The Barbero-Immirzi Field and the Nieh–Yan Topological Invariant

By George Shiber Thursday, October 20, 2016Friday, October 26, 2018 In Loop Quantum Gravity, Math, Mathematical Physics, Mathematics, Physics Barbero-Immirzi Parameter, Hilbert-Palatini Action

Why the need for the Barbero-Immirzi field? Let me briefly explain. We saw that LQG in the Holst formulation faces the serious problem that unless the Barbero-Immirzi parameter is promoted to a field, the three-dimensional action with the Barbero-Immirzi parameter:

$\begin{array}{l}S\left[ {e,x,\omega } \right] = {\int_{{M_3}} {\rm{d}} ^3}x{\varepsilon ^{\mu \nu \rho }}\left( {\frac{1}{2}} \right.{\varepsilon _{IJKL}}\\{x^I}e_\mu ^JF_{\nu \rho }^{KL}\left. { + \frac{1}{\gamma }{x^I}e_\mu ^J{F_{\nu \rho }}_{IJ}} \right)\end{array}$

cannot admit a 4-D uplifting of the reduced 3-D gauge-free spacetime compactified action:

$\begin{array}{*{20}{l}}{{S^{{\rm{Red}}}} = - \int_{{S^1}} {\rm{d}} {x^3}\int_{{M_3}} {{{\rm{d}}^3}} x{\varepsilon ^{\mu \nu \rho }}\left( {\frac{1}{2}} \right.}\\{{\varepsilon _{IJKL}}e_3^Ie_\mu ^JF_{\nu \rho }^{KL}\left. { + \frac{1}{\gamma }e_3^Ie_\mu ^J{F_{\nu \rho }}_{IJ}} \right)}\end{array}$

to the 4-D Holst-action:

$\begin{array}{*{20}{l}}{{S_{4D}}\left[ {e,\omega } \right] = \int_{{{\tilde M}_4}} {{{\rm{d}}^4}} x{\varepsilon ^{\mu \nu \rho \sigma }}\left( {\frac{1}{2}} \right.{\varepsilon _{IJKL}}}\\{e_\mu ^Ie_\nu ^JF_{\rho \sigma }^{KL}\left. { + \frac{1}{\gamma }e_\mu ^Ie_\nu ^J{F_{\rho \sigma }}_{IJ}} \right)}\end{array}$

and that is because the total 3-D action with the Barbero-Immirzi parameter:

is invariant under rescaling symmetry and translational symmetry, which destroy the time-gauge accessibility of the theory and 4-D-uplifting. Let us see whether and how promoting the Barbero-Immirzi parameter to a field and using the Nieh–Yan topological invariant can ameliorate our crises. In Lagrangian Holst theory, a Hilbert–Palatini action can always be generalized to contain the Holst term and promotes the Barbero–Immirzi parameter to a field via:

$\begin{array}{l}{S^{{\gamma _f}}} = \int_{{M_4}} {dtL = - \frac{1}{2}} \int_{{M_4}} {{d^4}} x{\,^{(4)}}e\\e_a^\mu e_b^\nu {\left( {{R^{ab}}} \right._{\mu \nu }} - \frac{{{\gamma _f}}}{2}{\varepsilon ^{ab}}_{cd}\left. {{R^{cd}}_{\mu \nu }} \right)\end{array}$

with:

$^{(4)}e \equiv \det \left( {e_a^\mu } \right)$

the determinant of the LQG-tetrad, and:

${R^{ab}}_{\mu \nu }: = 2{\partial _{\left[ {_\mu {\omega ^{ab}}_\nu } \right]}} + 2\omega _{c\left[ {_\mu {\omega ^{ab}}_\nu } \right]}^\alpha$

being the Riemannian curvature corresponding to:

$\omega _\mu ^{ab}$

Clearly,

${S^{{\gamma _f}}}$

is not equivalent to a Hilbert-Palatini action, since the first Cartan equation is affected by the BI-field: a torsion trace contributes depending on the derivative of the BI-field. Letting

${D_\mu }$

be the covariant Lorentz spin-valued connection-derivative, with spin-connection

${\omega ^{ab}}$

and

$T_{\mu \nu }^a$

the torsion tensor.

The Bianchi cyclic equation is then:

$R_{b\left[ {_\mu {\nu ^{eb}}_\rho } \right]}^a = {D_{\left[ {_\mu T_{\nu \rho }^\alpha } \right]}} \ne 0$

and so the Holst term does not vanish, thus

${\gamma _f}$

is a constant.

Now, given the trace vector:

$T_\mu ^{Tr} \equiv T_{\mu \nu }^\nu$

and the identity:

$q_{e\nu }^\nu = 0 = {\varepsilon ^{\mu \nu \rho \sigma }}{q_{\mu \nu \rho }}$

it follows that:

${S^{{\gamma _f}}}$

reduces to:

$\begin{array}{l}{S^{{\gamma _f}}} = - \frac{1}{2}\int {{d^4}{\,^{(4)}}} e\left[ {e_a^\mu } \right.e_b^\nu {\overline {{R_{\mu \nu }}} ^{ab}}\\ + \frac{{{\gamma _f}}}{2}{{\bar \nabla }_\mu }{S^\mu } + \frac{1}{{24}}{S_\mu }{S^\mu } - \\\frac{2}{3}{T_\mu }{T^\mu } + \frac{{{\gamma _f}}}{3}{T_\mu }{S^\mu } + \frac{1}{2}{q_{\mu \nu \rho }}{q^{\mu \nu \rho }}\\ + \frac{{{\gamma _f}}}{2}{\varepsilon _{\mu \nu e\sigma }}{q_\tau }^{\mu \rho }\left. {{q^{\tau \nu \sigma }}} \right]\end{array}$

with

${S_\mu } \equiv {\varepsilon _{\nu \rho \sigma \mu }}{T^{\nu \rho \sigma }}$

and

${\bar \nabla _\mu }$

is the torsion-less metric-compatible covariant derivative. By solving,

${\gamma _f}$

induces contortion spin-connections, and hence:

${S^{{\gamma _f}}}$

generalizes to:

$\begin{array}{l}S_{NY}^{{\gamma _f}} = - \frac{1}{2}\int {{d^4}} x{\,^{(4)}}e\,e_a^\mu e_b^\nu {R_{\mu \nu }}^{ab} - \\\frac{1}{4}\int {{d^4}} x{\,^{(4)}}e{\gamma _f}\left( {{\eta _{ab}}} \right.T_{\mu \nu }^aT_{\rho \sigma }^a - \\e_a^\mu e_b^\nu \varepsilon _{cd}^{ab}\left. {{R_{\mu \nu }}^{cd}} \right)\end{array}$

Thus, the second integral is the Nieh-Yan topological invariant and connects to the Holst term, yielding

$\begin{array}{l}^\dagger S_{NY}^{{\gamma _f}} = - \frac{1}{2}\int {{d^4}} x{\,^{(4)}}e\left[ {e_a^\mu } \right.e_b^\nu {\overline {{R_{\mu \nu }}} ^{ab}}\\ + \frac{{{\gamma _f}}}{2}{{\bar \nabla }_\mu }{S^\mu } + \frac{1}{{24}}{S_\mu }{S^\mu } - \frac{1}{3}{T_\mu }{T^\mu }\\ + \frac{1}{2}{q_{\mu \nu \rho }}\left. {{q^{_{\mu \nu \rho }}}} \right]\end{array}$

Now, one varies the action with respect to the irreducible components of:

$\left\{ {\begin{array}{*{20}{c}}{{S^\mu }}\\{{T^\nu }}\\{{q^{\rho \sigma \tau }}}\end{array}} \right.$

to obtain:

$\left\{ {\begin{array}{*{20}{c}}{{\partial _\mu }{\gamma _f} - \frac{1}{6} - {S_\mu } = 0}\\{{T_\mu } = 0}\\{{q_{\mu \nu \rho }} = 0}\end{array}} \right.$

Inserting into:

$^\dagger S_{NY}^{{\gamma _f}}$

one gets the effective action:

$\begin{array}{l}S_{eff}^{{\gamma _f}} = - \frac{1}{2}\int {{d^4}} x{\,^{(4)}}ee_a^\mu e_b^\nu {\overline {{R_{\mu \iota }}} ^{ab}} + \\\frac{3}{4}\int {{d^4}} x{\,^{(4)}}e{\partial _\alpha }{\gamma _f}{\partial ^\alpha }{\gamma _f}\end{array}$

giving us an equivalence with the Hilbert-Palatini torsion-free action and thus solving the gauge-free accessibility problem as well as the 4-D uplifting problem caused by invariance under rescaling symmetry and translational symmetry: the proof is straightforward.

Since the phase-space has symplectic structure:

$\left\{ {K_\alpha ^i\left( {t,x} \right),E_j^\gamma \left( {t,x'} \right)} \right\} = \delta _\alpha ^\gamma \delta _j^i\delta \left( {x,x'} \right)$

and

$\left\{ {{\gamma _f}\left( {t,x} \right),\prod \left( {t,x'} \right)} \right\} = \delta \left( {x,x'} \right)$

It thus follows that the total BI-field Hamiltonian:

${H_{tot}} = \int {{d^3}} x\left( {{ \wedge ^i}} \right.{\tilde R_i} + {N^\alpha }\widetilde {{H_\alpha }} + \left. {N\widetilde H} \right)$

with

$\left\{ {\begin{array}{*{20}{c}}{{ \wedge ^i}}\\{{N^\alpha }}\\N\end{array}} \right.$

the Lagrange multipliers, obeys:

${\partial _{{o_t}}}{\gamma _f}^\phi \left\{ {{H_{tot}},{\gamma _f}^\phi } \right\}$

where

$\left\{ {\;,\;} \right\}$

is the Poisson bracket satisfying:

$\begin{array}{l}\left\{ {E_i^a\left( x \right),A_b^i\left( x \right)} \right\} = \delta _b^a\delta _j^i\left( {x,y} \right) = \\\left\{ {\pi _i^\alpha \left( x \right),\omega _b^{(3)i}\left( y \right)} \right\}\end{array}$