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

$[P_\mu,P_\nu]=0$,

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

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

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).

Lopey Tall
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  • Related/possible duplicates: https://physics.stackexchange.com/q/46324/50583, https://physics.stackexchange.com/q/126978/50583, https://physics.stackexchange.com/q/346793/50583 – ACuriousMind Feb 10 '20 at 16:35
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    "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. – ACuriousMind Feb 10 '20 at 16:36
  • 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. – R. Rankin Feb 12 '20 at 07:58
  • @ACuriousMind gives a succinct summary of this in terms of bundles in the his third link above (the second point in his answer). – R. Rankin Feb 12 '20 at 08:03

1 Answers1

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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..!

Nikita
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  • Thank you for the answer! Can you tell me where this expression for the covariant derivation comes from? $$\nabla_m \phi^a = \partial_m \phi^a -\frac{i}{2}\omega_m^{;;;cd}(S_{cd})^a_{;b} \phi^b $$ In my experience, the covariant derivative in gauge theory takes the form $$ (D_\mu) \nabla_\mu= \partial_\mu - \delta (B_\mu) = \partial_\mu - B_\mu{}^A T_A $$ where $B_\mu$ is the gauge field and $T_A$ is the symmetry transformation. – Lopey Tall Feb 12 '20 at 19:55
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    This derivative takes form, that you present! $B$ is $\omega$ and $S$ is $T$. Is it clear? – Nikita Feb 12 '20 at 20:25
  • I suppose it's not clear to me why the symmetry transformation if only the Lorentz transformations. Oh wait! Your next expression deals with this, notably, the Poincare spacetime translations take the form $\partial_\mu$. Thus, $$ (D_\mu) \nabla_\mu= \partial_\mu - \delta (B_\mu) = \partial_\mu - B_\mu{}^A T_A = \partial_\mu - e_\mu{}^A P_A - \omega_\mu{}^{CD} M_{CD} $$ – Lopey Tall Feb 12 '20 at 20:37
  • A bit confused on the next step to take in the above expression. I'm tempted to use this expression for the global Poincaré symmetry transformation, $$ \delta(a,\lambda)\phi=[a^\mu\partial_\mu-\frac{1}{2}\lambda^{\mu\nu}M_{\mu\nu}]\phi $$ And use this to advance to the next equality, $$ \partial_\mu - e_\mu{}^A P_A - \omega_\mu{}^{CD} M_{CD} = \partial_\mu - e_\mu{}^A \partial_A + \frac{1}{2}\lambda^{\mu\nu}\omega_\mu{}^{CD} M_{CD}$$

    where I know I've done something disgusting in the last line but I'm just brainstorming. Thoughts?

     – Lopey Tall Feb 12 '20 at 21:01
  • I updated answer. Sorry for lot of misleading statements... – Nikita Feb 12 '20 at 21:48
  • Thank you for your help with this! Can you explain why the P and M Poincaré and Lorentz transformations respectively have flat indices a,b,... as opposed to curved $\mu,\nu,...$? (other than to make the summed/saturated indices of the terms look pretty?) – Lopey Tall Feb 13 '20 at 15:55
  • Because this transformation acts as usual Lorentz transformations only locally, in special flat coordinates. The reason is that Poincaré symmetry is symmetry of flat space. – Nikita Feb 13 '20 at 16:26
  • Following up on my confusion about flat versus curved indices in the above comment, can you tell me your thoughts on this? Source, Freedman-Van Proeyen's Supergravity. "In particular, since the general coordinate transformation of a covariant quantity should not involve a derivative of the parameter $\xi^\mu$, a covariant quantity must be a world scalar. Hence, the covariant derivative that we will use is $D_a\phi$ as defined in (11.63), rather than $D_\mu \phi$." Where the motivation/derivation of 11.63 follows. – Lopey Tall Feb 13 '20 at 16:38
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    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? – Lopey Tall Feb 13 '20 at 16:44
  • This will give the same answer due to $\nabla_m e_n^{;;a} 0 $. I my post situation is more clear and concrete that in Freedman's book.. Do you understand my logic? Can you check all equations? – Nikita Feb 13 '20 at 21:47
  • Not seeing it yet, but I will work on it! Also, why does the third term in the covariant derivative have a factor of 1/2? – Lopey Tall Feb 14 '20 at 10:25
  • Also, can you clarify how to write the curvatures? I have the formula for the curvatures as $$ R_{\mu\nu}{}^A = 2\partial_{[\mu} B_{\nu]}{}^A + B_\nu{}^C B_\mu{}^B f_{BC}{}^A $$ where in this case we have, $$ R_{\mu\nu}^A (e_\mu{}^A) = 2\partial_{[\mu} e_{\nu]}{}^A + ... $$ and $$ R_{\mu\nu}^{AB} (\omega_\mu{}^{AB})=2\partial_{[\mu} \omega_{\nu]}{}^{AB} + ... $$

    I believe the $f_{BC}^A$ are 1 from the Poincaré algebra (?) but what are the gauge fields here? Should it be two more of the same, $ee$ and $\omega \omega$ or mixed $e \omega$?

     – Lopey Tall Feb 14 '20 at 12:29
  • The thesis I mention above, [2], lists them as $$ R_{\mu\nu}^{A} (e_\mu{}^{A})=2\partial_{[\mu} e_{\nu]}{}^{A} - 2 \omega_{[\mu}{}^A{}B e{\nu]}{}^B $$ and $$ R_{\mu\nu}^{AB} (\omega_\mu{}^{AB})=2\partial_{[\mu} \omega_{\nu]}{}^{AB} - 2 \omega_{[\mu}{}^A{}C \omega{\nu]}{}^{CB}. $$

    [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).

    Can you help clarify where these come from?

     – Lopey Tall Feb 14 '20 at 12:32
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    Factor 1/2 due to antisymetry of $M_{[ab]}$. – Nikita Feb 14 '20 at 19:18
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    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. – Nikita Feb 14 '20 at 19:20
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    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! – Lopey Tall Feb 15 '20 at 14:34
  • Oh my! haha we have been chatting so long I thought this was all in the comments of my question ;) accepted! – Lopey Tall Feb 15 '20 at 18:37