## Poincaré disk: metric and isometric action

Hi!

I'm trying to give a few examples of symmetric manifolds. In the article "Introduction to Symmetric Spaces and Their Compactification" Lizhen Ji mentions the Poincaré disk as a symmetric space in the following way:

$D = \{z \in \mathbb C | |z| < 1\}$

with metric

$ds^2 = \frac{|dz|^2}{(1-|z|^2)^2}$.

First question: I don't know this "ds˛" notation and I wasn't able to figure out how to convert it into the more familiar notation $g_p(x, y) = ...$. Is this possible? Last semester we defined a Riemannian metric on the Poincaré disk as follows:

$g_p(x, y) = \frac{4\langle x, y\rangle}{(1-||p||^2)^2}$

Is this the same metric?

The second thing:

Ji says that the group

$SU(1, 1) = \{\begin{pmatrix}a && b \\ \bar b && \bar a\end{pmatrix} | a, b \in \mathbb C, |a|^2-|b|^2 = 1\}$

acts isometrically and transitively on D by setting

$\begin{pmatrix}a && b \\ \bar b && \bar a\end{pmatrix}z = \frac{az+b}{\bar b z+\bar a}$

But he doesn't prove this and instead says "This follows by a direct computation".

I would like to do this computation, and as I see this I need to show:

$g_{Az}(A_*x, A_*y) = g_z(x, y)$

But then again I would need to know the metric explicitly and also unfortunately I couldn't figure out how to compute the derivative $A_*x$ for this group action :(

I would be very grateful for some help with this. Thank you in advance!
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 Quote by Sajet with metric $ds^2 = \frac{|dz|^2}{(1-|z|^2)^2}$. First question: I don't know this "ds˛" notation and I wasn't able to figure out how to convert it into the more familiar notation $g_p(x, y) = ...$. Is this possible? Last semester we defined a Riemannian metric on the Poincaré disk as follows: $g_p(x, y) = \frac{4\langle x, y\rangle}{(1-||p||^2)^2}$ Is this the same metric?
You should be able to show that, up to numerical factors, these are equivalent. Namely,

$$ds^2 = \frac{1}{4} g_z(dz,d\bar{z}).$$

 The second thing: Ji says that the group $SU(1, 1) = \{\begin{pmatrix}a && b \\ \bar b && \bar a\end{pmatrix} | a, b \in \mathbb C, |a|^2-|b|^2 = 1\}$ acts isometrically and transitively on D by setting $\begin{pmatrix}a && b \\ \bar b && \bar a\end{pmatrix}z = \frac{az+b}{\bar b z+\bar a}$ But he doesn't prove this and instead says "This follows by a direct computation". I would like to do this computation, and as I see this I need to show: $g_{Az}(A_*x, A_*y) = g_z(x, y)$ But then again I would need to know the metric explicitly and also unfortunately I couldn't figure out how to compute the derivative $A_*x$ for this group action :( I would be very grateful for some help with this. Thank you in advance!
If you verify that

$$dz' = \frac{dz}{(\bar{b} z + \bar{a})^2}$$

(which is presumably related to your ##A_*##) and

$$1-|z'|^2 = \frac{1-|z|^2}{|\bar{b} z + \bar{a}|^2},$$

you can show that the metric tensor is invariant.
 Thank you for your response. I'm sorry but I still couldn't figure this ds˛ notation out and I was hoping I could somehow get around it by using the other notation. So I tried the following: $g_p(x, y) = \frac{}{(1-|p|^2)^2}$ which would be identical to Ji's definition of ds˛ according to what you said at the beginning. I tried to calculate $A_*x$ using: $\frac{d}{dt}_{t=0} \frac{a\gamma(t)+b}{\bar b\gamma(t)+\bar a}$ for some curve $\gamma(0) = p, \dot\gamma(0) = x$ which gave me $\frac{|a|^2+|b|^2}{(\bar b p+\bar a)^2}x = \frac{2|a|^2-1}{(\bar b p+\bar a)^2} x$ (since |a|˛-|b|˛ = 1) (Is this what the bottom of your post is referring to by dz'?) Assuming this is correct, I would have to show: $\frac{}{(1-|p|^2)^2} = g_{Ap}(A_*x, A*y) = \frac{<\frac{2|a|^2-1}{(\bar b p+\bar a)^2}x, \frac{2|a|^2-1}{(\bar b p+\bar a)^2}y>}{(1-|\frac{ap+b}{\bar b p+\bar a}|^2)^2}= (\frac{2|a|^2-1}{(\bar b p+\bar a)^2})^2\frac{}{(1-|\frac{ap+b}{\bar b p+\bar a}|^2)^2}$ But I don't get any further from here since I don't how to simplify $|\frac{ap+b}{\bar b p+\bar a}|^2$. Is this even correct up to this point?

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## Poincaré disk: metric and isometric action

 Quote by Sajet Thank you for your response. I'm sorry but I still couldn't figure this ds˛ notation out and I was hoping I could somehow get around it by using the other notation. So I tried the following: $g_p(x, y) = \frac{}{(1-|p|^2)^2}$ which would be identical to Ji's definition of ds˛ according to what you said at the beginning. I tried to calculate $A_*x$ using: $\frac{d}{dt}_{t=0} \frac{a\gamma(t)+b}{\bar b\gamma(t)+\bar a}$ for some curve $\gamma(0) = p, \dot\gamma(0) = x$ which gave me $\frac{|a|^2+|b|^2}{(\bar b p+\bar a)^2}x = \frac{2|a|^2-1}{(\bar b p+\bar a)^2} x$ (since |a|˛-|b|˛ = 1) (Is this what the bottom of your post is referring to by dz'?)
I find

$$\frac{d}{dt}_{t=0} \frac{a\gamma(t)+b}{\bar b\gamma(t)+\bar a}=\frac{|a|^2-|b|^2}{(\bar b p+\bar a)^2}x$$

This is essentially the same computation as for ##dz'##.

 Assuming this is correct, I would have to show: $\frac{}{(1-|p|^2)^2} = g_{Ap}(A_*x, A*y) = \frac{<\frac{2|a|^2-1}{(\bar b p+\bar a)^2}x, \frac{2|a|^2-1}{(\bar b p+\bar a)^2}y>}{(1-|\frac{ap+b}{\bar b p+\bar a}|^2)^2}= (\frac{2|a|^2-1}{(\bar b p+\bar a)^2})^2\frac{}{(1-|\frac{ap+b}{\bar b p+\bar a}|^2)^2}$ But I don't get any further from here since I don't how to simplify $|\frac{ap+b}{\bar b p+\bar a}|^2$. Is this even correct up to this point?
You haven't used the fact that the coordinates are complex, so presumably you should have ##(A_*x, A_*\bar{y})## or maybe ##(A_*x, A_*\bar{y})+(A_*\bar{x}, A_*y)## in the metric. With the appropriate expression, the factors cancel as in the computation I mentioned.
 Thanks. You're right, I made a mistake in the first part. |a|˛ - |b|˛, and thus 1, is correct. With regard to the second part, I'm not really used to complex manifolds... I thought, x and y were real since they are in the tangent space. Would it be possible to transfer the entire thing to real coordinates? Meaning $D = \{p \in \mathbb R^2 | |p| < 1\}$ and then take SO(1, +) (?) as the isometry group? Or would some mathematical insight be lost when doing that? Or maybe I should just stick to a simpler example...

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 Quote by Sajet Thanks. You're right, I made a mistake in the first part. |a|˛ - |b|˛, and thus 1, is correct. With regard to the second part, I'm not really used to complex manifolds... I thought, x and y were real since they are in the tangent space. Would it be possible to transfer the entire thing to real coordinates? Meaning $D = \{p \in \mathbb R^2 | |p| < 1\}$ and then take SO(1, +) (?) as the isometry group? Or would some mathematical insight be lost when doing that? Or maybe I should just stick to a simpler example...
$$x_0^2 - x_1^2 -x_2^2=1$$