Question

I quote:

... a (3+1)
-dimensional spacetime exhibiting spatial spherical symmetry, namely, a manifold with the three-dimensional special orthogonal group SO(3)
(representing rotations in three-dimensional Euclidean space) as its group of symmetries. The three generators of the action of SO(3)
on the spacetime are the following:

J1J2J3=x2∂3−x3∂2=−sinφ∂θ−cotθcosφ∂φ ,=x3∂1−x1∂3=cosφ∂θ−cotθsinφ∂φ ,=x1∂2−x2∂1=∂φ ,

where xi
are Cartesian coordinates (with spatial indices i,j,…=1,2,3
), and r∈[0,+[infinity])
, θ∈[0,π]
, φ∈[0,2π)
are spherical coordinates. The transformation between Cartesian and spherical coordinates is given by the usual expressions

x1=rsinθcosφ ,x2=rsinθsinφ ,x3=rcosθ .

The generators Ji
satisfy the commutation relations

[Ji,Jj]=εijkJk ,

where εijk
is the Levi-Civita symbol. The generators of symmetries are also known as Killing vectors
since they satisfy the Killing equation
.

My question is how to define generators Ji
in case of a (2+1)-dimensional space, where the spatial manifold is a two-sphere. Such spacetime exhibits spherical symmetry but how to define Ji
generators and Killing algebra without the third spatial dimension.

Answer 1

In a (2+1)-dimensional space with a two-sphere as the spatial manifold, the generators Ji can still be defined, but there will be fewer of them.

Step-by-step explanation:

In a (2+1)-dimensional space where the spatial manifold is a two-sphere, the generators Ji can still be defined, but there will be fewer of them compared to the (3+1)-dimensional case.

In a (2+1)-dimensional space with a two-sphere as the spatial manifold, the generators Ji can still be defined, but there will be fewer of them.

This is because the three generators correspond to the three dimensions in the spatial manifold. In a two-sphere, there are only two dimensions, so there would be only two generators. These generators can be defined by taking linear combinations of the coordinate vectors in the spherical coordinate system.