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Poincaré disk: metric and isometric action

  1. Jun 21, 2012 #1
    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:

    [itex]D = \{z \in \mathbb C | |z| < 1\}[/itex]

    with metric

    [itex]ds^2 = \frac{|dz|^2}{(1-|z|^2)^2}[/itex].

    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 [itex]g_p(x, y) = ...[/itex]. Is this possible? Last semester we defined a Riemannian metric on the Poincaré disk as follows:

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

    Is this the same metric?

    The second thing:

    Ji says that the group

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

    acts isometrically and transitively on D by setting

    [itex]\begin{pmatrix}a && b \\ \bar b && \bar a\end{pmatrix}z = \frac{az+b}{\bar b z+\bar a}[/itex]

    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:

    [itex]g_{Az}(A_*x, A_*y) = g_z(x, y)[/itex]

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

    I would be very grateful for some help with this. Thank you in advance!
     
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  3. Jun 21, 2012 #2

    fzero

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    You should be able to show that, up to numerical factors, these are equivalent. Namely,

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

    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.
     
  4. Jun 22, 2012 #3
    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:

    [itex]g_p(x, y) = \frac{<x,y>}{(1-|p|^2)^2}[/itex]

    which would be identical to Ji's definition of ds² according to what you said at the beginning.

    I tried to calculate [itex]A_*x[/itex] using:

    [itex]\frac{d}{dt}_{t=0} \frac{a\gamma(t)+b}{\bar b\gamma(t)+\bar a}[/itex]

    for some curve [itex]\gamma(0) = p, \dot\gamma(0) = x[/itex] which gave me

    [itex]\frac{|a|^2+|b|^2}{(\bar b p+\bar a)^2}x = \frac{2|a|^2-1}{(\bar b p+\bar a)^2} x[/itex] (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:

    [itex]\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}[/itex]

    But I don't get any further from here since I don't how to simplify [itex]|\frac{ap+b}{\bar b p+\bar a}|^2[/itex].

    Is this even correct up to this point?
     
  5. Jun 22, 2012 #4

    fzero

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    I find

    [tex]\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[/tex]

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

    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.
     
    Last edited: Jun 22, 2012
  6. Jun 22, 2012 #5
    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

    [itex]D = \{p \in \mathbb R^2 | |p| < 1\}[/itex]

    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...
     
  7. Jun 22, 2012 #6

    fzero

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    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.
     
  8. Jun 24, 2012 #7
    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.
     
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