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Indexes: \(\alpha-\omega\) = spacetime curved, \(a-h,o-z\) = spatial curved, \(A-Z\) = spacetime flat, \(i-n\) = spatial flat
spatial curved indexes are raised and lowered by the spatial metric \( \gamma_{ij} \)
spatial flat indexes are raised and lowered by the spatial components of the Minkowski metric, i.e. \( \delta_{ij} \)

Extrinsic curvature in triad representation
${K_a}^i = K_{a\mu} e^{\mu i}$
$= K_{at} e^{ti} + K_{ab} e^{bi}$
$= K_{at} \cdot 0 + K_{ab} e^{bi}$
$= K_{ab} e^{bi}$
$= K_{ab} \cdot (^3e)^{bi}$

Levi-Civita covariant relation between 3D and 4D:
$\epsilon_{0ijk} = \epsilon_{ijk}$ by definition

Levi-Civita contravariant relation between 3D and 4D:
$-\epsilon^{0ijk} = \epsilon_{0ijk}$ because $\epsilon^{IJKL} = \sigma \epsilon_{IJKL} = -\epsilon_{IJKL}$ for signature $\sigma = -1$
$= \epsilon_{ijk}$ by converting the permutation tensor from 4D to 3D
$= \epsilon^{ijk}$ by raising by $\eta^{ij} = \delta^{ij}$

dual of the spin connection in triad:
${\Gamma^i}_a = \star {\omega_a}^{0i} = -{\star}{\omega_a}^{i0} $
${\Gamma^i}_a = {1\over2} {\epsilon^{0i}}_{jk} {\omega_a}^{jk}$
${\Gamma^i}_a = -{1\over2} {\epsilon^i}_{jk} {\omega_a}^{jk}$
Though some sources define this as ${\Gamma^i}_a = {\epsilon^i}_{jk} {\omega_a}^{jk}$
TODO more on this, i.e. from Kiefer

timelike component of the self-dual spin connection:
${A_a}^i = {A_a}^{i0}$
$= {1\over2} {\omega_a}^{i0} - {i\over4} {\epsilon^{i0}}_{KL} {\omega_a}^{KL}$
$= {1\over2} {\omega_a}^{i0} + {i\over4} {\epsilon^{0i}}_{jk} {\omega_a}^{jk}$ by permuting $\epsilon^{IJKL}$
$= {1\over2} {K_a}^i - {i\over4} {\epsilon^i}_{jk} {\omega_a}^{jk}$ by definition $-\epsilon^{0ijk} = \epsilon^{ijk}$
$= {1\over2} ({K_a}^i + i {\Gamma^i}_a)$
$= {1\over2} ({K_a}^i + \beta {\Gamma^i}_a)$
...for Barbero-Immirzi parameter $\beta=i$

Seems the original Ashtekar variables are derived from a definition of self-duals that neglects the ${1\over2}$ scalar and everyone ran with it after that.
So the original papers' self-dual spin connections double in magnitude every time the dual is taken.

$n_I = {e^\mu}_I n_\mu$
$ = {e^t}_I n_t + {e^j}_I n_j $
$ = {e^t}_I (-\alpha) + {e^j}_I \cdot 0$
$ = -\alpha \overset{I(i)\downarrow}{\left[\matrix{ {e^t}_0 \\ {e^t}_i }\right]} $ $ = -\alpha \overset{I(i)\downarrow}{\left[\matrix{1/\alpha \\ 0 }\right]} $ $ = \overset{I(i)\downarrow}{\left[\matrix{-1 \\ 0 }\right]} $ $ = -\delta^0_I$

$\star {A_a}^{i0} = i {A_a}^{i0}$ by self-duality
$\star {A_a}^{i0} = {1\over 2} {\epsilon^{i0}}_{jk} {A_a}^{jk}$ by dual definition
by combining the two definitions...
$ i {A_a}^{i0} = {1\over 2} {\epsilon^{i0}}_{jk} {A_a}^{jk}$
$ {A_a}^{i0} = -i {1\over 2} {\epsilon^{i0}}_{jk} {A_a}^{jk}$ by multiplying both sides by $-i$
$ {A_a}^{i0} = i {1\over 2} {\epsilon^{0i}}_{jk} {A_a}^{jk}$ by permutation of $\epsilon^{0ijk}$
$ {A_a}^{i0} = -i {1\over 2} {\epsilon^i}_{jk} {A_a}^{jk}$ by above $\epsilon^{0ijk} = -\epsilon^{ijk}$
$ {A_a}^i = -i {1\over 2} {\epsilon^i}_{jk} {A_a}^{jk}$ by ${A_a}^i = {A_a}^{i0}$

${A_a}^{jk} = -i \star {A_a}^{jk}$
$ = -{i\over2} {\epsilon^{jk}}_{LM} {A_a}^{LM}$
$ = -{i\over2} {\epsilon^{jk}}_{0i} {A_a}^{0i} - {i\over2} {\epsilon^{jk}}_{i0} {A_a}^{i0} $
$ = -i {\epsilon^{jk}}_{i0} {A_a}^{i0} $
$ = -i {\epsilon^{jk}}_{i0} {A_a}^i $
$ = i {{\epsilon_0}^{jk}}_i {A_a}^i $
${A_a}^{jk} = i {\epsilon^{jk}}_i {A_a}^i $

verify:
$ {A_a}^i = -i {1\over2} {\epsilon^i}_{jk} {A_a}^{jk}$
$ {A_a}^i = -i {1\over2} {\epsilon^i}_{jk} (i {\epsilon^{jk}}_l {A_a}^l)$
$ {A_a}^i = {1\over2} \epsilon^{jki} \epsilon_{jkl} {A_a}^l$
$ {A_a}^i = \delta^i_l {A_a}^l$ for $\epsilon^{ijk} \epsilon_{ijl} = 2 \delta^i_l$
$ {A_a}^i = {A_a}^i$

timelike+spatial component of self-dual of curvature tensor:
${F_{ab}}^{i0} = \partial_a {A_b}^{i0} - \partial_b {A_a}^{i0} + {A_a}^{ik} {A_{bk}}^0 - {A_b}^{ik} {A_{ak}}^0$
$= \partial_a {A_b}^i - \partial_b {A_a}^i + {A_a}^{ik} A_{bk} - {A_b}^{ik} A_{ak}$
$ = \partial_a {A_b}^i - \partial_b {A_a}^i + i {\epsilon^{ik}}_j {A_a}^j A_{bk} - i {\epsilon^{ik}}_j {A_b}^j A_{ak}$ using ${A_a}^{ij} = i {\epsilon^{ij}}_k {A_a}^k$
$ = \partial_a {A_b}^i - \partial_b {A_a}^i - i {\epsilon^i}_{jk} {A_a}^j {A_b}^k - i {\epsilon^i}_{jk} {A_a}^j {A_b}^k$
$ = \partial_a {A_b}^i - \partial_b {A_a}^i - 2 i {\epsilon^i}_{jk} {A_a}^j {A_b}^k$

self-dual covariant spin connection derivative of timelike self-dual spin connection:
$\mathcal{D}_b (t^a {A_a}^i) = \mathcal{D}_b (t^a {A_a}^{i0})$ by ${A_a}^i = {A_a}^{i0}$
$= \partial_b (t^a {A_a}^{i0}) + {{A_b}^i}_J t^a {A_a}^{J0} + {{A_b}^0}_J t^a {A_a}^{iJ}$ by self-dual covariant spin-connection def
$= \partial_b (t^a {A_a}^i) + {A_b}^{ij} t^a A_{aj} - A_{bj} t^a {A_a}^{ij}$ by excluding ${A_a}^{00} = 0$ from sums and index gymnastics
$= \partial_b (t^a {A_a}^i) + i {\epsilon^{ij}}_k {A_b}^k t^a A_{aj} - A_{bj} t^a i {\epsilon^{ij}}_k {A_a}^k$ by ${A_a}^{ij} = i {\epsilon^{ij}}_k {A_a}^k$
$= \partial_b (t^a {A_a}^i) + i t^a {\epsilon^i}_{jk} {A_a}^j {A_b}^k + i t^a {\epsilon^i}_{jk} {A_a}^j {A_b}^k$ by permuting the second ${\epsilon^{ij}}_k$, switching $j$ and $k$, and some more index gymnastics
$= \partial_b t^a {A_a}^i + t^a \partial_b {A_a}^i + 2 i t^a {\epsilon^i}_{jk} {A_a}^j {A_b}^k$ by distributing the $\partial_b$
$\mathcal{D}_b (t^a {A_a}^i) = \partial_b t^a {A_a}^i + t^a \partial_b {A_a}^i + 2 i t^a {\epsilon^i}_{jk} {A_a}^j {A_b}^k$

combine to find...
$\mathcal{L}_\vec{t} {A_b}^i - \mathcal{D}_b (t^a {A_a}^i)$
...using $\mathcal{L}_\vec{t} {A_b}^i = t^a \partial_a {A_b}^i + {A_a}^i \partial_b t^a$...
$ = t^a \partial_a {A_b}^i + {A_a}^i \partial_b t^a - \mathcal{D}_b (t^a {A_a}^i)$
...using $\mathcal{D}_b (t^a {A_a}^i) = \partial_b t^a {A_a}^i + t^a \partial_b {A_a}^i + 2 i t^a {\epsilon^i}_{jk} {A_a}^j {A_b}^k$...
$ = t^a \partial_a {A_b}^i + {A_a}^i \partial_b t^a - \partial_b t^a {A_a}^i - t^a \partial_b {A_a}^i - 2 i t^a {\epsilon^i}_{jk} {A_a}^j {A_b}^k$
$ = t^a \partial_a {A_b}^i - t^a \partial_b {A_a}^i - 2 i t^a {\epsilon^i}_{jk} {A_a}^j {A_b}^k$
$ = t^a (\partial_a {A_b}^i - \partial_b {A_a}^i - 2 i {\epsilon^i}_{jk} {A_a}^j {A_b}^k)$
$\mathcal{L}_\vec{t} {A_b}^i - \mathcal{D}_b (t^a {A_a}^i) = t^a {F_{ab}}^i$

---

Self-dual Palatini Action:
$ S = \int e {e^a}_I {e^b}_J {F_{ab}}^{IJ} dx^4 $
TODO show $ \delta S / \delta {e^a}_I = 0 $ for $ {^+R_{ab}} - {1\over 2} g_{ab} {^+R} = 0 $ for $^+R_{ab}$ is the self-dual of $ R_{ab} $?

Redefine $ t^a, \alpha, \beta^a, n^a $ to be complex
Same as before, $ {\gamma_b}^a = \delta^a_b + n^a n_b $ and $ \delta^a_b = {\gamma_b}^a - n^a n_b $
Let $ {E^a}_I $ be the projected vierbein defined by $ {E^a}_I = {\gamma_b}^a {e^b}_I $
so $ {e^a}_I = \delta^a_b {e^b}_I = ({\gamma_b}^a - n^a n_b) {e^b}_I = {\gamma_b}^a {e^b}_I - n^a n_b {e^b}_I = {E^a}_I - n^a n_b {e^b}_I $
TODO is ${E^a}_I$ lowered by $g_{ab}$ or $\gamma_{ab}$?
Then
$ S = \int e {e^a}_I {e^b}_J {F_{ab}}^{IJ} dx^4 $
$ = \int e ({E^a}_I - n^a n_c {e^c}_I) ({E^b}_J - n^b n_d {e^d}_J) {F_{ab}}^{IJ} dx^4 $
$ = \int e ({E^a}_I {E^b}_J - n^a n_c {e^c}_I {E^b}_J - n^b n_d {e^d}_J {E^a}_I + n^a n_c n^b n_d {e^c}_I {e^d}_J) {F_{ab}}^{IJ} dx^4 $
...using symmetry of $ {F_{ab}}^{IJ} = -{F_{ba}}^{IJ} = -{F_{ab}}^{JI} $...
$ S = \int e ({E^a}_I {E^b}_J - 2 {E^a}_I {e^c}_J n^b n_c) {F_{ab}}^{IJ} dx^4 $

Let $ {\tilde{E}^a}_I = \sqrt\gamma {E^a}_I $
...using $ e = \alpha \sqrt\gamma $
$ S = \int \alpha \sqrt\gamma ({E^a}_I {E^b}_J - 2 {E^a}_I {e^c}_J n^b n_c) {F_{ab}}^{IJ} dx^4 $
$ S = \int \alpha ( {1\over\sqrt\gamma} {\tilde{E}^a}_I {\tilde{E}^b}_J - 2 {\tilde{E}^a}_I {e^c}_J n^b n_c) {F_{ab}}^{IJ} dx^4 $
...using $\star {F_{ab}}^{IJ} = {1\over 2} {\epsilon^{IJ}}_{KL} {F_{ab}}^{KL} = i {F_{ab}}^{IJ} $...
$ S = \int -i {1\over 2} {1\over\sqrt\gamma} \alpha {\tilde{E}^a}_I {\tilde{E}^b}_J {\epsilon^{IJ}}_{MN} {F_{ab}}^{MN} - 2 \alpha {\tilde{E}^a}_I {e^c}_J n^b n_c {F_{ab}}^{IJ} dx^4 $
$ S = \int -i {1\over 2} {1\over\sqrt\gamma} \alpha {\tilde{E}^a}_I {\tilde{E}^b}_J {\epsilon^{IJ}}_{MN} {F_{ab}}^{MN} - 2 \alpha {\tilde{E}^a}_I n^b n_J {F_{ab}}^{IJ} dx^4 $
...using ${\epsilon^{IJ}}_{MN} {F_{ab}}^{MN}$ with ${\epsilon^{IJ}}_{M0} {F_{ab}}^{M0} + {\epsilon^{IJ}}_{0M} {F_{ab}}^{0M}$...i.e. the values for which $N=0$ plus the values for which $N \ne 0$ (substiting the $M$ for $N$)
$ S = \int -i {1\over 2} {1\over\sqrt\gamma} \alpha {\tilde{E}^a}_I {\tilde{E}^b}_J ({\epsilon^{IJ}}_{M0} {F_{ab}}^{M0} + {\epsilon^{IJ}}_{0M} {F_{ab}}^{0M}) - 2 \alpha {\tilde{E}^a}_I n^b n_J {F_{ab}}^{IJ} dx^4 $
...using ${\epsilon^{IJ}}_{M0} = -{\epsilon^{IJ}}_{0M}$ and ${F_{ab}}^{M0} = -{F_{ab}}^{0M}$...
$ S = \int -i {1\over\sqrt\gamma} \alpha {\tilde{E}^a}_I {\tilde{E}^b}_J {\epsilon^{IJ}}_{M0} {F_{ab}}^{M0} - 2 \alpha {\tilde{E}^a}_I n^b n_J {F_{ab}}^{IJ} dx^4 $
...using $n_I = -\delta^0_I$...
$ S = \int -i {\alpha\over\sqrt\gamma} {\tilde{E}^a}_I {\tilde{E}^b}_J {\epsilon^{IJ}}_{M0} {F_{ab}}^{M0} + 2 \alpha {\tilde{E}^a}_I n^b {F_{ab}}^{I0} dx^4 $

Now we only need to consider the 3D hypersurface.
All other values for which $IJK=0$ are zero
$ S = \int -i {\alpha\over\sqrt\gamma} {\tilde{E}^a}_i {\tilde{E}^b}_j {\epsilon^{ij}}_k {F_{ab}}^{k0} + 2 \alpha {\tilde{E}^a}_i n^b {F_{ab}}^{i0} dx^4 $
Let $ {F_{ab}}^I = {F_{ab}}^{I0} $
$ S = \int -i {\alpha\over\sqrt\gamma} {\tilde{E}^a}_i {\tilde{E}^b}_j {\epsilon^{ij}}_k {F_{ab}}^k + 2 \alpha {\tilde{E}^a}_i n^b {F_{ab}}^i dx^4 $
...using $\alpha n^a = t^a - \beta^a$...
$ S = \int -i {\alpha\over\sqrt\gamma} {\tilde{E}^a}_i {\tilde{E}^b}_j {\epsilon^{ij}}_k {F_{ab}}^k + 2 (t^b - \beta^b) {\tilde{E}^a}_i {F_{ab}}^i dx^4 $
...switch $a$s and $b$s in rightmost term...
$ S = \int -i {\alpha\over\sqrt\gamma} {\tilde{E}^a}_i {\tilde{E}^b}_j {\epsilon^{ij}}_k {F_{ab}}^k + 2 (t^a - \beta^a) {\tilde{E}^b}_i {F_{ba}}^i dx^4 $
...using ${F_{ab}}^{IJ} = -{F_{ba}}^{IJ}$...
$ S = \int -i {\alpha\over\sqrt\gamma} {\tilde{E}^a}_i {\tilde{E}^b}_j {\epsilon^{ij}}_k {F_{ab}}^k - 2 (t^a - \beta^a) {\tilde{E}^b}_i {F_{ab}}^i dx^4 $
...distribute rightmost term...
$ S = \int -i {\alpha\over\sqrt\gamma} {\tilde{E}^a}_i {\tilde{E}^b}_j {\epsilon^{ij}}_k {F_{ab}}^k - 2 t^a {\tilde{E}^b}_i {F_{ab}}^i + 2 \beta^a {\tilde{E}^b}_i {F_{ab}}^i dx^4 $
...using $t^a {F_{ab}}^i = \mathcal{L}_\vec{t} {A_b}^i - \mathcal{D}_b (t^a {A_a}^i)$ ...
$ S = \int - i {\alpha\over\sqrt\gamma} {\tilde{E}^a}_i {\tilde{E}^b}_j {\epsilon^{ij}}_k {F_{ab}}^k - 2 {\tilde{E}^b}_i (\mathcal{L}_\vec{t} {A_b}^i - \mathcal{D}_b (t^a {A_a}^i)) + 2 \beta^a {\tilde{E}^b}_i {F_{ab}}^i dx^4 $
$ S = \int - i {\alpha\over\sqrt\gamma} {\epsilon^{ij}}_k {\tilde{E}^a}_i {\tilde{E}^b}_j {F_{ab}}^k - 2 {\tilde{E}^b}_i \mathcal{L}_\vec{t} {A_b}^i + 2 {\tilde{E}^b}_i \mathcal{D}_b (t^a {A_a}^i) + 2 \beta^a {\tilde{E}^b}_i {F_{ab}}^i dx^4 $
...integrate by parts to find $\int\int_\Sigma {\tilde{E}^b}_i \mathcal{D}_b (t^a {A_a}^i) dx^3 dt = \int \left( \int_{\partial\Sigma} t^a {A_a}^i {\tilde{E}^b}_i ... \sigma_b dx^3 - \int_\Sigma t^a {A_a}^i \mathcal{D}_b {\tilde{E}^b}_i dx^3 \right) dt$
Pullin ditches the $t^a {A_a}^i \mathcal{D}_b {\tilde{E}^b}_i$ term entirely.
Wiki then ditches the total integral term and sticks with $\int {\tilde{E}^b}_i \mathcal{D}_b (t^a {A_a}^i) dx^4 = -\int t^a {A_a}^i \mathcal{D}_b {\tilde{E}^b}_i dx^4$
Ashtekar has what looks like the total integral and some extra terms.
Ashtekar's results also match up with mine, up to a factor of ${i\over2}$, which is suspiciously the same as the amount Wiki shows the Poisson brackets scaled by.
$ S = \int - i {\alpha\over\sqrt\gamma} {\epsilon^{ij}}_k {\tilde{E}^a}_i {\tilde{E}^b}_j {F_{ab}}^k - 2 {\tilde{E}^a}_i \mathcal{L}_\vec{t} {A_a}^i - 2 t^a {A_a}^i \mathcal{D}_b {\tilde{E}^b}_i + 2 \beta^a {\tilde{E}^b}_i {F_{ab}}^i dx^4 $
...move first time last...
$ S = \int - 2 {\tilde{E}^a}_i \mathcal{L}_\vec{t} {A_a}^i - 2 t^a {A_a}^i \mathcal{D}_b {\tilde{E}^b}_i + 2 \beta^a {\tilde{E}^b}_i {F_{ab}}^i - i {\alpha\over\sqrt\gamma} {\epsilon^{ij}}_k {\tilde{E}^a}_i {\tilde{E}^b}_j {F_{ab}}^k dx^4 $
...factor out 2...
$ S = 2 \int - {\tilde{E}^a}_i \mathcal{L}_\vec{t} {A_a}^i - 2 t^a {A_a}^i \mathcal{D}_b {\tilde{E}^b}_i + \beta^a {\tilde{E}^b}_i {F_{ab}}^i - {i\over2} {\alpha\over\sqrt\gamma} {\epsilon^{ij}}_k {\tilde{E}^a}_i {\tilde{E}^b}_j {F_{ab}}^k dx^4 $
Now I have some sign errors with Wiki and some different sign errors with Pullin...
TODO show how this infers $p \dot{q} = {\tilde{E}^a}_i \mathcal{L}_\vec{t}$, which infers $\{ {A_a}^i(\vec{x}), {\tilde{E}^b}_j(\vec{y}) \} = c \delta_a^b \delta^i_j \delta^3(\vec{x}-\vec{y})$ for $c$ some value preferably only made up of $1$'s, $2$'s, and $i$'s

For $0 = \delta S / \delta (t^a {A_a}^i) = \mathcal{D}_a {\tilde{E}^a}_i = \mathcal{G}_i$ we find $\mathcal{G}_i = 0$
For $0 = \delta S / \delta \beta^b = {F_{ab}}^i {\tilde{E}^b}_i = \mathcal{V}_a$ we find $\mathcal{V}_a = 0$
For $0 = \delta S / \delta \alpha = {\epsilon^{ij}}_k {\tilde{E}^a}_i {\tilde{E}^b}_j {F_{ab}}^k = \mathcal{S}$ for $\mathcal{S} = 0$

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