- #1

yyoon@fas.harvard.edu

connection A in Ashtekar variables?

Best,

Youngsub

- Mathematica
- Thread starter yyoon@fas.harvard.edu
- Start date

- #1

yyoon@fas.harvard.edu

connection A in Ashtekar variables?

Best,

Youngsub

- #2

torre@cc.usu.edu

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

- #3

Rock Brentwood

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.

- Replies
- 8

- Views
- 50K

- Replies
- 6

- Views
- 407

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 5K

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 3

- Views
- 6K

- Replies
- 8

- Views
- 2K

- Replies
- 5

- Views
- 4K

- Replies
- 3

- Views
- 4K

- Replies
- 1

- Views
- 2K