I Killing vectors in isotropic space-times

["Don't panic!"]

I've been reading up on Killing vectors, and have got on to the topics of homogeneous, isotropic and maximally symmetric space-times. I've read that for an isotropic spacetime, one can construct a set of Killing vector fields $K^{(i)}$, such that, at some point $p\in M$ (where $M$ is the space-time manifold), $K^{(i)\mu}(p)=0$ and $\nabla_{\nu}K^{(i)\mu}(p)\neq 0$. Intuitively, I understand that we must have $K^{(i)\mu}(p)=0$, since isotropy is a rotational symmetry, and so we don't want the Killing vectors to translate the points, we want them to rotate about them. However, I don't understand why we require $\nabla_{\nu}K^{(i)\mu}(p)\neq 0$. Why must the covariant derivative of each Killing vector be arbitrary? I read that it has something to do with $\nabla_{\nu}K^{(i)\mu}(p)$ generating "rotations" of neighbouring points about $p$, but this is unclear to me.

 

[PeterDonis]

I understand that we must have $K^{(i)\mu}(p)=0$, since isotropy is a rotational symmetry, and so we don't want the Killing vectors to translate the points, we want them to rotate about them.

That's not quite what the condition $K^{(i)\mu}(p)=0$ is saying. It is saying that any rotation must leave at least one point (the point $p$) unchanged; hence, the action of the Killing vector on that point vanishes.

I don't understand why we require $\nabla_{\nu}K^{(i)\mu}(p)\neq 0$.

Because if it were zero, the action of the KVF on points neighboring $p$ would also vanish, but that's not what a rotation would do.

 

["Don't panic!"]

That's not quite what the condition K(i)μ(p)=0K(i)μ(p)=0K^{(i)\mu}(p)=0 is saying. It is saying that any rotation must leave at least one point (the point ppp) unchanged; hence, the action of the Killing vector on that point vanishes.

Ah ok. So the Killing vector will move about all other points, but will keep the point p, about which you are applying it, fixed. It must vanish because $K^{\mu}$ generates space-time translations, i.e. $x'^{\mu}=x^{\mu}+\epsilon K^{\mu}$, right? (And so, at p, we must have $x'^{\mu}=x^{\mu}$).

Because if it were zero, the action of the KVF on points neighboring ppp would also vanish, but that's not what a rotation would do.

So intuitively, is this condition stating that at neighbouring points, the action of the KVF is non-zero (i.e. its components are non-zero), such that these points will move around relative to the fixed point p, i.e. a rotation?

 

[PeterDonis]

It must vanish because $K^{\mu}$ generates space-time translations, i.e. $x'^{\mu}=x^{\mu}+\epsilon K^{\mu}$, right?

No. A spacetime translation would not leave any points unchanged. The Killing vector we are discussing generates rotations.

is this condition stating that at neighbouring points, the action of the KVF is non-zero (i.e. its components are non-zero), such that these points will move around relative to the fixed point p, i.e. a rotation?

Yes.

 

["Don't panic!"]

No. A spacetime translation would not leave any points unchanged. The Killing vector we are discussing generates rotations.

What is the intuition for why $K^{\mu}=0$ at a point p, such that it is unchanged?

 

[PeterDonis]

What is the intuition for why $K^{\mu}=0$ at a point p, such that it is unchanged

Heuristically, the Killing vector at any point "points" in the direction in which the transformation induced by the Killing vector field will "move" the point. So if the point is left unchanged by the transformation, it doesn't get "moved" anywhere, and the Killing vector must vanish since there is nowhere for it to "point".

 

["Don't panic!"]

Heuristically, the Killing vector at any point "points" in the direction in which the transformation induced by the Killing vector field will "move" the point. So if the point is left unchanged by the transformation, it doesn't get "moved" anywhere, and the Killing vector must vanish since there is nowhere for it to "point".

But is the transformation generated by the KVF of the form $x’^{\mu}=x^{\mu}+\epsilon K^{\mu}$? If so, it makes sense to me that $K^{\mu}=0$ at p, as then it won’t be transformed.

 

[PeterDonis]

is the transformation generated by the KVF of the form $x’^{\mu}=x^{\mu}+\epsilon K^{\mu}$?

Infinitesimally, yes, you can view it that way. But only infinitesimally.

 

["Don't panic!"]

Infinitesimally, yes, you can view it that way. But only infinitesimally.

Yes, sorry. I should have stated that I meant infinitesimally. So is it correct to say that, at the infinitesimal level, the reason why the KVF must vanish at the point p is because we require that it remains unchanged, i.e. $x’^{\mu}=x^{\mu}$?

 

[Orodruin]

But is the transformation generated by the KVF of the form $x’^{\mu}=x^{\mu}+\epsilon K^{\mu}$?

Just to interject that this is true for the flow generated by any vector field by definition.

Yes, sorry. I should have stated that I meant infinitesimally. So is it correct to say that, at the infinitesimal level, the reason why the KVF must vanish at the point p is because we require that it remains unchanged, i.e. $x’^{\mu}=x^{\mu}$?

Yes. Any point $p$ for which $K(p) = 0$ is a fixed point under the flow of $K$. Again, this is regardless of whether $K$ is a Killing field or not.

 

["Don't panic!"]

Any point ppp for which K(p)=0K(p)=0K(p) = 0 is a fixed point under the flow of KKK. Again, this is regardless of whether KKK is a Killing field or not.

Thanks for the info. In the case where K is a killing field, is the interpretation of $K(p)=0$ at $p$, that the point p remains fixed, i.e. its coordinates remain unchanged?

 

[Orodruin]

Why would it change just because it is a Killing field?

 

["Don't panic!"]

Why would it change just because it is a Killing field?

You’re right, I got myself mixed up there.