I'm reading Griffiths and Harris and just want to check I'm interpreting a passage correctly.
Let $M$ be a $n$ dimensional complex manifold.
We define $T_p^\mathbb{R}M$ as the tangent space of the underlying smooth manifold, and define $T_pM:=T_p^\mathbb{R}M\otimes \mathbb{C}$. We then define $T_p'M=\langle\frac{\partial}{\partial z_i} \rangle_{i=1}^n$ where $\frac{\partial}{\partial z_i} =\frac{1}{2}\left(\frac{\partial}{\partial x_i} -i\frac{\partial}{\partial y_i}\right) $. When then have $T_pM=T_p'M\oplus\overline{T_p'M}$.
Now we have a natural isomorphism of real vector spaces $T_p^\mathbb{R}M\to T_p'M$ defined by the composition
$$T_p^\mathbb{R}M\hookrightarrow T_pM\stackrel{\pi}{\to}T_p'M$$
I see that one could define this, but it seems a bit strange to consider this map when we consider what it does to our elements:
$$\frac{\partial}{\partial x_i}\mapsto \frac{\partial}{\partial x_i}\otimes 1=\frac{\partial}{\partial z_i}+\frac{\partial}{\partial\bar{z_i}}\stackrel{\pi}{\to}\frac{\partial}{\partial z_i}$$
$$\frac{\partial}{\partial y_i}\mapsto \frac{\partial}{\partial y_i}\otimes 1=i\frac{\partial}{\partial z_i}-i\frac{\partial}{\partial \bar{z_i}}\stackrel{\pi}{\to}i\frac{\partial}{\partial z_i}$$
So under this $\mathbb{R}$ linear isomorphism we map $\frac{\partial}{\partial x_i}$ to $\frac{\partial}{\partial z_i}$ and $\frac{\partial}{\partial y_i}$ to $i\frac{\partial}{\partial z_i}$. Of course we can do this, but why is this natural? I get that is it natural in the sense that we get this map essentially from the definition of complexification of a vector space.
But I mean moreso why it is natural from a more analytic point of view when we think about our tangent vectors as derivations. Or for example when we have a Hermitian inner product $g$ on $T_p'M$, we define the inner product on $T_p^\mathbb{R}M$ as $Re(g)$, where we also use the identification above, so for example
$$Re(g)(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j})=g(\frac{\partial }{\partial z_i},i\frac{\partial }{\partial z_j})+\overline{g(\frac{\partial }{\partial z_i},i\frac{\partial }{\partial z_j})}$$
I can see how this definition is correct in the sense of giving us an inner product on $T_p^\mathbb{R}M$, but I feel that I'm missing some understanding as to why this is a definition that makes sense, which in turn makes me suspect that there's something I'm not getting.
So my question is whether what I've written above is correct in the sense that this is what Griffiths and Harris meant. And then if it is, if someone could maybe motivate some of this.
Best Answer
Reviving an old post here.
As I am sure Griffiths and Harris mention in the book, the bundle $T'M$ (or $T^{1,0}M$ in other books) is supposed to be the holomorphic tangent bundle. One way to specify what that is to say that it should act on holomorphic functions, in other words it should consist of derivations mapping holomorphic functions back to holomorphic functions.
Now note that given a holomorphic function $f$ on an open subset of $\mathbb{C}^n$, we have $$ \frac{\partial{f}}{\partial{x_j}}=\frac{\partial{f}}{\partial{z_j}}, \quad \frac{\partial{f}}{\partial{y_j}}=i\frac{\partial{f}}{\partial{z_j}}$$where $\frac{\partial{}}{\partial z_j}$ can either be given by the formula you mention in your post, or simply as the partial differentiation symbol in the $j$-th direction (which makes sence, because the limit in the partial derivative exists by the virtue of $f$ being holomorphic). In this sense it is a pretty natural identification.