What is the transitivity of the complex orthogonal group on generalized spheres?

In summary, the conversation discusses the action of the complex (special) orthogonal group on \mathbb{C}^n and whether it is transitive on each sphere, similar to the real case. It is noted that the sum of squares does not define a norm in the complex case, making it difficult to prove transitivity. The zero sphere is also discussed, which is not connected in 2 dimensions but appears to be two copies of \mathbb{C}^{n-1} glued together at their own zero spheres in higher dimensions. It is suggested that the action of SO(n,C) is transitive on each connected component, but reflections in O(n,C) are needed to achieve transitivity between components.
  • #1
StatusX
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I'm wondering about the action of the complex (special) orthogonal group on [itex]\mathbb{C}^n[/itex]. 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?
 
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  • #2
Are you sure it's the special orthogonal group you're interested in, and not the special unitary group?
 
  • #3
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
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
StatusX said:
I'm wondering about the action of the complex (special) orthogonal group on [itex]\mathbb{C}^n[/itex]. 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.
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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?

But the action of the whole orthogonal group, O(n,C), is transitive, yes?
 
  • #6
OrderOfThings said:
But the action of the whole orthogonal group, O(n,C), is transitive, yes?

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:

[tex] (z_1, z_2,...,z_{n-1}, \pm i \sqrt{ {z_1}^2 + ... + {z_{n-1}}^2 } ) [/tex]

In 2 dimensions, this is not connected (unless you include zero), and in higher dimensions, it seems to consist of two copies of [itex]\mathbb{C}^{n-1}[/itex] 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
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

[tex]\left(\begin{matrix}1 & 0 \\ 0 & -1\end{matrix}\right)[/tex]

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

FAQ: What is the transitivity of the complex orthogonal group on generalized spheres?

1. What is the Complex Orthogonal Group?

The Complex Orthogonal Group, denoted as O(n,C), is a mathematical group consisting of all n-by-n unitary matrices with determinant equal to ±1. It is a subgroup of the general linear group GL(n,C) and is an important object in the study of linear algebra and group theory.

2. What are the properties of the Complex Orthogonal Group?

Some key properties of the Complex Orthogonal Group include orthogonality, unitarity, and preserving lengths and angles. It also has a non-compact Lie group structure and is isomorphic to the special unitary group SU(n).

3. How is the Complex Orthogonal Group used in scientific research?

The Complex Orthogonal Group has many applications in physics, engineering, and computer science. It is used in quantum mechanics to describe the symmetries of physical systems, in signal processing to analyze signals in the frequency domain, and in computer graphics to represent rotations and reflections of 3D objects.

4. What is the relationship between the Complex Orthogonal Group and other groups?

The Complex Orthogonal Group is closely related to other groups such as the real orthogonal group O(n) and the complex special orthogonal group SO(n,C). It is also a subgroup of the complex general linear group GL(n,C) and shares many properties with these other groups.

5. Are there any real-world examples of the Complex Orthogonal Group?

Yes, there are many real-world examples of the Complex Orthogonal Group. For instance, in crystallography, the symmetry of a crystal can be described by the rotations and reflections of its unit cell, which form elements of the Complex Orthogonal Group. It is also used in robotics to represent the movements and rotations of robotic arms.

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