Let the Poincaré algebra be given without any factors of i as


$[M_{\rho \sigma},P_\mu]=\eta_{\sigma\mu}P_\rho-\eta_{\rho\mu}P_\sigma$,


where $P$ are the generators of the translational symmetries of the Poincaré group, $M$ are the generators of the Lorentz rotations and boosts, and $\eta$ is the spacetime metric 1.

The author of this thesis 2 writes on pages 20 and 21, that infinitesimal transformations take the form,

$\delta \bullet = \xi^A P_A$

where recall that the $P_A$ translation generator is given by $\partial_\mu$. This holds for the bullet denoting an arbitrary field.

When discussing the first step in gauging the Poincare algebra — assigning a gauge field to each generator — the author states that the transformation rules for these gauge fields can be derived from the above expression and the structure constants of the algebra.

From here he states that we can "define a connection $A_\mu$ that takes values in the adjoint of the gauge group."

$A_\mu = E_\mu^A P_A +\frac{1}{2} \Omega_\mu{}^{AB} M_{AB}$

where $E_\mu^A$ and $\Omega_\mu{}^{AB}$ will eventually be realized as the vielbein and spin connection.

Where did this come from? Can anyone provide me some more clear motivation for constructing such an object? My only idea is comparing it to the following expression for the global Poincaré transformation acting on a scalar field,

$\delta(a,\lambda)\phi(x)=[a^\mu \partial_\mu - \frac{1}{2} \lambda^{\mu\nu}M_{\mu\nu}]\phi(x)=[a^\mu P_\mu - \frac{1}{2} \lambda^{\mu\nu}M_{\mu\nu}]\phi(x)$

where it seems the construction of $A_\mu$ was done by considering the "vielbein and spin connection" as parameters for the Poincaré and Lorentz symmetry transformations.

I have addition questions after this is resolved, but I will either edit the question later to include more, or post another question.

1 D. Freedman and A. Van Proeyen, Supergravity, Textbook (2012).

2 T. Zojer, Non-relativistic supergravity in three space-time dimensions, Van Swinderen Institute for Particle Physics and Gravity at Rijksuniversiteit Groningen, PhD Thesis (2016).

  • $\begingroup$ Related/possible duplicates: physics.stackexchange.com/q/46324/50583, physics.stackexchange.com/q/126978/50583, physics.stackexchange.com/q/346793/50583 $\endgroup$
    – ACuriousMind
    Commented Feb 10, 2020 at 16:35
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    $\begingroup$ "I have addition questions after this is resolved, but I will either edit the question later to include more, or post another question." - please never edit a question to add more subquestions after answers have been given, as this invalidates existing answers. Ask a new question if you have a new question. $\endgroup$
    – ACuriousMind
    Commented Feb 10, 2020 at 16:36
  • $\begingroup$ To motivate that connection you need to consider nothing more than the covariant derivative of your tetrad field, to solve for your connection. you end up getting commutators of gamma matrices, which are lorentz algebra valued objects. The translation group valued terms come about similarly. $\endgroup$
    – R. Rankin
    Commented Feb 12, 2020 at 7:58
  • $\begingroup$ @ACuriousMind gives a succinct summary of this in terms of bundles in the his third link above (the second point in his answer). $\endgroup$
    – R. Rankin
    Commented Feb 12, 2020 at 8:03

1 Answer 1


Main idea to introduce gauge field for every generator, is to provide invariance under some group of transformation, in your case under group of diffeomorphism and local Lorentz transformations (local version of global Poincaré group). This logic is very similar to gauge invariance, where we introduce gauge field and covariant derivative to provide invariance under local gauge transformations. But now gauge group is Poincaré group.

1) Gauge symmetry with generators $T^t$:

$$ \nabla_m = \partial_m - i A_m^t T^t $$

Poincaré group with generators $P_a$ and $M_{ab}$:

$$ \nabla_m = \partial_m -i e_m^{\;a}P_a -\frac{i}{2}\omega_m^{\;\;\;cd}M_{cd} $$

2) Now we can directly calculate commutator:

$$ [\nabla_m, \nabla_n] = -i R_{mn}^{\;\;\;a}P_a -\frac{i}{2}R_{mn}^{\;\;\;ab}M_{ab} $$

$R_{mn}^{\;\;\;a}$ is curvature (or field strength), that corresponds to $P_a$, $R_{mn}^{\;\;\;ab}$ curvature for $M_{ab}$. They expressed only in terms $e_m^{\;\;a}$ and $\omega_m^{\;\;cd}$. They will be identified with tetrad and spin connection.

3) Gauge transformations for $e_m^{\;\;a}$ and $\omega_m^{\;\;cd}$ follow from requirement:

$$ \phi^\prime = h \phi $$ $$ (\nabla_m\phi)^\prime = \nabla_m^\prime\phi^\prime = h \nabla_m\phi \;\;\;\;\;\Rightarrow \;\;\;\;\; \nabla_m^\prime = h \nabla_m h^{-1} $$ $$ h = e^{i a^a P_a + i\omega^{ab}M_{ab}} $$ Here $\phi$ is some field, $h$ is element of Poincaré group.

From here you can found laws of transformation for $e_m^{\;\;a}$ and $\omega_m^{\;\;cd}$. From this transformations you can argue, that this fields transform as tetrad and spin connection.

Note, that curvatures transforms homogeneously $R^\prime = hRh^{-1}$.

4) In gravity, spin connection is expressed in terms of tetrad. So we need covariantly constraint geometry: $$ R_{mn}^{\;\;\;a} =0 $$

From this equation one can express $\omega_m^{\;\;\;cd}$ in terms of $e_m^{\;a}$.

One can also check vielbein postulate: $$ \nabla_n e_m^{\;\;a} = 0 $$

5) From tetrads one can construct metric:

$$ g_{mn}(x) = e_m^{\;a}(x) e_n^{\;b}(x) \eta_{ab}$$

Indeces $a,b, \dots$ are Lorentz indeces, and transforms onle under Lorentz transformations.

Indeces $m,n, \dots$ are world indeces, and transforms only under diffeomorphism transformations.

Using tetrad one can convert Lorentz indeces to curved indeces:

$$ \xi_m = e_m^{\;\;a}\xi_a $$

From Riemannian tensor $R_{mn}^{\;\;\;ab}$ one can construct standard gravity actions.

Please, Ask questions, I will try to clarify answer..!

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    $\begingroup$ This derivative takes form, that you present! $B$ is $\omega$ and $S$ is $T$. Is it clear? $\endgroup$
    – Nikita
    Commented Feb 12, 2020 at 20:25
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    $\begingroup$ The authors replace general coordinates transformations $\delta_{gct}$ with covariant general coordinates transformations $\delta_{cgct}$ because scalar fields transforming under internal symmetry transformations do not transform covariantly, $\delta(\xi) \phi^i = \xi^\mu \partial_\mu \phi$. So they transform $\delta_{gct} \rightarrow \delta_{cgct} = \xi^\mu \partial_\mu \phi + (\xi^\mu A_\mu{}^A)t_A{}^i{}_j \phi^j = \xi^\mu D_\mu \phi = \xi^a D_a \phi$. Where they have defined $D_a \phi$ as $e_a{}^\mu D_\mu \phi.$ Does this mean I ought to be calculating $[D_a,D_b]$ to find curvatures? $\endgroup$
    – Lopey Tall
    Commented Feb 13, 2020 at 16:44
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    $\begingroup$ Factor 1/2 due to antisymetry of $M_{[ab]}$. $\endgroup$
    – Nikita
    Commented Feb 14, 2020 at 19:18
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    $\begingroup$ Curvatures, that you present here are right. You can easily calculate them from commutator of covariant derivatives:$$[\nabla_m, \nabla_n] = -i R_{mn}^{\;\;\;a}P_a -\frac{i}{2}R_{mn}^{\;\;\;ab}M_{ab}$$ In calculation you need use Poincaré algebra relations. $\endgroup$
    – Nikita
    Commented Feb 14, 2020 at 19:20
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    $\begingroup$ Ahh brilliant! :D It can practically be read off from the algebra I begin the post with: $[P,P] = 0 \rightarrow$ no vielbein squared term. I have only more question about why both curvatures don't each get the $\omega$e AND $\omega^2$ term, but I finally understand your comment about the commutator of the covariant derivatives, and so I will work on this on my own first. Thank you so much for your diligent help with this! $\endgroup$
    – Lopey Tall
    Commented Feb 15, 2020 at 14:34

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