Can General Relativity Accommodate Spaces Without Killing Vectors?

[Tio Barnabe]

By requiring the inner product in two points $x$ and $x'$ having metrics $g$ and $g'$ to be invariant, i.e. $g'(x') = g(x)$, one is lead to the Killing equation. Does general relativity forbiddes spaces where the Killing equation cannot be satisfied?

It seems obvious that we want conserved quantities in our theories. But, is there a way around in which we can consider a space-time having no Killing Vectors at all?

 

[PeroK]

By requiring the inner product in two points $x$ and $x'$ having metrics $g$ and $g'$ to be invariant, i.e. $g'(x') = g(x)$, one is lead to the Killing equation. Does general relativity forbiddes spaces where the Killing equation cannot be satisfied?

It seems obvious that we want conserved quantities in our theories. But, is there a way around in which we can consider a space-time having no Killing Vectors at all?

A Killing vector represents a symmetry in the metric. If the metric has no symmetries then there are no Killing vectors.

Although the gravity affecting the Earth's solar orbit is nearly symmetric, if you include all the factors, then at a certain level of accuracy it won't be symmetric at all.

 

[PeterDonis]

By requiring the inner product in two points $x$ and $x'$ having metrics $g$ and $g'$ to be invariant, i.e.$g'(x') = g(x)$, one is lead to the Killing equation.

How so?

Does general relativity forbiddes spaces where the Killing equation cannot be satisfied?

Certainly not.

 

[PAllen]

Inner product, so far as I know, is an operation on two vectors producing a scalar. Your usage does not appear to jive with this. Please clarify

 

[Tio Barnabe]

How so?

Please clarify

I just considered that the two points $x$ and $x'$ were vectors themselves, in which my notation $g(x)$ means the inner product of $x$ with itself (similarly for $g'(x')$). Couldn't I do that?

 

[PAllen]

I just considered that the two points $x$ and $x'$ were vectors themselves, in which my notation $g(x)$ means the inner product of $x$ with itself (similarly for $g'(x')$). Couldn't I do that?

Points in a manifold are not vectors. Vectors live in the tangent space to a manifold at a given point. In flat space, position vectors do happen to form a vector space, but not if there is any curvature. Also, for flat space, you can choose to treat the space as it’s own tangent space, but again you cannot if there is any curvature.

 

[PeterDonis]

I just considered that the two points $x$ and $x'$ were vectors themselves

Which, as @PAllen has pointed out, is incorrect. So this entire thread appears to be based on a mistaken premise in your OP.

Thread closed.