How to construct generators and Lie Algebra for Lorentz group?

Question

I'm trying to figure out Lorentz group in 2+1. First of all, I'd like to think the special orthgonal group as a combination of rotation and boost in space. Then I construct it as below. First rotation part: $$ R(\theta)= \begin{pmatrix} cos\theta& -sin\theta & 0 \\ sin\theta & cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ I assume $\theta$ is infinitesimally small and $\theta \sim \epsilon$ then $cos\theta \sim 1$ and $sin\theta \sim \theta$ $$ R(\theta)=\begin{pmatrix} 1 & 0 & 0 \\ 0& 1&0 \\ 0&0&1\end{pmatrix} -\epsilon \begin{pmatrix} 0&1&0 \\ -1&0&0\\ 0&0&0\end{pmatrix} $$ I have just one parameter $\theta$ so I suppose I have one generator as well from rotations. And it is $$ J=\begin{pmatrix} 0&i&0 \\ -i&0&0\\ 0&0&0\end{pmatrix} $$ and I construct the boost part as $$ T(\phi)=\begin{pmatrix} cosh\phi&0&sinh\phi\\0&1&0\\sinh\phi&0&cosh\phi\end{pmatrix} $$ $$ T(\theta)= \begin{pmatrix} 1 & 0 & 0 \\ 0& 1&0 \\ 0&0&1\end{pmatrix} - \epsilon\begin{pmatrix}0&0&1\\0&0&0\\ 1&0&0\end{pmatrix} $$ and the generator comes from translation is $$ K=\begin{pmatrix}0&0&i\\0&0&0\\ i&0&0\end{pmatrix} $$ Apologies for poor notation and terminology. But I think I need to have $\frac{n(n-1)}{2}$ generator right? If yes I have one more but how can I get it? And how can I construct Lie Algebra then?
NOTE:I also posted this question on Math.SE but I'm afraid here is more proper. I am not looking for a way of any other isomorphic groups to construct this LG. I have not much knowledge about homomorphism, isomorphism etc.. Thanks.

Answer 1

$SO(1,\,2)$, with two spatial dimensions, only has rotations in one plane - that of all space. So if our co-ordinates are $(t,\,x,\,y)$ (spatial co-ords last), its unique (up to a real scale factor of course) rotational Lie algebra generator must be:

$$R=\left(\begin{array}{ccc}0&0&0\\0&0&-1\\0&1&0\end{array}\right)$$

where we've put the unit $I=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$ which exponentiates to the 2D rotation $e^{I\,\theta}=\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)$ in the lower right corner.

Likewise, we put the generator $K=\left(\begin{array}{cc}0&+1\\+1&0\end{array}\right)$ of 1D boosts in the $x$ direction, which exponentiates to $e^{\eta\,K}=\left(\begin{array}{cc}\cosh\eta&\sinh\eta\\\sinh\eta&\cosh\eta\end{array}\right)$ in the upper right corner to get the generator of boosts in the $x$ direction:

$$B_x=\left(\begin{array}{ccc}0&1&0\\1&0&0\\0&0&0\end{array}\right)$$

Do the same for boosts in the $y$ direction in the first / third rows/columns which correspond to the time and y co-ordinates:

$$B_y=\left(\begin{array}{ccc}0&0&1\\0&0&0\\1&0&0\end{array}\right)$$

and then readily show that the linear space $\mathfrak{l}=\langle\{B_x,\,B_y,\,R\}\rangle$ spanned by $\{B_x,\,B_y,\,R\}$ is closed under the Lie bracket, with bracket relationships:

$$[B_x,\,B_y]=-R;\quad[B_x,\,R]=-Ly;\quad[B_y,\,R]=+Lx$$

Therefore, under the Lie correspondence, $\mathfrak{L}=\bigcup\limits_{k=1}^\infty\,\exp(\mathfrak{l})^k$ is the identity connected component $SO^+(1,\,2)$ of $SO(1,\,2)$: the Lie group of all boosts and rotations in the $X\wedge Y$ plane.

The generator of boosts in a direction at angle $\theta$ relative to the $x$ axis is:

$$B_\theta = \mathrm{Ad}(e^{I\,\theta})\,B_x = e^{I\,\theta}\,B_x\,e^{-I\,\theta} = \left(\begin{array}{ccc}0&\cos\theta&\sin\theta\\\cos\theta&0&0\\\sin\theta&0&0\end{array}\right)$$

exponentiating to the general boost $\beta(\theta\,\eta) = \exp(\eta\,B_\theta)$, which works out to:

$$\beta(\theta,\,\eta)=\left( \begin{array}{ccc} \cosh (\eta ) & \cos (\theta ) \sinh (\eta ) & \sin (\theta ) \sinh (\eta ) \\ \cos (\theta ) \sinh (\eta ) & \cosh (\eta ) \cos ^2(\theta )+\sin ^2(\theta ) & \cos (\theta ) (\cosh (\eta )-1) \sin (\theta ) \\ \sin (\theta ) \sinh (\eta ) & \cos (\theta ) (\cosh (\eta )-1) \sin (\theta ) & \cos ^2(\theta )+\cosh (\eta ) \sin ^2(\theta ) \\ \end{array} \right)$$

Alternatively, you can write a general boost as $\exp(\eta_x\,B_x+\eta_y\,B_y)$ with the result:

$$\exp(\eta_x\,B_x+\eta_y\,B_y)=\left( \begin{array}{ccc} \cosh \left(\sqrt{\eta_x^2+\eta_y^2}\right) & \frac{\eta_x \sinh \left(\sqrt{\eta_x^2+\eta_y^2}\right)}{\sqrt{\eta_x^2+\eta_y^2}} & \frac{\eta_y \sinh \left(\sqrt{\eta_x^2+\eta_y^2}\right)}{\sqrt{\eta_x^2+\eta_y^2}} \\ \frac{\eta_x \sinh \left(\sqrt{\eta_x^2+\eta_y^2}\right)}{\sqrt{\eta_x^2+\eta_y^2}} & \frac{\cosh \left(\sqrt{\eta_x^2+\eta_y^2}\right) \eta_x^2+\eta_y^2}{\eta_x^2+\eta_y^2} & \frac{\eta_x \eta_y \left(\cosh \left(\sqrt{\eta_x^2+\eta_y^2}\right)-1\right)}{\eta_x^2+\eta_y^2} \\ \frac{\eta_y \sinh \left(\sqrt{\eta_x^2+\eta_y^2}\right)}{\sqrt{\eta_x^2+\eta_y^2}} & \frac{\eta_x \eta_y \left(\cosh \left(\sqrt{\eta_x^2+\eta_y^2}\right)-1\right)}{\eta_x^2+\eta_y^2} & \frac{\eta_x^2+\eta_y^2 \cosh \left(\sqrt{\eta_x^2+\eta_y^2}\right)}{\eta_x^2+\eta_y^2} \\ \end{array} \right)$$

Unlike the $SO(1,\,3)$, which has two connected components, there are no unity determinant matrices which conserve the pseudonorm $t^2 - x^2 -y^2$ which are outside $SO^+(1,\,2)$. So in fact $SO^+(1,\,2)\cong SO(1,\,2)$ whereas $SO^+(1,\,3)\not\cong SO(1,\,3)$.

Witness the following Mathematica session, to help you with these calculations:

In[1]:= Bx = {{0, 1, 0}, {1, 0, 0}, {0, 0, 0}};
In[2]:= By = {{0, 0, 1}, {0, 0, 0}, {1, 0, 0}};
In[3]:= R = {{0, 0, 0}, {0, 0, -1}, {0, 1, 0}};

In[4]:= Lie[X_, Y_] := X.Y - Y.X

In[5]:= Lie[Bx, By] + R
Out[5]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}

In[6]:= Lie[Bx, R] + By
Out[6]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}

In[7]:= Lie[By, R] - Bx
Out[7]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}

In[8]:= MatrixExp[theta R].Bx.MatrixExp[-theta R] // Simplify
Out[8]= {{0, Cos[theta], Sin[theta]}, {Cos[theta], 0, 0}, {Sin[theta],0, 0}}

In[9]:= ExpToTrig[MatrixExp[eta MatrixExp[theta R].Bx.MatrixExp[-theta R]]] // Simplify

Out[9]= {{Cosh[eta], Cos[theta] Sinh[eta], 
  Sin[theta] Sinh[eta]}, {Cos[theta] Sinh[eta], 
  Cos[theta]^2 Cosh[eta] + Sin[theta]^2, 
  Cos[theta] (-1 + Cosh[eta]) Sin[theta]}, {Sin[theta] Sinh[eta], 
  Cos[theta] (-1 + Cosh[eta]) Sin[theta], 
  Cos[theta]^2 + Cosh[eta] Sin[theta]^2}}

In[10]:= ExpToTrig[MatrixExp[etax Bx + etay By]] // Simplify

Out[10]= {{Cosh[Sqrt[etax^2 + etay^2]], (
  etax Sinh[Sqrt[etax^2 + etay^2]])/Sqrt[etax^2 + etay^2], (
  etay Sinh[Sqrt[etax^2 + etay^2]])/Sqrt[etax^2 + etay^2]}, {(
  etax Sinh[Sqrt[etax^2 + etay^2]])/Sqrt[etax^2 + etay^2], (
  etay^2 + etax^2 Cosh[Sqrt[etax^2 + etay^2]])/(etax^2 + etay^2), (
  etax etay (-1 + Cosh[Sqrt[etax^2 + etay^2]]))/(etax^2 + etay^2)}, {(
  etay Sinh[Sqrt[etax^2 + etay^2]])/Sqrt[etax^2 + etay^2], (
  etax etay (-1 + Cosh[Sqrt[etax^2 + etay^2]]))/(etax^2 + etay^2), (
  etax^2 + etay^2 Cosh[Sqrt[etax^2 + etay^2]])/(etax^2 + etay^2)}}

Answer 2

You should have two boost generators. You have constructed one for boost in the $x$ direction, but there is also one for boost in $y$.