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The generators of the translation group $T(4)$ are given below:

$P_0 \equiv -i \begin{pmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix};$ $P_1 \equiv -i \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix};$ $P_2 \equiv -i \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix};$ $P_3 \equiv -i \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix};$

with $P_\mu = g_{\mu\nu}P^\nu$ with metric sign convention: $(+,-,-,-)$.

Is it correct to use the contravariant notation for the generators initially ?

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    $\begingroup$ Shouldn't matter what you start with. $\endgroup$ Commented Jul 22, 2014 at 0:11

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The generators are covariant vectors in space-time. Following Ome, in order to represent the translation generators by matrices, space-time is a 4-d projective space where points are rays $x^{i}\in V_{5}$ with $i=0,1,2,3,4$. Suppose Alice's coords are $x^{i}$ and Bob's are $x'^{i}$ and Alice is boosted along Bob's positive x axis with a small boost parameter $\eta$. The boost is, $$ x'^{0}=x^{0}+\eta x^{1}\\ x'^{1}=x^{1}+\eta x^{0} $$ which implies that the boost is the Lie algebra element, $$ K^{i}_{\ j}=\delta^{i}_{0}\delta^{1}_{j}+\delta^{i}_{1}\delta^{0}_{j} $$ because, $$ x'^{i}=x^{i}+\eta K^{i}_{\ j}x^{j} \ . $$ Ome's translations are, $$ [P_{k}]^{i}_{\ j}=\delta^{i}_{k}\delta^{4}_{j} $$ where the $(-i)$ factor has been omitted because everything is classical at present. Using the matrix commutator for the Lie bracket, $$ [P_{1},K]=-P_{0} $$ The response of $P_{1}$ to the boost is, $$ P'_{1}=P_{1}+\eta[P_{1},K]=P_{1}-\eta P_{0} \ . $$ Now compare this equation with the response of the contravariant space-time coordinate $x^{1}$ in the second equation at the top; when we think of $P_{1}$ as a vector in space-time, it is not transforming as a contravariant vector. It's easy to see that it is transforming as a covariant vector in $V_{5}$. So, the translation generators are covariant vectors in space-time $P_{i}\in\tilde{V}_{5}$.

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  • $\begingroup$ Superb answer. I started treating it as the contravariant and messed up. $\endgroup$ Commented Aug 5, 2014 at 15:57
  • $\begingroup$ Thanks, when you posted the question, I was confused about this, having also assumed the translation generators were contravariant. $\endgroup$ Commented Aug 5, 2014 at 19:35

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