Is local chiral symmetry qualitatively the same as gauge symmetries?

Question

I am confused by the role that local chiral symmetry plays in chiral perturbation theory. For the case of chiral QCD with three quark flavors, the Lagrangian is invariant under *global* $SU(3)_L\times{}SU(3)_R$ transformations on $U$

$$U\to{}g_RUg_L^{\dagger}\qquad{}g_L\in{}SU(3)_L;g_R\in{}SU(3)_R$$

where following http://arxiv.org/pdf/hep-ph/9502366v1.pdf $U$ is the matrix used to parametrize the pions, kaons and the eta (page 6 of the link)

$$U=e^{i\sqrt{2}\Phi/f}$$
$$\Phi=
\begin{pmatrix}
\frac{1}{\sqrt{2}}\pi^0+\frac{1}{\sqrt{6}}\eta&\pi^+&K^+\\
\pi^-&-\frac{1}{\sqrt{2}}\pi^0+\frac{1}{\sqrt{6}}\eta&K^0\\
K^-&\bar{K}^0&-\frac{2}{\sqrt{6}}\eta
\end{pmatrix}$$
where $f$ is just a constant with mass dimensions.

now, we can account for weak and electromagnetic interactions of the pions, kaons and etas by making the global chiral symmetry *local*. Just as in gauge theories covariant derivatives must be used now instead of partial derivatives (page 8)

$D_{\mu}=\partial_{\mu}U-ir_{\mu}U+il_{\mu}U$

where inside $r_{\mu}$ and $l_{\mu}$ we have the photon field, the $W$ bosons...(page 7)

$$r_{\mu}=eQA_{\mu}+\ldots$$
$$l_{\mu}=eA_{\mu}+\frac{e}{\sqrt{2}\sin\theta_W}(W_{\mu}^{\dagger}T_++h.c.)+\ldots$$

where $Q$ is the quark mass matrix

$$Q=\frac{1}{3}diag(2,-1,-1)$$

and $T_+$ is a matrix containning the relevant Cabibbo-Kobayashi-Maskawa factors
$$T_+=
\begin{pmatrix}
0&V_{ud}&V_{us}\\
0&0&0\\
0&0&0
\end{pmatrix}$$

This looks somewhat familiar as what is done in gauge theories but there are things that puzzle me. In gague theories we have as much gauge bosons as dimensions the Lie algebra of the gauge group has. For the case of $SU(3)_L\times{}SU(3)_R$ we ought to have 8+8=16 such gauge bosons! where are they? Moreover, if we consider chiral perturbation theory for two quarks and not three the gauge group would be $SU(2)_L\times{}SU(2)_R$. Nonetheless we can take into account just as much interactions as we did for the three quark case, even though the algebra has now 3+3=6 dimensions only!

All this leads me to think that local chiral symmetry is not a gauge symmetry, but I would like somebody to make this statement more precise. I want someone to clarify the role of local symmetries in chiral perturbation theory and why they are not like gauge symmetries.

Answer 1

1) there is NO local chiral symmetry,  chiral symmetry is global, so it is different with gauge symmetry definitely. Since it is global, there is no 16 gauge bosons.

2) this global symmetry \(SU(3)_L\times SU(3)_R\) is broken into \(SU(3)_V\),  then 8 pseudoscalar massless bosons appears, according to Goldstone's theorem,  which are pions, kaons and \eta boson, and their mass tells you ChPT is only an effective theory.