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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?

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To start, let's recall the definition of a hermitian form on a complex vector space.

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 $M$.

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  • $\begingroup$ 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 $\endgroup$
    – Louie
    Commented Jun 12, 2023 at 7:12

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