# (Alm) Complex Structures and things being holomorphic

I’m learning some complex geometry and this post is about some basics I re-organized for myself.  Most of the material is coming from chapter 1 Huybrechts book ‘Complex Geometry.’

Let $M$ be a complex manifold.  It is in particular a real manifold of even dimension.  Not every even dimensional real manifold is a complex manifold; I believe $S^4$ is an example.  This lead to the question:

When can an even dimensional real manifold $M$ be made into a complex manifold?

The short answer to this question is: whenever $M$ has a complex structure.  A complex structure is an almost complex structure such that the local charts are holomorphic.

Again $M$ is a real manifold and $E$ is a real vector bundle on $M$.  A complex structure on the vector bundle $E$ is a vector bundle endomorphism $J \colon E \to E$ such that $J^2 = -1$.  Note any real bundle admitting a complex structure must have even fiber dimension basically because it can be made into a $\mathbb{C}$ vector space.

An almost complex structure on $M$ is a complex structure on the tangent bundle $TM$.

1. Using this idea, for any even dimensional manifold the local $\mathbb{R}^{2n}$ charts can be made into $\mathbb{C}^n$ charts.

Locally $M$ is $\mathbb{R}^{2n}$ say with basis $e_1,..., e_{2n}$.   Relabel $x_i = e_{2i}$ and $y_i = e_{2i+1}$.   Define a map via $J(x_i) = y_i$ and $J(y_i) = -x_i$.  Note $J \in GL(2n, \mathbb{R})$ and $J^2 = -1$.  Together $(\mathbb{R}^{2n}, J)$ has the structure of a $\mathbb{C}$-vector space via $(a+bi)v = av+bJ(v)$.  So general theory says $(\mathbb{R}^{2n}, J)$ $\cong \mathbb{C}^n$.  The isomorphism is exactly

$\mathbb{C}^n \to (\mathbb{R}^{2n}, J)$

$(a_1 +ib_1,....,a_n+ib_n) \mapsto \sum_{i=1}^n a_ix_i +b_iy_i$

2. For $U \subset \mathbb{C}^n$ there is a natural complex structure.

Note $TU$ is trivial, i.e. $TU = U \times \mathbb{C}^n$.  So define $J$ via $J(u, v) = (u, iv)$.  In real coordinates $\frac{\partial}{\partial x_i}, \frac{\partial}{\partial y_i}$ this is

$J(\frac{\partial}{\partial x_i}) = \frac{\partial}{\partial y_i}$

$J(\frac{\partial}{\partial y_i}) = -\frac{\partial}{\partial x_i}$

this is just the above example in slightly different notation.

3. A very important aspect of bundles with complex structures is that they can be tensored with $\mathbb{C}$.  The significance is that each fiber is a vector space $V \otimes_{\mathbb{R}} \mathbb{C}$ with two complex structures:

$J(v \otimes (a+bi)) = J(v)\otimes (a+bi)$

$i(v \otimes (a+bi)) = v \otimes (-b +ai)$

This is very important because it gives rise to the decomposition

$V \otimes \mathbb{C} = V^{1,0} \oplus V^{0,1}$      (0)

$V^{1,0} = \{ v^{1,0}:= v\otimes 1 - J(v)\otimes i| v \in V\}$

$V^{0,1} = \{v^{0,1}:=v\otimes 1 +J(v)\otimes i|v \in V\}$

note $Jv^{1,0} = iv^{1,0}$ and $Jw^{0,1} = -iw^{0,1}$ .  In effect the complex structure $J$ naturally picks out a subspace that behaves $\mathbb{C}$ linearly and another that behaves $\mathbb{C}$ anti-linearly, analogous to $z, \bar z$.

This decomposition allows me to make sense of the notion of $(p,q)$ forms and the operators $d, \partial, \overline{\partial}$.

Thinking of the tangent bundle, i.e $V = T_xM$, the decomposition passes to the dual vector space $V^*_{\mathbb{C}}$.  Consequently I can declare $(p,q)$ forms to be

$V^{p,q} := \wedge^p V^{*1,0}\otimes_{\mathbb{C}}\wedge^qV^{*0,1}$

Its a linear algebra fact that

$\wedge^nV^* = \oplus_{p+q = n}V^{p,q}$  (1)

Recall there is the exterior derivative on sections of the exterior powers of the cotangent bundle $d \colon \Gamma(\wedge^p T^*M) \to \Gamma(\wedge^{p+1} T^*M)$ (I’m using $\Gamma$ to denote just smooth sections)

In local real coordinates $r_1,..., r_{2n}$ of $M$ a section is $f(r_i)dr_{i_1}\wedge ... \wedge dr_{i_p} = f(r_i)dr_I$ and

$d(f(r_i)dr_I) = \sum_i \frac{\partial f}{\partial r_i}dr_i\wedge dr_I$

Define  $A^kM = \Gamma(\wedge^k T^*_{\mathbb{C}}M)$ and define $A^{p,q}$ similarly.  $\mathbb{C}$-linear extension gives a map $d \colon A^k \to A^{k+1}$ which locally is

$d(f(r_j)dr_I + ig(r_j)dr_K) = d(f(r_j)dr_I) +id(g(r_j)dr_K)$

Since I can project to any of the subspaces in (1) I can define $\partial$ by the composition

$\partial \colon A^{p,q} \xrightarrow{d} A^{p+q+1}\to A^{p+1,q}$

and similarly for $\overline{\partial}$.  I’m use to it being true that $d = \partial + \overline{\partial}$ but this only happens for nice almost complex structures.  A not so nice one is presented below.

Notice applying the above decomposition for the case of $TU\otimes \mathbb{C}$ gives

$\frac{1}{2}\frac{\partial}{\partial x_i}^{1,0} = \frac{1}{2}\frac{\partial}{\partial x_i} -i\frac{1}{2}\frac{\partial}{\partial y_i}$ $= \frac{\partial}{\partial z_i}$  (2)

The 1/2 is in there so that it works out $dz(\frac{\partial}{\partial z}) = 1$ where $dz = dx + idy$

### Holomorphicity

Let $f \colon U \to V$ be a smooth map where $U \subset \mathbb{C}^n$ and $V \subset \mathbb{C}^m$ both open.  Let $J_U, J_V$ be the complex structures on $U,V$.  Let $Df \colon TU \to TV$ be the differential of the map.  Then $f$ is holomorphic if $Df \circ J_U = J_V \circ Df$.

This can be motivated from the 1-dim case.  In this case $f$ holomorphic means $\frac{\partial f}{\partial \bar z} = 0$.  In real coordinates this is the same as Cauchy Riemann equations:

The logic is $\frac{\partial f}{\partial \bar z} = 0$ iff the Cauchy-Riemann equations iff $Df T^{1,0}\mathbb{C} \subset T^{1,0}\mathbb{C}$ iff $f$ is holomorphic in sense of the definition above.

In more cumbersome language a function $f = f(z,\bar z) = f(x,y)$ is a map of (real) manifolds:

$f\colon \mathbb{C} \to \mathbb{C}$

$x+iy \mapsto f(x,y) = u(x,y) + iv(x,y)$

$(x,y) \mapsto (u(x,y), v(x,y))$

The maps on (real) tangent spaces is given by the Jacobian matrix $Df = \bigl(\begin{smallmatrix} u_x & u_y\\ v_x & v_y \end{smallmatrix}\bigr)$ i.e. $\frac{\partial}{\partial x} \mapsto u_x \frac{\partial}{\partial u} + v_x \frac{\partial}{\partial v}$.  For the sake of notation write this as $Df (1,0) = (u_x,v_x)$

Assuming the Cauchy-Riemann equations (for the third line):

$Df((1,0)^{1,0}) = Df[(1,0)\otimes 1 - (0,1) \otimes i]$

$= (u_x,v_x)\otimes 1 - (u_y,v_y)\otimes i$

$= (u_x,v_x)\otimes 1 - (-v_x,u_x)\otimes i = (u_x,v_x)^{1,0}$

where I note that $i(u_x,v_x) = (-v_x,u_x)$.  This argument is reversible.  Thus $\frac{\partial f}{\partial \bar z} = 0$ is equivalent to $Df T^{1,0} \subset T^{1,0}$ and similarly for $T^{0,1}$, i.e. preserves the decomposition (0).  The last implication in covered in the general case.

Now going to the general case $f \colon U \to V$ I see if $Df \circ J_U = J_V \circ Df$ then

$Df(v^{1,0}) = Df(v\otimes 1 -J_V(v)\otimes i)$

$= Df(v)\otimes 1 - J_U(Df(v))\otimes i = Df(v)^{1,0}$

Conversely, if $Df$ preserves (0) then

$Df(v^{1,0}) = Df(v)\otimes 1 - Df(J_Uv)\otimes i = \sum_{k = 1}^l w_k\otimes 1 - J_Vw_k\otimes i$

but noting that $V \to V^{1,0}$ given by $v \mapsto v\otimes 1 - Jv\otimes i$ is an isomoprhism implies in the above that $l=1$ and $Df(J_Uv) = J_VDf(v)$ as desired.  So $Df \circ J_U = J_V \circ Df$ is a good way to generalize holomorphicity.

An almost complex structure that doesn’t give a complex structure.

Look at $\mathbb{C}^n$ and use $\frac{\partial}{\partial x_i}, \frac{\partial}{\partial y_i}$ for the tangent space.  Define

$J(\frac{\partial}{\partial x_i}) = -\frac{\partial}{\partial y_i}$

$J(\frac{\partial}{\partial y_i}) = \frac{\partial}{\partial x_i}$

It can be checked if we map $\mathbb{C}^n$ to itself via the identity and given one of them this funky almost complex structure then calling one $U$ and the other $V$, then $Df \circ J_U = -J_V \circ Df$.

In normal complex coordinates i’m used to $d(x+iy) = dz$, but in the above formalism $idx = -dy$ so $dx +idy = dx\otimes 1 - idx\otimes i$ i.e

$d(x+iy) = dx^{0,1}$

but with the funky almost complex structure $J(dx) = dy$ so $d(x+iy) = dx\otimes 1 +J(dx)\otimes i = dx^{0,1}$.  So in perhaps poor notation this is something like

$d(x+iy) = d\bar z$

All sorts of bad things happen with this almost complex structure.  For example if you look at $x_2dx_1^{1,0} \in A^{1,0}\mathbb{C}^2$ and apply $d$ then you’ll get a component in $A^{0,2}$, i.e. $d \ne \partial + \overline{\partial}$.

In fact $d = \partial +\overline{\partial}$ is another characterization of the $Df \circ J_U = J_V \circ Df$ but this is I think difficult to show and is basically covered by a theorem of Newlander and Nierenberg.

In the situation of a manifold $M$ with almost complex structure on a vector bundle $E$, I can also consider the extra structure of having an inner product $\langle , \rangle$ on each fiber.

Some terminology. $\langle , \rangle$ is compatible with $J$ if $\langle Jv, Jw\rangle = \langle v,w \rangle$.  The fundamental form associated to $\langle , \rangle, J$ is $\omega(,) = \langle J(), \rangle$.  Together these determine a hermetian form $H(,) = \langle , \rangle \pm J\omega(,) = \omega(,J())\pm J\omega(,)$.  Its clear any two of $\langle , \rangle, \omega, J$ determine the third.

Turns out $\omega$ is a real valued 2 form, i.e. $\omega \in \wedge^2 V^*$ and lies in $\wedge^{1,1}V^*_{\mathbb{C}}$.  Notice the importance between differentiating whether or not I tensored with $\mathbb{C}$.  This is probably a good thing to check in local coordinates.  I wont do it now.

This post is already too long, but I’ll mention in passing $\omega$ defines an operator called the Lefshcetz operator $L$ on the exterior algebra $\wedge^* V^*_{\mathbb{C}}$ via wedging with $\omega$.  There is a dual operator of $L$ with respect to $\langle , \rangle$ called $\Lambda$, i.e. $\langle \Lambda( ), \rangle = \langle ,L() \rangle$.  With $H:= [L,\Lambda]$ these guys give a representation of $sl_2$.