(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


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.


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