I Spherically symmetric manifold

[cianfa72]

TL;DR Summary

About the properties of Killing Vector Fields for spherically symmetric manifold

A spherically symmetric manifold has, by definition, a set of 3 independent Killing Vector Fields (KVFs) satisfying: $$\begin{align}[R,S] &=T \nonumber \\ [S,T] &=R \nonumber \\ [T,R] &=S \nonumber \end{align}$$
These 3 KVFs define a linear subspace of the (infinite dimensional) vector space over $\mathbb R$ of the vector fields defined on the manifold $M$. Indeed any linear combination with constant real coefficients of the 3 KVFs is a KVF and, from the properties of the commutator $[~.,.]$ it follows that the span of $\{ T,R,S \}$ closes w.r.t. it as well.

My question: taken 3 independent linear combinations of $\{ T,R,S \}$, say $\{X,Y,Z\}$, do they satisfy the three conditions above as well ?

 

[Orodruin]

They will not generally satisfy [X,Y] = Z, [Y,Z] = X, [Z,X] = Y. This is trivial to see, just let them be a multiple of 2 times the original set.

 

[cianfa72]

They will not generally satisfy [X,Y] = Z, [Y,Z] = X, [Z,X] = Y. This is trivial to see, just let them be a multiple of 2 times the original set.

Any linear combination of $\{ T,R,S \}$, say $X = aT + bR + cS$, is a KVF that is a generator of rotations about a different axis.

span$\{ T,R,S \}$ is a sub Lie algebra of dimension 3 of the Lie algebra of the set of all KVFs. So one can pick a different basis $\{X,Y,Z\}$ within the span$\{ T,R,S \}$, however the conditions in post#1 are no longer true for them.

That said, does the definition of spherically symmetric manifold (e.g. spacetime) actually demand for the existence of 3 linear independent KVFs such that the conditions in post#1 hold true ?

 

[Orodruin]

Yes.

If you had XYZ you could always create RST as the appropriately normalised basis.

 

[fresh_42]

$\left\{R,S,T\right\}$ represent the cross product in $\mathbb{R}^3.$ The corresponding Lie algebra is of type $A_1,$ the Lie algebra $\mathfrak{sl}(2)$ with a standard basis $\{H,X,Y\}$ where
$$
[H,X]=2X\, , \,[H,Y]=-2Y\, , \,[X,Y]=H.
$$
However, here you need a complex transformation to transform $\{R,S,T\}$ into $\{H,X,Y\}.$ You also need a factor $-2i$ if you use Pauli-matrices. Here are some remarks on it:
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/

 

[cianfa72]

If you had XYZ you could always create RST as the appropriately normalised basis.

As far as I know, a Lie algebra is characterized by the structure constants. Fixed a basis $\{ X_i \}$ in the underlying vector space they are given by $$[X_i,X_j] = C_{ij}^k X_k$$
Do the structure constants $C_{ij}^k$ change when picking a different set of basis elements ?

 

[martinbn]

As far as I know, a Lie algebra is characterized by the structure constants. Fixed a basis $\{ X_i \}$ in the underlying vector space they are given by $$[X_i,X_j] = C_{ij}^k X_k$$
Do the structure constants $C_{ij}^k$ change when picking a different set of basis elements ?

He answered that in post #2. Double all basis elements. It gives you a new basis, with different structure constants.

 

[fresh_42]

As far as I know, a Lie algebra is characterized by the structure constants. Fixed a basis $\{ X_i \}$ in the underlying vector space they are given by $$[X_i,X_j] = C_{ij}^k X_k$$
Do the structure constants $C_{ij}^k$ change when picking a different set of basis elements ?

$X\longmapsto cX$ produces a factor $c^2$ on the product side of the equation but you can compensate only one factor $c$ on the other side of the equation. Now choose $c$ as any number $\neq \pm 1.$

$\mathbb{F}_2$ is a strange case when it comes to Lie algebras and is usually ruled out.

 

[cianfa72]

He answered that in post #2. Double all basis elements. It gives you a new basis, with different structure constants.

Ok, so which is the fundamental/distinctive property of the sub Lie algebra of the rotational KVFs of a spherically symmetric manifold ?

 

[martinbn]

Ok, so which is the fundamental/distinctive property of the sub Lie algebra of the rotational KVFs of a spherically symmetric manifold ?

Being able to pick a basis with those commutation relations doesn't mean that any basis has to have them.

 

[cianfa72]

Being able to pick a basis with those commutation relations doesn't mean that any basis has to have them.

Ah ok, so the fundamental property/feature is that such a sub Lie algebra admits as basis a set of 3 elements (KVFs) with those commutation relations.

In other words: if one gets a set of 3 independent KVFs with given commutation relations, in order to ascertain that their span generates the sub Lie algebra associated to a spherically symmetric manifold, one must check that there exist independent linear combinations of them having the relevant commutation relations as in post#1.