Instanton Representation of Plebanski Gravity. The Classical Theory

Abstract

This paper is a self-contained introduction to the instanton representation of Plebanski gravity (IRPG), a formulation of General Relativity (GR) where the basic variables are a spacetime gauge connection and a three by three matrix valued in the Lie algebra of so(3,C). We present a classical analysis of the IRPG from various perspectives, noting some of its interesting features and motivations.

Appendix A: Verification of the Hodge Duality Condition in 3 + 1 Form

We will now show that (26), re-written here for completeness

$$\begin{array}{@{}rcl@{}} F^{a}_{0i}-\left(\epsilon_{ijk}N^{k}+\beta\underline{N}h_{ij}\right){B^{j}_{a}}=0, \end{array} $$

(96)

is indeed the 3 + 1 form of the statement of Hodge duality of the field strength $F^{a}_{\mu \nu }$ with respect to the metric g _{μ ν} , whose covariant form in 3 + 1 form is

$$\begin{array}{@{}rcl@{}} g^{00}=-{1 \over {N^{2}}};~~g^{0i}={{N^{i}} \over {N^{2}}};~~g^{ij}=h^{ij}-{{N^{i}N^{j}} \over {N^{2}}}. \end{array} $$

(97)

To show this, we will derive the Hodge self-duality condition for Yang–Mills theory in curved spacetime, using the 3 + 1 decomposition (97). The Hodge self-duality condition for $F^{a}_{\mu \nu }$ can be written in the form

$$\begin{array}{@{}rcl@{}} \sqrt{-g}g^{\mu\rho}g^{\nu\sigma}F^{a}_{\rho\sigma}=-{\beta \over 2}\epsilon^{\mu\nu\rho\sigma}F^{a}_{\rho\sigma}, \end{array} $$

(98)

where β is a numerical constant to be fixed by a consistency condition. It will also be convenient to use ${B^{i}_{a}}={1 \over 2}\epsilon ^{ijk}F^{a}_{jk}$ for the spatial part of the field strength $F^{a}_{\mu \nu }$. Expanding (98) and using $F^{a}_{00}=0$, we have

$$\begin{array}{@{}rcl@{}} N\sqrt{h}\left(\left(g^{\mu0}g^{\nu{j}}-g^{\nu0}g^{\mu{j}}\right)F^{a}_{0j}+g^{\mu{i}}g^{\nu{j}}\epsilon_{ijk}{B^{k}_{a}}\right) =-{\beta \over 2}\left(2\epsilon^{\mu\nu{0}i}F^{a}_{0i}+\epsilon^{\mu\nu{ij}}\epsilon_{ijm}{B^{m}_{a}}\right). \end{array} $$

(99)

We will now examine the components of (99) in turn. Note that the time-time component μ=0,ν=0 yields 0=0, which is trivially satisfied. So we may move on to components involving the spatial indices.

1.1 Space-Time Components

Moving on to the μ=0,ν=k component of (99), we have

$$\begin{array}{@{}rcl@{}} N\sqrt{h}\left(\left(g^{00}g^{kj}-g^{k0}g^{0j}\right)F^{a}_{0j}+g^{0i}g^{kj}\epsilon_{ijm}{B^{m}_{a}}\right)=-\beta{B}^{k}_{a} \end{array} $$

(100)

Inserting the metric components from (97) into (100), we have

$$\begin{array}{@{}rcl@{}} N\sqrt{h}\left(\left(-{1 \over {N^{2}}}\right)\left(h^{kj}-{{N^{k}N^{j}} \over {N^{2}}}\right)-\left({{N^{k}N^{j}} \over {N^{4}}}\right)\right)F^{a}_{0j}\\ +\left.N\sqrt{h}\left({{N^{i}} \over {N^{2}}}\right)\left(h^{kj}-{{N^{k}N^{j}} \over {N^{2}}}\right)\epsilon_{ijm}{B^{m}_{a}}\right)=-\beta{B}^{k}_{a}. \end{array} $$

(101)

Cancelling off the terms multipying $F^{a}_{0j}$ which are quadratic in N ⁱ, we have

$$\begin{array}{@{}rcl@{}} -\left({{\sqrt{h}} \over N}\right)h^{kj}\left(F^{a}_{0j}-\epsilon_{jmi}{B^{m}_{a}}N^{i}\right)=-\beta{B}^{k}_{a}. \end{array} $$

(102)

Multiplying (102) by $\underline {N}=N/\sqrt {h}$ and by h _{l k} , this yields

$$\begin{array}{@{}rcl@{}} F^{a}_{0l}-\epsilon_{lmi}{B^{m}_{a}}N^{i}-\beta\underline{N}h_{lk}{B^{k}_{a}}=0. \end{array} $$

(103)

Equation (103) is the same as (96), which confirms that we are on the right track. Since we must verify Hodge duality on all components, then we must show that (103) is consistent with the condition of Hodge duality with respect to the spatial components of (99).

1.2 Space-Space Components

Moving on to the space-space component we substitute μ=m and ν=n in (99), which yields

$$\begin{array}{@{}rcl@{}} N\sqrt{h}\left(\left(g^{m0}g^{nj}-g^{n0}g^{mj}\right)F^{a}_{0j}+g^{mi}g^{nj}\epsilon_{ijk}{B^{k}_{a}}\right)=-\beta\epsilon^{mn0j}F^{a}_{0j}. \end{array} $$

(104)

Inserting the metric components from (97) into (104), we have

$$\begin{array}{@{}rcl@{}} N\sqrt{h}\left[\left.\left({{N^{m}} \over {N^{2}}}\right)\left(h^{nj}-{{N^{n}N^{j}} \over {N^{2}}}\right)-\left({{N^{n}} \over {N^{2}}}\right)\left(h^{mj}-{{N^{m}N^{j}} \over {N^{2}}}\right)\right)F^{a}_{0j}\right.\\ +\left.\left(h^{mi}-{{N^{m}N^{i}} \over {N^{2}}}\right)\left(h^{nj}-{{N^{n}N^{j}} \over {N^{2}}}\right)\epsilon_{ijk}{B^{k}_{a}}\right]=-\beta\epsilon^{0mnj}F^{a}_{0j}. \end{array} $$

(105)

Equation (105), upon vanishing of the term quartic in the shift vector N ⁱ, yields

$$\begin{array}{@{}rcl@{}} {{\sqrt{h}} \over N}\left(h^{nj}N^{m}-h^{mj}N^{n}\right)F^{a}_{0j}+N\sqrt{h}h^{mi}h^{nj}\epsilon_{ijk}{B^{k}_{a}}\\ -{{\sqrt{h}} \over N}\left(h^{mi}N^{n}N^{j}+h^{nj}N^{m}N^{i}\right)\epsilon_{ijk}{B^{k}_{a}}=-\beta\epsilon^{0mnj}F^{a}_{0j}. \end{array} $$

(106)

From the third term on the left hand side of (106), we have the following relation upon relabelling indices $i\leftrightarrow {j}$ on the first term in brackets

$$\begin{array}{@{}rcl@{}} -h^{mi}N^{n}N^{j}\epsilon_{ijk}{B^{k}_{a}}-h^{nj}N^{m}N^{i}\epsilon_{ijk}{B^{k}_{a}}&=&-h^{mj}N^{n}N^{i}\epsilon_{jik}{B^{k}_{a}}-h^{nj}N^{m}N^{i}\epsilon_{ijk}{B^{k}_{a}}\\ &=&\epsilon_{ijk}\left(h^{mj}N^{n}-h^{nj}N^{m}\right)N^{i}{B^{k}_{a}}. \end{array} $$

(107)

Note that the combination h ^nj N ^m−h ^mj N ⁿ on the right hand side of (107) is the same term multiplying $F^{a}_{0j}$ in the left hand side of (106). Using this fact, then (106) can be written as

$$\begin{array}{@{}rcl@{}} {{\sqrt{h}} \over N}\left[\left(h^{nj}N^{m}-h^{mj}N^{n}\right)\left(F^{a}_{0j}-\epsilon_{jki}{B^{k}_{a}}N^{i}\right)\right]+\underline{N}\epsilon^{mnl}h_{lk}{B^{k}_{a}}=-\beta\epsilon^{mnj}F^{a}_{0j} \end{array} $$

(108)

where 𝜖 ^0mnj=𝜖 ^mnj. Using $F^{a}_{0j}-\epsilon _{jki}{B^{k}_{a}}N^{i}=\beta \underline {N}h_{jk}{B^{k}_{a}}$ from (103) in (108), then we have

$$\begin{array}{@{}rcl@{}} {{\sqrt{h}} \over N}\left(-h^{mj}N^{n}+h^{nj}N^{m}\right)\beta\underline{N}h_{jk}{B^{k}_{a}}+\underline{N}\epsilon^{mnl}h_{lk}{B^{k}_{a}}=-\beta\epsilon^{mnj}F^{a}_{0j}. \end{array} $$

(109)

This simplifies to

$$\begin{array}{@{}rcl@{}} \beta\left({\delta^{n}_{k}}N^{m}-{\delta^{m}_{k}}N^{n}\right){B^{k}_{a}}+\underline{N}\epsilon^{mnl}h_{lk}{B^{k}_{a}}=-\beta\epsilon^{mnj}F^{a}_{0j}\\ \longrightarrow-\beta\left(\epsilon^{mnj}F^{a}_{0j}-{B^{m}_{a}}N^{n}+{B^{n}_{a}}N^{m}\right)=\underline{N}\epsilon^{mnj}h_{jk}{B^{k}_{a}}. \end{array} $$

(110)

Contracting (110) with 𝜖 _{m n l} and dividing by 2, we obtain the relation

$$\begin{array}{@{}rcl@{}} F^{a}_{0l}-\epsilon_{lmn}{B^{m}_{a}}N^{n}+{1 \over \beta}\underline{N}h_{lk}{B^{k}_{a}}=0. \end{array} $$

(111)

To avoid a contradiction, consistency of (111) with (103) implies that β ²=−1, or that β=±i.

The final result is

$$\begin{array}{@{}rcl@{}} F^{a}_{0i}=\left(\epsilon_{ijk}N^{k}+\beta\underline{N}h_{ij}\right){B^{j}_{a}} \end{array} $$

(112)

where β=±i, which is the 3 + 1 decomposition of the Hodge duality condition of the GYM field strength $F^{a}_{\mu \nu }$ with respect to the metric g _{μ ν} which it couples to.

The results of this appendix are based on the Lorentzian signature case. To extend these results to spacetimes of Euclidean signature, one needs only to perform a Wick rotation $N\rightarrow -iN$, and all the analogous results of this paper carry through.

The fields N and N ⁱ are auxiliary fields and β is a numerical constant which depends on the signature of spacetime.¹⁰ Additionally, the operator D _μ is the S O(3) covariant derivative, which acts as

$$\begin{array}{@{}rcl@{}} D_{\mu}V_{a}&=&\partial_{\mu}V_{a}+f_{abc}A^{b}_{\mu}V_{c};\\ D_{\mu}m_{ae}&=&\partial_{\mu}m_{ae}+A^{b}_{\mu}\left(f_{abc}m_{ce}+f_{ebc}m_{ae}\right) \end{array} $$

(113)

on S O(3)-valued 3-vectors V _c and second-rank tensors m _{a e} .