1. I need the complete proof for commutation relation of the Lorentz group generators.

  2. The proof of Lorentz algebra using this commutation relation.


The Lorentz group is the group of matrices that conserve the quadratic form:

$$\mathscr{Q}(X,\,Y) = X^T\,\eta\,Y\tag{1}$$

where here $X$ and $Y$ are $1\times 4$ column vectors, the $4\times 4$ group member matrices act on these from the left and $\eta$ is the Minkowski (pseuso) metric. Therefore, $\Lambda\in O(1,\,3)$ if and only if:

$$\mathscr{Q}(\Lambda\,X,\,\Lambda\,Y) = X^T\,\Lambda^T\,\eta\,\Lambda\,Y=X^T\,\eta\,Y;\;\forall\,X,\,Y\in\mathbb{R}^{1+3}\tag{2}$$

Exercise 1: Show that (2) can only hold for for any $\Lambda(s) = \exp(s\, L)\in O(1,\,3)\,\forall\,s\in\mathbb{R}$ for a given member $L$ of the Lie algebra of $O(1,\,3)$ (here written as a $4\times 4$ matrix) if and only if:

$$L^T\,\eta + \eta\,L = 0\tag{3}$$

This then is the defining relationship you must use for the Lie algebra members.

Exercise 2: Show that $L$ fulfills (3) if and only if $L$ is of the form:

$$L = \left(\begin{array}{cccc}0&\eta_x&\eta_y&\eta_z\\\eta_x&0&-\theta_z&\theta_y\\\eta_y&\theta_z&0&-\theta_x\\\eta_z&-\theta_y&\theta_x&0\end{array}\right)\tag{4}$$

where all the $\eta_j,\theta_j\in\mathbb{R}$. The Mathematica code:

In[1]:= L={{L11,L12,L13,L14},{L21,L22,L23,L24},{L31,L32,L33,L44},{L41,L42,L43,L44}};
In[2]:= Solve[Transpose[L].DiagonalMatrix[{1,-1,-1,-1}]+DiagonalMatrix[{1,-1,-1,-1}].L==0,Flatten[L]]
Out[2]= {{L11->0,L21->L12,L22->0,L31->L13,L32->-L23,L33->0,L44->0,L41->L14,L42->-L24,L43->0}}

will help here.

Thus you can write a general Lie algebra member as the following superposition of the six linearly independent matrices:

$$L = \sum\limits_{i\in\{x,\,y,\,z\}}(\eta_i\, K_i +\theta_i\,J_i)\tag{5}$$

where the $J_i$ are the "infinitessimal" rotations and span the Lie algebra of the rotation group $SO(3)$ and the $K_i$ are the "infinitessimal" boosts.

Exercise 3: Find the required commutation relationships using the definitions of $J_i,\,K_i$ that follow from (5).

An interesting aside:

Exercise: Using bilinearity of the form in (1), and considering each of the entities $\mathscr{Q}(X,\,X)$, $\mathscr{Q}(Y,\,Y)$, ,$\mathscr{Q}(X+Y,\,X+Y)$ and $\mathscr{Q}(X,\,Y)$ show that the condition (2) holds if and only if the seemingly weaker condition:

$$\mathscr{Q}(\Lambda\,X,\,\Lambda\,X) = X^T\,\Lambda^T\,\eta\,\Lambda\,X=X^T\,\eta\,X;\;\forall\,X\in\mathbb{R}^{1+3}\tag{6}$$


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