Proof of Killing Vectors Commutator Theorem

  • Thread starter Thread starter Altabeh
  • Start date Start date
  • Tags Tags
    Vectors
Click For Summary
SUMMARY

The discussion centers on the proof of the theorem stating that if A^{{\mu }} and B^{{\mu }} are Killing vectors, then their commutator C^{{\mu }}=[A,B]^{\mu} is also a Killing vector. The participants reference the metric g_{\mu \nu} and the properties of Killing vectors, specifically their covariant derivatives. The key takeaway is the relationship between the commutator of two Killing vectors and the preservation of the metric under their transformations.

PREREQUISITES
  • Understanding of Killing vectors in differential geometry
  • Familiarity with covariant derivatives and their notation
  • Knowledge of the metric tensor g_{\mu \nu}
  • Basic principles of Lie algebras and commutators
NEXT STEPS
  • Study the properties of Killing vectors in Riemannian geometry
  • Learn about the implications of the commutator in Lie algebra
  • Explore the derivation of the covariant derivative of Killing vectors
  • Investigate applications of Killing vectors in general relativity
USEFUL FOR

The discussion is beneficial for mathematicians, physicists, and students specializing in differential geometry and general relativity, particularly those interested in the properties of symmetries in spacetime.

Altabeh
Messages
657
Reaction score
0
I was wondering if you could help me with the proof of the following theorem.

If [tex]A^{{\mu }}[/tex] and [tex]B^{{\mu }}[/tex] are Killing vectors, then so is their commutator [tex]C^{{\mu }}=[A,B]^{\mu}[/tex].

Thanks in advance
 
Physics news on Phys.org
What have you tried?
 
diazona said:
What have you tried?

Assuming [tex]g_{\mu \nu}[/tex] as our metric, I can write

[tex]g_{\mu \alpha}A^{\alpha}_{;\nu}=-g_{\alpha\nu}A^{\alpha}_{;\mu}[/tex] and
[tex]g_{\mu \alpha}B^{\alpha}_{;\nu}=-g_{\alpha\nu}B^{\alpha}_{;\mu}[/tex].

And I can correlate these two to each other, but I'm afraid about the commutator. I just need a clue as to how the commutator is written in terms of [tex]A^{\mu}[/tex] and [tex]B^{\mu}[/tex].
 

Similar threads

  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
0
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K