# Deriving the Poisson bracket relation of the Ashtekar variables

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 $\delta_{(a}^c \delta_{b)}^d$]

$\delta_a^c \delta_b^d = \matrix{\matrix{&b\\a&c \backslash d}&\matrix{1&1&2&2\\1&2&1&2} \\ \matrix{1&1\\1&2\\2&1\\2&2} & \left[\matrix{1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1}\right]}$

$\delta_b^c \delta_a^d = \matrix{\matrix{&b\\a&c \backslash d}&\matrix{1&1&2&2\\1&2&1&2} \\ \matrix{1&1\\1&2\\2&1\\2&2} & \left[\matrix{1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1}\right]}$

$\delta_{(a}^c \delta_{b)}^d = \matrix{\matrix{&b\\a&c \backslash d}&\matrix{1&1&2&2\\1&2&1&2} \\ \matrix{1&1\\1&2\\2&1\\2&2} & \left[\matrix{1&0&0&1\over2\\0&0&1\over2&0\\0&1\over2&0&0\\1\over2&0&0&1}\right]}$

• You didn't symmetrize the LHS of Eq (4), which leads to the discrepency. If you match the downstairs indices, you end up with (10) and hence (13). Oct 21, 2015 at 20:17
• Yes the symmetry shows that $e \cdot e$ is symmetric, because this equals $\gamma$. But it says nothing about $\delta e \cdot e$. Oct 22, 2015 at 0:42
• no, look at Eq (4), it should read $\delta^{a}_{(c}\delta^{b}_{d)} = \partial\gamma_{cd}/\partial\gamma_{ab}$, which fixes your problem. Oct 22, 2015 at 2:34
• Even if you symmetrize the $cd$ on the deltas, it doesn't help. Without it we're showing a matrix equals a symmetrized matrix. With it we're showing a symmetrized matrix equals a symmetrized matrix. What you need to solve this is to show one matrix (not just the symmetric portion) equals another matrix (not just the symmetric portion). Just because $A^T + A = B^T + B$ doesn't mean $A = B$. And even with $A = B^T + B$, you can't get an exact solution for $B$. Oct 22, 2015 at 12:51
• But look, the only reason you have the symmetry problem is because you arbitrarily introduced it into Eq (3). If you instead rewrite it as $\gamma_{ab} = {e_{a}}^{i}{e_{b}}^{j}\delta_{ij}$ without explicitly symmetrizing the RHS, you're golden. (Your reasoning and counter-example is quite excellent, though.) Oct 22, 2015 at 15:47

You don't need to calculate anything explicitly.

$$H(q,\pi)$$ is related to $$H(E^i_a(q,\pi), K^a_i(q,\pi))$$ by a canonical transformation (see Perez 2004 Eq. (8) and (16)):

$$H(q,\pi)=H(E^i_a(q,\pi),K^a_i(q,\pi))-\frac{\partial F}{\partial t}$$

where $$F$$ is a linear function of lapse $$N$$ and shift $$N_a$$. A theorem exists (see Hand and Finch p.217) that states that

$$\left\{G,K\right\}_{p,q}=\left\{G,K\right\}_{P,Q}$$ where $$p,q$$ and $$P,Q$$ are canonical variables related by a canonical transformations. Thus, if $$\left\{q_{ab}(x),\pi_{cd}(y)\right\}=\delta_{(a|c}\delta_{b)d}\delta(x,y)$$

then automatically

$$\left\{E^i_a(x),K^b_j(y)\right\}=\delta^i_j\delta^b_a\delta(x,y)$$

• I assume that your $K$ is the same as the $A$ above? Jun 7, 2021 at 13:56
• No. But the answer above also works for $A^a_i$, where $K^a_i$ is related to $A^a_i$ by $K^a_i=A^a_i-\Gamma^a_i[E]$ where $\Gamma^a_i[E]$ is the connection compatible with the densitized triad $E^i_a$ (see Rovelli 2010 CUP). The same applies for $A$ because once again the canonical variables $A^a_i$ and $E^i_a$ are related to the metric $q_{ab}$ and its conjugate $\pi_{ab}$ by just another canonical transformation. Jun 7, 2021 at 18:13

Why not use $\gamma_{cd} = {e_c}^j {e_d}^k \delta_{jk}$, differentiate this with respect to $\gamma_{ab}$ and by the chain rule deduce the result? That is, we know $$\frac{\partial\gamma_{cd}}{\partial\gamma_{ab}}=\delta^{ab}_{cd}$$ but we also know $$\frac{\partial\gamma_{cd}}{\partial\gamma_{ab}}=\frac{\partial({e_c}^j {e_d}^k \delta_{jk})}{\partial\gamma_{ab}}\sim 2e\delta\frac{\partial e}{\partial\gamma}$$ Set these two expressions equal, and you've got your result, right?

Addendum. After some further thought, this might be the wrong question. Or at least posed badly.

You would want to rework the ADM formalism with tetrads (which has been known since the '70s). This would be the first step.

The next step would be trying to embed the "Gauss-constraint surface" of the Ashtekar phase space (i.e., the subspace specified by the Gauss constraint vanishing) into the ADM-tetrad phase space.

Even then, I'm not so sure this would necessarily work: the Ashtekar variable approach is "Palatini" in style. You might want to peruse this paper for some reference: Peter Peldan, Actions for Gravity, with Generalizations: A Review.

• Thanks for the tip! That looks like the approach I've got listed. The $\approx$ is bugging me -- I'm looking for something with a $=$. Oct 20, 2015 at 17:09
• @thenumbernine It's just the product rule: $\delta_{jk}({e_{c}}^{j} (\partial {e_{d}}^{k}/\partial_{ab}) + {e_{d}}^{k} (\partial {e_{c}}^{j}/\partial_{ab}))$, then set it equal to $\delta^{(a}_{c}\delta^{b)}_{d}$, and you're done. Oct 20, 2015 at 21:36
• This way works only if the tensor $e\delta e$ is symmetric -- I wrote out my work in the comments. Is there an easy proof that $e\delta e$ is symmetric? I'll start messing with that now... Oct 21, 2015 at 11:51
• The original question wasn't how to differentiate the square root of a matrix (as it has wildly deviated into). It was how to solve the Poisson brackets of Ashtekar variables. This Addendum is a much better candidate for an answer. Thanks for the link -- I will check it out and see if I can get something from this. Oct 31, 2015 at 10:05
• This paper still only states fundamental Poisson brackets for each of the different actions it describes, without explaining how they are chosen, as the Romano, Giulini, and Pullin papers I reference under "my rationality for this assumption" in the question. Maybe I should've asked how fundamental Poisson brackets are chosen in the first place, because that seems to be done in these papers rather than deriving one set from another set. Nov 1, 2015 at 6:29