I am trying understand if there is a relation between two formulations of the spontaneous symmetry breaking.


The first is provide by Derdzinski in his book "Geometry of the standard model of elementary particles" in which we have a vector bundle $(E, M, \mathbb{C}^2)$ with a usual inner product in each fiber, and the Higgs field is identified with a global section $\phi: M \rightarrow E$ with constant norm $|\phi| = v$. Then, we consider the subbundle (a line bundle) generated by $\phi^\perp$ as a eletromagnetic bundle.


The second approach is the natural approach in terms of the principal bundles. Given a principal bundle $(P, M, G= SU(2) \times U(1))$, we consider the eletromagnetic as a subbundle which is given by a global section $h: M \rightarrow P/U(1)$ as guaranteed by the Kobayashi & Namizu page 57-58 in their book "Foundations of differential geometry vol. 1". $P/U(1)$ is a bundle with typical fiber $G/U(1) \simeq SU(2)$.


My problem is that the line bundle $\phi^\perp$ (it's frame bundle) don't seem to have nothing related to the principal bundle of the second case. The better relation that I get see is: as $\phi$ belongs to orbit $Gv$ for all $x\in M$, which is diffeomorphic to $G/ U(1) \simeq SU(2)$, then we can use the global section $\phi$ to obtain a global section of the bundle $P/U(1)$. But, I don't get relate the $\phi^\perp$ subbundle.



> 
> I believe there is a deep structure behind these things. Both have
> further results that seem to be correct. Can anyone help me?
> 
> 
> 



Relationship between two bundles approaches of spontaneous symmetry breaking
Before continuing, let me make some algebraic observations.


1. We can view $U(1)$ as a subgroup of $U(2)$ via the injective homomorphism $\iota : U(1) \to U(2)$ given by
$$
 \forall z \in U(1), \quad \iota(z) := \begin{pmatrix} 1&0\\0&z \end{pmatrix};
$$
moreover, for all orthonormal $\{v,w\} \subset \mathbb{C}^2$, so that $[v\vert w] \in U(2)$, we have
$$
 \forall z \in U(1), \quad [v\vert w]\cdot \iota(z) = [v \vert zw],
$$
so that the map $(g \mapsto (g\_{11},g\_{21})^T) : U(2) \to S^3 \subset \mathbb{C}^2$ descends to a left $U(2)$-equivariant diffeomorphism $F : U(2)/U(1) \to S^3$.
2. We can view $U(1) \times SU(2)$ as a double cover of $U(2)$ via the surjective homomorphism $\mu : U(1) \times SU(2) \to U(2)$ given by
$$
 \forall (z,g) \in U(1) \times SU(2), \quad \mu(z,g) := zg,
$$
where $\ker\mu = \langle (-1,-I\_2) \rangle \cong \mathbb{Z}\_2$. As we’ll soon see, you’ll actually have to work with $U(2) \cong (U(1) \times SU(2))/\mathbb{Z}\_2$ instead of $U(1) \times SU(2)$, but note that the induced Lie algebra homomorphism $\mu\_\ast : \mathfrak{u}(1) \oplus \mathfrak{su}(2) \to \mathfrak{u}(2)$ is an isomorphism.1


Let $M$ be a smooth manifold and let $E \to M$ be a Hermitian vector bundle of rank $2$, which is precisely your data $(E,M,\mathbb{C}^2)$. Let $P \xrightarrow{\pi} M$ be the orthonormal frame bundle of the Hermitian vector bundle $E$, which is a principal $U(2)$-bundle; a further reduction of the structure group to $U(1) \times SU(2)$ would be nontrivial extra structure that we don’t actually need. Modulo this subtlety, you’re correct, one can use the canonical isomorphisms
$$
 P \times\_{U(2)} \mathbb{C}^2 \to E, \quad P \times\_{U(2)} (U(2)/U(1)) \to P/U(1)
$$
and the resulting identifications of global sections with equivariant maps to construct an explicit bijection between the following sets (which may or may not be empty):


* the set of global sections $\phi$ of $E$ with $\lvert \phi \rvert = 1$;
* the set of all global sections of the associated fibre bundle $P/U(1) \to M$.


Given a global section $\phi$ of $E$ with $\lvert \phi \rvert = 1$, I claim that the reduction of the structure group of $P$ induced by the corresponding global section of $P/U(1)$ recovers the circle bundle of the line bundle $\phi^\perp$.


Now, suppose that $\phi$ is a global section of $E$ with $\lvert \phi \rvert = 1$; let $\tilde{\phi} : P \to \mathbb{C}^2$ be the corresponding $U(2)$-equivariant map, so that global section $\psi$ of $P/U(1)$ induced by $\phi$ corresponds to the $U(2)$-equivariant map $\tilde{\psi} : P \to U(2)/U(1)$ given by
$$
 \forall p \in P,\quad \tilde{\psi}(p) := F^{-1} \circ \tilde{\phi}(p) = \begin{pmatrix} \tilde{\phi}(p)\_1 & -\overline{\tilde{\phi}(p)\_2} \\ \tilde{\phi}(p)\_2 & \overline{\tilde{\phi}(p)\_1} \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & U(1) \end{pmatrix}.
$$
The section $\psi$ of $P/U(1) \to M$ yields a reduction of the structure group of $P$ to $U(1)$, i.e., the principal $U(1)$-bundle
$$
 \tilde{P} := \{ p \in P \,\mid\, p \cdot U(1) = \psi(\pi(p))\}
$$
satisfying $P \cong \tilde{P} \times\_{U(1)} U(2)$ as principal $U(2)$-bundles on $M$. By unpacking definitions and canonical isomorphisms and applying observation 1, one can check that
$$
 \forall x \in M, \quad \tilde{P}\_x := \{(\phi\_x,w) \, \mid \, w \in \phi\_x^\perp, \, \lvert w \rvert = 1\}.
$$
But now, this is clearly isomorphic as a principal $U(1)$-bundle over $M$ to the circle bundle
$$
 Q := \{ w \in \phi^\perp \mid \lvert w \rvert = 1\}
$$
of $\phi^\perp$ via the isomorphism $f : Q \to \tilde{P}$ given by
$$
 \forall x \in M, \, \forall w \in Q\_x, \quad f(w) := (\phi\_x,w).
$$




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1 Compare the Standard Model, where the famed group $U(1) \times SU(2) \times SU(3)$ is actually a $6$-fold cover of the true structure group $(U(1) \times SU(2) \times SU(3))/\mathbb{Z}\_6$. Since the covering homomorphism of Lie groups induces an explicit isomorphism of Lie algebras, one can formally work with the group $U(1) \times SU(2) \times SU(3)$ at the price of introducing fractional charges.


