Best way to check for anisotropy given a metric tensor Carroll gives the definition of isotropy at a point as given vector $V$ and $W$ in $T_{p}M$, there is some isometry that can push $V$ forward such that it ends up parallel to $W$. I understand what this is saying, but if I'm given the line element for a spacetime in some coordinate system, what is the best practical way to check if the spacetime has anisotropies?
 A: Firstly, it should be made clear that being isotropic is a very special and rare property. (A spacetime can never be truly isotropic because no isometry can map spacelike vectors to timelike vectors, for example, so I'll talk about "space" being isotropic). There are very few spaces isotropic around every point, only very few spaces will even be isotropic about any point and these will typically be one or two special points. As a rule, if you have an isotropic space you constructed it that way, or there is some special reason why it must be isotropic, and it won't happen "by accident".
With that said: the problem you need to solve is to find all the isometries. The best way to do this is to find the infinitesimal isometries (the Lie algebra of the isometry Lie group, if that language is helpful), which are Killing vector fields: mapping each point a fixed distance along the integral curves of a Killing vector field gives you an isometry.
Another way of phrasing this is that the Lie derivative of the metric with respect to a Killing field must vanish. This gives the Killing equations, which can be expressed in various ways:
$\displaystyle 
(\mathcal{L}_\xi g)_{a b}=\nabla_a \xi_b + \nabla_b \xi_a=\xi^c \partial_c g_{ab}+g_{cb}\partial_a \xi^c+g_{ac}\partial_b\xi^c=0
$
Now in general this is going to be pretty tricky to solve. But in practice there are a few tricks that will help out:


*

*There are often very obvious Killing vectors: in particular, if the metric components are all independent of a coordinate ($t$, say), then the derivative in that direction ($\partial_t$ here) will be a Killing field. This is immediate from the third form of the Killing equation, and intuitive: translating the coordinate ($t\to t+\Delta$) does nothing to the metric.

*There may be other isometries that you can spot by inspection if you think about what the metric means. For example there may be a part looking like flat space or a sphere,  with the familiar isometries associated with that, or there may be some way of rescaling coordinates by a factor that leaves the metric invariant.

*If you have found any two Killing fields $\xi$ and $\eta$, the group structure may teach you something. Because the composition of two isometries is another isometry, there is a corresponding statement for infinitesimal isometries. This will be familiar if you know about Lie groups and Lie algebras: the commutator $[\xi,\eta]$ must also be a Killing field. It may be that the two isometries commute, in which case you get zero, but often you will not.


So, if you thing you have found all the possible Killing fields (hard to know!), and hence symmetries, how do we work out whether the metric is isotopic at a point? The isometries relevant for this question are those that leave the point invariant, which means that the associated Killing fields must vanish there. You could rule out isotropy if there aren't enough such independent fields: see the comments. If there are enough, you need to check how they act on the vectors, so you can map from any one unit vector to any other.
