TITLE: Complexified tangent space
QUESTION [26 upvotes]: Let $M$ be a complex manifold of dimension $n$ and $p\in M$. So $M$ can be viewed as a real manifold of dimension $2n$ and we can consider the usual real tangent space at $p$, $T_{\mathbb{R},p}(M)$, that is the space of $\mathbb{R}$ linear derivations on the $\mathbb{R}$-algebra of germs of $C^\infty$ functions in a neighbourhood of $p$, and if we write $z_i=x_i+y_i$, then
$T_{\mathbb{R},p}(M)=\mathbb{R}\big\{\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_i}\big\}$.
So if, $v\in T_{\mathbb{R},p}(M)$, then we can write $v=\sum_{i=1}^{n}a_i\frac{\partial}{\partial x_i}|_p+\sum_{i=1}^{n}b_i\frac{\partial}{\partial y_i}|_p$, where $a_i,b_i\in\mathbb{R}$. Is that right?
Next, we define the complexified tangent space to $M$ at $p$, $T_{\mathbb{C},p}(M)$ to be $T_{\mathbb{R},p}(M)\otimes_\mathbb{R}\mathbb{C}$. The book then says elements of $T_{\mathbb{C},p}(M)$ can be realized as $\mathbb{C}$ linear derivations in the ring of complex valued $C^\infty$ functions on $M$ around $p$. I don't know how I can realize this. For if we choose an element $(v\otimes z_0)\in T_{\mathbb{C},p}(M)$, how do we act it on complex valued functions? Initially I thought that if $f$ is a complex valued function in a neighbourhood of $p$, I can define $(v\otimes z_0)(f) = z_0 v(f)$. But this doesn't make sense because, $v$ is a real tangent vector, further $(v\otimes z_0)$ has to be $\mathbb{C}$-linear and should be a derivation. So how can I see this?

REPLY [28 votes]: "The" book is right: $T_{\mathbb{R},p}(M)\otimes_\mathbb{R}\mathbb{C}$ can be identified with the complex linear vector space  of $\mathbb C$-linear derivations $C^\infty_{M,p,\mathbb C}\to \mathbb C$.
Indeed,  given the real derivation $v\in T_{\mathbb{R},p}(M)$ the elementary tensor  $v\otimes z\in T_{\mathbb{R},p}(M)\otimes_\mathbb{R}\mathbb{C}$ acts on $f+ig\in C^\infty_{M,p,\mathbb C}$ (=germs of smooth complex-valued functions defined near $p$) by the formula $$ (v\otimes z) (f+ig)  =z\cdot [v(f)+iv(g)]$$ This action is a $\mathbb C$-linear derivation $  C^\infty_{M,p,\mathbb C}\to \mathbb C          $ and all such  complex derivations  are uniquely obtained from $ T_{\mathbb{R},p}(M)\otimes_\mathbb{R}\mathbb{C}$ .
To sum up in a formula: $$Der (C^\infty_{M,p,\mathbb C}\to \mathbb C)=T_{\mathbb{R},p}(M)\otimes_\mathbb{R}\mathbb{C}=   Der_\mathbb R (C^\infty_{M,p,\mathbb R}\to \mathbb R) \otimes_\mathbb R \mathbb C         $$
Did you notice that the the complex structure on $M$ is irrelevant?
 No?
I'm not surprised: this fact is almost never mentioned in books or lectures.
The complex structure on $M$ gives rise to a canonical direct sum decomposition $T_{\mathbb{R},p}(M)\otimes_\mathbb{R}\mathbb{C}=T^{1,0}\oplus T^{0,1}$, where $T^{0,1}$ consists of  derivations killing  germs of holomorphic functions but that is another (long and interesting!) story.