latentcorpse
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Is it true to say that most higher dimensional spacetimes have symmetries amongst their n extra dimensions?
Thanks.
Thanks.
latentcorpse said:Is it true to say that most higher dimensional spacetimes have symmetries amongst their n extra dimensions?
Thanks.
fzero said: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 said: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.