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Let $X$ be a complex manifold with a Hermitian metric $h$. Locally, we can consider the space $T^{1,0}X$ with basis $\displaystyle\left\{ \frac{\partial}{\partial z_j} \right\}$ and the space $T^{0,1}X$ with basis $\displaystyle\left\{ \frac{\partial}{\partial \overline{z_j}} \right\}$. From what I understand, for any indexes $k,l=1 \ldots n$, we have that $\displaystyle\frac{\partial}{\partial z_k}$ and $\displaystyle\frac{\partial}{\partial \overline{z_k}}$ are orthogonal with respect to $h$.

We can write $$h = \sum_{i,j=1}^n h_{ij} dz_i \otimes d\overline{z_j}.$$

Now note that $$h\left(\frac{\partial}{\partial z_k}, \frac{\partial}{\partial \overline{z_l}}\right) = \sum_{i,j=1}^n h_{ij} dz_i \otimes d\overline{z_j}\left(\frac{\partial}{\partial z_k}, \frac{\partial}{\partial \overline{z_l}}\right) = $$

$$ = h_{kl} dz_k \otimes d\overline{z_l}\left(\frac{\partial}{\partial z_k}, \frac{\partial}{\partial \overline{z_l}}\right) = h_{kl} dz_k \left(\frac{\partial}{\partial z_k} \right) d\overline{z_l} \left(\frac{\partial}{\partial \overline{z_l}}\right) = h_{kl}.$$

The answer is not zero, as I thought it should be. Worse yet, if one try define a norm using $\| \cdot \| = h(\cdot, \cdot)$, then $\displaystyle\left\|\frac{\partial}{\partial z_k}\right\| = \left\|\frac{\partial}{\partial \overline{z_k}}\right\| = 0$, which makes no sense. I don't know what is incorrect in my understanding. I need some clarification on this topic.

Thank you very much for your help.

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2 Answers 2

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Your question is essentially one of Hermitian linear algebra. The following little essay is essentially lifted out of Huybrechts' book "Complex Geometry." This is an elaboration on Ted Shifrin's answer, which I'm mostly writing for my own clarification (sorry).


Let $(V, J, g)$ be a real vector space with a complex structure $J$ and a positive-definite inner product $g$ that are compatible. By "compatible," we mean that $g(Jv, Jw) = g(v,w)$ for all $v,w \in V$.

Define an $\mathbb{R}$-bilinear form $\omega \colon V \times V \to \mathbb{R}$ by $\omega(v,w) := g(Jv, w)$. One can check that $\omega$ is alternating and non-degenerate, hence is a symplectic $2$-form.

Define an $\mathbb{R}$-bilinear form $h \colon V \times V \to \mathbb{C}$ by $h := g - i\omega$. One can check that:

  • $h(Jv,w) = i\,h(v,w)$.
  • $h(w,v) = \overline{h(v,w)}$.
  • $h(v,v) = g(v,v) > 0$ for $v \in V$, $v \neq 0$.

In other words, $h$ is a positive-definite hermitian form on $(V,J)$. Thus far, everything has happened on the vector space $V$.


Let $V^{\mathbb{C}}$ be the complexification of $V$. Let $J \colon V^{\mathbb{C}} \to V^{\mathbb{C}}$ be the $\mathbb{C}$-linear extension of $J \colon V \to V$, meaning that $J(iv) = iJ(v)$ for all $v \in V^{\mathbb{C}}$. Thus, there two different complex structures on $V^{\mathbb{C}}$, namely $J$ and $i$.

Since $J^2 = -\text{Id}$, the map $J \colon V^{\mathbb{C}} \to V^{\mathbb{C}}$ has eigenvalues $\pm i$. We denote the eigenspaces by \begin{align*} V^{1,0} & = \{v \in V^{\mathbb{C}} \colon J(v) = iv\} \\ V^{0,1} & = \{v \in V^{\mathbb{C}} \colon J(v) = -iv\}. \end{align*} One can see that $V^{\mathbb{C}} = V^{1,0} \oplus V^{0,1}$ by writing $v \in V^{\mathbb{C}}$ as $$v = \textstyle \frac{1}{2}(v - iJ(v)) + \frac{1}{2}(v + iJ(v)).$$ One can also check that for the following two compositions \begin{align*} \pi^{1,0} \colon (V,J) \hookrightarrow V^{\mathbb{C}} & \to (V^{1,0}, i) & \pi^{0,1} \colon (V,J) \hookrightarrow V^{\mathbb{C}} & \to (V^{0,1}, i) \\ \pi^{1,0}(v) & = \textstyle \frac{1}{2}(v - iJ(v)) & \pi^{0,1}(v) &= \textstyle \frac{1}{2}(v + iJ(v)), \end{align*} the map $\pi^{1,0}$ is a $\mathbb{C}$-linear isomorphism, while $\pi^{0,1}$ is a $\mathbb{C}$-antilinear isomorphism.


There are two ways one might like to extend the $\mathbb{R}$-bilinear form $g \colon V \times V \to \mathbb{R}$ to the complexification $V^{\mathbb{C}}$.

One way is to extend $g$ to a $\mathbb{C}$-bilinear form $g^{\text{bil}} \colon V^{\mathbb{C}} \times V^{\mathbb{C}} \to \mathbb{C}$ via (for $v,w \in V$ and $\alpha, \beta \in \mathbb{C}$) $$g^{\text{bil}}(\alpha v, \beta w) := \alpha \beta\, g(v,w)$$ The other way is to extend $g$ to a sesquilinear form $g^{\text{ses}} \colon V^{\mathbb{C}} \times V^{\mathbb{C}} \to \mathbb{C}$ via $$g^{\text{ses}}(\alpha v, \beta w) := \alpha \overline{\beta}\, g(v,w).$$ One can check that for $v,w \in V^{\mathbb{C}}$:

  • $g^{\text{ses}}(iv, w) = i\,g^{\text{ses}}(v,w)$.
  • $g^{\text{ses}}(w,v) = \overline{g^{\text{ses}}(v,w)}$
  • $g^{\text{ses}}(v,v) > 0$ for $v \in V^{\mathbb{C}}$, $v \neq 0$.

In other words, $g^{\text{ses}}$ is a positive-definite Hermitian form on $(V^{\mathbb{C}}, i)$.


Note that $g^{\text{bil}}$ and $g^{\text{ses}}$ are defined on $V^{\mathbb{C}}$, whereas $h$ is defined on $(V,J)$. The upshot is now:

Fact: Use the setup above.

(a) For $v,w \in V$, we have: \begin{align*} g^{\text{ses}}(\pi^{1,0}(v), \pi^{1,0}(w)) & = \textstyle\frac{1}{2}h(v,w) \tag{1} \\ g^{\text{ses}}(\pi^{1,0}(v), \pi^{0,1}(w)) & = 0 \tag{2} \\ g^{\text{ses}}(\pi^{0,1}(v), \pi^{0,1}(w)) & = \textstyle\frac{1}{2}\overline{h(v,w)}. \tag{3} \end{align*} Equation (2) says that the decomposition $V^{\mathbb{C}} = V^{1,0} \oplus V^{0,1}$ is $g^{\text{ses}}$-orthogonal. Equation (1) says that after the isomorphism $\pi^{1,0} \colon (V,J) \xrightarrow{\cong} (V^{1,0}, i)$, we have $\left.g^{\text{ses}}\right|_{V^{1,0}} = \frac{1}{2}h$.

(b) For $v,w \in V$, we have: \begin{align*} g^{\text{bil}}(\pi^{1,0}(v), \pi^{1,0}(w)) & = 0 \tag{4} \\ g^{\text{bil}}(\pi^{1,0}(v), \pi^{0,1}(w)) & = \textstyle\frac{1}{2}h(v,w) \tag{5} \\ g^{\text{bil}}(\pi^{0,1}(v), \pi^{0,1}(w)) & = 0. \tag{6} \end{align*} Equation (5) says that the $\mathbb{C}$-bilinear form $g^{\text{bil}}$ restricts to $V^{1,0} \times V^{0,1} \to \mathbb{C}$ to be the sesquilinear form $h$, after applying the isomorphisms $\pi^{1,0}$ and $\pi^{0,1}$.

The proof is a calculation using the definitions.

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  • $\begingroup$ Thank you for your detailed answer. I have some question. $g^\mathbb{C}$ is really $\mathbb{R}$-linear? Because I thought it would be $\mathbb{C}$-linear. At last, by your conclusion looks like I should consider $h$ as a bilinear map from $T^{1,0}X \times T^{1,0}X$ to $\mathbb{C}$, not $T^{1,0}X \times T^{0,1}X$ to $\mathbb{C}$. Is that correct? $\endgroup$
    – Integral
    Commented Dec 7, 2017 at 1:03
  • $\begingroup$ My maps $g^{\mathbb{C}} \colon V^{\mathbb{C}} \times V^{\mathbb{C}} \to \mathbb{C}$ and $h \colon V \times V \to \mathbb{C}$ are both sesquilinear and $\mathbb{R}$-bilinear, but not $\mathbb{C}$-bilinear. In particular, we can regard $h$ as a sesquilinear map $V^{1,0} \times V^{1,0} \to \mathbb{C}$. (The words "$\mathbb{R}$-linear" and "$\mathbb{C}$-linear" do not make much sense for maps $W \times W \to \mathbb{C}$: one should use "$\mathbb{R}$-bilinear" or "$\mathbb{C}$-bilinear" or "sesquilinear.") $\endgroup$ Commented Dec 7, 2017 at 1:58
  • $\begingroup$ You are right, I really meant to say bilinear, not linear, my mistake. Regarding h as a sesquilinear map $V^{1,0} \times V^{1,0} \to \mathbb{C}$ in the manifold context, this means the correct definition locally is $$h\left( \frac{\partial}{\partial z_k}, \frac{\partial}{\partial z_k} \right) = \sum_{i,j=1}^n h_{ij} dz_i \otimes d\overline{z_j} \left( \frac{\partial}{\partial z_k}, \frac{\partial}{\partial \overline{z_k}} \right) ?$$ $\endgroup$
    – Integral
    Commented Dec 7, 2017 at 11:29
  • $\begingroup$ I mean, I have to use $\left( \frac{\partial}{\partial z_k}, \frac{\partial}{\partial z_k} \right)$ as argument to $h$ and shift to $\left( \frac{\partial}{\partial z_k}, \frac{\partial}{\partial \overline{z_k}} \right)$ when making calculations. $\endgroup$
    – Integral
    Commented Dec 7, 2017 at 11:29
  • $\begingroup$ You need change in operand. Because you are using a $\otimes_{\mathbb{R}}$ product. $h\left( \frac{\partial}{\partial z_k} , \frac{\partial}{\partial \bar{z}_k} \right)$ and don't change again inside calculations. $\endgroup$ Commented Jun 9, 2022 at 0:22
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The best way to think about this is that the hermitian inner product is defined on the holomorphic tangent space (not the complexified tangent space), and you should interpret $d\bar z_j(v)$ as $\overline{dz_j(v)}$.

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