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  Differential Geometry of the Higgs Field

+ 3 like - 0 dislike

I am trying to understand the mathematical structure of the Higgs fields and the Spontaneous Symmetry Breaking according to the spirit of the Nakahara`s book  "Geometry, Topology and Physics"  and the paper in the same spirit by  G.Sardanashvily  "Geometry of Classical Higgs Fields" (http://arxiv.org/pdf/hep-th/0510168.pdf).  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 (Please look an example of this at http://www.physicsoverflow.org/32653/differential-geometry-of-the-standard-model).   In this sense I think that the proposition 5.6, the proposition 6.4  and the proposition 7.4 in chapter II of  Kobayashi-Nomizu give a very important foundation of the mathematical structure of the Higgs field and the Spontaneous Symmetry Breaking  from the point of view of the modern differential geometry. 

Proposition 5.6.  The structure group $G$ of a principal bundle  $P(M, G)$ is reducible to a closed subgroup $H$ of $G$ if and only if the associated bundle $E(M, G/H, G, P)$  admits  a global cross section $\sigma :  M  \rightarrow  (E = P/H)$.

Proposition 6.4.  Let  $Q(M, H)$ be a principal subbundle of the principal bundle $P(M, G)$,  where $H$ is a Lie.subgroup of $G$. Assume that the Lie algebra $g$ of $G$ admits a subspace $m$ such that $g = m \oplus h$ (direct sum).and  $ad (H)(m)  \subset m$, where $h$ is the Lie algebra of $H$. For every connection  one-form $\omega_P$ in the principal bundle $P$,  the $h$-component $\omega_Q$ of $\omega_P$ restricted to the principal subbundle $Q$ is a connection one-form in $Q$.

Proposition 7.4.  Let $P(M, G)$ be  a principal bundle and $E(M, G/H, G, P)$ the associated bundle with standard fibre $G/H$, where $H$ is a closed subgroup of $G$. Let $\sigma :  M  \rightarrow  E$ be a global cross section and $Q ( M, H)$ the reduced subbundle of $P(M, G)$  corresponding to the global cross section $\sigma$ according with  Proposition 5.6. Then a connection $\Gamma_P$ in $P$ is reducible to a connection  $\Gamma_Q$ in $Q$ if and only the global cross section $\sigma $ is parallel with respect to  $\Gamma_P$ .

My questions are: 

1. How to prove the propositions 5.6, 6.4 and 7.4?

2. How to apply the propositions 5.6, 6.4 and 7.4  to the Higgs fields and the Spontaneous Symmetry Breaking ?

asked Jul 30, 2015 in Theoretical Physics by juancho (1,130 points) [ revision history ]
edited Jul 31, 2015 by juancho

2 Answers

+ 3 like - 0 dislike

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

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

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 Jul 31, 2015 by juancho (1,130 points) [ revision history ]
edited Jul 31, 2015 by juancho
+ 2 like - 0 dislike

Application of the proposition 5.6 to the Higgs field and the Spontaneous Symmetry Breaking:

From the physical point of view the proposition 5.6 says that there is a spontaneous symmetry breaking from \(G\) to \(H\)  if an only if exists a Higgs field. In proposition 5.6, the Yang-Mills  bundle is the principal bundle \(P(M,G)\); the Higgs bundle is the associated bundle \(E(M, G/H, G, P)\); the Higgs field is the  global cross section  \(\sigma\).

In General Relativity, the metric $g$ plays the role of the Higgs field and then the metric produces a spontaneous symmetry breaking from $GL_4^+(\mathbb{R})$ to $SO(1,3)$  (http://arxiv.org/pdf/1212.6702.pdf).

More formally,  The structure group $GL_4^+(\mathbb{R})$ of the principal bundle  $P(M,GL_4^+(\mathbb{R}))$  over a space-time manifold $M$ is reduced to the closed subgroup $SO(1,3)$ of $GL_4^+(\mathbb{R})$  by the metric $g$ which is a global cross section of the form

$$g :  M  \rightarrow  (E = P/SO(1,3))$$ where $E$ is the associated bundle $E(M,GL_4^+(\mathbb{R})/SO(1,3),GL_4^+(\mathbb{R}),P) $.

answered Jul 30, 2015 by juancho (1,130 points) [ revision history ]
edited Aug 1, 2015 by juancho

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