Poincaré disk: metric and isometric action

[Sajet]

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 "ds²" 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!

 

[fzero]

with metric

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

First question: I don't know this "ds²" 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.

 

[Sajet]

Thank you for your response. I'm sorry but I still couldn't figure this ds² notation out and I was hoping I could somehow get around it by using the other notation. So I tried the following:

$g_p(x, y) = \frac{<x,y>}{(1-|p|^2)^2}$

which would be identical to Ji's definition of ds² 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|²-|b|² = 1)

(Is this what the bottom of your post is referring to by dz'?)

Assuming this is correct, I would have to show:

$\frac{<x, y>}{(1-|p|^2)^2} = g_{Ap}(A_*x, A*y) = \frac{<\frac{2|a|^2-1}{(\bar b p+\bar a)^2}x, \frac{2|a|^2-1}{(\bar b p+\bar a)^2}y>}{(1-|\frac{ap+b}{\bar b p+\bar a}|^2)^2}= (\frac{2|a|^2-1}{(\bar b p+\bar a)^2})^2\frac{<x, y>}{(1-|\frac{ap+b}{\bar b p+\bar a}|^2)^2}$

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?

 

[fzero]

Thank you for your response. I'm sorry but I still couldn't figure this ds² notation out and I was hoping I could somehow get around it by using the other notation. So I tried the following:

$g_p(x, y) = \frac{<x,y>}{(1-|p|^2)^2}$

which would be identical to Ji's definition of ds² 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|²-|b|² = 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:

$\frac{<x, y>}{(1-|p|^2)^2} = g_{Ap}(A_*x, A*y) = \frac{<\frac{2|a|^2-1}{(\bar b p+\bar a)^2}x, \frac{2|a|^2-1}{(\bar b p+\bar a)^2}y>}{(1-|\frac{ap+b}{\bar b p+\bar a}|^2)^2}= (\frac{2|a|^2-1}{(\bar b p+\bar a)^2})^2\frac{<x, y>}{(1-|\frac{ap+b}{\bar b p+\bar a}|^2)^2}$

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.

 

[Sajet]

Thanks. You're right, I made a mistake in the first part. |a|² - |b|², 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...

 

[fzero]

Thanks. You're right, I made a mistake in the first part. |a|² - |b|², 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...

In the real case, we have $SL(2,\mathbb{R})$, which is isomorphic to $SU(1,1)$ and $SO^+(1,2)$. However, the $SO^+(1,2)$ actually acts on coordinates $(x_0,x_1,x_2)$ which are constrained to lie on the hyperboloid

$$x_0^2 - x_1^2 -x_2^2=1$$

with $x_0>0$.

I'm sure once the details are worked out, everything would fit together. But it's much simpler to use the complex coordinate.

 

[Sajet]

Thanks for all your help! I haven't figured it out completely but I get the basic idea now and I'm sure I'll get there soon.