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.(adsbygoogle = window.adsbygoogle || []).push({});

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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# Complex orthogonal group

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