Pullbacks respect order of forms

Question

Consider a holomorphic map $f:M\to N$ between two complex manifolds. This gives rise to a pullback $f^*:\Omega^k(N)\to\Omega^k(M)$, that is, a complex linear map between the complex $k$-forms on $N$ and the complex $k$-forms on $M$. We furthermore have decompositions $$\Omega^k(N)=\bigoplus_{p,q\geq 0,\,p+q=k}\Omega^{p,q}(N),\quad \Omega^k(M)=\bigoplus_{p,q\geq 0,\,p+q=k}\Omega^{p,q}(M),$$ and my goal is to show that restricting $f^*$ gives rise to a map $f^*:\Omega^{p,q}(N)\to\Omega^{p,q}(M)$.

Working in local coordinates $z_1,\dots,z_n$ on $N$, we have that $dz_1,\dots,dz_n\in\Omega^{1,0}$ and $\overline z_1,\dots,\overline z_n\in\Omega^{0,1}$. One easily sees that the general statement follows from proving that

$$f^*(dz_i)\in\Omega^{1,0},\quad f^*\left(d\overline{z_i}\right)\in\Omega^{0,1},$$ or equivalently, that $$d(z_i\circ f)\in\Omega^{1,0},\quad d\left(\overline{z_i}\circ f\right)\in\Omega^{0,1}.$$

I am unsure how to finish the argument from here. Is the simplification described above even necessary?

Answer 1

Let $L:(V,I)\to (W,J)$ be a linear map of complex vector spaces. This means that $L\circ I=J\circ L$. Let $\eta\in\wedge^{1,0}W^*$. Then by definition $\eta(Jw)=i\eta(w)$. Now, $(L^*\eta)(Iv)=\eta(L\circ Iv)=\eta(J\circ Lv)=i\eta(Lv)$. Hence, $L^*\eta\in \wedge^{1,0}V^*$. I am using the characterisation of $\wedge^{1,0}V^*$ as the $+i$-eigenspace of $I$, whereas $\wedge^{0,1}V^*$ is the $-i$-eigenspace of $I$.

Now you apply this pointwise, to the differential of $f:M\to N$, and you get the result. Explicitly, let $(M,I)$ and $(N,J)$ be complex manifolds. Let $f:M\to N$ be holomorphic. This means that $df\circ I=J\circ df$. Let $\eta\in\Omega^{1,0}(N)$. This means that $\eta(Jv)=i\eta(v)$. Hence, $(f^*\eta)(Iw)=\eta(df(Iw))=\eta(J\circ df(w))=i\eta(df(w))=i(f^*\eta)(w)$, so $f^*\eta\in\Omega^{1,0}(M)$.

Answer 2

A complex $r$-form on a complex manifold $X$ is in general a local section of the vector bundle $$\bigwedge^r(T_a^*X_0)_{\mathbb C}=\bigoplus_{p+q=r}\left(\bigwedge^r(T_a^*X_0)_{\mathbb C}\right)^{p,q},$$ so, given $V\subset M$ and a chart $(U,z)$, it's an element of $A^r(V)=\bigoplus_{p+q=r}A^{p,q}(V)$ where $$A^{p,q}(V)=\{\omega\in A^r(V)|\omega=\sum_{|I|=p,|J|=q}\omega_{I,J}dz_I\wedge d\bar z_J\text{ on }U\cap V\text{ and }I,J\text{ are multi-indices}\}.$$ The differental on $A^{p,q}$ is given by a map $d=\partial+\bar\partial:A^{p,q}\to A^{p+1,q}\oplus A^{p,q+1}$ and we can also define projection maps $\pi^{p,q}:A^r\to A^{p,q}$, hence $\partial=\pi^{p+1,q}\circ d$ and $\bar \partial=d\circ \pi^{p,q+1}$.
If we now have an holomorphic map $f:X\to Y$ between complex manifolds one has:

1. $f^*A^{p,q}(Y)\subseteq A^{p,q}(X)$; If we have $A^{p,q}(V)\ni \omega=\sum_{I,J} \omega_{IJ}dz_I\wedge d\bar z_J\implies f^*\omega=\sum_{I,J}\omega_{IJ}\circ f \underbrace{d(z_I\circ f)}_{\in A^{p,0}}\wedge d\underbrace{(\bar z_J\circ f)}_{\in A^{0,q}}$
2. $f^*\circ\pi_Y^{p,q}=\pi^{p,q}_X\circ f^*$;
3. $f^*\circ\partial_Y=\partial_X\circ f^*$;