# Hermitian form on the manifold $\mathbb{C}^n$

I'm trying to verify why the Hermitian form on $$\mathbb{C}^n$$ is given by $$\langle u,v \rangle = \sum_{i}u_i\overline{v}_i = \sum_{i} dz_i \otimes d\overline{z}_i(u,v)$$ and the last equality is bothering me. If $$u,v \in T_p\mathbb{C}^n$$, I can't find a representation for the vectors with respect to the basis for $$T_p\mathbb{C}^n$$. I've found that $$T_p\mathbb{C}^n = T^{1,0}_p\mathbb{C}^n \oplus T^{0,1}_p\mathbb{C}^n$$ where $$T^{1,0}_p\mathbb{C}^n$$ and $$T^{0,1}_p\mathbb{C}^n$$ are the eigenspaces given by $$i$$ and $$-i$$ respectively. There is an issue here also since this decomposition is usually done when the manifold has an almost complex structure so can we assume that $$\mathbb{C}^n$$ naturally carries one?

A hermitian form is first of all a complex bilinear map $$h:V\times \bar{V}\to \mathbb{C}$$. Recall that if $$V$$ is a complex vector space, $$\bar{V}$$ is the complex vector space with complex structure given by $$-i$$. This means that $$h$$ can be thought of as a bilinear map $$V\times V\to \mathbb{C}$$ when $$V$$ is thought of as a real vector space. The fact that this map is complex bilinear from $$V\times\bar{V}$$ means that it is $$\mathbb{C}$$ linear in the first input and $$\mathbb{C}$$ anti linear in the second component.
The space of all bilinear maps $$V\times \bar{V}\to \mathbb{C}$$ is given by $$V^*\otimes_{\mathbb{C}} (\bar{V})^*$$ where $$V^*$$ is the space of complex linear maps $$V\to \mathbb{C}$$. If we fix a basis of $$V$$, $$e_1,\cdots, e_n$$ then we have a basis of $$V^*$$ given by the dual basis ($$e^i: V\to \mathbb{C}$$ is the unique collection of complex linear maps with $$\langle e^i, e_j\rangle=\delta^i_j$$). From the collection $$\{ e^i\}$$ we can yield a basis for $$(\bar{V})^*$$ by defining $$\bar{e}^i(v):= \overline{\langle e^i, v\rangle}$$. In the case where $$V=T_p\mathbb{C}^n\cong \mathbb{C}^n$$ with the standard complex structure and $$e_i=\frac{\partial}{\partial z^i}$$, the corresponding dual bases for $$V^*$$ and $$\bar{V}^*$$ are given by $$dz^i$$ and $$d\bar{z}^i$$ respectively. See that $$d\bar{z}^i\left(v^j\frac{\partial}{\partial z^j}\right)=\bar{v}^j$$ so that the expression you have written $$h=\sum_{i} dz^i\otimes d\bar{z}^i$$ yields the desired result.
It is very important to distinguish between the tangent space of a complex manifold and its complexified tangent space. If $$V$$ is a real vector space, we may form its complexification by tensoring with $$\mathbb{C}$$. $$V_{\mathbb{C}}:= V\otimes_{\mathbb{R}}\mathbb{C}$$. A choice of complex structure on $$V$$ (an endomorphism $$J: V\to V$$ with $$J^2=-1$$) allows us to split $$V_{\mathbb{C}}$$ into eigenspaces of $$J$$. The $$i$$ eigenspace of $$V_{\mathbb{C}}$$ is canonically isomorphic (as a complex vector space) to $$V$$ and the $$-i$$ eigenspace is canonically isomorphic (as a complex vector space) to $$\bar{V}$$, and hence $$V_{\mathbb{C}}\cong V\oplus \bar{V}$$. In differential geometry, if $$M$$ is given an almost complex structure we denote $$T_pM_{\mathbb{C}}= T^{1,0}_pM\oplus T^{0,1}_pM$$ and there are really two copies of $$T_pM$$ inside of $$T_pM_{\mathbb{C}}$$. Since these pointwise constructions are made without choice, one can apply them pointwise to the entirety of $$TM$$ for an almost complex manifold $$M$$.
• Thanks for the great answer! I would still be curious how do you express the elements in $T_pM_{\mathbb{C}}= T^{1,0}_pM\oplus T^{0,1}_pM$. Are they of the form $$v =\sum v^{1,0}_i \frac{\partial}{\partial z_i} + v^{0,1}_i \frac{\partial}{\partial \overline{z}_i}?$$ @j-v-gaiter Jun 12, 2023 at 7:12