Exploring Isometry: Questions on Symmetry and Transformation in Manifolds

[synoe]

Isometry is the symmetry s.t.
$$g^\prime_{\mu\nu}(x)=g_{\mu\nu}(x)$$
under the transformation $x^\mu\to x^{\prime\mu}(x)$. This means under infinitesimal transformation
$$x^\mu\to x^\mu+\epsilon \xi^\mu(x)$$
where $\epsilon$ is any infinitesimal constant, the vector field $\xi^\mu(x)$ satisfies Killing equation:
$$\nabla_\mu\xi_\nu(x)+\nabla_\nu\xi_\mu(x)=0.$$

I have questions about above description.
1. Why the infinitesimal parameter $\epsilon$ is constant and $\xi^\mu$ is depend on $x$? The most general form seems to be $\epsilon(x)\xi^\mu(x)$.

2.Is this transformation global or local? It's global symmetry because $\epsilon$ is constant?

3.I have seen the description like "isometry means the metric is independent on the coordinate of the isometry direction." That is, if the isometry transformation is $x^0\to x^0+\epsilon\xi(x)$, $g_{\mu\nu}=g_{\mu\nu}(x^1,\dots)$ or $\partial_0g_{\mu\nu}=0$. But I think this is true only when the Killing equation have a constant solution. The statement is true in more general?

 

[Ben Niehoff]

Isometry is the symmetry s.t.
$$g^\prime_{\mu\nu}(x)=g_{\mu\nu}(x)$$
under the transformation $x^\mu\to x^{\prime\mu}(x)$.

Not quite. If you have a diffeomorphism $\phi : M \to M$, then $\phi$ is an isometry when

$$(\phi^* g'_{\mu\nu}) (\phi^{-1}(x')) = g_{\mu\nu}(x).$$
That is, you must pull back $g'_{\mu\nu}$ along $\phi$.

This means under infinitesimal transformation
$$x^\mu\to x^\mu+\epsilon \xi^\mu(x)$$
where $\epsilon$ is any infinitesimal constant, the vector field $\xi^\mu(x)$ satisfies Killing equation:
$$\nabla_\mu\xi_\nu(x)+\nabla_\nu\xi_\mu(x)=0.$$

I have questions about above description.
1. Why the infinitesimal parameter $\epsilon$ is constant and $\xi^\mu$ is depend on $x$? The most general form seems to be $\epsilon(x)\xi^\mu(x)$.

$\xi^\mu(x)$ is already an arbitrary function, so there is no need to make the infinitesimal parameter also a function.

2.Is this transformation global or local? It's global symmetry because $\epsilon$ is constant?

Most mathematical texts are referring to global properties unless they say otherwise. That means if your manifold cannot be covered by a single chart, you need to concern yourself with transition functions to move between charts.

Often a Killing vector will vanish at some points on the manifold, so that although you do have an isometry, it does not necessarily "move" every point. For example, if you rotate a sphere, the north and south poles do not change.

3.I have seen the description like "isometry means the metric is independent on the coordinate of the isometry direction." That is, if the isometry transformation is $x^0\to x^0+\epsilon\xi(x)$, $g_{\mu\nu}=g_{\mu\nu}(x^1,\dots)$ or $\partial_0g_{\mu\nu}=0$. But I think this is true only when the Killing equation have a constant solution. The statement is true in more general?

That's a bit sloppy.

It is true that if you have a continuous isometry $\xi^\mu$, it is possible to choose coordinates such that $\xi = \partial_\varphi$ for some coordinate $\varphi$. However, if you have multiple isometries, it is not necessarily true that you can do this for all of them simultaneously. Again, the sphere is a good example. The sphere has three rotational isometries, but you can only "coordinatize" one of them at a time.

In general, you can choose adapted coordinates for any set of isometries that mutually commute (I believe this comes from the Frobenius theorem).

 

[synoe]

Thank you, BenNiehoff

Could you show the statement,

It is true that if you have a continuous isometry ξμ\xi^\mu, it is possible to choose coordinates such that ξ=∂φ\xi = \partial_\varphi for some coordinate φ\varphi.

I interpreted this means the Killing equation have a constant solution like $\xi^\mu=(1,0,0,\dots)$ if the manifold have a single continuous isometry. Is it true?

Most mathematical texts are referring to global properties unless they say otherwise. That means if your manifold cannot be covered by a single chart, you need to concern yourself with transition functions to move between charts.

Can we consider "gauged" isometries? If we make naively $\epsilon$ to be a function $\epsilon(x)$, there is no change by absorbing $x$-dependence to $\xi^\mu(x)$.