# Definition of complex differential forms of bidegree $(p,q)$

Let $M$ be a complex manifold of dimension $n$. It means that $M$ is a real smooth smooth manifold of dimension $2n$. Suppose that the real tangent bundle of $M$ has a local basis: $$\left\{\frac{\partial }{\partial x_1},\ldots,\frac{\partial}{\partial x_n},\frac{\partial }{\partial y_1},\ldots,\frac{\partial}{\partial y_n}\right\}$$ and the real cotangent bundle has local basis $$\left\{dx_1,\ldots,dx_n,dy_1,\ldots dy_n\right\}$$

Then we put $$\frac{\partial}{\partial z_j}:=\frac{1}{2}\left(\frac{\partial }{\partial x_j}-i \frac{\partial }{\partial y_j}\right)$$ $$\frac{\partial}{\partial \bar z_j}:=\frac{1}{2}\left(\frac{\partial }{\partial x_j}+i \frac{\partial }{\partial y_j}\right)$$ $$dz_j:=dx_j+idy_j$$ $$d\bar z_j:=dx_j-idy_j$$

Now consider the complexified cotangent bundle $(T^\ast M)_{\mathbb C}$, it has a decomposition: $$(T^\ast M)_\mathbb C:=T^\ast M^{1,0}\oplus T^\ast M^{0.1}$$

where

$$T^\ast M^{(1,0)}=\left<dz_j:j=1,\ldots n\right>$$

$$T^\ast M^{(1,0)}=\left<d\bar z_j:j=1,\ldots n\right>\,.$$

At this point one defines the algebra of differential $(p,q)$-forms on $M$ as: $$\bigwedge^{p,q}M:=\bigwedge^{p}T^\ast M^{1,0}\otimes \bigwedge^{q}T^\ast M^{0,1}$$

So locally any $(p,q)$-differential form can be written as

$$\omega=\sum_{i_1<\ldots<i_p} \alpha_{i_1,\ldots,i_p}dz_{i_1}\wedge\ldots\wedge dz_{i_p}\otimes \sum_{j_1<\ldots<j_q}\beta_{i_1,\ldots,i_q}d\bar z_{j_1}\wedge\ldots\wedge d\bar z_{j_q}\,.$$

So why in every textbook a $(p,q)$-dffierential form is written simply as: $$\omega=\sum \alpha_{i_1,\ldots,i_p}dz_{i_1}\wedge\ldots\wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge\ldots\wedge d\bar z_{j_q}\;\;?$$ Where is the tensor product?

## 1 Answer

If $V$ and $W$ are complex vector spaces, then you have canonical isomorphisms $$\bigoplus_{j+k=m} \bigwedge^j V \otimes \bigwedge^k W \to \bigwedge^m V \oplus W$$ given on each direct summand $\wedge^j V \otimes \wedge^k W$ with $j+k=m$ by $$(v_1 \wedge \cdots \wedge v_j) \otimes (w_1 \wedge \cdots w_k) \mapsto v_1 \wedge \cdots \wedge v_j \wedge w_1 \wedge \cdots \wedge w_k.$$ Applied fibrewise in your case, this gives isomorphisms of complex vector bundles $$\bigoplus_{p+q=m} \bigwedge^p T^\ast M^{(1,0)} \otimes \bigwedge^q T^\ast M^{(0,1)} \to \bigwedge^m T^\ast M_{\mathbb{C}};$$ it's completely standard (and generally convenient) to identify $$\bigwedge^{p,q} T^\ast M := \bigwedge^p T^\ast M^{(1,0)} \otimes \bigwedge^q T^\ast M^{(0,1)}$$ with its image in $\wedge^{p+q}(TM)_{\mathbb{C}}$ under this map, which is what you're seeing whenever textbooks leave out the tensor product signs you're asking about.

• Thank you! I was aware of the isomorphism as a "given result". I didn't know the map on each summand... Now it is all clear. – manifold Sep 21 '16 at 21:51
• Note that you have to skew-symmetrize the tensor product to get something that's totally skew-symmetric. The tensor product itself is not. – Ted Shifrin Sep 23 '16 at 1:01
• @ted shifrin could you expand a little bit your comment? – manifold Sep 24 '16 at 19:38
• @manifold: For example, consider a form of type $(1,1)$ on $\Bbb C$, for simplicity. You wanted to write it as $\alpha dz\otimes d\bar z$; although this is a $2$-tensor, it is not a $2$-form, because it is not anti-symmetric. You either have to skew-symmetrize, writing instead $\alpha (dz\otimes d\bar z + d\bar z\otimes dz)$ or have to pass to the quotient of the tensor algebra that gives you the exterior algebra, "identifying" $dz\otimes d\bar z$ to $dz\wedge d\bar z$. But, again, as it stands, it is not a $2$-form. As Branimir wrote, this identification is "taken for granted." – Ted Shifrin Sep 24 '16 at 19:51
• ok so basically if we have $(v_1\wedge\ldots\wedge v_p)\otimes (w_1\wedge\ldots\wedge w_q)\in \bigwedge^p V\otimes\bigwedge^q W$ I pass to the quotient, and I construct the usual vector space $\bigwedge^p V\wedge\bigwedge^q W$. – manifold Sep 24 '16 at 19:55