# Some Comments on Linear Complex Structures via an Example

Consider a real vector space ${V}$ generated by ${\partial_ x}$ and ${\partial_ y}$. There is an obvious identification ${L:V\rightarrow\mathbb C}$ of ${V}$ with the complex plane ${\mathbb C}$ such that ${L(\partial_ x) = 1}$ and ${L(\partial_ y) = i}$. Define a linear complex structure on ${V}$ by setting ${J(\partial_ x) = \partial_ y}$ and ${J(\partial_ y)=-\partial_ x}$. With the identification mentioned above, since ${\mathbb C}$ is a complex vector space, ${V}$ can be viewed as a complex vector space, too. Furthermore, the action of ${J}$ can be viewed as multiplication by ${i}$ on ${V}$ but we will see below why this view does not extend further.

Next, we complexify ${V}$ by taking a tensor product with ${\mathbb C}$ over ${\mathbb R}$. We know that (real) dimension of ${V_{\mathbb C} = V\otimes \mathbb C}$ is ${4}$ and it is generated by ${\partial_ x\otimes 1, \partial_ y \otimes 1, \partial_ x \otimes i}$ and ${\partial_ y \otimes i}$. We can view ${V_{\mathbb C}}$ as a complex vector space and, for notational simplicity, write ${v = v \otimes 1}$ and ${iv = v \otimes i}$. Note that over the complex numbers ${V_{\mathbb C}}$ is ${2}$ dimensional and generated by ${\partial_ x}$ and ${\partial_ y}$. However, these are not the “natural” bases to work with as we wil see. Next, we extend (complexify) ${J:V\rightarrow V}$ to get ${J_{\mathbb C}:V_{\mathbb C}\rightarrow V_{\mathbb C}}$ which we will still denote by ${J}$ for notational simplicity. Let ${\partial_ z = \frac{1}{2}(\partial_ x - i \partial_ y)}$ and ${\partial_ {\bar z} = \frac{1}{2}(\partial_ x + i\partial_ y)}$. Now, we see that

$\displaystyle \begin{array}{rcl} J(\partial_ z) &=& \frac{1}{2}\left( J(\partial_ x) -i J(\partial_ y) \right) \\ &=& \frac{1}{2}\left( \partial_ y +i \partial_ x \right) \\ &=& i\frac{1}{2}\left(\partial_ x - i \partial_ y \right) \\ &=& i \partial_ z \end{array}$

and also,

$\displaystyle \begin{array}{rcl} J(\partial_ {\bar z}) &=& \frac{1}{2}\left( J(\partial_ x) +i J(\partial_ y) \right) \\ &=& \frac{1}{2}\left( \partial_ y -i \partial_ x \right) \\ &=& -i\frac{1}{2}\left(\partial_ x + i \partial_ y \right) \\ &=& -i \partial_ {\bar z}. \end{array}$

This means that ${\partial_ z}$ is an eigenvector of ${J}$ corresponding to the eigenvalue ${i}$. Similarly, ${\partial_ {\bar z}}$ is an eigenvector corresponding to the eigenvalue ${-i}$. So, the set ${\left\{ \partial_ z, \partial_ {\bar z} \right\}}$ is an eigenbasis for ${J}$ and it gives us an eigenspace decomposition of ${V_{\mathbb C}}$. Computing ${J}$, using this basis, is clearly more convenient and hence, this is a “natural” choice as a basis. Furthermore, from this viewpoint, it is also clear why the action of ${J}$ cannot be viewed as multiplication by ${i}$ any more.