# tangent vector on a complex manifold

Let $x_1,x_2, \ldots, x_n$ be local coordinates on a manifolds $M$. One can interpret $\frac{\partial}{\partial x_i}(p)$ as a tangent vetor to a curve with constant $x_j$ (where $j \neq i$). What is the interpretation of $\frac{\partial}{\partial z_i}(p)$ on a complex manifold in local coordinates? I'll be glad for any references.

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I feel like the standard texts are a little brief explaining this. I was confused about it for a really long time. This is going to be a fairly long answer.

First up, lets start with a smooth manifold $M$ with dimension $n$. Let $\mathscr{S}$ be the sheaf of smooth functions on $M$. More concretely, $\mathscr{S}(U)$ is the $\mathbb{R}$-algebra of smooth functions $U \to \mathbb{R}$. We are going to define another sheaf on $M$. Let $\mathscr{C}$ be the sheaf of complex valued smooth functions on $M$. More concretely, $\mathscr{C}(U)$ is the $\mathbb{C}$-algebra of smooth functions $U \to \mathbb{C}$, where by smooth, I mean that each component function is smooth. It is a standard result from differential geometry that vector fields on $U \subseteq M$ are the same as $\mathbb{R}$-linear maps $D : \mathscr{S}(U) \to \mathscr{S}(U)$ satisfying $D(fg) = f D(g) + g D(f)$. These are called $\mathbb{R}$-linear derivations.

It is natural to ask the following question: Are the $\mathbb{C}$-linear derivations $\mathscr{C}(U) \to \mathscr{C}(U)$ smooth sections of some vector bundle over $U$? The answer is yes! Here is the idea (to make this rigorous quite a lot more work is required. Inpaticular, at some point you need to muck around with partitions of unity). Firstly, if $D: \mathscr{S}(U) \to \mathscr{S}(U)$ is an $\mathbb{R}$-linear derivation, then $D(\alpha + i \beta) = D \alpha + i D \beta$ defines a $\mathbb{C}$-linear derivation $\mathscr{C}(U) \to \mathscr{C}(U)$. Now assume that $U$ is small, $x_1, \dots, x_n$ are coordinates on $U$ and $D : \mathscr{C}(U) \to \mathscr{C}(U)$ is a $\mathbb{C}$-derivation. Taylor's theorem tells us that $$D (\alpha + \beta i) = \sum c_j \frac{\partial (\alpha + \beta i)}{\partial x_j} \quad c_j \in \mathscr{C}(U).$$ This is exactly a smooth section of $$TM \otimes_{\mathbb R} \mathbb{C} = \coprod_{p \in M} T_p \otimes_{\mathbb R} \mathbb{C}$$ Before we start talking about complex manifolds, notice that we still have the non degenerate pairing $$TM \otimes_{\mathbb{R}} \mathbb{C} \times TM^* \otimes_{\mathbb{R}} \mathbb{C} \to \mathscr{C}$$ defined by $( \partial / \partial x_i , dx_j) \mapsto \delta_{ij}$. This allows us to define the differential of $f \in \mathscr{C}(U)$ by $(\partial / \partial x_i, df) = \partial f / \partial x_i$.

OK, Now let $X$ be a complex manifold with complex dimension $n$. Let $z_j = x_j + i y_j$ be holomorphic coordinates on $U \subseteq X$. Then both $z_j$ and $\overline{z_j}$ are in $\mathscr{C}(U)$. Inpaticular, we have that $dz_j = d x_j + i d y_j$ and $d \overline{z_j} = dx_j - i dy_j$. From the equations $2 dx_j = dz_j + d \overline{z_j}$ and $2i dy_j = dz_j - d \overline{z_j}$, it follows that the sections of $TM^* \otimes_{\mathbb{R}} \mathbb{C}$ over $U$ are given by $$\sum_j f_j dz_j + g_j d \overline{z_j} \quad f_j,g_j \in \mathscr{C}(U)$$ Under the non-degenerate pairing, the dual basis to $dz_j, d \overline{z_j}$ is given by $$\frac{\partial}{\partial z_j} = 1/2 \left( \frac{\partial}{\partial x_j} - i \frac{\partial}{\partial y_j} \right)$$ $$\frac{\partial}{\partial \overline{z_j}} = 1/2 \left( \frac{\partial}{\partial x_j} + i \frac{\partial}{\partial y_j} \right)$$ This is how I think about $\partial / \partial z_j$.

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DBr's answer is very good. Here is another way to think about it:

In a smooth manifold, as the OP says, one way to think of $\dfrac{\partial}{\partial {x_i}}$ at $p$ is as taking the derivative along the curve $x_j =$ constant for $j \neq i$. In other words, we get a map $\gamma$ from an interval $I$ around $0$ to the manifold $M$, taking $0$ to $p$, and taking $t$ to $(0,\ldots, t, \ldots, 0)$ (assuming coords. are chosen so that $p$ is at the origin), where $t$ appears in the $i$th place and all other coords. are $0$. We can compose a function $f$ with $\gamma$, and then differentiate $f\circ \gamma$ w.r.t. $t$. This gives the value of $\dfrac{\partial f}{\partial x_i}$ at $p$.

Similarly, on a complex manifold with local coords. $z_i$, we may define a map $\gamma$ in exactly the same way, but now taking a n.h. $U$ of $0$ in the complex plane to $M$ (so $t$ is now a complex variable rather than a real one). If $f$ is a holomorphic function on $M$, then the composite $f\circ \gamma$ is a holomorphic function on $x$, and we may differentiate it w.r.t. $t$. This gives $\dfrac{\partial f}{\partial z_i}$ at $p$.

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