Differential Forms on a complex manifold

Question

Given a complex manifold $M$, we can complexify its wedge product of tangent cotangent bundles. Consider $\wedge^{k}T^{\ast}M \otimes \mathbb C$, and we know that $T^{\ast}M \otimes \mathbb C$ decomposes into $T^{\ast}M^{(1,0)} \oplus T^{\ast}M^{(0,1)}$.

Then by the commutativity of wedge product and direcsum, we conclude that $\wedge^{k}T^{\ast}M = \bigoplus_{p+q = k} \wedge^{p}T^{\ast}M^{(1,0)} \otimes \wedge^{q}T^{\ast}M^{(0,1)}$.

A $(p,q)$ form is a section of $\wedge^{p}T^{\ast}M^{(1,0)} \otimes \wedge^{q}T^{\ast}M^{(0,1)}$, but why does a typical element of (p,q) form looks like $dz_{k_1} \wedge dz_{k_p} \ldots \wedge d\bar{z_{t_1}} \ldots \wedge d \bar{z_{t_1}}$ in local coordinates? Where does the tensor product go?

Answer 1

You wrote:$$\Lambda^{k}T^{\ast}M = \bigoplus_{p+q = k} \Lambda^{p}{(T^{1,0})}^*M \otimes \Lambda^{q}{(T^{0,1})}^*M$$

This is wrong, the correct identity is $$(\Lambda^{k}T^{\ast}M)\otimes \mathbb{C} = \bigoplus_{p+q = k} \Lambda^{p}{(T^{1,0})}^*M \wedge \Lambda^{q}{(T^{0,1})}^*M$$

You derive this identity like so: $$\begin{align} (\Lambda^{k}T^*M) \otimes \mathbb{C} &= \Lambda^{k} (T^*M \otimes \mathbb{C})\\ & = \Lambda^{k} ({(T^{1,0})}^*M \oplus {(T^{0,1})}^*M)\\ &= \bigoplus_{p+q = k} \Lambda^{p}{(T^{1,0})}^*M \wedge \Lambda^{q}{(T^{0,1})}^*M \end{align} $$ The tensor product didn't go anywhere because it wasn't there to begin with: it was already gone as soon as you wrote $T^*M \otimes \mathbb{C} = {(T^{1,0})}^*M \oplus {(T^{0,1})}^*M$.

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EDIT: As TedShifrin points out in the comment, most texts write a tensor product (as you did) instead of a wedge there. I suppose that in general $V \wedge W$ does not make sense when $V$ and $W$ are different vector spaces but $V \otimes W$ does, which probably explains the choice of "most texts". But they're wrong: $dz \otimes d\bar{z}$ is not the same as $dz \wedge d\bar{z}$. The point is that $dz \wedge d\bar{z}$ does make sense, because $dz$ and $d\bar{z}$ are both one-forms (complexified). More generally, $\Lambda^{p}{(T^{1,0})}^*M$ and $\Lambda^{q}{(T^{0,1})}^*M$ are not the same vector space but they are both subspaces of the complexified exterior algebra $(\Lambda T^*M) \otimes \mathbb{C}$, so their wedge product is a well-defined subspace of $(\Lambda T^*M) \otimes \mathbb{C}$.

Answer 2

From the identity $T^{\ast}M\otimes \mathbb{C} = T^{\ast}M^{(1,0)} \oplus T^{\ast}M^{(0,1)}$ and with fixed local frames $\{dz_1 , \dots, dz_n \}$ and $\{d\bar z_1 , \dots, d\bar z_n \}$ for $T^{\ast}M^{(1,0)}$ and $T^{\ast}M^{(0,1)}$ respectively, we can construct the bundle $$ \bigwedge^\ast (T^{\ast}M\otimes \mathbb{C}) $$ wich is has a frame given by the wedge products of $\{dz_1 , \dots, dz_n, d\bar z_1 , \dots, d\bar z_n \}$.

Here no tensor product will apear when you pick an element of type $(p,q)$.

The formula for the exterior power of a direct sum $$ \wedge^{k}(T^{\ast}M\otimes \mathbb{C}) \, {\Huge \simeq} \bigoplus_{p+q = k} \wedge^{p}T^{\ast}M^{(1,0)} \otimes \wedge^{q}T^{\ast}M^{(0,1)} $$ is correct. The missing thing is an identification. Here a $(p,q)$ form is a section $$ \eta =\sum\limits_{|I|=p, |J|=q} a_{IJ}dz_I\otimes d\bar z_{J} \in \Gamma( \wedge^{p}T^{\ast}M^{(1,0)} \otimes \wedge^{q}T^{\ast}M^{(0,1)}) $$ The identification is $$ \eta \longmapsto \sum\limits_{|I|=p, |J|=q} a_{IJ}dz_I \wedge d\bar z_{J} \in \Gamma( \wedge^{p+q}(T^{\ast}M\otimes \mathbb{C})). $$ Note that this is well defined and bijective (onto its image, $\wedge^{p,q}(T^{\ast}M\otimes \mathbb{C})$) as you can arrange the terms so that the $dz_j$ come ahead of the $d\bar z_i$.