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As we know the Complexified Tangent Space of a Riemann Surface $M$ at a point $p$ is $T_{\mathbb{R},p}(M)\otimes_\mathbb{R}\mathbb{C}=\mathbb{C}\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\}= \mathbb{C}\{\frac{\partial}{\partial z},\frac{\partial}{\partial \bar z}\}$ , where $z=x+\sqrt{-1}y$ is a local chart around p. It can also be realised as the $\mathbb{C}$-vector space of the $\mathbb{C}$-linear derivations on the germ of complex valued smooth functions at $p$.

But we know that $\frac{\partial}{\partial \bar z}$ is $\mathbb{C}$ anti-linear. So, what am I missing here? Is it really all $\mathbb{C}$-linear derivations or just $\mathbb{R}$-linear derivations of complex-valued functions? A detailed well-explained solution is most welcome.

Partha
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    I think you should write $\frac{\partial}{\partial\bar{z}}=\frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})$, and then look at what it does to $f=u(x,y)+iv(x,y)$, and $(a+bi)f$. That might make you doubt the antilinear comment. – Steve D Oct 28 '19 at 13:27
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    What you have in mind is probably that $\frac{\partial}{\partial \overline{z}}$ is anti-holomorphic, rather than anti-linear? – Derived Cats Oct 28 '19 at 17:43

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I don't understand why you assume that $\frac{\partial}{\partial{\overline{z}}}$ is $\mathbb{C}$-antilinear. You know that it is a derivation and $$\frac{\partial}{\partial \overline{z}}(c) = 0$$ for every constant $c$ complex function on $M$ (what you expect to be the value of $\frac{\partial}{\partial \overline{z}}(i)$, anyways? Or how about $\frac{\partial}{\partial x}(i),\frac{\partial}{\partial y}(i)$? ). This implies (is equivalent to) $\mathbb{C}$-linearity by the calculation

$$\frac{\partial}{\partial \overline{z}}(cf) = c\frac{\partial}{\partial\overline{z}}(f) + f(p)\cdot \frac{\partial}{\partial \overline{z}}(c) = c\frac{\partial}{\partial\overline{z}}(f) + f(p)\cdot 0 = c\frac{\partial}{\partial\overline{z}}(f)$$

Actually you have identifications $$T_p(M)\otimes_{\mathbb{R}}\mathbb{C} = \mathrm{Der}_{\mathbb{R}}(\mathcal{O}_{M,p},\mathbb{C}) = \mathrm{Der}_{\mathbb{C}}\left(\mathcal{O}_{M,p}\otimes_{\mathbb{R}}\mathbb{C},\mathbb{C}\right)$$

It is a general result. Let $A$ be a $k$-algebra $M$ be an $A$-module and $L$ be a $k$-algebra (all algebras and rings commutative). Then there are isomorphisms of $L$-modules

$$\mathrm{Hom}_k(A,M)\otimes_kL\cong \mathrm{Hom}_k(A,M\otimes_kL) \cong \mathrm{Hom}_L(A\otimes_kL,M\otimes_kL)$$

They induce

$$\mathrm{Der}_k(A,M)\otimes_kL\cong \mathrm{Der}_k(A,M\otimes_kL) \cong \mathrm{Der}_L(A\otimes_kL,M\otimes_kL)$$

Now just substitute $A = \mathcal{O}_{M,p}$, $k= \mathbb{R}$ as a ring, $M = \mathbb{R}$ as $A=\mathcal{O}_{M,p}$-module with action $$\mathcal{O}_{M,P}\times \mathbb{R}\ni f\cdot r\mapsto f(p)\cdot r \in \mathbb{R}$$ and finally $L =\mathbb{C}$ as a $k=\mathbb{R}$-algebra.

Slup
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