New answers tagged

3

Apologies for not producing a most general answer for arbitrary Lie groups, (which you might tease with great effort out of WP ), but only a trail-map for your particular (charmed!) problem. I call it charmed because it should remind you of the Lorentz group, with a,b,c parameterizing Kx,Ky,Kz boosts and d,e,f the three J rotation angles. Decent treatments ...


1

To a given Lie algebra $\mathfrak g$ there is a unique group $\tilde G$, called the universal covering group, with the property of being simply connected. For example, the covering group of the algebra $\mathfrak{su}(2)$ is $SU(2)$. The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way ...


4

The vector $(0,0,0,v)$ is left invariant by the set of matrices of the form \begin{align*} M=\begin{bmatrix} R & \vec 0 \\ \vec 0^T & 1\end{bmatrix} \end{align*} where $\det(M)=\det(R)=1$ and $M^{-1}=M^T$ implies $R^{-1}=R^T$. By definition, $SO(3)$ is the group of 3 by 3 orthogonal matrices with determinant 1. In general, you need to know the Lie ...


1

I can't comment on string theory, but in quantum field theory the U(1), SU(2), etc symmetry groups are local gauge symmetries. They are not a symmetry of the spacetime in which the symmetry is formulated. So whether the spacetime is discrete or not makes no difference to the local gauge symmetry. As far as I know the physical significance of the local gauge ...


1

Let $\mathfrak{g,h}$ denote the respective Lie algebras of $G,H$. Note that $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{g}/\mathfrak{h}$ as vector spaces. Obviously, there is the adjoint representation of $\mathfrak{h}$ on the whole of $\mathfrak{g}$. Also, the adjoint action of $\mathfrak{h}$ on itself is a subrepresentation, since the algebras of Lie ...



Top 50 recent answers are included