I'm trying to figure out how to calculate the orthogonality of Ashtekar variables with respect to the ADM hypersurface metric and conjugate momentum.
$$\{{A_a}^i(x), {E^b}_j(y)\} = 8 \pi \beta \delta^i_j \delta^b_a \delta(x,y)$$
as is given in Kiefer's book on p.127 eqn 4.120.
I've been assuming the configuration variables of these Poisson brackets are the ADM hypersurface metric $\gamma_{ab}$ and conjugate momentum $\pi^{ab}$.
My rationality for this assumption:
Immediately after the equation I'm trying to solve, the book states that ${A_a}^i$ and ${E^b}_j$ will be the new configuration variable and canonical momentum. This implies that different configuration variables were previously being used.
The Poisson bracket $\{f,g\}$ applied to the ADM formalism is defined in terms of the spatial metric $\gamma_{ab}$ and conjugate momentum $\pi^{ab}$ as follows: $\{f,g\} = {{\partial f}\over{\partial \gamma_{ab}}} {{\partial g}\over{\partial \pi^{ab}}} - {{\partial f}\over{\partial \pi^{ab}}} {{\partial g}\over{\partial \gamma_{ab}}}$. This definition is found in Kiefer p.112 eqn.4.64, Romano page 14 eqn 2.33, Alcubierre p.81 eqn.2.7.14, Pullin p.6 eqn.13.
Romano has a similar statement, footnote 11 on page 26, but with ${A_a}^i$ and ${E^b}_j$ already chosen to be the configuration variables: $\{{A_a}^I(x), {E^b}_j(y)\} = \delta^b_a \delta^i_j \delta(x,y)$. Giulini also has a similar statement at p.26 eqn.5.23. Pullin has a similar statement shortly after eqn.23 on p.11.
What I've gathered so far:
Substituting $\{{A_a}^i(x), {E^b}_j(y)\}$ into the Poisson bracket definition gives us:
$${{\partial}\over{\partial \gamma_{ab}}} {A_a}^i(x) {{\partial}\over{\partial \pi^{ab}}} {E^b}_j(y) - {{\partial}\over{\partial \pi^{ab}}} {A_a}^i(x) {{\partial}\over{\partial \gamma_{ab}}} {E^b}_j(y)$$
Now ${A_a}^i = {A_a}^{i\hat{t}}$ is the timelike portion of the self-dual connection, and $${A_\alpha}^{IJ} = {^+}{\omega_\alpha}^{IJ} = {1\over2}({\omega_\alpha}^{IJ} + {i\over2} {\epsilon^{IJ}}_{KL} {\omega_\alpha}^{KL})$$ is the self-dual of the spin connection.
And the spin connection ${\omega_\alpha}^{IJ}$ is defined as the Minkowski coordinate connection to cancel $\nabla_\alpha {e_\mu}^I$, as $${{\omega_\alpha}^I}_J = {\Gamma^\mu}_{\nu\alpha} {e_\mu}^I {e^\nu}_J - {e^\mu}_J \partial_\alpha {e_\mu}^I$$
For ${\Gamma^\mu}_{\nu\alpha}$ the affine connection of the spacetime metric.
With all this said, I would think to chain rule ${{\partial {A_a}^i}\over{\partial \gamma_{cd}}} = {{\partial {A_a}^i}\over{\partial {e_f}^J}} {{\partial {e_f}^J}\over{\gamma_{cd}}}$ and then substitute the first term for the derivatives of the definitions of ${A_\alpha}^{IJ}$ and ${\omega_\alpha}^{IJ}$ above.
The second term stumps me though. For $\gamma_{cd} = {e_c}^j {e_d}^k \delta_{jk}$ how would you calculate ${{\partial {e_f}^i}\over{\partial \gamma_{cd}}}$?
My next thought is to simplify the inverse of the derivative: \begin{align} {{\partial \gamma_{cd}}\over{\partial {e_f}^i}} &= {\partial\over{\partial {e_f}^i}} ({e_c}^j {e_d}^k \delta_{jk})\\ &= ({\partial\over{\partial {e_f}^i}} {e_c}^j) {e_d}^k \delta_{jk} + {e_c}^j ({\partial\over{\partial {e_f}^i}} {e_d}^k) \delta_{jk}\\ &= \delta^j_i \delta^f_c {e_d}^k \delta_{jk} + {e_c}^j \delta^k_i \delta^f_d \delta_{jk}\\ &= \delta^f_c e_{di} + \delta^f_d e_{ci} \end{align} ...but how do you solve the inverse of a rank-4 tensor?
Is there a better approach?
Thanks.
Sources:
Kiefer, Claus. "Quantum Gravity."
Romano. "Geometrodynamics vs Connection Dynamics."
Giulini, "Ashtekar Variables in Classical General Relativity.
Alcubierre, Miguel. "Introduction to 3+1 Numerical Relativity."
Pullin, Jorge. "Knot Theory and Quantum Gravity in Loop Space: A Primer"
[edit: hint suggested by Alex Nelson]
(1) $\gamma_{ab} = {e_a}^i {e_b}^j \delta_{ij}$ <- metric def by vielbein
(2) $\gamma_{ab} = \gamma_{(ab)}$ <- metric symmetric
(3) $\gamma_{ab} = {e_{(a}}^i {e_{b)}}^j \delta_{ij}$ <- substitute (1) into (2)
(4) $\delta^a_c \delta^b_d = {{\partial \gamma_{cd}}\over{\partial \gamma_{ab}}}$ <- deriv of a var wrt itself
(5) $\delta^a_c \delta^b_d = {\partial\over{\partial \gamma_{ab}}} ({e_{(c}}^i {e_{d)}}^j \delta_{ij})$ <- substitute (3) into (4)
(6) $\delta^a_c \delta^b_d = \left( ({\partial\over{\partial \gamma_{ab}}} {e_{(c}}^i) {e_{d)}}^j + ({\partial\over{\partial \gamma_{ab}}} {e_{(d}}^j) {e_{c)}}^i \right) \delta_{ij}$ <- product rule
(7) $\delta^a_c \delta^b_d = 2({\partial\over{\partial \gamma_{ab}}} {e_{(c}}^i) {e_{d)}}^j \delta_{ij}$ <- symmetry lets us rearrange c & d indexes, then combine
(8) ${1\over2} \delta^a_c \delta^b_d = ({\partial\over{\partial \gamma_{ab}}} {e_{(c}}^i) e_{d)i}$ <- lower j to i, divide by two
... this doesn't give a solution of $({\partial\over{\partial \gamma_{ab}}} {e_c}^i) e_{di}$ (i.e. $\delta e \cdot e$), it gives a solution to only the symmetric part of $\delta e \cdot e$. If I wanted to reconstruct $\delta e \cdot e$ fully then I'm still missing the antisymmetric part of $\delta e \cdot e$. If I could get the antisymmetric part then I could multiply both sides by $e^{-1}$ and get a solution for $\delta e$.
This has an easy solution for $\delta e \cdot e = (\delta e \cdot e)^T$, but I don't believe I can make that assumption.
If $\delta e \cdot e = (\delta e \cdot e)^T$ were true ...
(9) $({\partial\over{\partial\gamma_{ab}}} {e_{(c}}^i) e_{d)i} = ({\partial\over{\partial\gamma_{ab}}} {e_c}^i) e_{di}$
... then you could say ...
(10) ${1\over2} \delta^a_c \delta^b_d = ({\partial\over{\partial \gamma_{ab}}} {e_c}^i) e_{di}$ <- substitute (9) into (8)
(11) ${1\over2} \delta^a_c \delta^b_d e^{dj} = ({\partial\over{\partial \gamma_{ab}}} {e_c}^i) e_{di} e^{dj}$ <- transform both sides by $e^{dj}$
(12) ${1\over2} \delta^a_c \delta^b_d e^{dj} = ({\partial\over{\partial \gamma_{ab}}} {e_c}^i) \delta^j_i$ <- $e$ times $e^{-1}$ is identity
(13) ${1\over2} \delta^a_c e^{bj} = {\partial\over{\partial \gamma_{ab}}} {e_c}^j$ <- viola
. . . so can you prove that $({\partial\over{\partial\gamma_{ab}}} {e_{(c}}^i) e_{d)i} = ({\partial\over{\partial\gamma_{ab}}} {e_c}^i) e_{di}$ ?
[edit: here's a counterproof to both the assumptions that $e\delta e$ is symmetric and that the derived partial based on that assumption is correct:]
Coordinate system $\{t, r\}$
vielbein: ${e_a}^i = a\downarrow\overset{i\rightarrow}{\left[\matrix{\sqrt{r}\over2&\sqrt{r}\over2\\0&r}\right]}$
vielbein inverse: ${e^a}_i = a\downarrow\overset{i\rightarrow}{\left[\matrix{\sqrt{2\over r}&0\\-{1\over r}&1\over r}\right]}$
metric: $\gamma_{ab} = {e_a}^i {e_b}^j \delta_{ij} = {e_a}^i e_{bi} = a\downarrow\overset{i\rightarrow}{\left[\matrix{\sqrt r\over2&\sqrt r\over2\\0&r}\right]} \cdot i\downarrow\overset{b\rightarrow}{\left[\matrix{\sqrt r\over2&0\\\sqrt r\over2&r}\right]} = a\downarrow\overset{b\rightarrow}{\left[\matrix{r&\sqrt{r^3\over2}\\\sqrt{r^3\over2}&r^2}\right]}$
$\gamma_{tt} = r$, so ${\partial\over\partial\gamma_{tt}} = {\partial\over\partial r}$
${\partial\over\partial\gamma_{tt}} {e_a}^i = {\partial\over\partial r} {e_a}^i = a\downarrow\overset{i\rightarrow}{\left[\matrix{1\over\sqrt{8r}&1\over\sqrt{8r}\\0&1}\right]}$
proof $e \delta e$ is not symmetric:
$({\partial\over\partial\gamma_{tt}} {e_a}^i) e_{bi} = a\downarrow\overset{i\rightarrow}{\left[\matrix{1\over\sqrt{8r}&1\over\sqrt{8r}\\0&1}\right]} \cdot i\downarrow\overset{b\rightarrow}{\left[\matrix{\sqrt{r\over2}&0\\\sqrt{r\over2}&r}\right]} = a\downarrow\overset{b\rightarrow}{\left[\matrix{1\over2&\sqrt{r\over8}\\\sqrt{r\over2}&r}\right]}$
$({\partial\over\partial\gamma_{tt}} {e_{(a}}^i) e_{b)i} = {1\over2}\left( ({\partial\over\partial\gamma_{tt}} {e_a}^i) e_{bi} + ({\partial\over\partial\gamma_{tt}} {e_b}^i) e_{ai} \right) $ $= {1\over2}\left( a\downarrow\overset{b\rightarrow}{\left[\matrix{1\over2&\sqrt{r\over8}\\\sqrt{r\over2}&r}\right]} + a\downarrow\overset{b\rightarrow}{\left[\matrix{1\over2&\sqrt{r\over2}\\\sqrt{r\over8}&r}\right]} \right) = a\downarrow\overset{b\rightarrow}{\left[\matrix{1\over2&{3\over4}\sqrt{r\over2}\\{3\over4}\sqrt{r\over2}&r}\right]}$
so $({\partial\over\partial\gamma_{tt}} {e_a}^i) e_{bi} \ne ({\partial\over\partial\gamma_{tt}} {e_{(a}}^i) e_{b)i}$
proof our derivative definition -- dependent on $e\delta e$ being symmetric -- is not true: ${1\over2}\delta_a^t e^{ti} = a\downarrow\overset{i\rightarrow}{\left[\matrix{1\over\sqrt{2r}&0\\0&0}\right]}$
${\partial\over\partial\gamma_{tt}} {e_a}^i = a\downarrow\overset{i\rightarrow}{\left[\matrix{1\over\sqrt{8r}&1\over\sqrt{8r}\\0&1}\right]}$
so ${1\over2} \delta_a^c e^{di} \ne {\partial\over\partial\gamma_{cd}} {e_a}^i$ for $c,d$ = $t,t$
[edit: some more thoughts]
Here's the problem:
${{\partial\gamma_{ab}}\over\partial\gamma_{cd}} = \delta^c_a \delta^d_b$
$\gamma_{ab} = \gamma_{(ab)}$
but
${{\partial\gamma_{(ab)}}\over\partial\gamma_{cd}} = \delta^c_{(a} \delta^d_{b)} \ne \delta^c_a \delta^d_b = {{\partial\gamma_{ab}}\over\partial\gamma_{cd}}$
Why? Just because $x = y$ does not mean $dx = dy$
Likewise, for $a \ne b$, despite $\gamma_{ab} = \gamma_{ba}$, it's still true that ${\partial\gamma_{ab}\over\partial\gamma_{ba}} = 0$
I remember getting in trouble for making this assumption when I first ran into (coincidentally) Poisson brackets in my first Quantum Mechanics class.
The counter-proof I offered relies on this assumption as well, so it is wrong as well as the suggestions so far offered.
With all that said, the solution to $\partial {e_a}^i \over \partial \gamma_{cd}$ is going to be a tensor ${T_a}^{icd}$ such that $2{T_{(ab)}}^{cd} = \delta^c_a \delta^d_b$
[edit: a 2D example of $\delta_a^c \delta_b^d$ vs $\delta_{(a}^c \delta_{b)}^d$]
$\delta_a^c \delta_b^d = \matrix{\matrix{&b\\a&c \backslash d}&\matrix{1&1&2&2\\1&2&1&2} \\ \matrix{1&1\\1&2\\2&1\\2&2} & \left[\matrix{1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1}\right]}$
$\delta_b^c \delta_a^d = \matrix{\matrix{&b\\a&c \backslash d}&\matrix{1&1&2&2\\1&2&1&2} \\ \matrix{1&1\\1&2\\2&1\\2&2} & \left[\matrix{1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1}\right]}$
$\delta_{(a}^c \delta_{b)}^d = \matrix{\matrix{&b\\a&c \backslash d}&\matrix{1&1&2&2\\1&2&1&2} \\ \matrix{1&1\\1&2\\2&1\\2&2} & \left[\matrix{1&0&0&1\over2\\0&0&1\over2&0\\0&1\over2&0&0\\1\over2&0&0&1}\right]}$
