1
$\begingroup$

The most sensible geometric interpretation of spinors that I've come across is that they encode projections in the Clifford algebra. So if $\mathbf A$ is a vector with components $A_i$ and $\psi$ is a spinor, then $\psi^\dagger A_i \sigma^i \psi = \mathrm{Tr} (A_i\sigma^i \psi \psi^\dagger)$ gives the component of $\mathbf A$ along the direction encoded by $\psi$.

Is there a geometric object analagous to spinors, which projects onto a bivector orientation rather than a vector direction? So if $S_{ij}$ were the components of a bivector and $\chi$ were such an object, $\chi^\dagger S_{ij} \frac12[\sigma^i,\sigma^j] \chi$ might give the component of the bivector $\mathbf S$ with orientation encoded by $\chi$. As in the case of spinors, $\frac{1}{2}(\chi_1+\chi_2)(\chi_1+\chi_2)^\dagger$ would be another such projection.

Or can this already be accomplished using spinors?

$\endgroup$
1
$\begingroup$

The geometric object which corresponds to spinors is the external bundle, the bundle of differential forms. The natural equation on such a bundle is the Dirac-Kähler equation.

This bundle is essentially used in lattice computations under the name "staggered fermions".

A problem with this geometric interpretation is that it has too many components. So, the complexified bundle $\Lambda(\mathbb{R}^4,\mathbb{C})$ would describe four Dirac fermions. In arXiv:0908.0591 (see also http://ilja-schmelzer.de/matter/ ) a splitting spacetime into space and time allows to reduce this to two Dirac fermions, which, then, can be interpreted as an electroweak pair.

$\endgroup$
  • $\begingroup$ Very interesting paper, though I'll have to take some more time with it. So the Dirac-Kähler equation describes a complex multivector field in 4d spacetime, with scalar, pseudoscalar, vector, pseudovector,... components. You've taken this equation for 3d Euclidean space and constructed two Dirac fermions from the 8 complex degrees of freedom that define the multivector? $\endgroup$ – rossng Sep 12 '16 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.