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Let $M$ be a complex manifold of dimension $n$. That is, in each point of $M$ we have a complex chart $\varphi:U\longrightarrow \mathbb{C}^n$, which locally parameterizes $U\subset M$ by means of $n$ complex coordinates $z^1$, ... , $z^n$. From a differentiable point of view, $\mathbb{C}^n\simeq\mathbb{R}^{2n}$, so $M$ has also a structure of (real) differentiable manifold of dimension $2n$, locally described by the coordinates $x^1$, $y^1$, ... , $x^n$, $y^n$, with $z^j=x^j+\mathrm{i} y^j$. Associated to this structure we have the real tangent space, that may be seen as the set of derivations acting on real-valued functions:

$$TM=\left\langle\frac{\partial}{\partial x^1},\frac{\partial}{\partial y^1},...,\frac{\partial}{\partial x^n},\frac{\partial}{\partial y^n}\right\rangle_\mathbb{R}$$

and its dual space:

$$T^*M=\langle\mathrm{d} x^1,\mathrm{d} y^1,...,\mathrm{d} x^n,\mathrm{d} y^n\rangle_\mathbb{R}$$

If we'd like to work with the set of derivations acting on complex-valued functions, there's only need to complexify the previous space:

$$TM_\mathbb{C}=\left\langle\frac{\partial}{\partial x^1},\mathrm{i}\frac{\partial}{\partial x^1},...,\frac{\partial}{\partial y^n},\mathrm{i}\frac{\partial}{\partial y^n}\right\rangle_\mathbb{R}=\left\langle\frac{\partial}{\partial x^1},\frac{\partial}{\partial y^1},...,\frac{\partial}{\partial x^n},\frac{\partial}{\partial y^n}\right\rangle_\mathbb{C}=TM\otimes\mathbb{C}$$

and likewise for the dual space:

$$T^*M_\mathbb{C}=\langle\mathrm{d} x^1,\mathrm{i}\mathrm{d} x^1,...,\mathrm{d} y^n,\mathrm{i}\mathrm{d} y^n\rangle_\mathbb{R}=\langle\mathrm{d} x^1,\mathrm{d} y^1,...,\mathrm{d} x^n,\mathrm{d} y^n\rangle_\mathbb{C}=T^*M\otimes\mathbb{C}$$

But for the last two spaces another base will be more useful

$$\frac{\partial}{\partial z^j}=\dfrac{1}{2}\left(\frac{\partial}{\partial x^j}-\mathrm{i}\frac{\partial}{\partial y^j}\right)\qquad \frac{\partial}{\partial \bar{z}^j}=\dfrac{1}{2}\left(\frac{\partial}{\partial x^j}+\mathrm{i}\frac{\partial}{\partial y^j}\right)$$

$$\mathrm{d} z^j=\mathrm{d} x^j+\mathrm{i}\mathrm{d} y^j\qquad \mathrm{d} \bar{z}^j=\mathrm{d} x^j-\mathrm{i}\mathrm{d} y^j$$

respect to with the following decompositions appear:

$$T^{1,0}M=\left\langle\frac{\partial}{\partial z^1},...,\frac{\partial}{\partial z^n}\right\rangle_\mathbb{C}\qquad T^{0,1}M=\left\langle\frac{\partial}{\partial \bar{z}^1},...,\frac{\partial}{\partial \bar{z}^n}\right\rangle_\mathbb{C}$$

$$(T^{1,0}M)^*=\Lambda^{1,0}M=\langle\mathrm{d} z^1,...,\mathrm{d} z^n\rangle_\mathbb{C}\qquad(T^{0,1}M)^*=\Lambda^{0,1}M=\langle\mathrm{d} \bar{z}^1,...,\mathrm{d} \bar{z}^n\rangle_\mathbb{C}$$

$$TM_\mathbb{C}=T^{1,0}M\oplus T^{0,1}M\qquad T^*M_\mathbb{C}=\Lambda^{1,0}M\oplus \Lambda^{0,1}M$$

These are well-defined independently of the charts, because the derivations $\left\{\dfrac{\partial}{\partial \bar{z}^1},...,\dfrac{\partial}{\partial \bar{z}^n}\right\}$ are annihilated on holomorphic functions. This decomposition reaches the space of forms:

$$\Omega^k(M)=\bigoplus_{p+q=k}\Lambda^{p,q}(M)$$

where $\Lambda^{p,q}(M)$ is generated by the forms with $p$ holomorphic terms and $q$ antiholomorphic terms $\omega(z)=\eta(z)\mathrm{d} z^{j_1}\wedge\cdots\wedge\mathrm{d} z^{j_p}\wedge\mathrm{d}\bar{z}^{k_1}\wedge\cdots\wedge \mathrm{d}\bar{z}^{k_q}$.

We also have the following differential operators:

$$\partial=\frac{\partial}{\partial z^j}\mathrm{d} z^j:\Lambda^{p,q}(M)\longrightarrow\Lambda^{p+1,q}(M)$$
$$\bar{\partial}=\frac{\partial}{\partial \bar{z}^j}\mathrm{d} \bar{z}^j:\Lambda^{p,q}(M)\longrightarrow\Lambda^{p,q+1}(M)$$

that is,

$$\partial(f\mathrm{d} z^I)=\frac{\partial f}{\partial z^j}\mathrm{d} z^j\wedge\mathrm{d} z^I$$
$$\bar{\partial}(f\mathrm{d} z^I)=\frac{\partial f}{\partial \bar{z}^j}\mathrm{d} \bar{z}^j\wedge\mathrm{d} z^I$$

These operators satisfy

$$\mathrm{d}=\partial+\bar{\partial}\qquad\partial\partial=0\qquad \bar{\partial}\bar{\partial}=0\qquad\partial\bar{\partial}=-\bar{\partial}\partial$$

Indeed,

$$\partial+\bar{\partial}=\frac{\partial}{\partial z^j}\mathrm{d} z^j+\frac{\partial}{\partial \bar{z}^j}\mathrm{d} \bar{z}^j=\frac{\partial}{\partial x^j}\mathrm{d} x^j+\frac{\partial}{\partial y^j}\mathrm{d} y^j=\mathrm{d}$$

$$0=\mathrm{d}^2=(\partial+\bar{\partial})(\partial+\bar{\partial})=\partial^2+\partial\bar{\partial}+ \bar{\partial}\partial+\bar{\partial}^2$$

$\partial^2$ adds two holomorphic terms, $\bar{\partial}^2$ add two antiholomorphic terms and $\partial\bar{\partial}+ \bar{\partial}\partial$ adds a term of each type, so each term has to be identically zero.

There exists a canonical isomorphism

$$\theta:TM\xrightarrow{\simeq} T^{1,0}M,\qquad\frac{\partial}{\partial x^j}\longmapsto\frac{\partial}{\partial z^j},\qquad\frac{\partial}{\partial y^j}\longmapsto\mathrm{i}\frac{\partial}{\partial z^j}$$

given by the equality of derivations with respect to holomorphic functions. Multiplying by $\mathrm{i}$ in $T^{1,0}M$ induces, together with $\theta$, the following linear map:

$$J:TM\longrightarrow TM,\qquad\frac{\partial}{\partial x^j}\longmapsto\frac{\partial}{\partial y^j},\qquad\frac{\partial}{\partial y^j}\longmapsto -\frac{\partial}{\partial x^j},\qquad J^2=-\text{id}$$

called almost complex structure.