# Complex orthogonal group

1. Feb 4, 2009

### StatusX

I'm wondering about the action of the complex (special) orthogonal group on $\mathbb{C}^n$. In the real case, we can use a (real) orthogonal matrix to rotate any (real) vector into some standard vector, say (a,0,0,...,0), where a>0 is equal to the norm of the vector. In other words, the action is transitive on each sphere.

I'm wondering if something similar holds in the complex case. Clearly, this action preserves the sum of the squares of the components of a vector, but this quantity is now a general complex number (possibly even zero). Still, we can define a generalized "sphere" of radius (squared) z, some complex z, which is the set of all vectors for which the sum of the squares of the components is z, and ask if the action is transitive on these spheres. Note the sphere corresponding to z=0 has no non-trivial representative of the form (a,0,0,...,0), so things are going to be a little different than the real case.

The fact that the sum of squares does not define a norm here gets in the way of repeating the obvious proof for the real case. I think I have a roundabout proof of transitivity for spheres of non-zero radius, but things are stranger when the radius is zero. For example, for n=2, the zero sphere is the set of points (z,iz) and (z,-iz), for z non-zero, which is not connected, and so clearly the action can't be transitive (since SO(n,C) is connected). Does anyone know what's going on here?

2. Feb 4, 2009

### Hurkyl

Staff Emeritus
Are you sure it's the special orthogonal group you're interested in, and not the special unitary group?

3. Feb 5, 2009

### StatusX

Yes, it's because it's the complexification of the real orthogonal group. I realize things would be basically the same as the real case if it was the unitary group I wanted.

4. Feb 5, 2009

### Hurkyl

Staff Emeritus
It's hard to visualize, but I bet the whole thing looks a lot like the Minkowski plane. Recall there that the orbits under the action of the Lorentz group are generally rectilinear hyperbolic arcs centered at the origin, and opening either up, right, down, or left. (And these are precisely the "circles" under the Minkowski metric)

However, you also have the degenerate cases of the two diagonal lines. I believe these constitute 5 more orbits: one orbit is the origin, and the other four are the rays.

5. Feb 5, 2009

### OrderOfThings

But the action of the whole orthogonal group, O(n,C), is transitive, yes?

6. Feb 5, 2009

### StatusX

I don't know, is it? Do you have a proof?

My idea was to show first that the action is transitive locally, that is, that you can find an infinitessimal orthogonal transformation to move you in any direction along the sphere. Explicitly, this involves taking any vector a, and a perpendicular vector b that represents the direction of motion, and finding an antisymmetric matrix A that takes a to b. A good choice is A = (b aT- a bT)/( aT a), but this only works for non-zero radius. Then it only remains to show the spheres are connected, which can be done with a proof similar to the real case: take a straight line path between two points on the sphere, make sure it avoids zero (acutally, in this case, make sure it avoids the zero sphere, which is a little harder, but still seems possible), and then project it onto the sphere. But this method seems completely unadaptable to the zero sphere.

As for what exactly the zero sphere is, it's clear it consists of vectors of the form:

$$(z_1, z_2,...,z_{n-1}, \pm i \sqrt{ {z_1}^2 + ... + {z_{n-1}}^2 } )$$

In 2 dimensions, this is not connected (unless you include zero), and in higher dimensions, it seems to consist of two copies of $\mathbb{C}^{n-1}$ glued together at their own zero spheres (since this is where the last coordinate is degenerate). It's interesting to note the similarity of this to the inductive definition of the ordinary sphere in terms of gluing along lower dimensional spheres, but I'm still a little lost on what these spaces look like.

7. Feb 5, 2009

### OrderOfThings

The action of SO(C,2) is transitive on each of the components (z,iz) and (z,-iz) separately. To move between the components you take the matrix

$$\left(\begin{matrix}1 & 0 \\ 0 & -1\end{matrix}\right)$$

which belongs to O(2,C).

A quick generalization of this would be that the action of SO(n,C) is transitive on each connected component. To also get transitivity between components you need the reflections in O(n,C).

8. Feb 5, 2009

### StatusX

I'm pretty sure it's only disconnected in 2 dimensions: for example, to get from $(a,b,i \sqrt{a^2+b^2})$ to $(a,b,-i \sqrt{a^2+b^2})$, use a path from (a,b) to a non-zero point (c,d) with $c^2+d^2=0$, which allows you to pass from one sheet to the other without hitting zero. But I'm still not sure about transitivity.