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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:

\begin{equation} \begin{aligned} J_1&=x^2\,\partial_3-x^3\,\partial_2=-\sin\varphi \,\partial_\theta-\cot\theta\cos\varphi \,\partial_\varphi\ ,\\ J_2&=x^3\,\partial_1-x^1\,\partial_3=\cos\varphi \,\partial_\theta-\cot\theta\sin\varphi \,\partial_\varphi\ ,\\ J_3&=x^1\,\partial_2-x^2\,\partial_1=\partial_\varphi\ , \end{aligned} \end{equation}

where $x^i$ are Cartesian coordinates (with spatial indices $i, j, \dotsc = 1, 2, 3$), and $r\in [0,+\infty)$, $\theta\in[0,\pi]$, $\varphi\in[0,2\pi)$ are spherical coordinates. The transformation between Cartesian and spherical coordinates is given by the usual expressions

\begin{equation} x^1=r \sin\theta \cos\varphi \ , \qquad x^2=r \sin\theta \sin\varphi \ , \qquad x^3=r \cos\theta \ . \end{equation}

The generators $J_i$ satisfy the commutation relations

\begin{equation} [J_i,J_j]=\varepsilon_{ijk}J_k\ , \end{equation}

where $\varepsilon_{ijk}$ is the Levi-Civita symbol. The generators of symmetries are also known as $\it{Killing~vectors}$ since they satisfy the $\it{Killing~equation}$.

My question is how to define generators $J_i$ 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 $J_i$ generators and Killing algebra without the third spatial dimension.

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    $\begingroup$ The expressions for the $J_i$ in spherical coordinates don't involve $r$, so the answer is... exactly the same. $\endgroup$
    – Javier
    Nov 8, 2023 at 15:59
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    $\begingroup$ @Javier If it would be so easy. In this context $r$ is equal $K^{-1/2}$, where $K$ is Gaussian curvature - an intrinsic property. There is no center there, as well as interior or exterior space. It is 'standalone' unbounded closed 2-dimensional universe. $\endgroup$
    – JanG
    Nov 8, 2023 at 16:58
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    $\begingroup$ @Javier And, there is no global coordinate system. The only possible rotation is about a point on the two-sphere. In such infinitesimally local coordinate system we have got only two orthogonal vectors. $\endgroup$
    – JanG
    Nov 8, 2023 at 17:13
  • $\begingroup$ @Javier Do you know any definition of Killing vectors on $S^2$ sphere without using the metric induced by embedding it into $R^3$ space? $\endgroup$
    – JanG
    Nov 9, 2023 at 11:34
  • $\begingroup$ @Javier I have found answer to the question I have just asked - an intrinsic definition of Killing vectors - and accept your answer, thanks! $\endgroup$
    – JanG
    Nov 9, 2023 at 11:47

1 Answer 1

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In 3D space the three vectors $J_i$ are already tangent to spheres. You can see this clearly from the fact that all three are orthogonal to $\partial_r$; this also implies that at each point they are not linearly independent! This is a common confusion: they are independent as vector fields, but not at each point.

All of this means that if your space is just a sphere $S^2$, the Killing vectors are exactly the same, since they don't have a radial component. Put spherical coordinates on the sphere and use the expressions you already have for the $J_i$, and you're done. The radius of the sphere doesn't matter - it affects the norms of the vectors, but not their expressions. And sure, this system doesn't technically cover the whole sphere, but neither do the usual spherical coordinates in 3D space. In practice this is not an issue since the coordinates are singular on a measure zero subset.

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    $\begingroup$ Interesting answer. Yes, they are all orthogonal to $\partial_r$ but, at the same time, I imagine that at each point the generator of rotation in a plane tangent to the 2-sphere is directed along $\partial_r$... this confuses me, I am probably too naive in my geometrical picture. What am I missing? How about considering something like $x J_1 + y J_2 + z J_3$ restricted to $r=cost$ as the unique generator of rotations in the 2-space? $\endgroup$
    – Quillo
    Nov 8, 2023 at 18:30
  • $\begingroup$ @Javier Oh, I see now that the three vectors $J_{i}$ lie within the two-sphere. The index $i$ labeled theses vectors and is not component index as in $\partial_{i}$. So, I have still three vectors and $SO(3)$ as group but around what point or axis is this rotation? $\endgroup$
    – JanG
    Nov 8, 2023 at 18:52
  • $\begingroup$ @Javier By the way, the $\partial_{r}$ does not exist. There is no such direction due to no ambient space. $\endgroup$
    – JanG
    Nov 8, 2023 at 18:56
  • $\begingroup$ I know that $\partial_r$ doesn't exist on the sphere, but it does exist in 3D space. I mentioned it to show that even in 3D space, the $J_i$ are tangent to spheres. $\endgroup$
    – Javier
    Nov 9, 2023 at 12:58
  • $\begingroup$ Regarding your first question, the generator of rotations around an axis doesn't point along the axis like an angular velocity. It's a vector field that rotates around the axis, telling each point of the sphere where to move. For example, the generator of rotations around the $z$ axis is $\partial_\varphi$, and it's a vector field over the whole sphere, not just a single vector. $\endgroup$
    – Javier
    Nov 9, 2023 at 12:59

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