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I know that for $U \subset \mathbb{C}^n$ and $p \in U$ we construct the holomorphic tangent space in the following way.

We take $T_pU$ the real $2n$-dimensional over $\mathbb{R}$ tangent space.

Now we complexify, i.e. take $T_pU \otimes\mathbb{C}$. We got $2n$-dimensional vector space over $\mathbb{C}$ which nicely decomposes into $T'_pU \oplus T''_pU$. So we take $T'_pU$ for the holomorphic tangent space.

Usually we take $\frac{\partial }{\partial z^i}, \frac{\partial }{\partial \bar z^i}$ for a basis in $T_pU \otimes \mathbb{C}$ so $T'_pU = span(\frac{\partial }{\partial z^i})$ and $T''_pU = span(\frac{\partial }{\partial \bar z^i})$. By the construction $\frac{\partial }{\partial z^i}$ and $\frac{\partial }{\partial \bar z^i}$ are linearly independent.

Algebraicly everything is totally fine but we want to think of the tangent space as a space of differentials with the standard basis $\frac{\partial }{\partial z^i}$. But $\frac{\partial }{\partial z^i}$ and $\frac{\partial }{\partial \bar z^i}$ are not linearly independent as a differentials! I mean, they are determined by their action on functions and they are not independent over $\mathbb{C}$ at this point.

Invincible
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    So what is your question? Of course, the complexified tangent space is not the same as the tangent space, so of course $\partial_{z_i}$ and $\partial_{\bar{z}_i}$ are linearly dependent as differential operators. – Moishe Kohan Oct 24 '19 at 22:56
  • It means that $\partial_{\bar z}$ are in the tangent space too with differential operator point of view. – Invincible Oct 24 '19 at 23:08
  • More precisely, there are (natural) isomorphisms $T^{1,0}M\to TM$ and $T^{0,1}M\to M$ sending vectors to certain differential operators; I think, this is what confuses you. I remember that this matter was discussed at MSE many times. It is useful to remember that "are the same" is not the same as "are isomorphic". – Moishe Kohan Oct 24 '19 at 23:12
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    How do you define the right hand side? – Invincible Oct 24 '19 at 23:22

1 Answers1

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If this confuses you, my suggestion is to use a separate notation, such as $\partial_{i}, \bar\partial_k$ (or $\partial_{\bar{k}}$), for basis elements of the spaces of sections of $T^{1,0}U$ and $T^{0,1}U$ (regarded as $C^\infty(U)$-modules) and the usual notation $$ \frac{\partial}{\partial z_i}, \frac{\partial}{\partial \bar{z}_k}$$ for the corresponding differential operators acting on functions $C^\infty (U)$. Then what you have are isomorphisms of vector bundles $$ \phi^{1,0}: T^{1,0}U \to TU, \phi^{0,1}: T^{0,1}U \to TU, $$ sending $$ \partial_{i}\mapsto \frac{\partial}{\partial z_i}, \bar\partial_k \mapsto \frac{\partial}{\partial \bar{z}_k}. $$ Then $\partial_{i}$ and $\bar\partial_{i}$ are linearly independent while their images are linearly dependent. After you get comfortable with this thing, you will not need the separate notation.

Moishe Kohan
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