differential form of charge for pi_4(S^3) or pi_4(S^2)

2010-11-13T12:00:35Z

How to write a 4-form of topological charge which would correspond to non-zero element of the homotopy group $\pi_4(S^3)$ or $\pi_4(S^2)$ (both are equal to $Z_2$) ? ?

An example of such a mapping (non-trivial), or even a homotopy of maps ($0\leq \tau\le 1$), is (or `localized' maps $R^4=H \to S^3$)

$f(h) = \frac{q-1}{q+1}$, where $q=h\ i\ \bar{h}+\tau\ i= -\bar{q}$ ; i.e. $\bar{f}=f^{-1}$ ;

here $h$ - quaternion, $q$ - imaginary quaternion, $f\in S^3$ - quaternion of unit length.

One can check that the framed 1-manifold $f^{-1}(-1)$ is non-trivial (see $\tau=1$).

Motivation: to describe topological charges (and quasi-charges) in the frame field theory (or Absolute Parallelism), see arXiv: gr-qc/0610076 .

This map is of great symmetry, in a sense like: $f(gh\bar{g})=gf\bar{g}=f(ghg)$, where $g=e^{i\alpha/2}$. Moreover, there is a more huge symmetry, a kind of gauge (gradient) symmetry, I quess:

$q(h)=q(h\ e^{i\alpha(h)})$ .

So, if one write quaternion $h$ as a pair of complex number (spinor), $h=a\ e^{i\phi} + j\ b\ e^{i\psi}$, so $q=i+(a^2-b^2)i -2abe^{i(\phi-\psi)}k$, and it is evident that the differential, combination $d\phi+d\psi$ in no way can arise in any differential form constructed from only $f$-field.

The cases $\pi_3(S^2)$ and $\pi_7(S^4)$ have the known solution relating to the Hopf invariant (if I am not mistaken) and having the form (pullback form)

$\alpha\wedge d\alpha$, where $d\alpha$ is the volume form on the sphere $S^2$ or $S^4$;

that is, for the first case, $d\alpha\propto{\rm Re}(\omega d\omega\wedge d\omega) $, where $\omega$ is a unit imaginary quaternion, the coordinates on the $S^2$; $\omega^2=-1$. (${\rm Tr}()$ suits (instead of ${\rm Re}()$ if the units are represented with the Pauli matrices, $i_p=i\sigma_p$.)

differential form of charge for pi_4(S^3) or pi_4(S^2) - MathOverflow

differential form of charge for pi_4(S^3) or pi_4(S^2)