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.


1 Answer 1


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
    Commented May 1, 2015 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 $g$. $\endgroup$ Commented May 1, 2015 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
    Commented May 1, 2015 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 $O(2n) / U(n)$. $\endgroup$ Commented May 1, 2015 at 14:00

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