# Complexified tangent space

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?

• What book are you alluding to? Commented Nov 22, 2013 at 21:39
• I am trying to read Griffiths and Harris, 0th chapter. Commented Nov 23, 2013 at 4:06
• Aah, thanks for the information, poorna. Commented Nov 23, 2013 at 8:34

"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.

• Oh thanks a lot! Now, my doubt is clarified! And no, I dint notice that the complex structure is irrelevant, I shall try to read up on that now. I, however saw that, $T_{\mathbb{C}, p} =T_{p'}(M)\oplus T_{p''}(M)$, that is the direct sum of holomorphic and anti-holomorphic tangent spaces. I don't know if what you mentioned is the same. Commented Nov 23, 2013 at 4:17
• Dear poorna, yes, Griffiths and Harris's decomposition is exactly the same as mine. The names of the direct summands are just different. Commented Nov 23, 2013 at 8:42
• Why "all such complex derivations are uniquely obtained from $T_{\mathbb{R},p}(M)\otimes_\mathbb{R}\mathbb{C}$"? Do we have to refere to basis of $T_{\mathbb{R},p}(M),$ cause I can't create inverse of map $$Der_\mathbb{R} (C^\infty_{M,p,\mathbb R}\to \mathbb R)\otimes_\mathbb{R}\mathbb{C}=T_{\mathbb{R},p}(M)\otimes_\mathbb{R}\mathbb{C} \rightarrow Der_\mathbb C (C^\infty_{M,p,\mathbb C}\to \mathbb C)$$ that you introduced. Commented Nov 11, 2015 at 20:40
• @FallenApart I guess your inverse maps works as follows. Let $v \in \operatorname{Der}_{\mathbb C}$. Now we have to create an element of $\operatorname{Der}_{\mathbb R} \otimes \mathbb C$ as inverse image. What about $\Re(v)\otimes 1 + \Im(v) \otimes i$? Commented Oct 1, 2019 at 9:32
• @GeorgesElencwajg Hello Georges, I have a question about the "holomorphic-anti-holomorphic decomposition", can one deduce that, for a complex Manifold $M$, its holomorphic tangent bundle $TM$(as a complex bundle) is "equal"(not just isomorphic) to $TX^{(1,0)}$, which is the holomorphic subbundle of the complexified bundle $(TM)_{\mathbb{R}}\otimes\mathbb{C}$ ?And for an almost complex Manifold, can one deduce the same thing? Commented Jun 29, 2022 at 3:32