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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.