I am currently dealing with the almost complex structures associated with a non-degenerate quadratic form. In particular, I am interested the size of the almost complex structures as a homogeneous space (if not, as a topological space) that is "compatible" with the quadratic form $Q$. For the sake of simplicity, we assume the base space is $\mathbb{R}^{2n}$ associated with a quadratic form $Q$ of signature $(+1, +1, ..., -1, -1)$ ($p$ positive signatures and $q$ negative signatures).

Key Question

How many almost complex structures $J\in GL_{2n}(\mathbb{R})$ satisfy $J^TQJ=Q$ (where $\cdot^T$ denotes transpose)? Also, when $\omega$ is a symplectic form, how many almost complex structures that $K^T\omega K=\omega$.

Current Progress

When $J$ is one such almost complex structure, $J^TQJ=Q\iff QJ$ is a symplectic form. So it is also to find how many $J$ that make $QJ$ a symplectic form.

As for the second part of this problem, this problem is well-studied when we imposed the restriction that $QJK$ is positive-definite. The reader may refer to Chapter II of Holomorphic curves in symplectic geometry by Michele Audin and Jacques Lafontaine. Yet when this restriction is lifted, I do not know how to handle this problem.


First, note that if a nondegenerate bilinear quadratic form is compatible with a complex structure, then the components $(p, q)$ of the signature must both be even, so we'll write $p = 2 p'$, $q = 2 q'$.

Now, the special orthogonal group $SO(Q) \cong SO(2p', 2q')$ of orientation-preserving orthogonal transformations preserving a quadratic form $Q$ on $\Bbb R^{2n}$ acts transitively on the set $C$ of complex structures $J$ compatible with $Q$ (one can show this readily by picking, for any two such complex structures $J, K$, bases of $\Bbb R^{2n}$ similarly adapted to $(Q, J)$ and $(Q, K)$).

Now, if we pick a compatible complex structure $J$ and write it in a convenient basis, it's easy to compute explicitly the subgroup of $SO(Q)$ that preserves $J$, called the unitary group, and we usually denote this group $U(p' , q')$. (It's also a standard exercise to show that $\dim U(p', q') = n^2$.) So, the homogeneous space of complex structures compatible with $Q$ is $$SO(2p, 2q) / U(p, q),$$ which has dimension $$\dim SO(2n) - \dim U(n) = \frac{1}{2}(2n)(2n - 1) - n^2 = n (n - 1).$$ This is not only a homogeneous space, but a symmetric space (this series of such spaces is denoted type DIII); it is compact iff $Q$ is definite.

  • $\begingroup$ Many thanks, Travis. Do you have any clue regarding the second part of the question, i.e, $K^T\omega K=\omega$ without restricting $\omega K$ being positive-definite? $\endgroup$
    – Wunderbar
    May 1 '15 at 1:04
  • $\begingroup$ One should be able to argue similarly, starting with the fact that the group preserving a symplectic form $\omega$ on $\Bbb R^{2n}$ is the symplectic group $Sp(n)$. Note that any complex structure compatible with $\omega$ determines a symmetric bilinear form $g := \omega(\, \cdot \, , J\, \cdot \,)$ (which is nondegenerate because $\omega$ and $J$ are), and the action of $Sp(n)$ on the space of compatible complex structures must preserve the signature of $g$. $\endgroup$ May 1 '15 at 4:25
  • $\begingroup$ I still do not see why we need the group which to act on $J$ to be orientation preserving. Moreover, it does not match my calculation of the most trivial case, in which $Q$ is the Euclidean inner product. In that case, I found the space is $O(2n)/U(n)$. Could you explain why and if in case I was wrong, what I have missed? $\endgroup$
    – Wunderbar
    May 1 '15 at 12:42
  • 1
    $\begingroup$ You're right that we don't need to require orientation preservation; I imposed the condition because it leads to a connected homogeneous space and the general case can be readily deduced from this one anyway. I think what's going on geometrically is this: Any complex structure induces an orientation, and we can identify $SO(p, q) / U(p, q)$ as the complex structures that induce a particular orientation. If we ignore orientation, we get twice as many complex structures, and these are parameterized by the (nonconnected) homogeneous space $O(2n) / U(n)$. $\endgroup$ May 1 '15 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.