Trying to understand the breaking down of the complexified tangent space

Question

I am reading some notes on the complexified tangent space. I don't understand how do we arrive at that $V^{(1,0)} = \{ X - iJX : X \in V\}$. I mean in one direction we have $JZ = J( v \otimes \alpha) = Jv \otimes_{\mathbb{R}} \alpha$. In the other direction we have $iZ = i (v \otimes_{\mathbb{R}} \alpha) = v \otimes_{\mathbb{R}} i\alpha$. Why is this the same as $X - iJX$ for $X \in V$.

Answer 1

Notice that each vector in $V_{\mathbb C}$ can be written uniquely as $X\otimes 1 + Y\otimes i$, for some $X,Y\in V$. Given $Z = X\otimes1 + Y\otimes i$ with $JZ = iZ$, we have $$ JX\otimes 1 + JY\otimes i = JZ = iZ = -Y\otimes1 + X\otimes i. $$ By the uniqueness of the representation we conclude $JX = -Y$ (and $JY = X$). Therefore, $Z = X\otimes1 - JX\otimes i$ (for some $X\in V$).

Conversely, given a vector of the form $Z = X\otimes 1 - JX\otimes i$, for some $X\in X$, we have $$ JZ = JX\otimes 1 -J^2X\otimes i = JX\otimes1 + X\otimes i = i\cdot (X\otimes 1 - JX\otimes i) = iZ. $$