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The generators of the Poincare group $P(1;3)$ are supposed to obey the following commutation relation to be verified:

$$\left[ M^{\mu\nu}, P^{\rho} \right] = i \left(g^{\nu\rho} P^{\mu} - g^{\mu\rho} P^{\nu} \right)$$

where $M^{\mu\nu}$ are the 6 generators of the Lorentz group and $P^\mu$ are the 4 generators of the four-dimensional translation group $T(4)$.

For $\mu = 3, \nu=1, \rho=0$ the LHS becomes: $ [M^{31},P^{0}] = M^{31}P^{0} - P^{0}M^{31}$.

Here $M^{31} = J^2 = -J_2= \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 \end{pmatrix}$ and $ P^0 = P_0 = -i \begin{pmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$.

My question is that how can I multiply $M^{31}$ and $P^0$ when they are $4\times4$ and $5\times 5$ matrices respectively?

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    $\begingroup$ In fact you need to complete the $M_{ab}$ generators with zeros to complete the 5th row and column. $\endgroup$
    – Dox
    Jul 22, 2014 at 17:35
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/127559/2451 $\endgroup$
    – Qmechanic ♦
    Jul 22, 2014 at 17:41
  • $\begingroup$ @Dox: I have considered: $(x^0, x^1, x^2, x^3) \rightarrow (x^0 + b^0, x^1+b^1, x^2+b^2, x^3+b^3)$. Then cast this transformation as a matrix multiplication and then used the formula to find the generators $P_\mu=g_{\mu\nu}P^\nu$. What else I could do to define the $P_\mu$ properly? $\endgroup$
    – rainman
    Jul 22, 2014 at 17:45
  • $\begingroup$ @Qmechanic: yes, I asked that question. I was in deep confusion! $\endgroup$
    – rainman
    Jul 22, 2014 at 17:51

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Consider $$M_{31} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \text{ and } P_0 = -i \begin{pmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix},$$

Then the commutator vanishes! As expected from $\left[ M_{31}, P_{0} \right] = i \left(g_{10} P_{3} - g_{30} P_{1} \right) = 0$.

If you take $$M_{01} = \begin{pmatrix} 0 & i & 0 & 0 & 0 \\ i & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \text{ and } P_0 = -i \begin{pmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix},$$ then $$ \left[M_{01},P_0\right] = -i \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} = P_1. $$

And so on!

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