Special Euclidean group

Problem: Show that the Lorentz transformation $W$ that keeps $k^{\mu}\equiv[1, 0,  0,  1]^T$ unchanged forms a two-dimensional Euclidean group SE(2).







Solution:

As shown in this post, the general solution of the Lorentz transformation $W$ that satisfies $W^{\mu}_{\,\,\,\,\nu}\,k^{\nu}=k^{\mu}$ can be parametrized in three parameters $\alpha,\beta,\theta$ as \begin{equation}W(\alpha, \beta, \theta)=\left[\begin{matrix} 1+(\alpha^2+\beta^2)/2  & -\alpha & -\beta & -(\alpha^2+\beta^2)/2  \\  -\alpha & 1 & 0 & \alpha \\ -\beta & 0 &  1 & \beta \\ (\alpha^2+\beta^2)/2 & -\alpha & -\beta & 1-(\alpha^2+\beta^2)/2 \end{matrix}\right]\left[\begin{matrix} 1 &  &  & \\  & \cos\theta & -\sin\theta & \\ & \sin\theta & \cos\theta  &   \\ & &  & 1\end{matrix}\right]\,.\tag{1}\end{equation}

To work out the Lie algebra, we expand $W(\alpha, \beta, \theta)$ in the first order of $\alpha, \beta, \theta$ as \begin{eqnarray}W(\alpha, \beta, \theta)&\approx&\left[\begin{matrix} 1  & -\alpha & -\beta & 0 \\  -\alpha & 1 & 0 & \alpha \\ -\beta & 0 &  1 & \beta \\  0 & -\alpha & -\beta & 1 \end{matrix}\right]\left[\begin{matrix} 1 &  &  & \\  & 1 & -\theta & \\ & \theta & 1  &   \\ & &  & 1\end{matrix}\right]\approx \left[\begin{matrix} 1  & -\alpha & -\beta & 0 \\  -\alpha & 1 & -\theta & \alpha \\ -\beta & \theta &  1 & \beta \\  0 & -\alpha & -\beta & 1 \end{matrix}\right]\\  \\&\equiv& I - i \alpha A -i\beta B - i\theta J\,,\end{eqnarray} where the generators are \begin{eqnarray}A\equiv\left[\begin{matrix} & -i & & \\ -i & & & i \\ & & & \\ & -i & & \end{matrix}\right]\,,\quad B\equiv\left[\begin{matrix} & & -i & \\ & & & \\ -i & & & i \\ & & -i & \end{matrix}\right]\,,\quad J\equiv\left[\begin{matrix} & & & \\ & & -i & \\ & i & & \\ & & & \end{matrix}\right] \,.\tag{2}\end{eqnarray} The generator $J$ represents a rotation of the x-y plane, which is obvious that keeps $[1, 0, 0, 1]^T$ unchanged. In contrast, the generators $A$ and $B$ are less obvious: $A$ represents a boost along x-direction plus a rotation of the x-z plane; $B$ represents a boost along y-direction plus a rotation of the y-z plane.

With the explicit matrix forms (2), one can verify that \begin{equation}[A,\,B] = 0\,,\quad [J,\,A]=i\,B\,,\quad [J,\,B]=-i\,A\,.\tag{3} \end{equation} As we will see below, the commutation relations (3) are exactly the Lie algebra of two-dimensional special Euclidean group SE(2).

SE(2) group

By definition, the two-dimensional special Euclidean group SE(2) consists of two-dimensional rotations plus translations. In terms of the homogenous coordinate $[x, y, 1]$, the transformation of SE(2) can be represented as \begin{equation}\left[\begin{matrix}x' \\ y' \\ 1\end{matrix}\right]=\left[\begin{matrix}\cos\theta & -\sin\theta & \alpha \\  \sin\theta & \cos\theta & \beta \\ 0 & 0 & 1\end{matrix}\right]\left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]\,,\end{equation} where $\theta$ is the rotation angle along the z-axis and $[\alpha,\beta]^T$ is the translation vector within the x-y plane. 

Expand the above transformation matrix to the first order of $\alpha, \beta, \theta$: \begin{equation}\left[\begin{matrix}\cos\theta & -\sin\theta & \alpha \\  \sin\theta & \cos\theta & \beta \\ 0 & 0 & 1\end{matrix}\right]\approx\left[\begin{matrix}1 & -\theta & \alpha \\  \theta & 1 & \beta \\ 0 & 0 & 1\end{matrix}\right]\equiv I - i \alpha A - i\beta B - i\theta J\,,\end{equation} where the generators are \begin{equation}A=\left[\begin{matrix}0 & 0 & i \\  0 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right]\,,\quad B=\left[\begin{matrix}0 & 0 & 0 \\  0 & 0 & i \\ 0 & 0 & 0\end{matrix}\right]\,,\quad J=\left[\begin{matrix}0 & -i & 0 \\  i & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right]\,.\tag{4} \end{equation} The generator $J$ represents a rotation of the x-y plane and $A\,(B)$ represents a translation along the x(y)-direction.

With the explicit matrix forms (4), one can work out the Lie algebra of SE(2): \begin{equation}[A,\,B] = 0\,,\quad [J,\,A]=i\,B\,,\quad [J,\,B]=-i\,A\,,\tag{5} \end{equation} which are the same as (3). 

Therefore, we can conclude that

  • The matrices (4) are the 3-dimensional representation of SE(2) group, and the parameters $\alpha,\beta$ are related a 2-dimensional translation.
  • The matrices (2) are the 4-dimensional representation of SE(2) group, and the parameters $\alpha,\beta$ are related to the Lorentz boosts followed by rotations.

Irreducible representation of SE(2)

To find out the irreducible representation of SE(2), we define a set of new operators \begin{equation}T^2\equiv A^2+B^2\,,\quad T_{\pm}\equiv A \pm i\,B\,,\end{equation} then the Lie algebra (5) becomes \begin{equation} [T^2, \,T_{\pm}] = 0\,,\quad [T_+,\,T_-]=0\,,\quad [T^2,\,J] = 0\,,\quad [J,\,T_{\pm}]=\pm T_{\pm}\,. \end{equation}

Since $T^2$ and $J$ commute, we can introduce a state $|t,j\rangle$ such that \begin{equation}T^2\,|t,j\rangle=t^2\,|t,j\rangle\,,\quad  J\,|t,j\rangle=j\,|t,j\rangle\,.\tag{6}\end{equation} The state $|t,j\rangle$ in (6) is the basis of the SE(2) irreducible representation. What about the state $T_{\pm}|t,j\rangle$? Consider \begin{eqnarray}T^2\,T_{\pm}|t,j\rangle&=&T_{\pm}\,T^2|t,j\rangle = T_{\pm}\,t^2|t,j\rangle = t^2\, T_{\pm}|t,j\rangle\,,\\ J\,T_{\pm}|t,j\rangle &=& T_{\pm}\,J|t,j\rangle \pm T_{\pm}|t,j\rangle=T_{\pm}\,j|t,j\rangle \pm T_{\pm}|t,j\rangle = \left(j\pm 1\right)\,T_{\pm}|t,j\rangle\,, \end{eqnarray} we can conclude that $T_{\pm}|t,j\rangle$ is the state $|t,j\pm 1\rangle$ up to a constant. The constant can be obtained by considering the inner product \begin{equation}\langle t,j|T^{\dagger}_{\pm}\,T_{\pm}|t,j\rangle= \langle t, j|\,T^2\,|t,j\rangle = t^2\langle t, j|t,j\rangle = t^2\,.\end{equation} As a result, we have \begin{equation}T_{\pm}\,|t,j\rangle=\pm\frac{t}{i}\,|t,j\pm 1\rangle\,,\end{equation} where the phase factor $\pm 1/i$ is chosen for other convenience.

The simplest case is the "degenerate" representation when $t=0$. In this case, we have \begin{equation}J\,|0,j\rangle = j\,|0, j\rangle\,,\quad T^2|0,j\rangle = T_{\pm}|0,j\rangle=A|0,j\rangle=B|0,j\rangle = 0\,.\end{equation} The degenerate representation of SE(2) is \begin{equation}\langle 0, j'| \exp\left[-i\alpha A - i\beta B - i\theta J\right]|0, j\rangle=e^{-i\,j\,\theta}\delta_{j',j}\,.\end{equation}

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