Here $a$, $b$,$c$, $d$ $u$, $v$ range from 0 to 4 and the metric $g^{ab}=\text{diag}(-1,1,1,1)$.

16 $4 \times 4$ matrices $M^{ab}$ are defined as follows: \begin{equation*} (M^{ab})_{uv} = -i(\delta^a_u\delta^b_v-\delta^b_u\delta^a_v) \end{equation*}

Then equivalently, \begin{equation*} (M^{ab})^{u}\;_{v} = -i(g^{ua}\delta^b_v-g^{ub}\delta^a_v) \end{equation*}

What is the meaning of these matrices? And these matrices are said to have the following commutation relation: \begin{equation*} [M^{ab}, M^{cd}]=i(g^{ac}M^{bd}-g^{ad}M^{bc}-g^{bc}M^{ad}+g^{bd}M^{ac}) \end{equation*}

How can I efficiently prove this relation? All I can think of is to compare the both side by components. Could anyone please help me with proving this relation?


The matrices $M^{ab}$ span a 4 dimensional representation of the Lorentz algebra $so(1,3)$. The commutation relation you want to prove is the defining property of the Lorentz algebra. This means that you can use these matrices to generate Lorentz transformations via exponantiation.

To prove this relation, I'm afraid you have to plug in the definitions and fight yourself through the indices.

  • $\begingroup$ You mean I have to compare the both sides by components as I wrote in the question? $\endgroup$ – Keith Mar 21 '18 at 8:29
  • $\begingroup$ I think $(M^{ab})^u_v $ expression would be more convenient for this. Right? $\endgroup$ – Keith Mar 21 '18 at 8:31
  • $\begingroup$ Yes, to both questions. $\endgroup$ – fermionator Mar 21 '18 at 8:47

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