K-Poincaré group

In physics and mathematics, the κ-Poincaré group, named after Henri Poincaré, is a quantum group, obtained by deformation of the Poincaré group into an Hopf algebra. It is generated by the elements $a^{\mu }$ and ${\Lambda ^{\mu }}_{\nu }$ with the usual constraint:

\eta ^{\rho \sigma }{\Lambda ^{\mu }}_{\rho }{\Lambda ^{\nu }}_{\sigma }=\eta ^{\mu \nu }~,

where $\eta ^{\mu \nu }$ is the Minkowskian metric:

\eta ^{\mu \nu }=\left({\begin{array}{cccc}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}}\right)~.

The commutation rules reads:

$[a_{j},a_{0}]=i\lambda a_{j}~,\;[a_{j},a_{k}]=0\,$
$[a^{\mu },{\Lambda ^{\rho }}_{\sigma }]=i\lambda \left\{\left({\Lambda ^{\rho }}_{0}-{\delta ^{\rho }}_{0}\right){\Lambda ^{\mu }}_{\sigma }-\left({\Lambda ^{\alpha }}_{\sigma }\eta _{\alpha 0}+\eta _{\sigma 0}\right)\eta ^{\rho \mu }\right\}\,$

In the (1 + 1)-dimensional case the commutation rules between $a^{\mu }$ and ${\Lambda ^{\mu }}_{\nu }$ are particularly simple. The Lorentz generator in this case is:

{\Lambda ^{\mu }}_{\nu }=\left({\begin{array}{cc}\cosh \tau &\sinh \tau \\\sinh \tau &\cosh \tau \end{array}}\right)\,

and the commutation rules reads:

$[a_{0},\left({\begin{array}{c}\cosh \tau \\\sinh \tau \end{array}}\right)]=i\lambda ~\sinh \tau \left({\begin{array}{c}\sinh \tau \\\cosh \tau \end{array}}\right)\,$
$[a_{1},\left({\begin{array}{c}\cosh \tau \\\sinh \tau \end{array}}\right)]=i\lambda \left(1-\cosh \tau \right)\left({\begin{array}{c}\sinh \tau \\\cosh \tau \end{array}}\right)\,$

The coproducts are classical, and encode the group composition law:

$\Delta a^{\mu }={\Lambda ^{\mu }}_{\nu }\otimes a^{\nu }+a^{\mu }\otimes 1\,$
$\Delta {\Lambda ^{\mu }}_{\nu }={\Lambda ^{\mu }}_{\rho }\otimes {\Lambda ^{\rho }}_{\nu }\,$

Also the antipodes and the counits are classical, and represent the group inversion law and the map to the identity:

$S(a^{\mu })=-{(\Lambda ^{-1})^{\mu }}_{\nu }a^{\nu }\,$
$S({\Lambda ^{\mu }}_{\nu })={(\Lambda ^{-1})^{\mu }}_{\nu }={\Lambda _{\nu }}^{\mu }\,$
$\varepsilon (a^{\mu })=0$
$\varepsilon ({\Lambda ^{\mu }}_{\nu })={\delta ^{\mu }}_{\nu }\,$