Revisiting loop quantum gravity with selfdual variables: classical theory

In signature $(-+++)$, the Hodge star $*$ is a skew involution, $*^2 = -1$. As a map $\Lambda^2(\mathbb{M})\rightarrow \Lambda^2(\mathbb{M})$ it is self-adjoint with respect to the natural inner product. To obtain a decomposition into eigenspaces, we have to work with the complexification $\Lambda^2(\mathbb{M})_{\mathbb{C}}$. On this space $*$ has eigenvalues ±i and the projectors onto the eigenspaces are

$$\begin{align} \require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}: = \frac{1\mp i\ast}{2}. \end{align} \tag{ 7 }$$

Working in a basis, we get the explicit expression

$$\begin{align} \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}}: = \frac{1}{2}\left({\delta^{[I}_K\delta^{J]}_L\mp\frac{i}{2}{\epsilon^{IJ}}_{KL}}\right), \end{align} \tag{ 8 }$$

acting on an object with two internal upper indices. From this we directly see that the projectors are symmetric under exchanging the first and second pair of indices, i.e.

$$\begin{align} \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}} = \require{mathtools}\prescript{\pm}{}{{P}{_{KL}{}^{IJ}}}. \end{align} \tag{ 9 }$$

Hence we have immediately the action on internal two-forms. It is easy to see that $\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}$ is self-dual/anti-self-dual in both pairs of indices:

$$\begin{align} \begin{aligned} \frac{1}{2}{\epsilon}{^{IJ}{}_{MN}}\require{mathtools}\prescript{\pm}{}{{P}{^{MN}{}_{KL}}} = \pm i \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}},\\ \frac{1}{2}{\epsilon}{_{KL}{}^{MN}}\require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{MN}}} = \pm i \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}}. \end{aligned} \end{align} \tag{ 10 }$$

Since $\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}$ are projections onto eigenspaces, one has

$$\begin{align} \begin{aligned} &\require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}}\require{mathtools}\prescript{\mp}{}{{P}{^{KL}{}_{MN}}} = 0,\\ &\require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}}\require{mathtools}\prescript{\pm}{}{{P}{^{KL}{}_{MN}}} = \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{MN}}},\\ &\require{mathtools}\prescript{+}{}{{P}{^{IJ}{}_{KL}}}+\require{mathtools}\prescript{-}{}{{P}{^{IJ}{}_{KL}}} = \delta^{[I}_K\delta^{J]}_L. \end{aligned} \end{align} \tag{ 11 }$$

So $\require{mathtools}\prescript{+}{}{{P}{^{}_{}}}$ and $\require{mathtools}\prescript{-}{}{{P}{^{}_{}}}$ are indeed orthogonal projectors.

In the following we will use the basis

$$\begin{align} {\left(L_{IJ}\right)}{^K{}_L} = \delta^K_{[I}\eta_{J]L} \end{align} \tag{ 12 }$$

of $\mathfrak{so}(3,1)_{\mathbb{C}}$. In this basis, the components for a Lie algebra element, e.g. the connection $\omega^I{}_J$, are simply ω^IJ .

Given this basis, we can now understand how the notion of (anti-)self-duality, as a concept on the internal Minkowski space, translates to $\mathfrak{so}(3,1)_{\mathbb{C}}$ matrices. Acting on the labels of the generator yields

$$\begin{align} \require{mathtools}\prescript{\pm}{}{{P}{_{IJ}{}^{MN}}}{\left(L_{MN}\right)}{^K{}_L} = \require{mathtools}\prescript{\pm}{}{{P}{_{IJ}{}^{MN}}}\delta^K_{[N}\eta_{M]L} = {\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}}{_{IJ}{}^K{}_L}. \end{align} \tag{ 13 }$$

Because of the symmetry of labels and matrix indices of the generators, i.e. ${(L_{IJ})}{_{KL}}$, this is exactly the same as acting with the projection on the matrix indices:

$$\begin{align} {\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}}{^K{}_{LM}{}^N}{\left(L_{IJ}\right)}{^M{}_N} = {\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}}{_{IJ}{}^K{}_L}. \end{align} \tag{ 14 }$$

Hence (anti-)self-dual projection of components makes the corresponding matrix (anti-)self-dual as well. As a consequence, the (anti-)self-dual generators are

$$\begin{align} {L^{\pm}}_{IJ}: = \require{mathtools}\prescript{\pm}{}{{P}{^{MN}{}_{IJ}}}L_{MN}, \end{align} \tag{ 15 }$$

with

$$\begin{align} {\left({L^{\pm}}_{IJ}\right)}{^K{}_L}: = {\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}}{_{IJ}{}^K{}_L}. \end{align} \tag{ 16 }$$

With this established, we can now use the last relation in (11) to split $a\in\mathfrak{so}(3,1)_{\mathbb{C}}$ into its self-dual and anti-self-dual parts

$$\begin{align} a = a^{IJ}L_{IJ} = a^{IJ}\left({\require{mathtools}\prescript{+}{}{{P}{^{KL}{}_{IJ}}}+\require{mathtools}\prescript{-}{}{{P}{^{KL}{}_{IJ}}}}\right)L^{KL} = a^{+IJ}{L^+}_{IJ}+a^{-IJ}{L^-}_{IJ}, \end{align} \tag{ 17 }$$

where both the components $a^{\pm IJ}: = \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}}a^{KL}$ and the generators ${L^{\pm}}_{IJ}$ are now (anti-)self-dual. As the projection $\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}$ is orthogonal, this decomposes $\mathfrak{so}(3,1)_{\mathbb{C}}$ into self-dual and anti-self-dual parts $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}: = \require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}\mathfrak{so}(3,1)_{\mathbb{C}}$, respectively:

$$\begin{align} \mathfrak{so}\left(3,1\right)_{\mathbb{C}}\cong\mathfrak{so}\left(3,1\right)_{\mathbb{C}}^+\oplus\mathfrak{so}\left(3,1\right)_{\mathbb{C}}^-. \end{align} \tag{ 18 }$$

Elements of $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ therefore are of the form

$$\begin{align} a^\pm = a^{IJ}\require{mathtools}\prescript{\pm}{}{{P}{_{IJ}{}^{MN}}}L_{MN} = a^{\pm IJ}{L^{\pm}}_{IJ}. \end{align} \tag{ 19 }$$

The next step is to analyse the commutation relation of $\mathfrak{so}(3,1)_{\mathbb{C}}$, in order to show that $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ are indeed Lie subalgebras. The commutation relation of $\mathfrak{so}(3,1)_{\mathbb{C}}$ is given by

$$\begin{align} \left[{L_{IJ}},{L_{KL}}\right] = {\,\mathring{f}}{_{IJKL}{}^{MN}}L_{MN}, \end{align} \tag{ 20 }$$

where in the basis (12), the structure constants ${\,\mathring{f}}{}{}$ are

$$\begin{align} {\,\mathring{f}}{_{IJKL}{}^{MN}} = 2\delta^{[M}_{[I}\eta_{J][K}\delta^{N]}_{L]}. \end{align} \tag{ 21 }$$

We now look at this commutation relation when $a,b\in\mathfrak{so}(3,1)_{\mathbb{C}}$ are decomposed into their (anti-)self-dual parts:

$$\begin{align} \begin{aligned} \left[{a},{b}\right]& = \left[{a^++a^-},{b^++b^-}\right] = \left[{a^+},{b^+}\right]+\left[{a^+},{b^-}\right]+\left[{a^-},{b^+}\right]+\left[{a^-},{b^-}\right]. \end{aligned} \end{align} \tag{ 22 }$$

In order to understand the action of the projection on this commutator, we have to understand both the purely (anti-)self-dual commutators and the mixed ones.

To this end we consider the adjoint action on $\mathfrak{so}(3,1)_{\mathbb{C}}$. Let still $a,b\in\mathfrak{so}(3,1)_{\mathbb{C}}$:

$$\begin{align} \pi\left(a\right)b: = \left[{a},{b}\right]. \end{align} \tag{ 23 }$$

We know $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}\subset\mathfrak{so}(3,1)_{\mathbb{C}}$ is invariant. Furthermore let $b^\pm\in\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$. Hence,

$$\begin{align} b\in\mathfrak{so}\left(3,1\right)^\pm_{\mathbb{C}} \Leftrightarrow \require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}b = b. \end{align} \tag{ 24 }$$

Now

$$\begin{align} \begin{aligned} \pi\left(a\right)b^\pm\in\mathfrak{so}\left(3,1\right)^\pm_{\mathbb{C}} &\Leftrightarrow \require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}\pi\left(a\right)b^\pm = \pi\left(a\right)b^\pm \\ &\Leftrightarrow\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}\left[{a},{b^\pm}\right] = \left[{a},{b^\pm}\right]\quad\forall a \in \mathfrak{so}\left(3,1\right)_{\mathbb{C}}, \forall b^\pm\in \mathfrak{so}\left(3,1\right)^\pm_{\mathbb{C}}. \end{aligned} \end{align} \tag{ 25 }$$

Restricting to $a^\pm\in\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$, this tells us that the commutators of (anti-)self-dual Lie algebra elements are indeed (anti-)self-dual, even before the actual projection. Restricting instead to $a^\mp$, the commutators $\left[{a^\mp},{b^\pm}\right]$, with what we just showed, have to both self-dual and anti-self-dual. As the (anti-)self-dual decomposition is an orthogonal one, these commutators have to vanish. Hence we are left with

$$\begin{align} \left[{a},{b}\right] = \left[{a^+},{b^+}\right]+\left[{a^-},{b^-}\right] \end{align} \tag{ 26 }$$

and projection yields

$$\begin{align} \require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}\left[{a},{b}\right] = \left[{a^\pm},{b^\pm}\right]. \end{align} \tag{ 27 }$$

This in turn shows that $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ are closed under the commutator and hence are Lie algebras on their own.

The structure constants of these Lie subalgebras can be determined from the ones of $\mathfrak{so}(3,1)_{\mathbb{C}}$ as follows. As all (anti-)self-dual Lie algebra elements are still in $\mathfrak{so}(3,1)_{\mathbb{C}}$, it holds that

$$\begin{align} \left[{a^\pm},{b^\pm}\right] = a^{\pm IJ}b^{\pm KL}{\,\mathring{f}}{_{IJKL}{}^{MN}}L^\pm_{MN}. \end{align} \tag{ 28 }$$

So at first glance, the structure constants of $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ seem to be the same as the ones of $\mathfrak{so}(3,1)_{\mathbb{C}}$. This however disregards the fact that all pairs of indices of the structure constants are (anti-)self-dually projected. The actual structure constants of $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ are therefore given by

$$\begin{align} \left[{L^\pm_{IJ}},{L^\pm_{KL}}\right] = {\,\mathring{f}\,}{^\pm{}_{IJKL}{}^{MN}}L^\pm_{MN} \end{align} \tag{ 29 }$$

with

$$\begin{align} {\,\mathring{f}\,}{^\pm{}_{IJKL}{}^{MN}}: = \require{mathtools}\prescript{\pm}{}{{P}{^{I^{\prime} J^{\prime}}{}_{IJ}}}\require{mathtools}\prescript{\pm}{}{{P}{^{K^{\prime} L^{\prime}}{}_{KL}}}{\,\mathring{f}}{_{I^{\prime} J^{\prime} K^{\prime} L^{\prime}}{}^{M^{\prime} N^{\prime}}}\require{mathtools}\prescript{\pm}{}{{P}{^{M^{\prime} N^{\prime}}{}_{MN}}}. \end{align} \tag{ 30 }$$

Using the explicit form of the structure constants in (30), the self-dual structure constants can be expressed solely in terms of contracted projections, namely

$$\begin{align} {\,\mathring{f}\,}{^\pm{}_{IJKL}{}^{MN}} = -2{\require{mathtools}\prescript{\pm}{}{{P}{_{}^{}}}}{_{IJ}{}^{K^{\prime}}{}_{L^{\prime}}}{\require{mathtools}\prescript{\pm}{}{{P}{_{}^{}}}}{_{KL}{}^{L^{\prime}}{}_{M^{\prime}}}{\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}}{^{MNM^{\prime}}{}_{K^{\prime}}} = -2\operatorname{tr}\left({{{L^\pm}}{_{IJ}}{{L^\pm}}{_{KL}}{{L^\pm}}{^{MN}}}\right), \end{align} \tag{ 31 }$$

where we employed the relation between projectors and the (anti-)self-dual generators.

Going back one step, we can look again at the projection of the $\mathfrak{so}(3,1)_{\mathbb{C}}$ commutator

$$\begin{align} \begin{aligned} &\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}\left[{a},{b}\right] = \left[{a^\pm},{b^\pm}\right] = \left[{\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}a},{\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}b}\right]\\ &\Leftrightarrow a^{IJ}b^{KL}{\,\mathring{f}}{_{IJKL}{}^{MN}}\require{mathtools}\prescript{\pm}{}{{P}{^{M^{\prime} N^{\prime}}{}_{MN}}}L_{M^{\prime} N^{\prime}} = a^{I^{\prime} J^{\prime}}b^{K^{\prime} L^{\prime}}\require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{I^{\prime} J^{\prime}}}}\require{mathtools}\prescript{\pm}{}{{P}{^{KL}{}_{K^{\prime} L^{\prime}}}}{\,\mathring{f}}{_{IJKL}{}^{MN}}L_{MN}. \end{aligned} \end{align} \tag{ 32 }$$

Here we treated the left hand side as the projection of the commutator and the right hand side as the commutator of the projections, explicitly. This now expresses the fact that the projectors are Lie algebra homomorphisms and that the adjoint representation of $\mathfrak{sl}(2,\mathbb{C})_{\mathbb{C}}$ is reducible. This yields the relation

$$\begin{align} {\,\mathring{f}}{_{IJKL}{}^{M^{\prime} N^{\prime} }}\require{mathtools}\prescript{\pm}{}{{P}{^{MN}{}_{M^{\prime} N^{\prime} }}} = \require{mathtools}\prescript{\pm}{}{{P}{^{I^{\prime} J^{\prime} }{}_{IJ}}}\require{mathtools}\prescript{\pm}{}{{P}{^{K^{\prime} L^{\prime} }{}_{KL}}}{\,\mathring{f}}{_{I^{\prime} J^{\prime} K^{\prime} L^{\prime} }{}^{MN}}. \end{align} \tag{ 33 }$$

Projecting the pair MN is trivial on the left hand side, as $(\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}})^2 = \require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}$, but gives the (anti-)self-dual structure constants on the right hand side. Doing this and using the total antisymmetry of ${\,\mathring{f}}{_{IJKLMN}}$ and ${\,\mathring{f}\,}{^\pm{}_{IJKLMN}}$ in the pairs IJ, KL and MN yields

$$\begin{align} {\,\mathring{f}\,}{^\pm{}_{IJKLMN}} = \require{mathtools}\prescript{\pm}{}{{P}{^{I^{\prime} J^{\prime} }{}_{IJ}}}{\,\mathring{f}}{_{I^{\prime} J^{\prime} KLMN}} = \require{mathtools}\prescript{\pm}{}{{P}{^{K^{\prime} L^{\prime} }{}_{KL}}}{\,\mathring{f}}{_{IJK^{\prime} L^{\prime} MN}} = \require{mathtools}\prescript{\pm}{}{{P}{^{M^{\prime} N^{\prime} }{}_{MN}}}{\,\mathring{f}}{_{IJKLM^{\prime} N^{\prime} }}. \end{align} \tag{ 34 }$$

Hence it is sufficient to project just one of the pairs of indices of the $\mathfrak{so}(3,1)_{\mathbb{C}}$ structure constants.

With the structure constants for both $\mathfrak{so}(3,1)_{\mathbb{C}}$ and $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$, we can compare the respective Cartan–Killing metrics. For $\mathfrak{so}(3,1)_{\mathbb{C}}$ it is given by

$$\begin{align} k_{IJKL} = 2\operatorname{tr}\left({L_{IJ}L_{KL}}\right) = 2\delta^M_{[I}\eta_{J]N}\delta^N_{[K}\eta_{L]M} = -2\eta_{I[K}\eta_{L]J}. \end{align} \tag{ 35 }$$

The same result is obtained in a more tedious fashion, when determining the Cartan–Killing metric from contraction of two structure constants according to $k_{IJKL} = {\,\mathring{f}}{_{IJMN}{}^{M^{\prime} N^{\prime} }}{\,\mathring{f}}{_{KLM^{\prime} N^{\prime} }{}^{MN}}$.

For $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$, however, we start from the structure constants and use (34):

$$\begin{align} {{k^\pm}}{_{IJKL}} = {\,\mathring{f}\,}{^\pm{}_{IJMN}{}^{M^{\prime} N^{\prime} }}{\,\mathring{f}\,}{^\pm{}_{KLM^{\prime} N^{\prime} }{}^{MN}} = \require{mathtools}\prescript{\pm}{}{{P}{^{I^{\prime} J^{\prime} }{}_{IJ}}}\require{mathtools}\prescript{\pm}{}{{P}{^{K^{\prime} L^{\prime} }{}_{KL}}}{\,\mathring{f}}{_{I^{\prime} J^{\prime} MN}{}^{M^{\prime} N^{\prime} }}{\,\mathring{f}}{_{K^{\prime} L^{\prime} M^{\prime} N^{\prime} }{}^{MN}}. \end{align} \tag{ 36 }$$

So the Cartan–Killing metric of $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ is indeed the projection of the Cartan–Killing metric of $\mathfrak{so}(3,1)_{\mathbb{C}}$, i.e.

$$\begin{align} {{k^\pm}}{_{IJKL}} = \require{mathtools}\prescript{\pm}{}{{P}{^{I^{\prime} J^{\prime} }{}_{IJ}}}\require{mathtools}\prescript{\pm}{}{{P}{^{K^{\prime} L^{\prime} }{}_{KL}}}k_{I^{\prime} J^{\prime} K^{\prime} L^{\prime} } = -2{\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}}{_{K^{\prime} L^{\prime} IJ}}\require{mathtools}\prescript{\pm}{}{{P}{^{K^{\prime} L^{\prime} }{}_{KL}}} = -2{\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}}{_{IJKL}}, \end{align} \tag{ 37 }$$

where in the last step the symmetry under exchange of first and second pair of indices was used. Note that the pair $I,J$ here is pulled down with Minkowski metrics. So in combination with the factor of −2, we can understand this as

$$\begin{align} {{k^\pm}}{_{IJKL}} = \require{mathtools}\prescript{\pm}{}{{P}{^{MN}{}_{IJ}}}k^\pm_{MNKL} = \require{mathtools}\prescript{\pm}{}{{P}{^{MN}{}_{KL}}}k^\pm_{IJMN} \end{align} \tag{ 38 }$$

and again, projecting one pair of indices is equivalent to projecting all pairs.

We want to make a remark about dimensions. The projection to the (anti-)self-dual subalgebra halves the dimension of $\mathfrak{so}(3,1)_{\mathbb{C}}$ from $6\,\mathbb{C}$ to $3\,\mathbb{C}$ for $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$, respectively. The structure constants of $\mathfrak{so}(3,1)_{\mathbb{C}}$ are the same as for $\mathfrak{so}(3,1)$ and the projection does not change the symmetries of ${\,\mathring{f}\,}{^\pm}$ as compared to ${\,\mathring{f}}{}$. Because of this and the six real dimensions of the (anti-)self-dual algebra, we see that indeed both $\mathfrak{so}(3,1)^+_{\mathbb{C}}$ and $\mathfrak{so}(3,1)^-_{\mathbb{C}}$ are isomorphic to $\mathfrak{so}(3,1)$ as real Lie algebras.

We also remark that, besides $\mathrm{so}(3,1)^{+}$ and $\mathrm{so}(3,1)^{-}$, the (real) subspace of real matrices fulfilling (2) furnishes a third subalgebra of $\mathrm{so}(3,1)_{\mathbb{C}}$. For $a\in \mathrm{so}(3,1)_{\mathbb{C}}$ one has

$$\begin{align} {\mathrm{Im}}\; a = 0 \quad \Longleftrightarrow \quad a^- = \overline{a^+}. \end{align} \tag{ 39 }$$

As we want to formulate self-dual GR in terms of an $\mathrm{SL}(2,\mathbb{C})$ connection, we need go over from the (anti-)self-dual Lorentz algebra to $\mathfrak{sl}(2,\mathbb{C})$. In order to fix the notation,

$$\begin{align} \mathfrak{sl}\left(2,\mathbb{C}\right) = \left\{a\in\mathrm{Mat}\left(2,\mathbb{C}\right)\,|\, \mathrm{tr}\left(a\right) = 0\right\} = \mathrm{span}_{\mathbb{C}}\left\{\tau_i \, | \, i = 1,2,3\right\}, \end{align} \tag{ 40 }$$

with $\{\tau_i\}$ the generators of $\mathfrak{sl}(2,\mathbb{C})$ that are also generators of $\mathfrak{su}(2)$.

Since $\mathfrak{so}(3,1)$ is isomorphic to $\mathfrak{sl}(2,\mathbb{C})$ (as a real algebra), so is $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$. Let the isomorphism be given by

$$\begin{align} {}^{\pm}P^{}_{}: \mathfrak{so}\left(3,1\right)^\pm_{\mathbb{C}} \rightarrow \mathfrak{sl}\left(2,\mathbb{C}\right), \end{align} \tag{ 41 }$$

where we have used the same symbol as for the (anti-)sef-dual projector. The two are in fact closely related as will become clear in a moment ³ .

Under this isomorphism $a^\pm\in\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ is mapped to

$$\begin{align} {\require{mathtools}\prescript{\pm}{}{P}^{}_{}}\left(a^\pm\right): = {\require{mathtools}\prescript{\pm}{}{P}^{i}_{IJ}}a^{\pm IJ}\tau_i \in\mathfrak{sl}\left(2,\mathbb{C}\right), \end{align} \tag{ 42 }$$

which defines its components ${\require{mathtools}\prescript{\pm}{}{P}^{i}_{IJ}}$ and $a^{\pm i}: = {\require{mathtools}\prescript{\pm}{}{P}^{i}_{IJ}}a^{\pm IJ}$. This is invertible, so conversely we can map $x\in\mathfrak{sl}(2,\mathbb{C})$ back to

$$\begin{align} \require{mathtools}\prescript{\pm}{}{P^{-1}}\left(x\right) = {\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{IJ}_{i}}x^i L^\pm_{IJ} \in\mathfrak{so}\left(3,1\right)^\pm_{\mathbb{C}}. \end{align} \tag{ 43 }$$

Combining these two, we can write

$$\begin{align} \begin{aligned} &a^{\pm IJ} = \left(P^{-1}\circ P\left(a\right)\right)^{IJ} = {\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{IJ}_{i}}{\require{mathtools}\prescript{\pm}{}{P}^{i}_{KL}}a^{\pm KL},\\ &x^{i} = \left(P\circ P^{-1}\left(x\right)\right)^i = {\require{mathtools}\prescript{\pm}{}{P}^{i}_{IJ}}{\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{IJ}_{j}}x^{j}. \end{aligned} \end{align} \tag{ 44 }$$

From this we can conclude the following relations when contracting the isomorphism components over their $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ and $\mathfrak{sl}(2,\mathbb{C})$ indices, respectively. Namely,

$$\begin{align} \begin{aligned} &{\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{IJ}_{i}}{\require{mathtools}\prescript{\pm}{}{P}^{i}_{KL}} = \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}},\\ &{\require{mathtools}\prescript{\pm}{}{P}^{i}_{IJ}}{\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{IJ}_{j}} = \delta^i_j. \end{aligned} \end{align} \tag{ 45 }$$

For the first relation, we do not get just $\delta^I_{[K}\delta^J_{L]}$ but the projector, as self-duality of the index pairs has to be maintained. The projector however acts as $\mathbb{I}$ on self-dual objects ⁴ .

Next, we want to show, that the isomorphism does indeed map the respective Cartan–Killing metric onto each other. For $x,y\in\mathfrak{sl}(2,\mathbb{C})$ the commutation relation is given by

$$\begin{align} \left[{x},{y}\right] = x^iy\,^j{\mathring{\epsilon}}{_{ij}{}^k}\tau_k. \end{align} \tag{ 46 }$$

Acting with the isomorphism on the commutator of $a^\pm,b^\pm\in\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ and employing (45) yields

$$\begin{align} \begin{aligned} \require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}\left(\left[{a^\pm},{b^\pm}\right]\right)& = {\require{mathtools}\prescript{\pm}{}{P}^{k}_{MN}}\left[{a^\pm},{b^\pm}\right]^{MN}\tau_k\\ & = a^{\pm IJ}b^{\pm KL}{\,\mathring{f}\,}{^\pm{}_{IJKL}{}^{MN}}{\require{mathtools}\prescript{\pm}{}{P}^{k}_{MN}}\tau_k\\ & = a^{\pm i}b^{\pm j}{\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{IJ}_{i}}{\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{KL}_{j}}{\,\mathring{f}\,}{^\pm{}_{IJKL}{}^{MN}}{\require{mathtools}\prescript{\pm}{}{P}^{k}_{MN}}\tau_k. \end{aligned} \end{align} \tag{ 47 }$$

From this we conclude the expected relation of structure constants

$$\begin{align} {\mathring{\epsilon}}{_{ij}{}^k} = {\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{IJ}_{i}}{\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{KL}_{j}}{\,\mathring{f}\,}{^\pm{}_{IJKL}{}^{MN}}{\require{mathtools}\prescript{\pm}{}{P}^{k}_{MN}}. \end{align} \tag{ 48 }$$

Using once again (45), we obtain the desired relation between the Cartan–Killing metrics,

$$\begin{align} \begin{aligned} k_{ij}& = {\mathring{\epsilon}}{_{im}{}^n}{\mathring{\epsilon}}{_{jn}{}^m} = {\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{II^{\prime} }_{i}}{\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{JJ^{\prime} }_{j}}{\,\mathring{f}\,}{^\pm{}_{II^{\prime} KL}{}^{MN}}{\,\mathring{f}\,}{^\pm{}_{JJ^{\prime} MN}{}^{KL}}\\ & = {\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{II^{\prime} }_{i}}{\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{JJ^{\prime} }_{j}}k^\pm_{II^{\prime} JJ^{\prime} }, \end{aligned} \end{align} \tag{ 49 }$$

as expected.

Solving this for one of the inverse isomorphism components, we get

$$\begin{align} {\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{IJ}_{i}} = {\require{mathtools}\prescript{\pm}{}{P}^{\,j}_{KL}} k_{ij} k^{\pm IJKL}. \end{align} \tag{ 50 }$$

We leave this relation as it is, because it is valid without reference to specific choices of bases. Further we do not want to move indices with the Cartan–Killing metrics, but with internal ones, i.e. the Minkowski metric and a yet to be specified internal three metric, acting at $\mathfrak{sl}(2,\mathbb{C})$ indices.

Throughout the rest of this article, however, the generators of $\mathfrak{sl}(2,\mathbb{C})$ are given by

$$\begin{align} \tau_j = \frac{\sigma_j}{2i}, \end{align} \tag{ 51 }$$

with the Pauli matrices σ_j . Hence the structure constants are just the Levi-Civita symbols ${\epsilon}{_{ijk}}$ where one index is raised with the Euclidean metric ⁵ :

$$\begin{align} {\mathring{\epsilon}}{_{ij}{}^k} = {\epsilon}{_{ijm}}\delta^{mk}. \end{align} \tag{ 52 }$$

Consequently, the explicit form of the Cartan–Killing metric is

$$\begin{align} k_{ij} = {\epsilon}{_{ikm}}\delta^{ml}{\epsilon}{_{jln}}\delta^{nk} = -2\delta_{ij} \end{align} \tag{ 53 }$$

and is also defined with respect to the Euclidean metric. With this prefactor matching the one of $\eta_{I[K}\eta_{L]J}$ in k_IJKL , (50) becomes

$$\begin{align} {\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{IJ}_{i}} = {\require{mathtools}\prescript{\pm}{}{P}^{\,j}_{KL}} \delta_{ij} \eta^{IK}\eta^{LJ}, \end{align} \tag{ 54 }$$

where we absorbed (anti-)self-duality completely into the contraction with the isomorphism components. As this now involves the metric of the internal Minkowski space, associated with $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$, it is already at this point suggestive that the metric of the internal space, associated with $\mathfrak{sl}(2,\mathbb{C})$, is the Euclidean metric. We will come back to this later.

Given our choice of basis, we can determine explicit expressions for the isomorphism components from (45), up to a sign $s = \pm 1$, that is unrelated to (anti-)self-duality ⁶ . The relation between structure constants (48) uniquely fixes this sign and we end up with

$$\begin{align} \begin{aligned} {\require{mathtools}\prescript{\pm}{}{P}^{\,j}_{KL}}& = \begin{cases} \mp\frac{i}{2}\delta^j_L & K = 0, L = 1,2,3 \\ \frac{1}{2}{\epsilon}{\,^j{}_{KL}} & K,L = 1,2,3 \end{cases} ,\\ {\left(\require{mathtools}\prescript{\pm}{}{P^{-1}}\right)^{KL}_{j}}& = \begin{cases} \pm\frac{i}{2}\delta_j^L & K = 0, L = 1,2,3 \\ \frac{1}{2}{\epsilon}{_j{}^{KL}} & K,L = 1,2,3 \end{cases}. \end{aligned} \end{align} \tag{ 55 }$$

Index positions have to be understood with respect to the Euclidean metric. All other components, except the ones obtained from antisymmetry in the Lorentz indices, are vanishing.

With the isomorphism of Lie algebras (41) are able to map an (anti-)self-dual Lorentz connection to an $\mathfrak{sl}(2,\mathbb{C})$-valued field. However, we have to make sure that this object is indeed a connection as well. In order to see this, we need to have the correct behaviour under gauge transformation with respect to $\mathrm{SL}(2,\mathbb{C})$.

At first we have to specify the transformation behaviour of the complex Lorentz connection ω, which we started with. Recall that the complex Lorentz group

$$\begin{align} \mathrm{SO}\left(3,1;\mathbb{C}\right) = \left\{{M\in\mathrm{Mat}\left(4,\mathbb{C}\right)\, |\, M^\mathrm{T}\eta M = \eta, \text{det}\left(M\right) = 1}\right\} \end{align} \tag{ 56 }$$

is connected [36]. Therefore every element can be expressed by a product of exponentials of its Lie algebra, which is the complexified Lorentz Lie algebra $\mathfrak{so}(3,1)_{\mathbb{C}}$. Given $g\in\mathrm{SO}(3,1;\mathbb{C})$, we can decompose the corresponding Lie algebra elements $a_i = a^+_i+a^-_i$ into its commuting self-dual and anti-self-dual parts. This yields

$$\begin{align} g = \prod_i e^{a_i} = \prod_i e^{a^+_i+a^-_i} = \prod_i e^{a^+_i}\prod_j e^{a^-_j} = :g^+g^-. \end{align} \tag{ 57 }$$

The group elements $g^\pm = \prod_i\text{exp}{a^\pm_i}$ are commuting. This motivates to define

$$\begin{align} G_\pm: = \left\{{g_\pm\in\mathrm{SO}\left(3,1;\mathbb{C}\right)\bigg| g_\pm = \prod_i e^{a^\pm_i}, a^\pm_i\in \mathfrak{so}\left(3,1\right)^\pm_{\mathbb{C}} }\right\}. \end{align} \tag{ 58 }$$

So $G_\pm$ are all the elements of $\mathrm{SO}(3,1;\mathbb{C})$ which can be associated to self-dual or anti-self-dual Lie algebra elements, respectively. They inherit multiplication, inverse elements and the neutral element from $\mathrm{SO}(3,1;\mathbb{C})$. Therefore they are commuting subgroups of $\mathrm{SO}(3,1;\mathbb{C})$. By construction they are connected.

We want to clarify how $G_\pm$ are related to $\mathrm{SL}(2,\mathbb{C})$. Since they arise as exponentials of a Lie algebra isomorphic to $\mathrm{SL}(2,\mathbb{C})$, they are the operators of a four representation of $\mathrm{SL}(2,\mathbb{C})$. The Casimir elements are

$$\begin{align} \begin{aligned} C^\pm_1 & = k^{\,IJKL}L^\pm_{IJ}L^\pm_{KL}, \\ C^\pm_2 & = k^{\,IJKL}L^\pm_{IJ}\left(\ast L^\pm\right)_{KL} = \pm iC^\pm_{1}, \end{aligned} \end{align} \tag{ 59 }$$

where we used the (anti-)self-duality of the generators. Using the fact that the (anti-)self-dual generators are given by the projectors, see (16), and the relation between Cartan–Killing and Minkowski metric, it is easy to determine

$$\begin{align} k^{\,IJKL}{\left(L^\pm_{IJ}\right)}{^M{}_{M^{\prime} }}{\left(L^\pm_{KL}\right)}{^{M^{\prime} }{}_N} = -\frac{1}{2}\require{mathtools}\prescript{\mp}{}{{P}{^{MM^{\prime} }{}_{NM^{\prime} }}} = \frac{1}{4}\left({\delta^{[M}_{N}\delta^{M^{\prime}]}_{M^{\prime} }-\frac{i}{2}{\epsilon}{^{MM^{\prime} }{}_{NM^{\prime} }}}\right) = \frac{3}{8}\delta^M_N. \end{align} \tag{ 60 }$$

In the last step the trace of the Levi-Civita symbol vanishes. With this, the two Casimir elements are given by

$$\begin{align} \begin{aligned} {{C^\pm_1}}{^M{}_N} & = \frac{3}{8}\delta^M_N, \\ {{C_2^\pm}}{^M{}_N} & = \pm i\frac{3}{8}\delta^M_N. \end{aligned} \end{align} \tag{ 61 }$$

Next we introduce

$$\begin{align} {{\left(K^\pm_i\right)}}{^M{}_N}: = {{\left(L^\pm_{0i}\right)}}{^M{}_N} = {\require{mathtools}\prescript{\pm}{}{{P}{^{}_{}}}}{_{0i}{}^M{}_N} \end{align} \tag{ 62 }$$

and

$$\begin{align} {{\left(J^\pm_i\right)}}{^M{}_N}: = {\left(\ast L^\pm_{0i}\right)}{^M{}_N} = \frac{1}{2}{\epsilon}{_{0i}{}^{jk}}{{\left(L^\pm_{jk}\right)}}{^M{}_N} = \pm i{{\left(L^\pm_{0i}\right)}}{^M{}_N} = \pm i{{\left(K^\pm_i\right)}}{^M{}_N}. \end{align} \tag{ 63 }$$

Using this decomposition of the (anti-)self-dual basis, it is straightforward to see that we can express the Casimir elements in the additional forms

$$\begin{align} \begin{aligned} C^\pm_1 & = K_\pm^2-J_\pm^2 = 2K_\pm^2 = -2J_\pm^2, \\ C_2^\pm & = 2K_\pm\cdot J_\pm = \pm 2iK_\pm^2 = \mp 2iJ_\pm^2. \end{aligned} \end{align} \tag{ 64 }$$

Here all products are with respect to the Euclidean three-metric. In comparison to the form of standard Casimirs of the Lorentz algebra, this suggests that $K^\pm$ and $J^\pm$ have indeed an interpretation as generators of boosts and rotations, respectively, even if they have been obtained from the (anti-)self-dual generators ⁷ .

This can however be verified when determining the commutation relations for $K^\pm$ and $J^\pm$. Using the commutation relation of (anti-)self-dual generators and the relation between $K^\pm$ and $J^\pm$, we find

$$\begin{align} \begin{aligned} &\left[{K^\pm_i},{K^\pm_j}\right] = \pm \frac{i}{2} {\epsilon}{_{ij}{}^k} K^\pm_k = \frac{1}{2}{\epsilon}{_{ij}{}^k}J^\pm_k,\\ &\left[{J^\pm_i},{J^\pm_j}\right] = \left(\pm i\right)^2 \left[{K^\pm_i},{K^\pm_j}\right] = -\frac{1}{2}{\epsilon}{_{ij}{}^k}J^\pm_k,\\ &\left[{K^\pm_i},{J^\pm_j}\right] = \pm i \left[{K^\pm_i},{K^\pm_j}\right] = \pm \frac{i}{2} {\epsilon}{_{ij}{}^k}J^\pm_k = -\frac{1}{2} {\epsilon}{_{ij}{}^k}K^\pm_k. \end{aligned} \end{align} \tag{ 65 }$$

Hence $K_\pm$ and $J_\pm$ indeed satisfy the commutation relations for boosts and rotations. In this form, the relations are the same for both, self-dual and anti-self-dual algebra.

In the usual way we then recombine these into generators $M^\pm_j$ and $N^\pm_j$ of two commuting copies of $\mathfrak{sl}(2,\mathbb{C})$. It turns out that $M^+_j = J^+_j$ and $N^-_j = J^-_j$, while the other generators vanish. By relating $J^{\pm}_j$ to the actual spin operators, we can determine the representation label of the Casimir elements in (61) to be $j = \frac{1}{2}$. Therefore we have a multiplicity of two for this representation, in order to construct the four dimensional representation. We therefore find that $\mathfrak{so}(3,1)^+_{\mathbb{C}}$ corresponds to the

$$\begin{align} \pi_+ = \pi_{\left(\frac{1}{2},0\right)}\oplus\pi_{\left(\frac{1}{2},0\right)} \end{align} \tag{ 66 }$$

representation of $\mathfrak{sl}(2,\mathbb{C})$, while $\mathfrak{so}(3,1)^-_{\mathbb{C}}$ corresponds to the

$$\begin{align} \pi_- = \pi_{\left(0,\frac{1}{2}\right)}\oplus\pi_{\left(0,\frac{1}{2}\right)} \end{align} \tag{ 67 }$$

representation. Consequently there is indeed a basis in which one of the corresponding group representations is of diagonal form. Let us say this is the case for $G_+$. Let $g^+ = \mathbb{I}_2\otimes g$ and $g^- = \bar{g}\otimes\mathbb{I}_2$. With this, we understand how $G_\pm$ are four dimensional representations of $\mathrm{SL}(2,\mathbb{C})$. Hence the decomposition (57), given a specific basis, can be understood as

$$\begin{align} g = g^+ g^- = \Pi_+\left(g_1\right)\Pi_-\left(g_1\right) = \left(\mathbb{I}_2\otimes g_1\right)\left(\bar{g}_2\otimes\mathbb{I}_2\right) = \bar{g}_2\otimes g_1. \end{align} \tag{ 68 }$$

In this tensor product form we immediately see that

$$\begin{align} \bar{g}_2\otimes g_1 = -\bar{g}_2\otimes \left(-g_1\right). \end{align} \tag{ 69 }$$

Hence the decomposition of $\mathrm{SO}(3,1;\mathbb{C})$ into $G_\pm$ is not unique, but takes the form

$$\begin{align} \mathrm{SO}\left(3,1;\mathbb{C}\right) = G_+\times G_-/H\cong\mathrm{SL}\left(2,\mathbb{C}\right)\times\mathrm{SL}\left(2,\mathbb{C}\right)/H, \end{align} \tag{ 70 }$$

with $H = \{\mathbb{I}_4,-\mathbb{I}_4\}$. This is a well known result, see e.g. [37], which we arrived at, using the (anti-)self-dual decomposition of the complexified Lorentz–Lie algebra.

In section 2.2.2 we saw how to express $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$ objects in terms of $\mathfrak{sl}(2,\mathbb{C})$. In this section we want to establish how the self-dual and anti-self-dual Lorentz connections are related to $\mathrm{SL}(2,\mathbb{C})$ connections.

The last section tells us that there is a basis in which we can represent the $\mathfrak{so}(3,1)_{\mathbb{C}}$-connection ω in the form

$$\begin{align} \omega = \mathbb{I}\otimes A_++\bar{A}_-\otimes\mathbb{I}, \end{align} \tag{ 71 }$$

which ensures the commutativity of the self-dual and anti-self-dual parts, represented by the $\mathfrak{sl}(2,\mathbb{C})$ elements $A_+$ and $A_-$, respectively. Understanding again g as $\bar{g}_-\otimes g_+$, the transformation behaviour of ω takes the form

$$\begin{align} \begin{aligned} \omega^{\prime} & = g\omega g^{-1} + g\mathrm{d}{g^{-1}} = \mathbb{I}\otimes\left(g_+A_+g_+^{-1}+g_+\mathrm{d}g_+^{-1}\right)+ \left(\bar{g}_-\bar{A}_-\bar{g}_-^{-1}+\bar{g}_-\mathrm{d}\bar{g}_-^{-1}\right)\otimes\mathbb{I}\\ & = \mathbb{I}\otimes A^{\prime} _++\bar{A}^{\prime} _-\otimes\mathbb{I}. \end{aligned} \end{align} \tag{ 72 }$$

Hence the (anti-)self-dual parts of the connection ω are not only connections on their own, but are indeed already $\mathfrak{sl}(2,\mathbb{C})$-connections.

We note that our arguments so far only establish the existence of the $\mathfrak{sl}(2,\mathbb{C})$-connections $\omega_\pm$ locally. To give a rigorous argument, one would have to establish the existence of the two $\mathrm{SL}(2,\mathbb{C})$-bundles and their connections globally. We see no obstruction to doing this, and might come back to it elsewhere. Remarkably, it seems that no additional global structure, such as a spin structure, is needed when starting from the theory with $\mathrm{SO}(3,1;\mathbb{C})$ as structure group. This is in contrast to the real Ashtekar–Barbero formulation [21], where a spin structure is needed to obtain the $\mathrm{SU}(2)$-formulation [38].

Using that the projectors sum up to the identity in (11), the complex Palatini action decomposes into a self-dual and an anti-self-dual part:

$$\begin{align} \begin{aligned} S_{\mathbb{C}}\left[\omega,e\right]& = \frac{\lambda}{\kappa}\int_{\mathcal{M}} \epsilon_{IJKL}\Sigma^{IJ}\!\wedge F^{KL}\\ & = \frac{\lambda}{\kappa}\int_{\mathcal{M}} \epsilon_{IJKL}\left({\require{mathtools}\prescript{+}{}{{P}{^{IJ}{}_{I^{\prime} J^{\prime} }}}+\require{mathtools}\prescript{-}{}{{P}{^{IJ}{}_{I^{\prime} J^{\prime} }}}}\right)\left({\require{mathtools}\prescript{+}{}{{P}{^{KL}{}_{K^{\prime} L^{\prime} }}}+\require{mathtools}\prescript{-}{}{{P}{^{KL}{}_{K^{\prime} L^{\prime} }}}}\right)\Sigma^{I^{\prime} J^{\prime} }\!\wedge F^{K^{\prime} L^{\prime} }\\ & = \left[t\right] \frac{\lambda}{\kappa}\int_{\mathcal{M}} \epsilon_{IJKL}\require{mathtools}\prescript{+}{}{{P}{^{IJ}{}_{I^{\prime} J^{\prime} }}}\require{mathtools}\prescript{+}{}{{P}{^{KL}{}_{K^{\prime} L^{\prime} }}}\Sigma^{I^{\prime} J^{\prime} }\!\wedge F^{K^{\prime} L^{\prime} }\\ & \quad +\frac{\lambda}{\kappa}\int_{\mathcal{M}} \epsilon_{IJKL}\require{mathtools}\prescript{-}{}{{P}{^{IJ}{}_{I^{\prime} J^{\prime} }}}\require{mathtools}\prescript{-}{}{{P}{^{KL}{}_{K^{\prime} L^{\prime} }}}\Sigma^{I^{\prime} J^{\prime} }\!\wedge F^{K^{\prime} L^{\prime} } \end{aligned} \end{align} \tag{ 73 }$$

Introducing the projected Plebanski two-form $\Sigma^{\pm IJ}$ and the (anti-)self-dual projection of the curvature curvature $F^{\pm IJ}$, the complex action decomposes into

$$\begin{align} S = S_++S_-. \end{align} \tag{ 74 }$$

The (anti-)self-dual actions $S_\pm$ are

$$\begin{align} S_\pm\left[\omega^{\left(\pm\right)},e\right] = \frac{\lambda}{\kappa}\int_{\mathcal{M}} \epsilon_{IJKL}\Sigma^{\left(\pm\right)IJ}\!\wedge F^{\left(\pm\right)KL}. \end{align} \tag{ 75 }$$

While the projection $\Sigma^{(\pm) IJ} = \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}}\Sigma^{KL}$ is straightforward, there is some more work required, in order to show that $F^{(\pm)IJ}$ is indeed the curvature of the (anti-)self-dual connection $\omega^{(\pm)IJ} = \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}}\omega^{KL}$. For the second term in (6) we find that

$$\begin{align} {\omega}{^I{}_M}\wedge{\omega}{^{MJ}} = \delta^{[I}_{[K}\eta_{L][M}\delta^{J]}_{N]}{\omega}{^{KL}}\wedge{\omega}{^{MN}} = \frac{1}{2}{\,\mathring{f}}_{KLMN}{}^{IJ}{\omega}{^{KL}}\wedge{\omega}{^{MN}}. \end{align} \tag{ 76 }$$

Thus

$$\begin{align} \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{I^{\prime} J^{\prime} }}} {\omega}{^{I^{\prime} }{}_M}\wedge{\omega}{^{MJ^{\prime} }} = {\omega}{^{\left(\pm\right)I}{}_M}\wedge{\omega}{^{\left(\pm\right)MJ}}, \end{align} \tag{ 77 }$$

where we used (34) in order to pull the projectors through the structure constant, which again in absorbed in the contraction of indices. So indeed from (6) we get

$$\begin{align} \require{mathtools}\prescript{\pm}{}{{P}{^{IJ}{}_{KL}}}F\left[\omega\right]^{KL} = F\left[\omega^{\left(\pm\right)}\right]^{IJ}. \end{align} \tag{ 78 }$$

We want to make a remark about the use of internal metrics here. Given the considerations about the Cartan–Killing metric of $\mathfrak{so}(3,1)^\pm_{\mathbb{C}}$, one can ask oneself, whether is possible to write down the Palatini action (75) referring to the Cartan–Killing metric only. This metric, however, is only able to move antisymmetric pairs of indices and compared to equation (6) we need to raise a single index of the curvature of the Lorentz connection. For this, however, we need the internal Minkowski metric.

In the following we will restrict ourselves to the self-dual part of the action only, because it yields a theory that reduces to Einstein gravity under certain reality conditions. Out of convenience, we will also drop the + indicating self-dual objects.

Next we look at the projection of the Plebanski two-form Σ^IJ , i.e. the self-dual part of $e^I\wedge e^J$. It is given by

$$\begin{align} \Sigma^{IJ} = {P}{^{IJ}{}_{KL}}e^K_\alpha e^L_\beta \mathop{\mathrm{d}x^\alpha}\!\wedge\mathop{\mathrm{d}y^\beta} = \frac{1}{2}\left({e^{[I}_\alpha e^{J]}_\beta-\frac{i}{2}{\epsilon^{IJ}}_{KL}e^K_\alpha e^L_\beta}\right)\mathop{\mathrm{d}x^\alpha}\!\wedge\mathop{\mathrm{d}y^\beta}. \end{align} \tag{ 79 }$$

From this we define the following ⁸

$$\begin{align} \Sigma^{IJ}_{\alpha\beta}: = \frac{1}{2}\left({e^{[I}_\alpha e^{J]}_\beta-\frac{i}{2}{\epsilon^{IJ}}_{KL}e^K_\alpha e^L_\beta}\right). \end{align} \tag{ 80 }$$

In order to further fix the notation, $\Sigma^{\alpha\beta}_{IJ}$, where indices where moved with the respective metrics, will be referred to as Plebanski bi-vector.

Since the frame fields are soldering forms between the internal and the tangential space, it is natural to ask how self-duality of internal indices transfers to tangential ones. To this end, we first express the determinant of the spacetime metric in terms of the frame fields. This yields the usual relation

$$\begin{align} \text{det}\left({g_{\alpha\beta}}\right) = -\text{det}\left({e^I_\alpha}\right)^2 \end{align} \tag{ 81 }$$

even though g_αβ and $e^I_\alpha$ are complex quantities now. Taking the square root we can write this as

$$\begin{align} \sqrt{-\text{det}\left({g_{\alpha\beta}}\right)} = s_e\text{det}\left({e^I_\alpha}\right), \end{align} \tag{ 82 }$$

where we introduced a sign function for the determinant of complex tetrads, i.e.

$$\begin{align} s_e: = \frac{\text{det}\left({e^I_a}\right)}{\sqrt{\text{det}\left({e^I_a}\right)^2}}. \end{align} \tag{ 83 }$$

It is real and takes values ±1. Equation (82) further shows that $\text{det}\left({e^I_a}\right)$ carries a density weight of +1.

In order to relate the Hodge dual with respect to internal and tangential indices, respectively, we need the following identity for the Levi-Civita symbol:

$$\begin{align} {\tilde{\epsilon}}{^{\alpha^{\prime} \beta^{\prime} \gamma^{\prime} \delta^{\prime} }}\underset{{\smash{\tilde{}}}}{\epsilon}{_{\alpha \beta \gamma \delta}} = -4! \delta^{\alpha^{\prime} }_{[\alpha}\delta^{\beta^{\prime} }_\beta\delta^{\gamma^{\prime} }_\gamma\delta^{\delta^{\prime} }_{\delta]}. \end{align} \tag{ 84 }$$

Here the tilde indicates the density weight of ±1 and the sign on the right hand side comes from all primed indices being raised with the Minkowski metric. Further, as the density weights cancel, this is also valid for the Levi-Civita tensor and with just internal indices. Therefore

$$\begin{align} {\epsilon}{_{IJKL}}e^I_\alpha e^J_\beta e^K_\gamma e^L_\delta & = {\epsilon}{_{IJKL}}e^I_{\alpha^{\prime} } e^J_{\beta^{\prime} } e^K_{\gamma^{\prime} } e^L_{\delta^{\prime} } \delta^{\alpha^{\prime} }_{[\alpha}\delta^{\beta^{\prime} }_\beta\delta^{\gamma^{\prime} }_\gamma\delta^{\delta^{\prime} }_{\delta]} \notag\\ & = -\frac{1}{4!} {\epsilon}{_{IJKL}} {\tilde{\epsilon}}{^{\alpha^{\prime} \beta^{\prime} \gamma^{\prime} \delta^{\prime} }} e^I_{\alpha^{\prime} } e^J_{\beta^{\prime} } e^K_{\gamma^{\prime} } e^L_{\delta^{\prime} } \underset{{\smash{\tilde{}}}}{\epsilon}{_{\alpha \beta \gamma \delta}}\notag \\ & = -\text{det}\left({e^I_\alpha}\right)\underset{{\smash{\tilde{}}}}{\epsilon}{_{\alpha \beta \gamma \delta}}. \end{align} \tag{ 85 }$$

So using (82) and cancelling the density weight of the Levi-Civita symbol, we find

$$\begin{align} {\epsilon}{_{IJKL}}e^I_\alpha e^J_\beta e^K_\gamma e^L_\delta = -s_e{\epsilon}{_{\alpha \beta \gamma \delta}}. \end{align} \tag{ 86 }$$

From this we can determine the action of the tangential Hodge star:

$$\begin{align} ({\ast_{\mathrm{ST}}\Sigma})^{IJ}_{\alpha\beta} & = \frac{1}{2}g_{\alpha\alpha^{\prime} }g_{\beta\beta^{\prime} }{\epsilon}{^{\alpha^{\prime} \beta^{\prime} \gamma\delta}}\frac{1}{2} \left({e^{[I}_\gamma e^{J]}_\delta-\frac{i}{2}{\epsilon^{IJ}}_{KL}e^K_\gamma e^L_\delta} \right)\notag\\ & = \frac{-s_e}{4}g_{\alpha\alpha^{\prime} }g_{\beta\beta^{\prime} } \left({-{\epsilon}{^{IJKL}} e^{\alpha^{\prime} }_K e^{\beta^{\prime} }_L - \frac{i}{2} {\epsilon}{^{IJ}{}_{KL}} {\epsilon}{^{MNKL}} e^{\alpha^{\prime} }_M e^{\beta^{\prime} }_N } \right)\notag\\ & = \frac{-s_e}{4}g_{\alpha\alpha^{\prime} }g_{\beta\beta^{\prime} }\left({-{\epsilon}{^{IJKL}} e^{\alpha^{\prime} }_K e^{\beta^{\prime} }_L + 2i \eta^{II^{\prime} }\eta^{JJ^{\prime} } e^{\alpha^{\prime} }_{[I^{\prime} }e^{\beta^{\prime} }_{J^{\prime} ]}}\right) \notag\\ & = -is_e \frac{1}{2}\left({e^{[I}_\gamma e^{J]}_\delta-\frac{i}{2}{\epsilon^{IJ}}_{KL}e^K_\gamma e^L_\delta}\right). \end{align} \tag{ 87 }$$

Note that we used the usual identities for the Levi-Civita symbol from the second to the third line. Therefore we have

$$\begin{align} \left({\ast_{\mathrm{ST}}\Sigma}\right)^{IJ}_{\alpha\beta} = -is_e \Sigma^{IJ}_{\alpha\beta} \end{align} \tag{ 88 }$$

and hence internal self-duality is only carried over up to a minus sign and depends on the orientation of the complex frame fields.

Before expressing the self-dual action in terms of $\mathfrak{sl}(2,\mathbb{C})$ variables, we rewrite it as follows. As a first step we use the self-duality of the Plebanski two-form:

$$\begin{align} S = \frac{\lambda}{\kappa}\int_{\mathcal{M}} \epsilon_{IJKL}\Sigma^{IJ}\!\wedge F^{KL} = \frac{2i\lambda}{\kappa}\int_{\mathcal{M}} \Sigma_{IJ}\!\wedge F^{IJ}. \end{align} \tag{ 89 }$$

As a second step, we use (84) to separate the four-form $\mathrm{d}{^4x} = \frac{1}{4!}\underset{{\smash{\tilde{}}}}{\epsilon}{_{\mu\nu\rho\sigma}}\mathrm{d}{x^\mu}\!\wedge\mathrm{d}{x^\nu}\!\wedge\mathrm{d}{x^\rho}\!\wedge\mathrm{d}{x^\sigma}$ from the components:

$$\begin{align} S = -\frac{2i\lambda}{\kappa}\int_{\mathcal{M}}\!\mathrm{d}{^4x} {\tilde{\epsilon}}{^{\alpha\beta\gamma\delta}} \Sigma_{IJ\alpha\beta}\frac{1}{2}F^{IJ}_{\gamma\delta}. \end{align} \tag{ 90 }$$

Further we employ (88) in order to get rid of the spacetime Levi-Civita symbol. This yields the final form of the self-dual action:

$$\begin{align} S = -\frac{2\lambda}{\kappa}\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,}\sqrt{-g}\, s_e\Sigma_{IJ}^{\alpha\beta} F^{IJ}_{\alpha\beta}. \end{align} \tag{ 91 }$$

The determinant factor $\sqrt{-g}$ arises due to the transition from the Levi-Civita density to its tensor counterpart, needed for the spacetime Hodge-$\ast$.

The next step is to go over from $\mathfrak{so}(3,1)^+_{\mathbb{C}}$- to $\mathfrak{sl}(2,\mathbb{C})$-variables.

In the contraction of internal indices in (91), there is an implicit self-dual projection. We can now use the first relation in (45) in order to rewrite this contraction in terms of $\mathfrak{sl}(2,\mathbb{C})$-indices:

$$\begin{align} \Sigma_{IJ}^{\alpha\beta}{P}{^{IJ}{}_{KL}} F^{KL}_{\alpha\beta} = \Sigma_{IJ}^{\alpha\beta} \left(P^{-1}\right)^{IJ}_{i}P^{i}_{KL} F^{KL}_{\alpha\beta}. \end{align} \tag{ 92 }$$

This lets us define the $\mathfrak{sl}(2,\mathbb{C})$-valued Plebanski bi-vector

$$\begin{align} \Sigma_i^{\alpha\beta}: = \left(P^{-1}\right)^{IJ}_{i}\Sigma_{IJ}^{\alpha\beta}. \end{align} \tag{ 93 }$$

Further we have the action of the isomorphism on the curvature components, that yields the curvature of the $\mathrm{SL}(2,\mathbb{C})$ connection

$$\begin{align} A^i_\alpha: = P^{i}_{IJ}\omega^{IJ}_\alpha, \end{align} \tag{ 94 }$$

which was introduced in section 2.2.4. We use the fact that the curvature components can be expressed in terms of the commutator of the $\mathfrak{so}(3,1)^+_{\mathbb{C}}$ connection and hence find

$$\begin{align} \begin{aligned} P^{i}_{KL} F^{KL}_{\alpha\beta}& = P^{i}_{KL}2\partial_{[\alpha}\omega^{KL}_{\beta]}+P^{i}_{KL}[{\omega_\alpha},{\omega_\beta}]^{KL}\\ & = 2\partial_{[\alpha}A_{\beta]}^i+[{A_\alpha},{A_{\beta}}]^i\\ & = 2\partial_{[\alpha}A_{\beta]}^i+{\mathring{\epsilon}}{_{jk}{}^i}A^j_\alpha A^k_\beta = : F^{i}_{\alpha\beta}[A]. \end{aligned} \end{align} \tag{ 95 }$$

This allows to recast the self-dual action, based on $\mathfrak{so}(3,1)^+_{\mathbb{C}}$-variables, in terms of $\mathfrak{sl}(2,\mathbb{C})$-valued $A^i_\alpha$ and $\Sigma^{\alpha\beta}_i$.

The next step is to perform a $3+1$-split of this action. In the usual way, see e.g. [19, 20, 33], we introduce a foliation

$$\begin{align} t^\alpha = N n^\alpha + N^\alpha,\quad N^\alpha n_\alpha = 0 \end{align} \tag{ 96 }$$

with respect to lapse function N, shift vector N^a and the normal vector of spatial slices n^a . We want to emphasise that this foliation is real, i.e. lapse, shift and normal are real objects. The spacetime manifold therefore splits into $\mathcal{M} = M\times\mathbb{R}$, where M is the spatial manifold. This gives rise to the spatial metric

$$\begin{align} q_{\alpha\beta} = g_{\alpha\beta}+n_\alpha n_\beta,\quad q_{\alpha\beta}n^\beta = 0. \end{align} \tag{ 97 }$$

Projections to the spatial manifold are done by

$$\begin{align} q^\alpha_\beta = \delta^\alpha_\beta+n^\alpha n_\beta \end{align} \tag{ 98 }$$

and hence a $3+1$ split can be performed by using the relation

$$\begin{align} \delta^\alpha_\beta = q^\alpha_\beta-n^\alpha n_\beta. \end{align} \tag{ 99 }$$

Regarding the metric determinant we have the following identity, which is still valid for the complex spacetime and spatial metrics:

$$\begin{align} \sqrt{-g} = N\sqrt{q}. \end{align} \tag{ 100 }$$

Inserting (99) in the contraction of spacetime indices in the action therefore yields

$$\begin{align} S & = -\frac{2\lambda}{\kappa}\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \sqrt{q} N s_e \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i}\left(q_\alpha^\gamma-n_\alpha n^\gamma\right)\left(q_\beta^\delta-n_\beta n^\delta\right) \notag\\ & = -\frac{2\lambda}{\kappa}\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \sqrt{q} N s_e \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i} \left(q_\alpha^\gamma q_\beta^\delta-q_\alpha^\gamma n_\beta n^\delta-q_\beta^\delta n_\alpha n^\gamma + n_\alpha n_\beta n^\gamma n^\delta\right)\notag\\ & = -\frac{2\lambda}{\kappa}\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \sqrt{q} N s_e \left({q_\alpha^\gamma q_\beta^\delta \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i} -2 q_\alpha^\gamma n_\beta n^\delta \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i}}\right) \notag\\ & = -\frac{2\lambda}{\kappa}\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \sqrt{q} s_e \left({N q_\alpha^\gamma q_\beta^\delta \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i} + 2 q_\alpha^\gamma n_\beta N^\delta \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i} -2 t^\delta q_\alpha^\gamma n_\beta \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i}}\right). \end{align} \tag{ 101 }$$

From the second to the third line, we used the antisymmetry of $\Sigma^{i}_{\alpha\beta} F_{i}^{\gamma\delta}$ to combine the middle two terms in the bracket and to cancel the symmetric combination of n's. From the third to the fourth line we used $Nn^\beta = t^\beta-N^\beta$. Using the easy to show relation $t^\alpha F^i_{\alpha\beta} = \mathop{\mathcal{L}_t}(A^i_\beta)-\mathop{\mathcal{D}_\beta} (t^\alpha A^i_\alpha)$, where $\mathcal{D}$ is the covariant derivative with respect to $A^i_a$, and taking the antisymmetry of F_αβ into account, we arrive at

$$\begin{align} S = \int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \Bigl( &-\frac{4\lambda\sqrt{q} s_e}{\kappa} q_\alpha^\gamma n_\beta \Sigma^{\alpha\beta}_i \mathop{\mathcal{L}_t}(A^i_\gamma) \end{align} \tag{ 102 }$$

$$\begin{align} &- \frac{2\lambda\sqrt{q} s_e}{\kappa} N q_\alpha^\gamma q_\beta^\delta \Sigma_{i}^{\alpha\beta} F^{i}_{\gamma \delta} \end{align} \tag{ 103 }$$

$$\begin{align} &- \frac{4\lambda\sqrt{q} s_e}{\kappa} N^\delta q_\alpha^\gamma n_\beta \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i} \end{align} \tag{ 104 }$$

$$\begin{align} &+ \frac{4\lambda\sqrt{q} s_e}{\kappa} q_\alpha^\gamma n_\beta \Sigma_{i}^{\alpha\beta} \mathop{D_\gamma} (t^\delta A^i_\delta)\Bigr). \end{align} \tag{ 105 }$$

From the first term, we will extract the canonically conjugated variables and the symplectic structures. The terms proportional to lapse and shift will be identified as Hamilton and diffeomorphism constraints, respectively, and the last term will form the Gauß constraint. We will look at the individual contributions separately in the next sections.

We start with the piece (102), from which we will get the symplectic structure and the actual canonical pair consisting of the Ashtekar connection $A^i_a$ and a corresponding electric field $E^a_i$.

Using $\mathop{\mathcal{L}_t}(q^\alpha_\beta) = 0$ and the spacetime self-duality of the Plebanski two-form we find that

$$\begin{align} -\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \frac{4\lambda\sqrt{q} s_e}{\kappa} q_\alpha^\gamma n_\beta \Sigma^{\alpha\beta}_i \mathop{\mathcal{L}_t}\left(A^i_\gamma\right) = -\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \frac{4\lambda\sqrt{q} s_e}{\kappa} \frac{i}{2 s_e} n_\beta {\epsilon}{^{\alpha\beta}{}_{\gamma\delta}} \Sigma^{\gamma\delta}_i \mathop{\mathcal{L}_t}\left(q_\alpha^\rho A^i_\rho\right). \end{align} \tag{ 106 }$$

Here we have $q_\alpha^\rho A^i_\rho$, the spatial projection of the spacetime Ashtekar connection. All open indices of $n_\beta {\epsilon}{^{\alpha\beta}{}_{\gamma\delta}} = -n_\beta {\epsilon}{^{\beta\alpha}{}_{\gamma\delta}}$ are implicitly spatially projected and so is therefore $\Sigma^{\gamma\delta}_i$. Hence we pull everything back to the spatial manifold and replace the spacetime Levi-Civita tensor by its spatial analogue ${\epsilon}{^{a}{}_{bc}}$. This yields

$$\begin{align} \int_{\mathbb{R}}\mathrm{d}{t}\int_M \mathrm{d}{^3x} \frac{2\lambda\sqrt{q} i}{\kappa} {\epsilon}{^{a}{}_{bc}} \Sigma^{bc}_i \mathop{\mathcal{L}_t}(A^i_a) = \int_{\mathbb{R}}\mathrm{d}{t}\int_M \mathrm{d}{^3x}\frac{2\lambda}{\kappa i} \tilde{E}^a_i \mathop{\mathcal{L}_t}(A^i_a), \end{align} \tag{ 107 }$$

where $A^i_a$ is the pull-back of $q_\alpha^\rho A^i_\rho$ to the spatial manifold. It is the spatial $\mathfrak{sl}(2,\mathbb{C})$-valued, hence complex, Ashtekar connection. Recalling from where we started, this is the pull-back of the self-dual part of a complex spacetime Lorentz connection, expressed in terms of $\mathfrak{sl}(2,\mathbb{C})$. In comparison, the real Ashtekar–Barbero connection is a purely spatial object and is not the pull-back of a spacetime connection ⁹ .

Further we introduced the electric field $\tilde{E}^a_i$, a density of weight one, corresponding to the Ashtekar connection. It is given by

$$\begin{align} \tilde{E}^a_i = -\sqrt{q}{\epsilon}{^a{}_{bc}}\Sigma^{bc}_i. \end{align} \tag{ 108 }$$

Now we can read off the symplectic structure and invert it for a Poisson relation

$$\begin{align} \left\{A^i_a\left(x\right),\tilde{E}^b_j\left(y\right)\right\} = \frac{i\kappa}{2\lambda}\delta^b_a\delta^i_j\delta\left(x,y\right). \end{align} \tag{ 109 }$$

This tells us that the canonically conjugated variables of self-dual GR are the $\mathfrak{sl}(2,\mathbb{C})$ connection $A^i_a$ and the corresponding electric field $E^a_i$ which is complex valued as well.

At this point we see the effect of rescaling the action with λ, which enters our Poisson relation. We will make use of two special cases for the quantisation in [39]: λ = 1 and $\lambda = i$. They result in two distinct Poisson relations:

$$\begin{align} \left\{A^i_a\left(x\right),\tilde{E}^b_j\left(y\right)\right\} = \frac{i\kappa}{2}\delta^b_a\delta^i_j\delta\left(x,y\right), \end{align} \tag{ 110 }$$

if λ = 1 or

$$\begin{align} \left\{A^i_a\left(x\right),\tilde{E}^b_j\left(y\right)\right\} = \frac{\kappa}{2}\delta^b_a\delta^i_j\delta\left(x,y\right) \end{align} \tag{ 111 }$$

if $\lambda = i$. The first case (110) is the Poisson relation that is expected when using self-dual variables, see for example [4, 33]. In the second case however, we enforce the Poisson relation as for real Ashtekar–Barbero variables. Nevertheless, connection and electric field in (111) are complex. In [10, 34] this kind of Poisson relation for complex variables appears in the context of using complexifier methods in order to construct a transformation from real to complex loop quantum gravity, which already incorporates the reality conditions. A further instance of the use of this Poisson bracket for complex variables can be found in [11], but under a different premise.

Before we go on and show that the electric field indeed act as triads of the spatial manifold, we need to analyse this complex Poisson relation in more detail. We do this in the next section.

Looking at the term (107) giving the symplectic potential, we get a kinematical phase space in which Ashtekar connection and the electric field are canonical coordinates. Both are a priori complex. The question arises whether it is possible to derive from (107) a symplectic structure for real and imaginary parts of the connection and the electric field.

We split $A^i_a = {A_{\mathrm{R}}}^i_a+i{A_{\mathrm{I}}}^i_a$ and $\tilde{E}^b_j = {\tilde{E}_{\mathrm{R}}}^b_j+i{\tilde{E}_{\mathrm{I}}}^b_j$ into the respective real and imaginary parts and insert this into (107):

$$\begin{align} \tilde{E}^a_i \mathcal{L}_t {A}^i_a = {\tilde{E}_{\mathrm{R}}}^a_i \mathcal{L}_t {A_{\mathrm{R}}}^i_a -{\tilde{E}_{\mathrm{I}}}^a_i \mathcal{L}_t {A_{\mathrm{I}}}^i_a+i{\tilde{E}_{\mathrm{I}}}^a_i \mathcal{L}_t {A_{\mathrm{R}}}^i_a +i{\tilde{E}_{\mathrm{R}}}^a_i \mathcal{L}_t {A_{\mathrm{I}}}^i_a. \end{align} \tag{ 112 }$$

From this we determine a pre-symplectic structure for the variations $\delta A^i_a$, $\delta\tilde{E}^a_i$ split into their respective real an imaginary parts $\delta^\mathrm{R}A^i_a$, $\delta^\mathrm{I}A^i_a$ and $\delta^\mathrm{R}\tilde{E}^a_i$, $\delta^\mathrm{I}\tilde{E}^a_i$, respectively. It is given by

$$\begin{align}\Omega\left(\delta_1^\mathrm{R},\delta_1^\mathrm{I},\delta_2^\mathrm{R},\delta_2^\mathrm{I}\right) = \int_\Sigma\mathrm{d}{^3x} \delta^\mathrm{T}_1\left(x\right)M\delta_2\left(x\right), \end{align} \tag{ 113 }$$

where $\delta_i^\mathrm{T}: = (\delta_i^\mathrm{R}A^1_1,\delta_i^\mathrm{I}A^1_1,\delta_i^\mathrm{R}\tilde{E}^1_1$, $\delta_i^\mathrm{I}\tilde{E}^1_1,\dots)$. The $12\times 12$-matrix M, characterising the presymplectic form is given by

$$\begin{align} M: = \mathbb{I}_3\otimes\begin{pmatrix} 0 & 0 & 1 & i \\ 0 & 0 & i & -1 \\ -1 & -i & 0 & 0 \\ -i & 1 & 0 & 0 \end{pmatrix}. \end{align} \tag{ 114 }$$

The corresponding Poisson relation for phase space functions F and G would therefore be of the form

$$\begin{align} \left\{F,G\right\} = \int_\Sigma\mathrm{d}{^3x}\, \left({\nabla F}\right)^\mathrm{T}\left(x\right)\, M^{-1}\,\nabla G\left(x\right), \end{align} \tag{ 115 }$$

where $\nabla F$ and $\nabla G$ are the collections of functional derivatives, according to

$$\begin{align} \nabla^\mathrm{T}: = \left({\frac{\delta}{\delta {A^\mathrm{R}}^1_1\left(x\right)},\frac{\delta}{\delta {A^\mathrm{I}}^1_1\left(x\right)},\frac{\delta}{\delta {\tilde{E}^\mathrm{R}}^1_1\left(x\right)},\frac{\delta}{\delta {\tilde{E}^\mathrm{I}}^1_1\left(x\right)}, \dots}\right). \end{align} \tag{ 116 }$$

As the determinant of the right tensor factor in (114) is clearly vanishing, the determinant of M is vanishing and it is not invertible. Hence the presymplectic form is not symplectic and it is not possible to invert it for a corresponding Poisson bracket for real and imaginary parts of connection and electric field.

Without adding additional Poisson brackets to the theory by hand, it is not possible to rewrite the self-dual Palatini action in terms of real quantities and we have to use complex Ashtekar variables [33] ¹⁰ . This further implies that the phase space of self-dual GR can only consist of functions which are holomorphic in both, $A^i_a$ and $\tilde{E}^a_i$, and that the Poisson brackets must be understood with respect to holomorphic derivatives. This is the way we want to proceed in the following.

There is also an immediate consequence for a possible implementation of reality conditions. These conditions would necessarily involve the basic fields and their complex conjugates, hence they would not be functions in the holomorphic phase space described above. Working in a holomorphic setup excludes the possibility to introduce reality conditions as additional constraints ¹¹ . They can only be implemented by hand, for example as adjointness relations.

One of the beautiful results of the Ashtekar formulation of gravity is that the electric fields serve as densitised triads for the spatial metric, with respect to an internal three metric. Up to now, we refrained from specifying an internal metric, used to move $\mathfrak{sl}(2,\mathbb{C})$ indices $i = 1,2,3$, because no such movements of indices took place.

Without an internal metric, the natural way of contracting internal indices is the Cartan–Killing metric of $\mathfrak{sl}(2,\mathbb{C})$. As the definition of electric fields includes the inverse isomorphism, the inverse Cartan–Killing metric k^ij of $\mathfrak{sl}(2,\mathbb{C})$ can be expressed by its $\mathfrak{so}(3,1)^+_{\mathbb{C}}$ counterpart k^IJKL according to (49). With (37) we therefore find

$$\begin{align} \tilde{E}^a_i \tilde{E}^b_j k^{ij} = \left(-\sqrt{q}\right)^2 {\epsilon}{^a{}_{cd}}{\epsilon}{^b{}_{ef}}k^{ij}\left(P^{-1}\right)^{IJ}_{i}\left(P^{-1}\right)^{KL}_{j}\Sigma^{cd}_{IJ}\Sigma^{ef}_{KL} = -\frac{1}{2}q {\epsilon}{^a{}_{cd}}{\epsilon}{^b{}_{ef}} P^{IJKL}\Sigma^{cd}_{IJ}\Sigma^{ef}_{KL}. \end{align} \tag{ 117 }$$

As the pairs of Lorentz indices are already self-dual, the projection act trivially and just moves one pair of indices. Hence we look at

$$\begin{align} {\epsilon}{^a{}_{cd}}{\epsilon}{^b{}_{ef}}\Sigma^{cd}_{IJ}\Sigma^{efIJ} & = \frac{1}{4}{\epsilon}{^a{}_{cd}}{\epsilon}{^b{}_{ef}}\left({e_{cI} e_{dJ}-\frac{i}{2}{\epsilon}_{IJKL}e^K_c e^L_d}\right)\left({e^{I}_e e^{J}_f-\frac{i}{2}{\epsilon}{^{IJ}{}_{MN}}e^M_e e^N_f}\right) \notag\\ & = \frac{1}{4}{\epsilon}{^a{}_{cd}}{\epsilon}{^b{}_{ef}} \left( e_{cI} e^I_e e_{dJ} e^J_f - \frac{i}{2} e_{cI}e_{dJ}{\epsilon}{^{IJ}{}_{MN}}e^M_e e^N_f \right.\notag\\ & \quad - \left.\frac{i}{2} {\epsilon}{_{IJKL}}e^K_c e^L_d e^I_e e^J_f - \frac{1}{4} {\epsilon}_{IJKL} e^K_c e^L_d {\epsilon}{^{IJ}{}_{MN}}e^M_e e^N_f \right). \end{align} \tag{ 118 }$$

Here the two middle terms vanish since ${\epsilon}{_{IJKL}}e^I_a e^J_b e^K_c e^L_d = {\epsilon}{^{abcd}} = 0$ because of four spatial indices. Contracting the Levi-Civita symbols further yields

$$\begin{align} \frac{1}{4}{\epsilon}{^a{}_{cd}}{\epsilon}{^b{}_{ef}}\left(e_{cI} e^I_e e_{dJ} e^J_f + \delta_{K[M}\delta_{N]K} e^K_c e^L_d e^M_e e^N_f \right) & = \frac{1}{4}{\epsilon}{^a{}_{cd}}{\epsilon}{^b{}_{ef}} ({q_{ce}q_{df}+q_{ce}q_{df}}) \notag\\ & = \frac{1}{2 q} {\tilde{\epsilon}}{^{acd}}{\tilde{\epsilon}}{^{bef}}q_{ce}q_{df} = q^{ab}. \end{align} \tag{ 119 }$$

So indeed this complex electric field encoding the self-dual degrees of freedom contains the spatial metric. Inserting this back in (117) and solving for the spatial metric yields

$$\begin{align} q q^{ab} = -2\tilde{E}^a_i \tilde{E}^b_j k^{ij}. \end{align} \tag{ 120 }$$

This in turn suggest that the (inverse) internal three metric, related to the spatial metric by densitised triads $\tilde{E}^a_i$, is indeed

$$\begin{align} -2k^{ij} = -2\left({-\frac{1}{2}}\right)\delta^{ij} = \delta^{ij}. \end{align} \tag{ 121 }$$

The internal metric is therefore given by the Euclidean three metric δ_ij , which we will use henceforth. This is in fact remarkable. In the derivation of Ashtekar–Barbero gravity from a real Palatini formulation, time-gauge is used to align the frame fields with the foliation. Consequently the structure group is reduced to $\mathrm{SU}(2)$. In fixing this gauge, the internal Minkowski metric is reduced to its spatial part, which is exactly the Euclidean metric. Here however, we end up with exactly the same internal three metric, but no time gauge has been performed. It is easy to see that the split of $\mathfrak{so}(3,1)_{\mathbb{C}}$ into its (anti-)self-dual components is incompatible with stabilising a gauge fixing similar to the time gauge. As we discuss the number of physical degrees of freedom later on, it turns out, we do not even need an additional gauge fixing.

Having established the relation

$$\begin{align} q q^{ab} = \tilde{E}^a_i \tilde{E}^b_j \delta^{ij}, \end{align} \tag{ 122 }$$

we see that the electric fields can be interpreted as densitised triads of the spatial manifold, even though they carry spacetime information. This is contained in the components $e^0_a$, that enters via the Plebanski bi-vector.

We can now use (122) in order to invert the electric field:

$$\begin{align} \left(\tilde{E}^{-1}\right)^i_a = \frac{1}{q}q_{ab}\delta^{ij}\tilde{E}^b_j = {\underset{{\smash{\tilde{}}}}{E}}{}^i_a, \end{align} \tag{ 123 }$$

which is a density of weight −1. The spatial metric therefore is given by

$$\begin{align} q_{ab} = q \delta_{ij}{\underset{{\smash{\tilde{}}}}{E}}{}^i_a{\underset{{\smash{\tilde{}}}}{E}}{}^j_b. \end{align} \tag{ 124 }$$

In order to fix the notation, we strip the electric fields of their density weight in order to get the actual triads

$$\begin{align} \begin{aligned} \Sigma^a_i& = \frac{1}{\sqrt{q}}\tilde{E}^a_i = -{\epsilon}{^a{}_{bc}}\Sigma^{bc}_i = -{\epsilon}{^a{}_{bc}}\left({2\left(P^{-1}\right)^{0j}_{i}e^b_0e^c_j+\left(P^{-1}\right)^{kl}_{i}e^b_ke^c_l}\right)\\ & = -{\epsilon}{^a{}_{bc}}\left({i e^b_0e^c_i+\frac{1}{2}{\epsilon}{_i{}^{kl}}e^b_ke^c_l}\right). \end{aligned} \end{align} \tag{ 125 }$$

Here we used the explicit form of the isomorphism in (55) and identified uppercase Lorentz indices that run from 1 to 3 with lowercase indices. We see how the spacetime components $e_0^a$ enter the triads in the first term, while the second term looks similar to the result in Ashtekar–Barbero gravity ¹² .

In this section we look at the three other term in the self-dual Palatini action and want to bring them in their standard form, while expressing them in terms of the canonical variables.

For the Hamilton constraint part (103) of the action, we see that the contraction between Plebanski two-form and curvature is fully projected and we can pull back the entire expression to the spatial manifold. Therefore

$$\begin{align} - \int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \frac{2\lambda\sqrt{q} s_e}{\kappa} N q_\alpha^\gamma q_\beta^\delta \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i}& = -\int_{\mathbb{R}}\mathrm{d}{t}\int_M \mathrm{d}{^3x}\frac{2\lambda\sqrt{q} s_e}{\kappa} N \Sigma_{i}^{ab} F_{ab}^{i} \notag \\ & = \int_{\mathbb{R}}\mathrm{d}{t}\int_M\mathrm{d}{^3x} \frac{\lambda N s_e}{\kappa}{\epsilon}{_c^{ab}}\tilde{E}^c_i F_{ab}^i. \end{align} \tag{ 126 }$$

In the last step we we inverted the spatially projected Plebanski bi-vector for the electric field:

$$\begin{align} \Sigma^{ab}_i = -\frac{1}{2\sqrt{q}}{\epsilon}{_c^{ab}}\tilde{E}^c_i. \end{align} \tag{ 127 }$$

Next we want to replace ${\epsilon}{_c^{ab}}\tilde{E}^c_i F_{ab}^i$ by ${\epsilon}{_i{}^{kl}}\tilde{E}^a_k\tilde{E}^b_l F_{ab}^i$. For the determinant of electric fields, it holds that

$$\begin{align} \underset{{\smash{\tilde{}}}}{\epsilon}{_{abc}}\tilde{E}^c_i = \text{det}\left(\tilde{E}\right){\epsilon}{_{ijk}}{\underset{{\smash{\tilde{}}}}{E}}{}^j_a{\underset{{\smash{\tilde{}}}}{E}}{}^k_b. \end{align} \tag{ 128 }$$

We multiply this by another factor $\text{det}(\tilde{E})$ and replace the spatial Levi-Civita density by the corresponding tensor, i.e.

$$\begin{align} \frac{\text{det}\left(\tilde{E}\right)}{\sqrt{q}}{\epsilon}{_{abc}}\tilde{E}^c_i = \text{det}\left(\tilde{E}\right)^2{\epsilon}{_{ijk}}{\underset{{\smash{\tilde{}}}}{E}}{}^j_a{\underset{{\smash{\tilde{}}}}{E}}{}^k_b. \end{align} \tag{ 129 }$$

The relation (122) further allows to relate the determinants of spatial metric q and electric fields $\text{det}(\tilde{E})$. Just taking the determinant on both sides yields

$$\begin{align} q^2 = \text{det}\left(\tilde{E}\right)^2. \end{align} \tag{ 130 }$$

Completely analogous to (83) for tetrads, we introduce s_E as the complex sign of $\text{det}(\tilde{E})$, and write

$$\begin{align} \text{det}\left(\tilde{E}\right) = s_E\sqrt{\text{det}\left(\tilde{E}\right)^2}. \end{align} \tag{ 131 }$$

Because of this we find

$$\begin{align} \frac{\text{det}\left(\tilde{E}\right)}{\sqrt{q}} = s_E \sqrt{\frac{\text{det}\left(\tilde{E}\right)^2}{q}} = s_E\sqrt{q}. \end{align} \tag{ 132 }$$

Hence, inverting the inverse triads on the right hand side, (129) takes the form

$$\begin{align} {\epsilon}{_c^{ab}}\tilde{E}^c_i = \frac{1}{\sqrt{q}s_E}{\epsilon}{_{i}{}^{jk}}\tilde{E}^a_j\tilde{E}^b_k. \end{align} \tag{ 133 }$$

Inserting this into (126) yields

$$\begin{align} \int_{\mathbb{R}}\mathrm{d}{t}\int_M\mathrm{d}{^3x}\frac{N\lambda s_e}{\kappa}{\epsilon}{_c^{ab}}\tilde{E}^c_i F_{ab}^i = \int_{\mathbb{R}}\mathrm{d}{t}\int_M \mathrm{d}{^3x} \frac{N\lambda s_e}{\sqrt{q} \kappa s_E} {\epsilon}{_{k}{}^{ij}}\tilde{E}^a_i\tilde{E}^b_j F_{ab}^k. \end{align} \tag{ 134 }$$

Hence we can read off the Hamilton constraint C

$$\begin{align} N C = -N\frac{\lambda s_e}{\sqrt{q} \kappa s_E} {\epsilon}{_{k}{}^{ij}}\tilde{E}^a_i\tilde{E}^b_j F_{ab}^k. \end{align} \tag{ 135 }$$

Comparing this to the Hamilton constraint of GR in Ashtekar–Barbero variables, there are two important aspects to remark. First the constraint here consists only of what is usually called the Euclidean–Hamilton constraint. The so called Lorentzian part of the Hamilton constraint of real GR in Ashtekar–Barbero variables is not present. This is the reason why GR in self-dual variables is expected to have simpler dynamics. The second aspect is the presence of $s_e/s_E$, which encodes the the orientation of both triads and tetrads, which are complex quantities. Therefore right from the beginning it is not clear what choosing an orientation of these would actually mean.

We go on with the diffeomorphism constraint part of the action, i.e. (104). Here we see that

$$\begin{align} N^\delta = N^\alpha\delta^\delta_\alpha = N^\alpha q^\delta_\alpha-N^\alpha n_\alpha n^\delta = N^\alpha q^\delta_\alpha \end{align} \tag{ 136 }$$

as $N^\alpha n_\alpha = 0$. The shift vector is indeed spatially projected. Hence, using again spacetime self-duality and performing the pull-back to the spatial manifold

$$\begin{align} -\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \frac{4\lambda\sqrt{q} s_e}{\kappa} N^\delta q_\alpha^\gamma n_\beta \Sigma_{i}^{\alpha\beta} F_{\gamma\delta}^{i} & = - \int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,}\frac{4\lambda\sqrt{q} s_e}{\kappa} N^\lambda q^\delta_\lambda \frac{i}{2s_e}n_\beta {\epsilon}{^{\alpha\beta}{}_{\rho\sigma}} \Sigma_{i}^{\rho\sigma} F_{\gamma\delta}^{i} \notag\\ & = \int_{\mathbb{R}}\mathrm{d}{t}\int_M \mathrm{d}{^3x} \frac{2\lambda\sqrt{q} i}{\kappa} N^a {\epsilon}{^{d}{}_{bc}} \Sigma_{i}^{bc} F_{da}^{i} \notag\\ & = \int_{\mathbb{R}}\mathrm{d}{t}\int_M \mathrm{d}{^3x} \frac{2\lambda i}{\kappa} N^a \tilde{E}^b_i F_{ab}^{i}, \end{align} \tag{ 137 }$$

where we again expressed everything by curvature and electric field. The diffeomorphism constraint C_a now reads

$$\begin{align} N^a C_a = \frac{2\lambda}{i\kappa} N^a \tilde{E}^b_i F_{ab}^{i}. \end{align} \tag{ 138 }$$

Likewise to the canonical pair, the dependence on the orientation of tetrads disappeared.

Finally we work out the Gauß constraint from (105). Once again employing spacetime self-duality and using that the open indices of $n_\beta{\epsilon}{^{\beta\alpha}{}_{\rho\sigma}}$ are already projected (making additional projections obsolete), we find

$$\begin{align} \begin{aligned} \int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \frac{4\lambda\sqrt{q} s_e}{\kappa} q_\alpha^\gamma n_\beta \Sigma_{i}^{\alpha\beta} \mathop{D_\gamma} \left(t^\delta A^i_\delta\right) & = -\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \frac{4\lambda\sqrt{q} s_e}{\kappa} \frac{i}{2 s_e} n_\beta {\epsilon}{^{\beta\gamma}{}_{\rho\sigma}} \Sigma_{i}^{\rho\sigma} \mathop{D_\gamma} \left(t^\delta A^i_\delta\right)\\ & = \int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \frac{2\lambda\sqrt{q} }{i\kappa} n_\beta {\epsilon}{^{\beta\gamma}{}_{\rho\sigma}} \Sigma_{i}^{\rho\sigma} \mathop{D_\gamma} \left(t^\delta A^i_\delta\right). \end{aligned} \end{align} \tag{ 139 }$$

Next we use the Leibniz rule on the covariant derivative. This produces the term

$$\begin{align} \mathop{D_\gamma} \left({ \frac{2\lambda\sqrt{q} }{i\kappa} n_\beta {\epsilon}{^{\beta\gamma}{}_{\rho\sigma}} \Sigma_{i}^{\rho\sigma} t^\delta A^i_\delta}\right), \end{align} \tag{ 140 }$$

which is the covariant derivative of a vector density of weight one. The covariant derivative can therefore be replaced by a partial derivative, we obtain a surface term and drop it. Hence we are left with

$$\begin{align} -\int_{\mathcal{M}} \!\mathrm{d}^{4}{x\,} \left(t^\delta A^i_\delta\right)\mathop{D_\gamma}\left({\frac{2\lambda\sqrt{q} }{i\kappa} n_\beta {\epsilon}{^{\beta\gamma}{}_{\rho\sigma}} \Sigma_{i}^{\rho\sigma}}\right) & = \int_{\mathbb{R}}\mathrm{d}{t}\int_M \mathrm{d}{^3x} \left(t^\delta A^i_\delta\right)\mathop{D_a}\left({-\frac{2\lambda\sqrt{q} }{i\kappa} {\epsilon}{^{a}{}_{bc}} \Sigma_{i}^{bc}}\right) \notag\\ & = \int_{\mathbb{R}}\mathrm{d}{t}\int_M \mathrm{d}{^3x} \left(t^\delta A^i_\delta\right) \frac{2\lambda }{i\kappa} \mathop{D_a}\left(\tilde{E}^a_i\right), \end{align} \tag{ 141 }$$

where we performed the pull-back and introduced the electric field. The derivative D_a is the spatial projection of D_α and hence the covariant derivative with respect to $A_a^i$. With $(A\cdot t)^i: = t^\delta A^i_\delta$ The Gauß constraint G_i reads

$$\begin{align} \left(A\cdot t\right)^i G_i = - \frac{2\lambda }{i\kappa} \mathop{D_a}\left(\tilde{E}^a_i\right). \end{align} \tag{ 142 }$$

It is easy to see that this generates indeed $\mathrm{SL}(2,\mathbb{C})$ gauge transformations.

Unlike the Hamilton constraint, the diffeomorphism and Gauß constraints are of exactly the same form as in the real theory, see [19, 20].

Putting all pieces of the action together, we therefore have

$$\begin{align} S = \int_{\mathbb{R}}\mathrm{d}{t}\int_M \mathrm{d}{^3x} \left({\frac{2\lambda}{\kappa i} \tilde{E}^a_i\mathop{\mathcal{L}_t}\left(A^i_a\right)-\left({NC+N^aC_a+\left(A\cdot t\right)^i G_i}\right)}\right). \end{align} \tag{ 143 }$$

Once again, we can comment on the consequences of rescaling the action by λ. As we just established, all the constraints are rescaled by a factor λ, while the Poisson bracket is rescaled by a factor $1/\lambda$. Therefore the equations of motion obtained from this action are unchanged as

$$\begin{align} \left\{\boldsymbol{\,\cdot\,},C_{\left(\lambda\right)}\right\}_{\left(\lambda\right)} = \frac{1}{\lambda}\left\{\boldsymbol{\,\cdot\,},\lambda C\right\} = \left\{\boldsymbol{\,\cdot\,},C\right\}, \end{align} \tag{ 144 }$$

for all types of constraints. Similarly, the form of the hypersurface deformation algebra is unaffected.

Having established all the constraints, we now want to see how many physical degrees of freedom from complexified GR are left. We will do this on the $\mathfrak{sl}(2,\mathbb{C})$ level of the formulation only and count complex degrees of freedom. The configuration variable $A^i_a$ encodes nine complex degrees of freedom. The Hamiltonian constraints reduces this number by one, diffeomorphism and Gauß constraint reduce by 3 respectively. Therefore we are left with two complex degrees of freedom. Expecting the reality conditions to halve this again, we end up with two physical and real degrees of freedom for gravity. As mentioned earlier, we do not need an additional gauge fixing (time gauge) in order to obtain the correct number of degrees of freedom.

Simply solving the Ashtekar connection for the extrinsic curvature, we can compute the Poisson bracket between $K\,^i_a$ and the electric field. We find

$$\begin{align} \left\{K\,^i_a\left(x\right),\tilde{E}^b_j\left(y\right)\right\} = \frac{1}{i}\left\{A^i_a\left(x\right)-\Gamma^i_a\left(x\right),\tilde{E}^b_j\left(y\right)\right\} = \frac{\kappa}{2\lambda}\delta^i_j\delta^a_b\delta\left(x,y\right), \end{align} \tag{ 181 }$$

since $\left\{\Gamma^i_a(x),\tilde{E}^b_j(y)\right\} = 0$. For both $K\,^i_a$ and $\tilde{E}^b_j$ real, and λ = 1 this perfectly reproduces the ADM Poisson relation

$$\begin{align} \left\{K\,^i_a\left(x\right),\tilde{E}^b_j\left(y\right)\right\} = \frac{\kappa}{2}\delta^i_j\delta^a_b\delta\left(x,y\right), \end{align} \tag{ 182 }$$

Further, as the right hand side—in the form of the Kronecker-deltas—is invariant under orthogonal transformations, any gauge transformation present is removable and the Poisson relation would be equivalent to the ADM relation.

For $\lambda = i$ instead, we have—independent of the presence of gauge transformations—an additional factor of i present:

$$\begin{align} \left\{K\,^i_a\left(x\right),\tilde{E}^b_j\left(y\right)\right\} = \frac{\kappa}{2i}\delta^i_j\delta^a_b\delta\left(x,y\right). \end{align} \tag{ 183 }$$

Once we work out the reality of the constraints, it will be clear that this change to the ADM relation is consistent with a change of ADM constraints in this case and nevertheless reproduces the dynamics of ADM.

We recall from section 2.3 the diffeomorphism constraint (138) and the explicit expression for the curvature of the Ashtekar connection (162). In the latter we replace $A^i_a$ by $\Gamma^i_a+i K\,^i_a$ and therefore find

$$\begin{align} \begin{aligned} C_a & = \frac{2\lambda}{i\kappa} \tilde{E}^b_i F_{ab}^{i} \\ & = \frac{2\lambda}{i\kappa} \tilde{E}^b_i \left({2 \partial_{[a}\left(\Gamma^i_{b]}+i K^i_{b]}\right) + {\epsilon}{_{kl}{}^i} \left({\Gamma^k_a\Gamma^l_b+i\Gamma^k_aK^l_b+iK^k_a\Gamma^l_b-K^l_aK^k_b}\right)}\right)\\ & = \frac{2\lambda}{i\kappa} \left({\tilde{E}^b_i R^i_{ab} +2i \tilde{E}^b_i \partial_{[a}K^i_{b]} + i {\epsilon}{_{kl}{}^i} \left({\Gamma^k_a K^l_b-\Gamma^k_b\Gamma^l_a}\right)+K^i_b G_i }\right). \end{aligned} \end{align} \tag{ 184 }$$

Here we identified the Gauß constraint, used the antisymmetry of the structure constant and combined all terms with the spin connection only in its curvature

$$\begin{align} R^i_{ab} = 2 \partial_{[a}\Gamma^i_{b]}+{\epsilon}{_{kl}{}^i} \Gamma^k_a\Gamma^l_b. \end{align} \tag{ 185 }$$

Similar to a real formulation, and e.g. as shown in [19], the contraction $\tilde{E}^b_i R^i_{ab}$ vanishes because of the Bianchi identity. Realising that the antisymmetrisation of spatial indices would cause Christoffel symbols to vanish, we can introduce the covariant derivative compatible with $\tilde{E}^a_i$ in the expression. Because of the compatibility, we can further move the electric field inside the derivative. Hence, supposing the Gauß constraint is vanishing, we find

$$\begin{align} C_a = \frac{4\lambda}{\kappa} \mathcal{D}_{[a}\left(K^i_{b]}\tilde{E}^b_i\right) = -\frac{2\lambda}{\kappa}\mathcal{D}_b\left({K\,^i_a\tilde{E}^b_i-\delta^b_aK^i_c\tilde{E}^c_i}\right). \end{align} \tag{ 186 }$$

This has exactly the form of the ADM diffeomorphism constraint and therefore perfectly reproduces it for λ = 1. Present gauge transformations cancel in the contraction of electric field and extrinsic curvature. In the case $\lambda = i$, the real constraint is rescaled by i, as expected. We come back to this after working out the Hamilton constraint.

Eventually we express the Hamilton constraint (135) by electric field and extrinsic curvature. From our treatment of the diffeomorphism constraint we already known what the curvature of the Ashtekar connection turns into. Only the contractions with electric fields are different. Hence

$$\begin{align} \begin{aligned} C & = -\frac{\lambda s_e}{\sqrt{q} \kappa s_E} {\epsilon}{_{k}{}^{ij}}\tilde{E}^a_i\tilde{E}^b_j F_{ab}^k\\ & = -\frac{\lambda s_e}{\sqrt{q} \kappa s_E} {\epsilon}{_{k}{}^{ij}}\tilde{E}^a_i\tilde{E}^b_j \left({R^k_{ab}+2i \mathcal{D}_{[a}K^k_{b]}- {\epsilon}{_{mn}{}^k}K^m_a K^n_b }\right). \end{aligned} \end{align} \tag{ 187 }$$

We look individually at the three contributions.

The first term we consider is the one including the curvature of $\Gamma^i_a$. A detailed calculation reveals that this contribution is in fact proportional to the Ricci curvature R of the spin connection. The result is

$$\begin{align} {\epsilon}{_{k}{}^{ij}}\tilde{E}^a_i\tilde{E}^b_j R^k_{ab} = -qR. \end{align} \tag{ 188 }$$

Next we show that the second term is related to the Gauß constraint. In order to realise this, we write out the antisymmetrisation in the covariant derivative and use for each term the corresponding electric field to form the Gauß constraint according to (180). This yields

$$\begin{align} \begin{aligned} 2i {\epsilon}{_{k}{}^{ij}}\tilde{E}^a_i\tilde{E}^b_j \mathcal{D}_{[a}K^k_{b]} & = i {\epsilon}{_{k}{}^{ij}}\tilde{E}^a_i\tilde{E}^b_j \left({\mathcal{D}_{a}K^k_{b}- \mathcal{D}_{b}K^k_{a}}\right) \\ & = \frac{i\kappa}{2\lambda} \left({-\delta^{il}\tilde{E}^a_i\mathcal{D}_aG_l - \delta^{jl}\tilde{E}^b_j \mathcal{D}_bG_l}\right)\\ & = -\frac{i\kappa}{\lambda} \tilde{E}^a_j\mathcal{D}_aG^j. \end{aligned} \end{align} \tag{ 189 }$$

On the Gauß constraint surface G^j vanishes everywhere and so does its derivative. Hence—irrespective of the value of λ—this term does not contribute to the Hamilton constraint.

We are now left with the contraction of two electric field and two curvature components each. This yields

$$\begin{align} {\epsilon}{_{k}{}^{ij}}{\epsilon}{_{mn}{}^k}\tilde{E}^a_i\tilde{E}^b_j K^m_a K^n_b = 2 K^{[m}_a K^{n]}_b \tilde{E}^a_m\tilde{E}^b_n = \left({\left(K^m_a \tilde{E}^a_m\right)\left(K^n_b \tilde{E}^b_n\right)-K^n_a \tilde{E}^b_n K^m_b \tilde{E}^a_m}\right). \end{align} \tag{ 190 }$$

Therefore, on the Gauß constraint surface, the Hamilton constraint in terms of extrinsic curvature and electric field turns into

$$\begin{align} C = \frac{\lambda s_e}{\sqrt{q} \kappa s_E}\left({qR+\left({\left(K^m_a \tilde{E}^a_m\right)\left(K^n_b \tilde{E}^b_n\right)-K^n_a \tilde{E}^b_n K^m_b \tilde{E}^a_m}\right)}\right). \end{align} \tag{ 191 }$$

Again, possible gauge transformations of electric field and extrinsic curvature cancel in the contractions and the Ricci scalar is a function of the spatial metric, which is also invariant under such transformations.

Recalling the diffeomorphism constraint (186), both—this and the Hamilton constraint (191)—are identical in form to the real ADM constraints. So assuming the reality of electric field and extrinsic curvature, according to the reality conditions, they perfectly match these real constraints up to the factor λ and the sign combination $s_e/s_E$.

The factor λ is of course in perfect agreement with the symplectic structure. For λ = 1 we match the real ADM formulation. For $\lambda = i$ the constraints are purely imaginary but so is the Poisson bracket. Hence the described physics does not change. This has therefore the—rather unintuitive—interpretation of real ADM gravity that is described by using a purely imaginary action.

The additional sign $s_e/s_E$ in the Hamilton constraint is the only remnant left from starting with a complex formulation of gravity. This does not change the hypersurface deformation algebra as well and therefore does not matter.

This connection of the self-dual holomorphic formulation of GR to its real ADM formulation ultimately shows the eligibility of the choice of reality conditions.