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  Differential Geometry of the Georgi–Glashow model

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I am trying to understand the mathematical structure of the Georgi–Glashow model according to the spirit of the Nakahara`s book  "Geometry, Topology and Physics" .  In few words the Nakahara`s spirit consists in to convert in physics the theorems in "Foundations of Differential Geometry Vol 1" by Kobayashi and Nomizu.   In this sense I think that the proposition 6.4 of Kobayashi-Nomizu  gives a very important foundation of the mathematical structure of the Georgi–Glashow model from the point of view of the modern differential geometry.

In particle physics, the Georgi–Glashow model is a particular grand unification theory (GUT) proposed by Howard Georgi and Sheldon Glashow in 1974. In this model the standard model gauge groups SU(3) × SU(2) × U(1) are combined into a single simple gauge group—SU(5). The unified group SU(5) is then thought to be spontaneously broken into the standard model subgroup below some very high energy scale called the grand unification scale.

From other side proposition 6.4 reads (page 18 on):

http://users.math.msu.edu/users/parker/GT/Kobayashi-Nomizu.pdf

Then, my questions are: 

1. How to prove the proposition 6.4?

2. How to apply the proposition 6.4 to the Georgi–Glashow model?

asked Aug 4 in Theoretical Physics by juancho (990 points) [ revision history ]
edited Aug 9 by juancho

@igael, thanks for your comment.  You are right but I want to have a proof for physicists with all the details and the same thing for the application of 6.4 to the Georgi–Glashow model.  All the best.

6.4 may also be used to check the construction of SU(3) from its known irreducible groups. But it doesn't help directly to choose between the possible embedding solutions while each can be invalidated in a lab or by known measures. Intuitively, SU(5) proton lifetime expectation is a good point for SO(10) and Susy-like theories. ( @Juancho, I fear only to spoil the nice questions by trivial or inappropriate comments while commenting refreshs the question rank :) TY ) .

@igael, your comments are very important and relevant for the questions.  Again, thanks for your new comment which is illuminating.  All the best.


 

yes, indeed SO(10), sorry

1 Answer

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Proof of the proposition 6.4.

First we prove that $\omega_Q((R_a)_* X) = ad(a^{-1})\omega_Q(X)$ where $a \in H$ and $X \in T_v(Q)$ with $v \in Q$.  Given that $\omega_P$ is a connection one-form in the principal bundle $P(M,G)$; given that $v \in P$ , $X \in T_v(P)$   and $a \in G$; then it is verified that

$$\omega_P((R_a)_* X) = ad(a^{-1})\omega_P(X)$$.

Given that the last expression is valid for all $a \in G$, it is also valid for $a \in H$ because $H$ is a subgroup of $G$. Then for $X \in T_v(Q)$   and $a \in H$; then it is verified that

$$\omega_P((R_a)_* X) = ad(a^{-1})\omega_P(X)$$.

Now, let  $\phi$ the $m$-component of  $\omega_P$ restricted to the subbundle $Q$, then it is possible to write $\omega_P = \omega_Q  + \phi$; and for hence we have that

$$(\omega_Q  + \phi)((R_a)_* X) = ad(a^{-1})(\omega_Q  + \phi)(X)$$

$$\omega_Q((R_a)_* X)  + \phi((R_a)_* X) = ad(a^{-1})(\omega_Q(X)  + \phi(X))$$

$$\omega_Q((R_a)_* X)  + \phi((R_a)_* X) = a(\omega_Q(X)  + \phi(X))a^{-1}$$

$$\omega_Q((R_a)_* X)  + \phi((R_a)_* X) = a\omega_Q(X)a^{-1}  + a\phi(X)a^{-1}$$

$$\omega_Q((R_a)_* X)  + \phi((R_a)_* X) = ad(a^{-1})\omega_Q(X) + ad(a^{-1})\phi(X)$$.

The generators for the subalgebra $h$ are denoted by $ \hat{h}_{\alpha}$ and the generators for $m$ are denoted $ \hat{m}_{\beta}$; then we have that

$$[\omega_Q((R_a)_* X)]^{\alpha}\hat{h}_{\alpha}  + [\phi((R_a)_* X) ]^{\beta}\hat{m}_{\beta}=$$

$$ad(a^{-1})([\omega_Q(X)]^{\alpha}\hat{h}_{\alpha} ) +ad(a^{-1})([\phi(X)]^{\beta}\hat{m}_{\beta}) $$

which is rewritten as

$$[\omega_Q((R_a)_* X)]^{\alpha}\hat{h}_{\alpha}  + [\phi((R_a)_* X) ]^{\beta}\hat{m}_{\beta}=$$ $$[\omega_Q(X)]^{\alpha}ad(a^{-1})(\hat{h}_{\alpha} ) + [\phi(X)]^{\beta}ad(a^{-1})(\hat{m}_{\beta}) $$

and then we have that

$$[\omega_Q((R_a)_* X)]^{\alpha}\hat{h}_{\alpha}  + [\phi((R_a)_* X) ]^{\beta}\hat{m}_{\beta}= [\omega_Q(X)]^{\alpha}E_{\alpha}^{\beta}\hat{h}_{\beta}  + [\phi(X)]^{\beta}C_{\beta}^{\gamma}\hat{m}_{\gamma} $$

where $E_{\alpha}^{\beta}$ and $C_{\beta}^{\gamma}$ are constants.; and the  Einstein summation convention was used.

Then, given that $\hat{h}_{\alpha}$ and $\hat{m}_{\beta}$ are linearly independent, we deduce that

$$[\omega_Q((R_a)_* X)]^{\alpha}\hat{h}_{\alpha}  = [\omega_Q(X)]^{\alpha}E_{\alpha}^{\beta}\hat{h}_{\beta}$$

$$\omega_Q((R_a)_* X)  = [\omega_Q(X)]^{\alpha}ad(a^{-1})(\hat{h}_{\alpha} )$$

$$\omega_Q((R_a)_* X)  = ad(a^{-1})([\omega_Q(X)]^{\alpha}\hat{h}_{\alpha} )$$

$$\omega_Q((R_a)_* X) = ad(a^{-1})\omega_Q(X)$$.

Second, we prove that $\omega_Q(A^*) = A$, where $A\in h$ and $A^*$ is the fundamental vector field corresponding to $A$. Then, given that $\omega_P(A^*)= A$ and $\phi(A^*)=0$  we have that

$$\omega_P = \omega_Q  + \phi$$

$$\omega_P(A^*) = \omega_Q( A^*)+ \phi(A^*)$$

$$A= \omega_Q( A^*)+ 0$$

$$A= \omega_Q( A^*)$$

answered 5 days ago by juancho (990 points) [ revision history ]
edited 5 days ago by juancho

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