Apologies, I know the title's a mouth-full.
We can consider the standard unit three-sphere $S^{3}$ as being embedded in $\mathbb{C}^{2}$ with coordinates:
$$\Phi=\begin{array}{c} \phi^{1}+i\phi^{2}\\ \phi^{3}+i\phi^{4} \end{array}=\begin{array}{c} z^{1}\\ z^{2} \end{array}$$
Where $\Phi^{\dagger}\Phi=1$ ( the dagger indicating the conjugate transpose). A metric $g_{\mu\nu}$ on $S^{3}$ may then be regarded as the induced metric:
$$g_{\mu\nu}=\frac{\partial\Phi^{\dagger}}{\partial x^{\mu}}\frac{\partial\Phi}{\partial x^{\nu}}=\left(\frac{\partial\bar{z}^{i}}{\partial x^{\mu}}\right)^{T}\frac{\partial z^{i}}{\partial x^{\nu}}$$
Consider now the Yamabe problem: we know that any compact, simply connected three-manifold $\tilde{M}$ is homeomorphic to the three-sphere (Poincare conjecture) which itself has constant scalar curvature. Knowing the Yamabe problem has solutions in this case for any such manifold $\tilde{M}$, there must exist a positive function $f(x^{\mu})$ such that the metric on $\tilde{M}$:
$$\tilde{g}_{\mu\nu}=f(x^{\mu})g_{\mu\nu}$$
I'm wondering about applying such a dilation to the original coordinates of the embedding? Can I then also achieve the metric on $\tilde{M}$ by saying there also then exists a dilation $\sigma$ such that:
$$\tilde{\Phi}=\sigma\Phi,\tilde{\Phi}^{\dagger}=\bar{\sigma}\Phi^{\dagger}$$
Such that $\tilde{\Phi}^{\dagger}\tilde{\Phi}=|\sigma|^{2}$ is our always positive function? Then the metric can be written and rearranged as:
$$\tilde{g}_{\mu\nu}=\frac{\partial\left(\sigma\Phi\right)^{\dagger}}{\partial x^{\mu}}\frac{\partial\left(\sigma\Phi\right)}{\partial x^{\nu}}=\left(\bar{\sigma}\Phi_{,\mu}^{\dagger}+\Phi^{\dagger}\bar{\sigma}_{,\mu}\right)\left(\Phi_{,\nu}\sigma+\Phi\sigma_{,\nu}\right)=\left(\partial_{\mu}\bar{\sigma}+\Phi_{,\mu}^{\dagger}\Phi\bar{\sigma}\right)\left(\partial_{\mu}\sigma+\Phi^{\dagger}\Phi_{,\nu}\sigma\right)$$
Which just looks like a covariant derivative! In this case we can write our metric as:
$$\tilde{g}_{\mu\nu}=\left(\nabla_{\mu}\bar{\sigma}\right)\left(\nabla_{\nu}\sigma\right)$$
defining: $R_{\mu}=\Phi^{\dagger}\overrightarrow{\partial}_{\mu}\Phi$ and $L_{\mu}=\Phi^{\dagger}\overleftarrow{\partial}_{\mu}\Phi$. This can be rewritten as: $$\tilde{g}_{\mu\nu}=(\partial_{\mu}\bar{\sigma}+R_{\mu}\bar{\sigma})(\partial_{\nu}\sigma+L_{\nu}\sigma)$$
If I did this procedure right, then, in the language of fiber bundles, what exactly is going on here? I know that the original coordinates $\Phi$ are an $SU(2)$ doublet, so can I say the final equation has an $su(2)$ valued connection? What about the left and right components of the connection, can these be viewed as chiral fields on $\tilde{M}$???
ADDENDUM: We can perform a “check” of sorts on this result by seeing how a coordinate transformation on the $\Phi$ coordinates changes our metric. Such a coordinate transformation $A$ is an element of the complex equivalent of $SO(2)$, namely $SU(2)$.
$$\Phi\rightarrow A\Phi$$
$$\Phi^{\dagger}=\left(A\Phi\right)^{\dagger}$$
So, now we have:
$$\Phi^{\dagger}\Phi=\left(A\Phi\right)^{\dagger}\left(A\Phi\right)$$
$$\tilde{g}_{\mu\nu}=\partial_{\mu}(\sigma A\Phi)^{\dagger}\partial_{\nu}(\sigma A\Phi)=\left(\sigma\Phi_{,\mu}^{\dagger}A^{\dagger}+\sigma\Phi^{\dagger}A_{,\mu}^{\dagger}+\Phi^{\dagger}A^{\dagger}\sigma_{,\mu}\right)\left(A\Phi_{,\nu}\sigma+A_{,\nu}\Phi\sigma+A\Phi\sigma_{,\nu}\right)$$
$$=\left(\partial_{\mu}\bar{\sigma}+\Phi_{,\mu}^{\dagger}\Phi\bar{\sigma}+A_{,\mu}^{\dagger}A\bar{\sigma}\right)\left(\partial_{\nu}\sigma+\Phi^{\dagger}\Phi_{,\nu}\sigma+A^{\dagger}A_{,\nu}\sigma\right)$$
Note that because $A\in su(2)$ we have that the last term in each factor has the form $g^{-1}dg$ (is a pure gauge transformation). This seems to match with our interpretation of each factor as a covariant derivative.
For the interested, I'm actually working on four dimensional compact Lorentzian manifolds embedded in $\mathbb{C}^{3}$ In which case, we get an $su(3)$ type connection.