where each vector field is either a vector field of type $(1,0)$ or $(0,1)$. In this restricted case, $\alpha$ can only evaluate to something non-zero when there are $p$ vector fields of type $(1,0)$ and $q$ of type $(0, 1)$. If $X$ is a vector field of type $(1, 0)$, then $JX = iX$, while if $Y$ is a vector field of type $(0, 1)$, we have $JY = -iY$. Therefore, if $X_1, \dots, X_p$ are vector fields of type $(1, 0)$ and $Y_1, \dots, Y_q$ are vector fields of type $(0, 1)$, then

\begin{align*} (J\alpha)(X_1, \dots, X_p, Y_1, \dots, Y_q) &= \alpha(JX_1, \dots, JX_p, JY_1, \dots, JY_q)\\ &= \alpha(iX_1, \dots, iX_p, -iY_1, \dots, -iY_q)\\ &= i^p(-i)^q\alpha(X_1, \dots, X_p, Y_1, \dots, Y_q)\\ &= i^{p-q}\alpha(X_1, \dots, X_p, Y_1, \dots, Y_q). \end{align*}

Therefore $J\alpha = i^{p-q}\alpha$ so

\begin{align*} -JdJ\alpha &= -Jd(i^{p-q}\alpha)\\ &= -i^{p-q}Jd\alpha\\ &= -i^{p-q}J(\partial\alpha + \bar{\partial}\alpha)\\ &= -i^{p-q}(J\partial\alpha + J\bar{\partial}\alpha)\\ &= -i^{p-q}(i^{(p+1)-q}\partial\alpha + i^{p-(q+1)}\bar{\partial}\alpha)\\ &= -i\partial\alpha + i\bar{\partial}\alpha\\ &= -i(\partial - \bar{\partial})\alpha. \end{align*}

That is $d^c = -JdJ = -i(\partial - \bar{\partial})$.

Michael Albanese
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