Do higher dimensional spacetimes have symmetries in their extra dimensions?

[latentcorpse]

Is it true to say that most higher dimensional spacetimes have symmetries amongst their n extra dimensions?

Thanks.

 

[fzero]

Is it true to say that most higher dimensional spacetimes have symmetries amongst their n extra dimensions?

Thanks.

Most manifolds have no isometries at all, so no. If you are talking about some Kaluza-Klein or string theory model that would try to reproduce standard model physics, then there are additional constraints on the allowed manifolds that require certain symmetries.

 

[latentcorpse]

Most manifolds have no isometries at all, so no. If you are talking about some Kaluza-Klein or string theory model that would try to reproduce standard model physics, then there are additional constraints on the allowed manifolds that require certain symmetries.

Yeah. Thanks.

Just to clear up what an isometry actually is though:

I know it is a symmetry transformation of the metric tensor field.
i.e. a map $\phi: M \rightarrow M$ such that $(\phi)_*g=g$

However, what does this actually mean?

Suppose the metric is invariant of t. We can see from Killing's equation that $\frac{\partial}{\partial t}$ will be a Killing vector field easily enough but what does this actually mean the isometry is?
Would the isometry be the map $\phi: M \rightarrow M ; t \mapsto t+c$ i.e. time translations?

Thanks.

 

[fzero]

Yeah. Thanks.

Just to clear up what an isometry actually is though:

I know it is a symmetry transformation of the metric tensor field.
i.e. a map $\phi: M \rightarrow M$ such that $(\phi)_*g=g$

However, what does this actually mean?

Suppose the metric is invariant of t. We can see from Killing's equation that $\frac{\partial}{\partial t}$ will be a Killing vector field easily enough but what does this actually mean the isometry is?
Would the isometry be the map $\phi: M \rightarrow M ; t \mapsto t+c$ i.e. time translations?

Thanks.

Yes, any isometry corresponds to the existence of a Killing vector field. From the Killing vector, we can define a conjugate coordinate, at least locally, for which the isometry corresponds to a translation.