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I Embeddings of a gauge group

  1. Apr 9, 2017 #1
    In the framework of Einstein-Yang-Mills (EYM) theory, suppose the following action:

    \begin{equation}S=\int\left({\kappa R + \alpha tr(F_{\mu \nu}F^{\mu \nu})d^4 x}\right)\,,\end{equation}

    where F is the gauge curvature associated with a non-abelian Lie group G and a gauge connection A. Then, it was established (P. B. Yasskin, Solutions for Gravity Coupled to Massless Gauge Fields, Phys. Rev. D 12, 2212 (1975)) that there exist embedded U(1) solutions (i.e. with a commutative gauge connection) to the EYM field equations. Furhermore,, it was also shown (G. 't Hooft, Magnetic Monopoles in Unified Gauge Theories, Nucl. Phys. B 79, 276 (1974)) that additional embedded solutions can exist due to the compact covering group of G.

    Then my question is about the way of distinguishing between the gauge connection associated with a general group G and the connection associated with a subgroup of G in some complicated cases. For example, suppose the Lorentz group SO(3,1) and the spin connection A in a Minkowski space-time:

    \begin{equation}A^{a b}\,_{\mu}=e^{a}\,_{\lambda}\,e^{b \rho}\,\Gamma^{\lambda}\,_{\rho \mu}+e^{a}\,_{\lambda}\,\partial_{\mu}\,e^{b \lambda}\,.\end{equation}

    I have computed the non-vanishing components of this connection in the Minkowski space-time and this is the outcome by using the usual coordinates:

    \begin{equation}A^{1 2}\,_{\theta}=1\,,A^{1 3}\,_{\theta}=sin(\theta)\,,A^{2 3}\,_{\phi}=cos(\theta)\,.\end{equation}

    Hence there are three independent components. Then it seems that these components satisfy the commutation laws of the SO(3) subgroup, so that my first question is why? If the isometry group of the minkowski space-time associated to the rotations is the SO(3,1) group, where are the additional components of the whole group SO(3,1)? Obviously, if I was written an abelian spin connection instead of a non-abelian one as above, then I had wrongly computed the mentioned components because of the isometry group of the space-time is a non-abelian group, then why the components of such a minkowski spin connection are associated with the SO(3) subgroup instead of the whole group SO(3,1)?

    Similarly, my second question is about the way of distinguishing in certain cases between a connection associated with the whole group G and the one associated with a subgroup S of G. For example, suppose the spin connection of the Schwarzschild space-time, then the computations give rise to the following components:

    \begin{equation}A^{0 1}\,_{t}=-\frac{m}{r^2}\,,A^{1 2}\,_{\theta}=\sqrt{1-\frac{2m}{r}}\,,A^{1 3}\,_{\theta}=sin(\theta)\sqrt{1-\frac{2m}{r}}\,,A^{2 3}\,_{\phi}=cos(\theta)\,.\end{equation}

    In this situation, an additional component arises in presence of this curved space-time and I am not sure if this type of connection is associated with the SO(3) group or with the SO(3,1), because of the presence of such an additional component...

    Thanks in advance.
     
  2. jcsd
  3. Apr 9, 2017 #2
    According to my second question, I think that a possible answer is to check the number of non-vanishing components of the regarded connection. In the case of the spin connection and the Schwarzschild space-time, there are four non-vanishing components and therefore it cannot describe the SO(3) subgroup whose dimensión is 3, but the SO(3,1) group of dimensión 6. Anyway, I would like to know if it is available any alternative and rigorous demonstration, also for general connections and different examples.

    Best regards.
     
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