preserving the metric

Question

I am a bit confused with what it means to preserve the Riemannian metric.

1. The link below says that SL2(R) action preserves the hyperbolic half plane metric. https://en.wikipedia.org/wiki/Poincar%C3%A9_metric#Metric_and_volume_element_on_the_Poincar.C3.A9_plane

2. The link below says that ANY coordinate transform preserves the metric. https://en.wikipedia.org/wiki/Metric_tensor#Invariance_of_arclength_under_coordinate_transformations

Is the SL2(R) action a change of variable (coordinate transform)? The result in the second link(2) seems too powerful...

how do you derive the result in the first link (1)? how would I go about showing an action such as vertical translation on H2 affects the Riemannian metric? Please clear up the confusion.

Answer 1

These two articles address different issues.

1. If $(M,g)$ is a Riemannian manifold then a diffeomorphism $f: M\to M$ is said to preserve $g$ (or be an isometry of $g$) if $f^*(g)=g$. In other words, for each $x\in M$ and any two tangent vectors $u, v\in T_xM$, we have $$ g(u,v)= g(df(u), df(v)). $$ (You may prefer to use the notation $<u,v>$ instead of $g(u,v)$.)

How do you check this property in practice is another matter.

2. The point of the second statement is that the notion of length of a curve depends only on the curve and on the metric and not on some local coordinates. In fact, the length of a curve $c: [0,1]\to M$ is defined in a coordinate-free form as $$ \int_0^1 \sqrt{<c'(t), c'(t)>} dt. $$

Answer 2

Both link 1 and 2 involve "preserving the metric" by isometries.

link 1 is special in that $ds^2=(dx^2+dy^2)/y^2= dzd\bar{z}/im(z) = dz'd\bar{z'}/im(z')= ((dx')^2+(dy')^2)/y^2 =(ds')^2$ have the same expression so $(ds)_z^2(u)= (ds')_z^2(u)$ for $u \in T_z(\mathbb{H})$

link 2 is a more general result; it says that in two dimensional M with any coordinates, $(ds)^2_p(u)$= $(ds')^2_{F(p)}(dF(u))$ $u \in T_pM$ for p=(x,y) \in M and $F(x,y)= (x',y')$,as mentioned in the other answer, .

We prove this for the case of changing cartesian coordinates to polar coordinates. We work out the case of $ds^2= dx^2+dy^2= dr^2+r^2d\theta^2= (ds')^2$. $F(x,y)= (r,\theta)$ using the chain rule. Notice a different basis in the domain must be used for $(ds')^2$ this was implicit in the change of covectors.

We find L, the linear map that changes the $d/dx, d/dy$ basis to the $d/dr$, $d/d\theta$ basis. $L(d/dx)= d/dr \cdot dr/dx + d/d\theta \cdot d \theta/dx$

$L(d/dy)= d/dr \cdot dr/dy + d/d\theta \cdot d \theta/dy$

thus L in matrix form is $ \left( \begin{array}{ccc} dr/dx & dr/dy \\ d\theta/dx & d\theta/dy \end{array} \right) $ = dF (in it's matrix form)

therefore: $ds^2_p(u)= (ds')^2_{F(p)}(dF(p)u)$ for $p \in \mathbb{R}^2$, $u \in T_p(\mathbb{R}^2)$

I believe that link 1 can be interpreted as link 2 for F(x,y)= (x',y') induced by G(z)= (az+b)/(cz+d) =z'$