Linearly Independent Killing Fields in n-D Manifold

In summary, the question is asking how many spacetime symmetries there are, and it appears that there are a maximum of (n+1)/2.
  • #1
Cexy
70
1
A question on a General Relativity exam that I have asks how many linearly independent Killing fields there can be in an n-dimensional manifold. I'm sure I've seen this question before and I think that the answer is n(n+1)/2, but I can't remember why!

Any help?
 
Last edited:
Physics news on Phys.org
  • #2
Cexy said:
A question on a General Relativity exam that I have asks how many linearly independent Killing fields there can be in an n-dimensional manifold. I'm sure I've seen this question before and I think that the answer is n(n+1)/2, but I can't remember why!

Any help?

The reason might might be something like this.

Killing's equation is anti-symmetric in 2 indices, which gives

[tex]
\left(
\begin{array}{cc}
n\\
2
\end{array}
\right)
[/tex]

degrees of freedom, plus [itex]n[/itex] initial degrees of freedom, one for each component of a killing vector.

Regards,
George
 
  • #3
Another way of understanding this number is considering the symmetries belonging to the Killing fields.

The simplest example of a maximally symmetric (n dim.) manifold is just Euclidian [tex]\mathbb{R}^n[/tex] (or Minkowski space in case of Lorentzian metric). Independent symmetries are given by the translations and rotations. There are n independent translations and you can convince yourself that there are n(n-1)/2 independent rotations (the dimension of SO(4)). So the total number of independent symmetries is [tex]n+n(n-1)/2=n(n+1)/2[/tex].

So any (Riemannian) manifold can have at most n(n+1)/2 symmetries, and at most the same number of independent Killing fields.

-- Timbuqtu
 
  • #4
Timbuqtu said:
Another way of understanding this number is considering the symmetries belonging to the Killing fields.

The simplest example of a maximally symmetric (n dim.) manifold is just Euclidian [tex]\mathbb{R}^n[/tex] (or Minkowski space in case of Lorentzian metric). Independent symmetries are given by the translations and rotations. There are n independent translations and you can convince yourself that there are n(n-1)/2 independent rotations (the dimension of SO(4)). So the total number of independent symmetries is [tex]n+n(n-1)/2=n(n+1)/2[/tex].

So any (Riemannian) manifold can have at most n(n+1)/2 symmetries, and at most the same number of independent Killing fields.

-- Timbuqtu

This is excellent. My only gripe is using the term "rotation" for an element of SO(n). Isometry would be better.
 
  • #5
Okay, thanks for the help guys. I suppose the question to ask now is: Why do Killing vectors have a one-to-one correspondence to spacetime symmetries?
 
  • #6
The killing equation [tex]\nabla_{(\mu}k_{\nu)}=0[/tex] is equivalent to the vanishing of the Lie-derivative of the metric along k, [tex]\mathcal{L}_k g_{\mu\nu}=0[/tex]. This last equation actually just tells you that the metric does not change if you pull it along the vector field k, so k represents a symmetry of the metric (and therefore a symmetry of the space time).

-- Timbuqtu
 
  • #7
Ah, fantastic. Cheers. :)
 

Related to Linearly Independent Killing Fields in n-D Manifold

1. What are Killing Fields in a manifold?

Killing Fields are vector fields on a manifold that preserve the metric of the manifold. They are named after Wilhelm Killing, a German mathematician who first studied these fields in the late 19th century.

2. How do you determine if Killing Fields are linearly independent?

Killing Fields are considered to be linearly independent if their linear combinations do not result in the zero vector. This can be determined by solving a system of equations using the Lie derivative, which measures the change in a vector field along another vector field.

3. What is the significance of linearly independent Killing Fields in a manifold?

Linearly independent Killing Fields are important in studying the symmetries of a manifold. They can be used to find conserved quantities along geodesics, which are curves that minimize distance on a manifold. They also play a crucial role in the construction of the geodesic equation in general relativity.

4. How are Killing Fields related to isometries?

Isometries are transformations that preserve distances on a manifold. Killing Fields are closely related to isometries, as they generate these transformations. In fact, a manifold is said to be homogeneous if it possesses a set of Killing Fields that span the tangent space at every point, which allows for the existence of isometries.

5. Can Killing Fields exist in non-Riemannian manifolds?

Yes, Killing Fields can exist in non-Riemannian manifolds as well. However, in these cases, the metric is not preserved by the Killing Fields. Instead, they preserve a different structure, such as a pseudo-Riemannian metric in the case of Lorentzian manifolds.

Similar threads

I Induced orientation on boundary of ##\mathbb{H}^n## in ##\mathbb{R}^n##
    • Differential Geometry
Replies
6
Views
525
A Hilbert spaces and kets "over" manifolds
    • Differential Geometry
Replies
10
Views
2K
Why is there maximally 1/2 n(n+1) killing vectors?
    • Differential Geometry
Replies
1
Views
4K
A Question about the derivation of the tangent vector on a manifold
    • Differential Geometry
Replies
9
Views
3K
I Are Coordinates on a Manifold Really Functions from R^n to R?
    • Differential Geometry
Replies
17
Views
3K
A Representing flux tubes as a pair of level surfaces in R^3
    • Differential Geometry
Replies
1
Views
2K
A Simple definition of Lie group
    • Differential Geometry
Replies
11
Views
2K
A Manifolds: local & global coordinate charts
    • Differential Geometry
2
Replies
40
Views
6K
A Pushforward of Smooth Vector Fields
    • Differential Geometry
Replies
1
Views
3K
Are left-invariant fields mapped onto the manifold Killing vectors?
    • Differential Geometry
Replies
7
Views
3K

Hot Threads

Recent Insights

Back
Top