Orthogonal complex structures

Suppose that $n = 2m$ is even, $V \isom \bR^{2m}$. Then $V$ can be identified with $\bC^m$, but not canonically.

Then also $g(Ju, v) = -g(Ju, J^2v) = -g(u, Jv)$, so that $J$ is skewsymmetric with respect to $g$: $J^t = -J$. Note that (b) says that $J^t J = 1$.

We can now make $V$ a $\bC$-module by setting $iv := Jv$, that is,

$$(\al + i\bt)v := \al v + \bt J v \words{for all} \al,\bt \in \bR.$$

We define a hermitian scalar product on $V$ by

$$\braket{u}{v}_J := g(u, v) + i g(Ju, v)$$

Note that $\braket{Ju}{v}_J = -i\braket{u}{v}_J$ and $\braket{u}{Jv}_J = +i\braket{u}{v}_J$ (check it!). We denote the resulting $m$-dimensional complex Hilbert space by $V_J$.

If $\{u_1,\dots,u_m\}$ is an orthonormal basis for $V_J := (V,\braket{\cdot}{\cdot}_J)$, then $\{u_1, Ju_1,\dots, u_m, Ju_m\}$ is an orthonormal oriented basis for $V$ (over $\bR$). The orientation may or may not be compatible with the given one on $V$.

If $J^2 = -1$, $J^t J = 1$ and if $h \in \OO(n) = \OO(V,g)$ is an orthogonal linear transformation, then $K := hJh^{-1}$ is also an orthogonal complex structure. In that case,
$$\begin{align*} \braket{hu}{hv}_K &= g(hu, hv) + i g((hJh^{-1})hu, hv) \\ &= g(u, v) + i g(Ju, v) = \braket{u}{v}_J, \end{align*}$$
so that $h\: V_J \to V_K$ is unitary. Thus $hJh^{-1} = J$ if and only if $h \in \rU(V_J) \isom \rU(m)$. In short: $\OO(n) = \OO(2m)$ acts transitively on $\sJ(V,g)$ with isotropy subgroups isomorphic to $\rU(m)$. Hence, as a manifold,

$$\sJ(V,g) \approx \OO(2m)/\rU(m).$$

Those $J$ which are compatible with orientation on $V$ form one component (of two), homeomorphic to $\SO(2m)/\rU(m)$.

We may complexify $V$ to get $V^\bC = V \ox_\bR \bC = V \oplus iV$. Take

$$W_J := \set{v - iJv \in V^\bC : v \in V} = \half(1 - iJ)V = P_J V.$$

This is an isotropic subspace for the symmetric billinear form $g^\bC$ on $V^\bC$.

$$g(u - iJu, v - iJv) = g(u, v) - ig(u, Jv) - ig(Ju, v) - g(Ju, Jv) = 0.$$

The conjugate subspace

$$\overline{W}_J = \set{v + iJv : v \in V} = \half(1 + iJ)V = P_J V$$

satisfies $W_J \oplus \overline{W}_J \isom V^\bC$, an orthogonal direct sum for the hermitian scalar product

$$\bbraket{w}{z} := 2 g(\bar w, z) \word{for} w, z \in V^\bC.$$

Note that $P_J^2 = P_J^\0$ and $P_J^\0 = P_J^*$ with respect to this product. We say that $W_J$ is a polarization of $V^\bC$. Also $P_J \: V_J \to W_J$ is an unitary isomorphism.

Conversely: given a splitting $V = W \oplus \overline{W}$, orthogonal with respect to $\bbraket{\cdot}{\cdot}$, write $w =: u - iv$ for $w \in W$, with $u,v \in V$; then $J_W \: u \mapsto v$ lies in $\sJ(V,g)$, and $W_{J_W} = W$ (exercise). Thus the correspondence $J \otto W_J$ is bijective.