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What is the mathematical relationship between Christoffel symbol and

  1. Apr 23, 2008 #1
    What is the mathematical relationship between Christoffel symbol and
    connection A in Ashtekar variables?

    Best,

    Youngsub
     
  2. jcsd
  3. Apr 29, 2008 #2
    Re: What is the mathematical relationship between Christoffel symboland

    On Apr 22, 3:39 pm, yy...@fas.harvard.edu wrote:
    > What is the mathematical relationship between Christoffel symbol and
    > connection A in Ashtekar variables?
    >
    > Best,
    >
    > Youngsub


    The Christoffel symbols define the Levi-Civita connection on various
    tensor bundles.
    The Ashtekar type of connection is the self-dual spin connection.
    Here is a pretty detailed explanation with formulas, etc.:

    http://arxiv.org/pdf/gr-qc/9312032v2

    charlie torre
     
  4. May 4, 2008 #3
    Re: What is the mathematical relationship between Christoffel symboland

    On Apr 22, 4:39 pm, yy...@fas.harvard.edu wrote:
    > What is the mathematical relationship between Christoffel symbol and
    > connection A in Ashtekar variables?
    >
    > Best,
    >
    > Youngsub


    Gamma = christoffel connection. K = contorsion (which can be expressed
    in terms of the torsion, but I won't do it here).
    The Einstein-Cartan connection is omega = Gamma + K.

    The appropriate question is what is the relation between *omega* and
    Ashtekhar?

    omega is a 1-form: omega^a_b = (Gamma^a_{cb} + K^a_{cb}) e^c. The
    indices (a,b,c) are generally for non-coordinate frames here.

    You can generalize Gamma to non-coordinate frames by writing
    Gamma^a_{cb} = e^a.(Del_c e_b); expressing the last item in terms of
    Christoffel coefficients, if you wish, by using the usual expression
    for the covariant derivative.
    (Del_c e_b)^m = e_c^n (Del_n e_b)^m = e_c^n (de_b^m/dx^n +
    Gamma^m_{nr} e_b^r)
    where (m,n,r) are coordinate indices.

    For an orthonormal frame the frame metric is g(e_a,e_b) = eta_{ab} =
    Minkowski metric and the connection, when its indices are lowered,
    becomes
    omega_{ab} = eta_{ac} omega^c_b.
    It is anti-symmetric; omega_{ab} = -omega_{ba}. There are 6
    independent connection 1-forms ... equalling the number of independent
    degrees of freedom in the Lorentz group. It's a Lorentz connection.

    The split of Lorentz SO(3,1) into a left and right helicity SU(2) is
    effectively what's done here. The "self-dual" part is the right
    helicity part, the "anti-self-dual" part is the left helicity part.
    Explicitly, in units where c = 1, the two parts are A_{01} =
    omega_{01} +/- i omega_{23}. I forget which goes with the +i and which
    with the -i.

    When the Immirzi coefficient beta is used in place of +/-i then
    something else is revealed to actually be going on underneath all the
    Ashtekhar wizardry, which gets to the real root of your question:
    "what is all this extra stuff doing and effecting, in relation to the
    familiar connection?"

    That is as follows. The action for the Einstein-Hilbert action stated
    in the language of differential forms is
    S_{EH} = integral A epsilon_{abcd} Omega^{ab} ^ e^c ^ e^d
    where A is some coefficient I'm not concerned with here, and
    Omega^{ab} is the curvature 2-form Omega^{ab} = 1/2 (R^{ab}_{cd} e^c ^
    e^d); and is related to little omega by the Cartan equation
    Omega^a_c = d(omega^a_c} + omega^a_b ^ omega^b_c.

    The "Ashtekhar" trick modifies omega to the following form
    omega^{ab} --> omega^{ab} + i/2 epsilon_{abcd} omega_{cd}.
    Taking the same Cartan equatio with respect to the modified omega
    leads to the following modified Omega as a result
    Omega^{ab} -> Omega^{ab} + i/2 epsilon_{abcd} omega_{cd}.

    The generalization with the Immirzi parameter replaces i by beta. The
    resulting action is
    S = integral A epsilon_{abcd} (Omega^{ab} + beta/2 epsilon^{abef}
    Omega_{ef}) ^ e^c ^ e^d
    which works out to
    S = S_{EH} + (A beta) integral Omega_{cd} ^ e^c ^ e^d.

    The last integral is a parity violating term. In retrospect, you'd
    expect to see it, since the splitting into self-dual vs. anti-self-
    dual is a split into right and left helicity parts. So the difference
    between the two terms is a difference that goes to its negative under
    parity reversal.

    In the absence of torsion the components R^{abcd} of Omega^{ab} are
    the oridinary Riemannian tensor components derived from the
    Christoffel symbols. Otherwise, more generally, R^{abcd} will be the
    curvature tensor associated with the Einstein-Cartan connection omega
    = Gamma + K.

    The last integral reduces to an expression proportional to
    integral epsilon_{mnrs} R_{mnrs} d^4 x
    in coordinate indices. That's 0 when R_{mnrs} is the Riemann tensor
    derived from Christoffel, since one has the cyclic identity
    R_{mnrs} + R_{mrsn} + R_{msnr} = 0.
    It's non-zero in the presence of torsion.

    In effect, beta is the coefficient for an additional parity-violating
    contribution to the Einstein-Hilbert action. The wizardry with the i's
    that Ashtekhar started out with originally is seen here to apparently
    have never been anything more than a red herring. When the smoke
    clears, all they're really doing, it appears, is simply going with the
    usual Einstein-Cartan connection, but with a non-Einstein-Hilbert
    action.

    In the absence of external sources will produce as one of its field
    equations that torsion is 0 and the action above will evaluate to that
    of the ordinary Einstein Hilbert action. But even there, the
    symplectic structure of the field theory (quantum AND classical) is
    dependent on what's in the Lagrangian -- even those parts that aren't
    involved in the dynamics. So the inclusion of a parity violating term
    will have an effect on the definition of what constitutes the
    conjugate momentum (and the definition of the Hamiltonian), even when
    beta itself is not seen in the dynamics.

    You can generalize everything and here the picture becomes really
    clear. Take ALL the possible algebraic combinations of frame 1-form
    (e^a) and 2-forms (T^a for torsion, Omega^a_b) and include them in the
    Lagrangian. The resulting action looks like this:
    S = integral (A L_A + B L_B + C L_C + D L_D + E L_E + F L_F)
    with the Lagrangian consisting of the 6 terms (ignoring constant
    coefficients)
    L_A = epsilon_{abcd} Omega^{ab} ^ e^c ^ e^d --- Einstein-Hilbert
    L_B = Omega_{ab} ^ e^a ^ e^b -- Parity violating part, as related
    above
    L_C = epsilon_{abcd} e^a ^ e^b ^ e^c ^ e^d -- Cosmological constant
    part
    L_D = epsilon_{abcd} Omega^{ab} ^ Omega^{cd}
    L_E = epsilon_{abcd} Omega^{ab} ^ Omega_{ab}
    L_F = epsilon_{abcd} T^a ^ T_a
    The torsion part L_F is equivalent (up to boundary term) to L_B. Only
    a combination of the coefficints B and F appear in the dynamics (B+ kF
    for some constant multiple k). The terms L_D, L_E are boundary terms,
    so that the coefficients D and E do not appear in the dynamics.

    But all 6 coefficients are crucually involved in the definition of the
    symplectic structure and Hamiltonian. In fact, if D and E are not both
    0, then the "electric field is the area" creed of Ashtekhar can be
    replaced by "the electric field is the CURVATURE (plus a multiple of
    the area)" -- thus making gravity a Yang-Mills theory for the Lorentz
    group. A similar consideration applies for the field conjugate to
    (e^a), which is then related to the torsion, itself, by the
    coefficient F. The nice thing about F is that you can use it to off-
    set the Immirzi coefficient (which up to multiples of A is just B) and
    make the Lagrangian parity symmetric.
     
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