What is the mathematical relationship between Christoffel symbol and

  • Mathematica
  • Thread starter yyoon@fas.harvard.edu
  • Start date
  • #1
What is the mathematical relationship between Christoffel symbol and
connection A in Ashtekar variables?



Answers and Replies

  • #2

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.:


charlie torre
  • #3

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

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,
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

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
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
L_C = epsilon_{abcd} e^a ^ e^b ^ e^c ^ e^d -- Cosmological constant
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.

Suggested for: What is the mathematical relationship between Christoffel symbol and

  • Last Post