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Conformal killing vector

  1. Mar 7, 2010 #1
    Hey guys, I'm working on Polchinski's string book, and I have a problem. Around page 152 he uses an identity I'm not sure how to prove. Essentially he wants to compute conformal killing vector fields. So we have the eq for a CKV:

    [tex]
    P_{ab}=\Delta_a \xi_b+ \Delta_b\xi_a- g_{ab}\Delta_c\xi^c=0
    [/tex]

    And what I need to prove is that, in general

    [tex]
    P_{ab}P^{ab}=\Delta_a \xi_b\Delta^a \xi^b-R \xi_a\xi^a,
    [/tex]

    Where R is the Ricci scalar. This identity might (or might not) only work when integrated over, ie
    [tex]

    \int d^2x \sqrt{g}P_{ab}P^{ab} = ...
    [/tex]
    I suspect this might be useful to get some integration by parts.

    My GR is a little rusty, so I'm not sure how to get it. I can expand all the Lie derivatives, collect some terms, etc, and I get:

    [tex]
    2\Delta_a \xi_b\Delta^a \xi^b+2(\xi_{b,a}-\Gamma^i_{ba}\xi_i)(\xi^a_{,b}+\Gamma^a_{jb}\xi^j)-g^{ab}\Delta_c\xi^c(\xi_{b,a}+\xi_{a,b}-2\Gamma^{i}_{ba}\xi_i)-g_{ab}\Delta_c\xi^c(\xi^b_{,a}+\xi^a_{,b}+\Gamma^b_{ia}\xi^i+\Gamma^a_{ib}\xi^i)
    [/tex]

    Am I on the right track? I just don't see how to get a Ricci scalar out of all those terms..For example I would need two terms like
    [tex]
    \Gamma^a_{bc}\Gamma^c_{de}-\Gamma^a_{ec}\Gamma^c_{db}
    [/tex]

    If anyone did a calculation like this before or can easily spot that it does work out, please let me know! I just need to know if I'm the right track...GR calculations seem to usually work out, but for this one I'm lacking a general sense of direction in manipulating the indices ...you need like a million contractions to get the Ricci scalar :) Thanks for any help!
     
  2. jcsd
  3. Mar 7, 2010 #2
    You can try to do get the equation

    [tex]P^{ab}P_{ab}=4\xi_{b;a}g^{ab}(1-\xi^c_{;c})+\xi^c_{;c}\xi^c_{;c}.[/tex]

    Can't you?

    ---------------------------------
    How do you define [tex]\Delta^c\xi_c[/tex]? That is so rare in GR contexts!
     
  4. Mar 7, 2010 #3
    Well I assume
    [tex]
    \Delta^c\xi_c=g^{ac}\Delta_a\xi_c
    [/tex]
    I mean that's the only thing I can imagine..otherwise I don't know what P^{ab} would be.

    I'm not sure how to get to your equation. If I write everything out I get:
    [tex]
    P_{ab}P^{ab}=(\Delta_a\xi_b+\Delta_b\xi_a)(\Delta^a\xi^b+\Delta^b\xi^a)+g_{ab}g^{ab}\Delta_c\xi^c\Delta_c\xi^c+ \text{etc}
    [/tex]
    and
    [tex]
    (\Delta_a\xi_b+\Delta_b\xi_a)(\Delta^a\xi^b+\Delta^b\xi^a)=2\Delta_a\xi_b(\Delta^a\xi^b+\Delta^b\xi^a)
    [/tex]

    [tex]
    g_{ab}g^{ab}\Delta_c\xi^c\Delta_c\xi^c= 2 \Delta_c\xi^c\Delta_c\xi^c
    [/tex]

    [tex]
    etc=-2\Delta_c\xi^cg^{ab}(\Delta_b\xi_a+\Delta_a\xi_b)
    [/tex]
    It doesn't seem to add up to what you have :-?
    For example, where would
    [tex]
    4\xi_{b;a}g^{ab}
    [/tex]
    come from?
     
  5. Mar 8, 2010 #4

    Stingray

    User Avatar
    Science Advisor

    That identity is only true when integrating by parts (and off by an overall factor of 2). First prove that
    [tex]
    \nabla_a \nabla_b \xi^b = 0
    [/tex]
    for a conformal Killing vector. You then integrate by parts and use Ricci's identity plus the fact that a Ricci tensor in 2 dimensions is entirely determined by the Ricci scalar (and the metric).

    I don't have Polchinski, so I'm curious. Is he trying to write down a variation principle for an "approximate conformal Killing vector?"
     
  6. Mar 8, 2010 #5

    Stingray

    User Avatar
    Science Advisor

    I should probably add some extra information: In these kinds of problems, it is almost never a good idea to expand things using Christoffel symbols. The equation defining a conformal Killing vector is actually independent of the choice of derivative operator if written as [itex]\mathcal{L}_\xi g_{ab} = g_{ab} \phi[/itex]. The Lie derivative expands to
    [tex]
    \mathcal{L}_\xi g_{ab} = \xi^c D_c g_{ab} + g_{ac} D_b \xi^c + g_{cb} D_a \xi^c
    [/tex]
    for any derivative operator [itex]D_a[/itex].

    In the two dimensions you're considering, [itex]\phi = \nabla_a \xi^a[/itex] for the metric-compatible derivative operator [itex]\nabla_a[/itex]. It is, of course, convenient to use this derivative when integrating by parts. You then use the general identity
    [tex]
    (\nabla_a \nabla_b - \nabla_b \nabla_a) \xi^c = - R_{abd}{}^{c} \xi^d
    [/tex]
    to eventually get the Ricci tensor and then the Ricci scalar.
     
  7. Mar 8, 2010 #6
    Ok I think I got the general idea now, thanks! I can get it to this form
    [tex]
    2\nabla_a\xi_b\nabla^a\xi^b+2\nabla_a\xi_b\nabla^b\xi^a-2\nabla_a\xi^a\nabla_b\xi^b
    [/tex]
    and I assume that now I should use integration by parts. However some details are still escaping me.

    For example, how does integrating by parts precisely work? Naively I assumed from the start it should somehow work like
    [tex]
    \int d^2 x \sqrt{g} \nabla_a\xi^b\nabla_b\xi^a=\nabla_a\xi^b\xi^a|-\int d^2 x \sqrt{g} \nabla_b\nabla_a\xi^b\xi^a
    [/tex]
    But is it really just that simple?

    And now is it immediately clear that the surface term is zero?

    And third, I'm not very sure how or why your expression
    [tex]\nabla_a \nabla_b \xi^b = 0[/tex],
    appears. Should this be the surface term and what I did above is wrong? :-?

    Unless I'm mistaken, I think Polchinski does something like that in some other chapter, but here his goal is to find the number of conformal killing vectors (k=dim ker h_{ab} )and number of metric moduli (\mu=dim ker f_{ab}^T) for a surface of genus g. It turns out that by the Riemann-Roch theorem
    \mu-k=-3\chi
    He needs this when computing a path integral for strings.
    You can find a better description of what is going on here if you're interested.
    http://www.phys.columbia.edu/~kabat/strings/Spring08/handout3.pdf
     
  8. Mar 8, 2010 #7
    Well, I don't know why Polchinski writes the equation for P_{ab} as [tex]P_{ab}=\Delta_a \xi_b+ \Delta_b\xi_a- g_{ab}\Delta_c\xi^c=0[/tex] which is incorrect. It must be

    [tex]P_{ab}=\Delta_a \xi_b+ \Delta_b\xi_a- \frac{1}{4}g_{ab}\Delta_c\xi^c=0[/tex]

    in a four dimensional spacetime (see http://en.wikipedia.org/wiki/Conformal_Killing_equation).

    If we are required to find the contravariant tensor [tex]T^{ab}[/tex] from its covariant version [tex]T_{cd}[/tex], simply use the rule

    [tex]T^{ab}=g^{ac}g^{bd}T_{cd}.[/tex]

    Now for P^{ab} one would get

    [tex]P^{ab}=g^{ac}g^{bd}P_{cd}=(\Delta_c \xi_d+ \Delta_d\xi_c- \frac{1}{4}g_{cd}\Delta_h\xi^h)g^{bd}g^{ac}.[/tex]

    AB
     
    Last edited: Mar 9, 2010
  9. Mar 8, 2010 #8

    Stingray

    User Avatar
    Science Advisor

    Integration by parts is that simple. If you have some volume [itex]W[/itex],
    [tex]
    \int_W \nabla_a A^a dV = \int_{\partial W} A^a dS_a,
    [/tex]
    where I'm using covariant volume elements.

    I don't really see why the surface terms should vanish, but I haven't thought about it. Maybe they are there, but ignored "for simplicity" or because they aren't needed for what's coming next (something annoyingly common in physics books).

    This shows up twice. You integrate two terms by parts. One goes away (up to a surface term) due to this identity and the other is reduced to the Ricci term.

    Altabeh: Polchinski is undoubtedly discussing string theory. From what negru has written, this is taking placing on the 2D manifold that is the string worldsheet.
     
  10. Mar 8, 2010 #9
    Hmm I still don't see it. Starting from
    [tex]2\nabla_a\xi_b\nabla^a\xi^b+2\nabla_a\xi_b\nabla^b \xi^a-2\nabla_a\xi^a\nabla_b\xi^b[/tex]

    which I got above, the first term is ok, I need one like that, so I integrate by parts the other two, and I get
    [tex]
    s.t. -2 \int \nabla^b\nabla_a\xi_b\ \xi^a - s.t. + 2 \int \nabla_b\nabla_a\xi^a\xi^b
    =2\int (-\nabla_b\nabla_a\xi^b\xi^a+\nabla_a\nabla_b\xi^b\xi^a) \text{etc}
    [/tex]
    and that's where I get my Ricci tensor. Isn't that how it's supposed to work?

    Altabeh, the formula should have a 2/d factor where you put that 1/4, I'm not sure what's going on. And yes I should've mentioned this was for d=2, sorry about that.
     
  11. Mar 8, 2010 #10

    Stingray

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    Science Advisor

    Yes, that looks right. I was making it more complicated than necessary.
     
  12. Mar 8, 2010 #11
    cool, thanks again!
     
  13. Mar 9, 2010 #12
    Yeah, I checked my notes and noticed that there is a "2" missing in the last term so for a 2D space this should reduce to what you exactly wrote.

    Got it!

    AB
     
    Last edited: Mar 9, 2010
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