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I am self learning GR. This is a rather long post but I needed to clarify few things about the effect of general coordinate transformations on the global symmetries of metric. Any comments, insights are much appreciated.

To be concrete let's consider that $g_{\mu\nu}$ represents an axially symmetric spacetime i.e a Kerr black hole. In Boyer-Lindquist coordinates ($t,\, r,\, \theta,\, \phi $) metric has no dependence on $t$ and $\phi$ (they are cyclic coordinates):

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-(1-\frac{2Mr}{r^2+a^2\cos^2{\theta}})\,dt^2-\frac{r^2+a^2\cos^2{\theta}}{r^2+2Mr +a^2} dr^2 +(r^2+a^2\cos^2{\theta})\, d\theta^2 \\ \,\,\,+ \left( r^2+a^2+ \frac{2Ma^2 r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right) \sin^2{\theta}\, d\phi^2 - \frac{2Ma r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}} dt\, d\phi $$

Consequently particle's energy $E_0$ and angular momentum $L_0$ are conserved. Suppose I make a coordinate transformation: $d \bar{x}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,dx^{\nu}$ where, if I am not mistaken, $\Lambda$ is a matrix, which is an element of the general linear group GL$\,(4,\,R)$ with a non-zero determinant. In the new coordinates metric may not have cyclic coordinates, a good example would be representation of the above metric in Kerr-Schild coordinates:

$$\begin{align} g_{\mu\nu} dx^{\mu}dx^{\nu}=&-d\bar{t}^2+dx^2+dy^2+dz^2 \\ &+ \frac{2Mr^3}{r^4+a^2z^2} \left[d\bar{t} + \frac{r(x \,dx+y \,dy)+a(x\,dy-y\,dx)}{r^2+a^2} + \frac{z\,dz}{r} \right]^2\\ &\text{where}\qquad r^4+ (x^2+y^2+z^2)\, r^2 -a^2 z^2=0 \end{align} $$

Now there is only one cyclic coordinate, $\bar{t}$. I presume we may in principle introduce new coordinates where none of the coordinates appear to be cyclic in functional form of $g_{\mu\nu}\, dx^{\mu}dx^{\nu}$. Suppose I make such a coordinate transformation. My questions are:

1. Does the new metric still have global symmetries i.e conserved quantities that correspond to $E_0$ and $L_0$?

2. If the answer to the first question is yes, then suppose I gave this new metric to someone without telling about the coordinate transformation and ask if there are any symmetries. Will she/he able to find symmetries by finding closed orbits in the geodesic flow?

To me the answer to both questions above seemed to be yes. I think that the integrability (in the Liouville sense) of the metric should not depend on one's definition of the coordinates. In other words, because of the global symmetries, we expect bounded geodesics around the black hole. In the old coordinates we may easily calculate such closed trajectories and according to my thinking closed geodesics exist (objects revolve around the black hole) regardless of how we label the coordinates.

But I could not be sure of this. To explain my confusion let me write:

$$ \frac{d x^{\mu}}{ds^2} + \Gamma^{\mu}_{\nu \lambda} \frac{d x^{\nu}}{ds} \frac{d x^{\lambda}}{ds}=0 $$

which gives the geodesic flow. The existence of the bounded trajectories here depends on the number of zeros and poles of the Christoffel symbol, and also on their relative locations. The thing is, one can mess with these parameters by making an arbitrary but an invertible coordinate transformation to the metric and therby change the flow. Also, perhaps somewhat related to this, it was pointed out here that a coordinate transformation, when viewed as a diffeomorphism, does not always map geodesics to geodesics unless it is an isometry.

3. So what is the correct way to think about this? Do all coordinate transformations/diffeomorphisms conserve global symmetries? or a subgroup of them?

I am self learning GR. This is a rather long post but I needed to clarify few things about the effect of general coordinate transformations on the global symmetries of metric. Any comments, insights are much appreciated.

To be concrete let's consider that $g_{\mu\nu}$ represents an axially symmetric spacetime i.e a Kerr black hole. In Boyer-Lindquist coordinates ($t,\, r,\, \theta,\, \phi $) metric has no dependence on $t$ and $\phi$ (they are cyclic coordinates):

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-(1-\frac{2Mr}{r^2+a^2\cos^2{\theta}})\,dt^2-\frac{r^2+a^2\cos^2{\theta}}{r^2+2Mr +a^2} dr^2 +(r^2+a^2\cos^2{\theta})\, d\theta^2 \\ \,\,\,+ \left( r^2+a^2+ \frac{2Ma^2 r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right) \sin^2{\theta}\, d\phi^2 - \frac{2Ma r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}} dt\, d\phi $$

Consequently particle's energy $E_0$ and angular momentum $L_0$ are conserved. Suppose I make a coordinate transformation: $d \bar{x}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,dx^{\nu}$ where, if I am not mistaken, $\Lambda$ is a matrix, which is an element of the general linear group GL$\,(4,\,R)$ with a non-zero determinant. In the new coordinates metric may not have cyclic coordinates, a good example would be representation of the above metric in Kerr-Schild coordinates:

$$\begin{align} g_{\mu\nu} dx^{\mu}dx^{\nu}=&-d\bar{t}^2+dx^2+dy^2+dz^2 \\ &+ \frac{2Mr^3}{r^4+a^2z^2} \left[d\bar{t} + \frac{r(x \,dx+y \,dy)+a(x\,dy-y\,dx)}{r^2+a^2} + \frac{z\,dz}{r} \right]^2\\ &\text{where}\qquad r^4+ (x^2+y^2+z^2)\, r^2 -a^2 z^2=0 \end{align} $$

Now there is only one cyclic coordinate, $\bar{t}$. I presume we may in principle introduce new coordinates where none of the coordinates appear to be cyclic in functional form of $g_{\mu\nu}\, dx^{\mu}dx^{\nu}$. Suppose I make such a coordinate transformation. My questions are:

1. Does the new metric still have global symmetries i.e conserved quantities that correspond to $E_0$ and $L_0$?

2. If the answer to the first question is yes, then suppose I gave this new metric to someone without telling about the coordinate transformation and ask if there are any symmetries. Will she/he able to find symmetries by finding closed orbits in the geodesic flow?

To me the answer to both questions above seemed to be yes. I think that the integrability (in the Liouville sense) of the metric should not depend on one's definition of the coordinates. In other words, because of the global symmetries, we expect bounded geodesics around the black hole. In the old coordinates we may easily calculate such closed trajectories and according to my thinking closed geodesics exist (objects revolve around the black hole) regardless of how we label the coordinates.

But I could not be sure of this. To explain my confusion let me write:

$$ \frac{d x^{\mu}}{ds^2} + \Gamma^{\mu}_{\nu \lambda} \frac{d x^{\nu}}{ds} \frac{d x^{\lambda}}{ds}=0 $$

which gives the geodesic flow. The existence of the bounded trajectories here depends on the number of zeros and poles of the Christoffel symbol, and also on their relative locations. The thing is, one can mess with these parameters by making an arbitrary but an invertible coordinate transformation to the metric and therby change the flow. Also, perhaps somewhat related to this, it was pointed out here that a coordinate transformation, when viewed as a diffeomorphism, does not always map geodesics to geodesics unless it is an isometry.

3. So what is the correct way to think about this? Do all coordinate transformations/diffeomorphisms conserve global symmetries? or a subgroup of them?

I am self learning GR. This is a rather long post but I needed to clarify few things about the effect of general coordinate transformations on the global symmetries of metric. Any comments, insights are much appreciated.

To be concrete let's consider that $g_{\mu\nu}$ represents an axially symmetric spacetime i.e a Kerr black hole. In Boyer-Lindquist coordinates ($t,\, r,\, \theta,\, \phi $) metric has no dependence on $t$ and $\phi$ (they are cyclic coordinates):

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-(1-\frac{2Mr}{r^2+a^2\cos^2{\theta}})\,dt^2-\frac{r^2+a^2\cos^2{\theta}}{r^2+2Mr +a^2} dr^2 +(r^2+a^2\cos^2{\theta})\, d\theta^2 \\ \,\,\,+ \left( r^2+a^2+ \frac{2Ma^2 r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right) \sin^2{\theta}\, d\phi^2 - \frac{2Ma r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}} dt\, d\phi $$

Consequently particle's energy $E_0$ and angular momentum $L_0$ are conserved. Suppose I make a coordinate transformation: $d \bar{x}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,dx^{\nu}$ where, if I am not mistaken, $\Lambda$ is a matrix, which is an element of the general linear group GL$\,(4,\,R)$ with a non-zero determinant. In the new coordinates metric may not have cyclic coordinates, a good example would be representation of the above metric in Kerr-Schild coordinates:

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-d\bar{t}^2+dx^2+dy^2+dz^2 + \frac{2Mr^3}{r^4+a^2z^2} \times\\ \qquad\qquad \left[d\bar{t} + \frac{r(x \,dx+y \,dy)+a(x\,dy-y\,dx)}{r^2+a^2} + \frac{z\,dz}{r} \right]^2\\ \text{where}\qquad r^4+ (x^2+y^2+z^2)\, r^2 -a^2 z^2=0 $$

Now there is only one cyclic coordinate, $\bar{t}$. I presume we may in principle introduce new coordinates where none of the coordinates appear to be cyclic in functional form of $g_{\mu\nu}\, dx^{\mu}dx^{\nu}$. Suppose I make such a coordinate transformation. My questions are

1. Does the new metric still have global symmetries i.e conserved quantities that correspond to $E_0$ and $L_0$  ?

2. If the answer to the first question is yes, then suppose I gave this new metric to someone without telling about the coordinate transformation and ask if there are any symmetries. Will she/he able to find symmetries by finding closed orbits in the geodesic flow  ?

To me the answer to both questions above seemed to be yes. I think that the integrability (in the Liouville sense) of the metric should not depend on one's definition of the coordinates. In other words, because of the global symmetries, we expect bounded geodesics around the black hole. In the old coordinates we may easily calculate such closed trajectories and according to my thinking closed geodesics exist (objects revolve around the black hole) regardless of how we label the coordinates.

But I could not be sure of this. To explain my confusion let me write:

$$ \frac{d x^{\mu}}{ds^2} + \Gamma^{\mu}_{\nu \lambda} \frac{d x^{\nu}}{ds} \frac{d x^{\lambda}}{ds}=0 $$

which gives the geodesic flow. The existence of the bounded trajectories here depends on the number of zeros and poles of the Christoffel symbol, and also on their relative locations. The thing is, one can mess with these parameters by making an arbitrary but an invertible coordinate transformation to the metric and therby change the flow. Also, perhaps somewhat related to this, it was pointed out here that a coordinate transformation, when viewed as a diffeomorphism, does not always map geodesics to geodesics unless it is an isometry.

So what is the correct way to think about this ? Do all coordinate transformations/diffeomorphisms conserve global symmetries ? or a subgroup of them ?

I am self learning GR. This is a rather long post but I needed to clarify few things about the effect of general coordinate transformations on the global symmetries of metric. Any comments, insights are much appreciated.

To be concrete let's consider that $g_{\mu\nu}$ represents an axially symmetric spacetime i.e a Kerr black hole. In Boyer-Lindquist coordinates ($t,\, r,\, \theta,\, \phi $) metric has no dependence on $t$ and $\phi$ (they are cyclic coordinates):

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-(1-\frac{2Mr}{r^2+a^2\cos^2{\theta}})\,dt^2-\frac{r^2+a^2\cos^2{\theta}}{r^2+2Mr +a^2} dr^2 +(r^2+a^2\cos^2{\theta})\, d\theta^2 \\ \,\,\,+ \left( r^2+a^2+ \frac{2Ma^2 r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right) \sin^2{\theta}\, d\phi^2 - \frac{2Ma r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}} dt\, d\phi $$

Consequently particle's energy $E_0$ and angular momentum $L_0$ are conserved. Suppose I make a coordinate transformation: $d \bar{x}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,dx^{\nu}$ where, if I am not mistaken, $\Lambda$ is a matrix, which is an element of the general linear group GL$\,(4,\,R)$ with a non-zero determinant. In the new coordinates metric may not have cyclic coordinates, a good example would be representation of the above metric in Kerr-Schild coordinates:

$$\begin{align} g_{\mu\nu} dx^{\mu}dx^{\nu}=&-d\bar{t}^2+dx^2+dy^2+dz^2 \\ &+ \frac{2Mr^3}{r^4+a^2z^2} \left[d\bar{t} + \frac{r(x \,dx+y \,dy)+a(x\,dy-y\,dx)}{r^2+a^2} + \frac{z\,dz}{r} \right]^2\\ &\text{where}\qquad r^4+ (x^2+y^2+z^2)\, r^2 -a^2 z^2=0 \end{align} $$

Now there is only one cyclic coordinate, $\bar{t}$. I presume we may in principle introduce new coordinates where none of the coordinates appear to be cyclic in functional form of $g_{\mu\nu}\, dx^{\mu}dx^{\nu}$. Suppose I make such a coordinate transformation. My questions are:

1. Does the new metric still have global symmetries i.e conserved quantities that correspond to $E_0$ and $L_0$?

2. If the answer to the first question is yes, then suppose I gave this new metric to someone without telling about the coordinate transformation and ask if there are any symmetries. Will she/he able to find symmetries by finding closed orbits in the geodesic flow?

To me the answer to both questions above seemed to be yes. I think that the integrability (in the Liouville sense) of the metric should not depend on one's definition of the coordinates. In other words, because of the global symmetries, we expect bounded geodesics around the black hole. In the old coordinates we may easily calculate such closed trajectories and according to my thinking closed geodesics exist (objects revolve around the black hole) regardless of how we label the coordinates.

But I could not be sure of this. To explain my confusion let me write:

$$ \frac{d x^{\mu}}{ds^2} + \Gamma^{\mu}_{\nu \lambda} \frac{d x^{\nu}}{ds} \frac{d x^{\lambda}}{ds}=0 $$

which gives the geodesic flow. The existence of the bounded trajectories here depends on the number of zeros and poles of the Christoffel symbol, and also on their relative locations. The thing is, one can mess with these parameters by making an arbitrary but an invertible coordinate transformation to the metric and therby change the flow. Also, perhaps somewhat related to this, it was pointed out here that a coordinate transformation, when viewed as a diffeomorphism, does not always map geodesics to geodesics unless it is an isometry.

3. So what is the correct way to think about this? Do all coordinate transformations/diffeomorphisms conserve global symmetries? or a subgroup of them?

I am self learning GR. This is a rather long post but I needed to clarify few things about the effect of general coordinate transformations on the global symmetries of metric. Any comments, insights are much appreciated.

To be concrete let's consider that $g_{\mu\nu}$ represents an axially symmetric spacetime i.e a Kerr black hole. In Boyer-Lindquist coordinates ($t,\, r,\, \theta,\, \phi $) metric has no dependence on $t$ and $\phi$ (they are cyclic coordinates):

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-(1-\frac{2Mr}{r^2+a^2\cos^2{\theta}})\,dt^2-\frac{r^2+a^2\cos^2{\theta}}{r^2+2Mr +a^2} dr^2 +(r^2+a^2\cos^2{\theta})\, d\theta^2 \\ \,\,\,+ \left( r^2+a^2+ \frac{2Ma^2 r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right) \sin^2{\theta}\, d\phi^2 - \frac{2Ma r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}} dt\, d\phi $$

Consequently particle's energy $E_0$ and angular momentum $L_0$ are conserved. Suppose I make a coordinate transformation: $d \bar{x}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,dx^{\nu}$ where, if I am not mistaken, $\Lambda$ is a matrix, which is an element of the general linear group GL$\,(4,\,R)$ with a non-zero determinant. In the new coordinates metric may not have cyclic coordinates, a good example would be representation of the above metric in Kerr-Schild coordinates:

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-d\bar{t}^2+dx^2+dy^2+dz^2 + \frac{2Mr^3}{r^4+a^2z^2} \times\\ \qquad\qquad \left[d\bar{t} + \frac{r(x \,dx+y \,dy)+a(x\,dy-y\,dx)}{r^2+a^2} + \frac{z\,dz}{r} \right]^2\\ \text{where}\qquad r^4+ (x^2+y^2+z^2)\, r^2 -a^2 z^2=0 $$

Now there is only one cyclic coordinate, $\bar{t}$. I presume we may in principle introduce new coordinates where none of the coordinates appear to be cyclic in functional form of $g_{\mu\nu}\, dx^{\mu}dx^{\nu}$. Suppose I make such a coordinate transformation. My questions are

1. Does the new metric still have global symmetries i.e conserved quantities that correspond to $E_0$ and $L_0$ ?

2. If the answer to the first question is yes, then suppose I gave this new metric to someone without telling about the coordinate transformation and ask if there are any symmetries. Will she/he able to find symmetries by finding closed orbits in the geodesic flow ?

To me the answer to both questions above seemed to be yes. I think that the integrability (in the Liouville sense) of the metric should not depend on one's definition of the coordinates. In other words, because of the global symmetries, we expect bounded geodesics around the black hole. In the old coordinates we may easily calculate such closed trajectories and according to my thinking closed geodesics exist (objects revolve around the black hole) regardless of how we label the coordinates.

But I could not be sure of this. To explain my confusion let me write:

$$ \frac{d x^{\mu}}{ds^2} + \Gamma^{\mu}_{\nu \lambda} \frac{d x^{\nu}}{ds} \frac{d x^{\lambda}}{ds} $$

which gives the geodesic flow. The existence of the bounded trajectories here depends on the number of zeros and poles of the Christoffel symbol, and also on their relative locations. The thing is, one can mess with these parameters by making an arbitrary but an invertible coordinate transformation to the metric and therby change the flow. Also, perhaps somewhat related to this, it was pointed out here that a coordinate transformation, when viewed as a diffeomorphism, does not always map geodesics to geodesics unless it is an isometry.

So what is the correct way to think about this ? Do all coordinate transformations/diffeomorphisms conserve global symmetries ? or a subgroup of them ?

I am self learning GR. This is a rather long post but I needed to clarify few things about the effect of general coordinate transformations on the global symmetries of metric. Any comments, insights are much appreciated.

To be concrete let's consider that $g_{\mu\nu}$ represents an axially symmetric spacetime i.e a Kerr black hole. In Boyer-Lindquist coordinates ($t,\, r,\, \theta,\, \phi $) metric has no dependence on $t$ and $\phi$ (they are cyclic coordinates):

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-(1-\frac{2Mr}{r^2+a^2\cos^2{\theta}})\,dt^2-\frac{r^2+a^2\cos^2{\theta}}{r^2+2Mr +a^2} dr^2 +(r^2+a^2\cos^2{\theta})\, d\theta^2 \\ \,\,\,+ \left( r^2+a^2+ \frac{2Ma^2 r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right) \sin^2{\theta}\, d\phi^2 - \frac{2Ma r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}} dt\, d\phi $$

Consequently particle's energy $E_0$ and angular momentum $L_0$ are conserved. Suppose I make a coordinate transformation: $d \bar{x}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,dx^{\nu}$ where, if I am not mistaken, $\Lambda$ is a matrix, which is an element of the general linear group GL$\,(4,\,R)$ with a non-zero determinant. In the new coordinates metric may not have cyclic coordinates, a good example would be representation of the above metric in Kerr-Schild coordinates:

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-d\bar{t}^2+dx^2+dy^2+dz^2 + \frac{2Mr^3}{r^4+a^2z^2} \times\\ \qquad\qquad \left[d\bar{t} + \frac{r(x \,dx+y \,dy)+a(x\,dy-y\,dx)}{r^2+a^2} + \frac{z\,dz}{r} \right]^2\\ \text{where}\qquad r^4+ (x^2+y^2+z^2)\, r^2 -a^2 z^2=0 $$

Now there is only one cyclic coordinate, $\bar{t}$. I presume we may in principle introduce new coordinates where none of the coordinates appear to be cyclic in functional form of $g_{\mu\nu}\, dx^{\mu}dx^{\nu}$. Suppose I make such a coordinate transformation. My questions are

1. Does the new metric still have global symmetries i.e conserved quantities that correspond to $E_0$ and $L_0$ ?

2. If the answer to the first question is yes, then suppose I gave this new metric to someone without telling about the coordinate transformation and ask if there are any symmetries. Will she/he able to find symmetries by finding closed orbits in the geodesic flow ?

To me the answer to both questions above seemed to be yes. I think that the integrability (in the Liouville sense) of the metric should not depend on one's definition of the coordinates. In other words, because of the global symmetries, we expect bounded geodesics around the black hole. In the old coordinates we may easily calculate such closed trajectories and according to my thinking closed geodesics exist (objects revolve around the black hole) regardless of how we label the coordinates.

But I could not be sure of this. To explain my confusion let me write:

$$ \frac{d x^{\mu}}{ds^2} + \Gamma^{\mu}_{\nu \lambda} \frac{d x^{\nu}}{ds} \frac{d x^{\lambda}}{ds}=0 $$

which gives the geodesic flow. The existence of the bounded trajectories here depends on the number of zeros and poles of the Christoffel symbol, and also on their relative locations. The thing is, one can mess with these parameters by making an arbitrary but an invertible coordinate transformation to the metric and therby change the flow. Also, perhaps somewhat related to this, it was pointed out here that a coordinate transformation, when viewed as a diffeomorphism, does not always map geodesics to geodesics unless it is an isometry.

So what is the correct way to think about this ? Do all coordinate transformations/diffeomorphisms conserve global symmetries ? or a subgroup of them ?