The generators ##\{ L^1, L^2 , L^3 , K^1 , K^2 , K^3 \}## of the Lorentz group satisfy the Lie algebra:(adsbygoogle = window.adsbygoogle || []).push({});

\begin{array}{l}

[L^i , L^j] = \epsilon^{ij}_{\;\; k} L^k \\

[L^i , K^j] = \epsilon^{ij}_{\;\; k} K^k \\

[K^i , K^j] = \epsilon^{ij}_{\;\; k} L^k

\end{array}

It has the Casimirs

[tex]

C_1 = \sum_i (K^i K^i - L^i L^i) , \qquad C_2 = \sum_i K^i L^i

[/tex]

I wish to prove that the Casimirs commute with all of the generators of the Lie algebra. It is easy to prove that ##[C_1 , L^j] = 0##, ##[C_2 , L^j]## and ##[C_2 , K^j] = 0##. However, I'm having more trouble proving ##[C_1 , K^j] = 0##. What I obtain, for example for ##j=1##, is:

[tex]

[C_1 , K^1] = 2 [L^2 , K^3]_+ - 2 [L^3 , K^2]_+

[/tex]

where ##[\cdot , \cdot]_+## is the anti-commutator. I'm not sure how to prove directly that this vanishes. However, there may be an indirect way of proving ##[C_1 , K^1] = 0##. First note:

\begin{array}{l}

[C_1 , C_2] = \sum_i [K^i K^i - L^i L^i , C_2] \\

\sum_i (K^i [K^i , C_2] + [K^i , C_2] K^i - L^i [L^i , C_2] - [L^i , C_2] L^i) \\

= 0

\end{array}

where we have used ##[C_2 , L^j]## and ##[C_2 , K^j] = 0##. Next write

\begin{array}{l}

0 = [C_1 , C_2] \\

= \sum_i [C_1 , K^i L^i] \\

= \sum_i ([C_1 , K^i] L^i + K^i [C_1 , L^i]) \\

= \sum_i [C_1 , K^i] L^i .

\end{array}

where we have used ##[C_1 , L^j] = 0##.

Is it possible to use this to prove ##[C_1 , K^1] = 0##?

I would prefer to prove first that the Casimirs commutate with all the generators first and then conclude the two Casimirs commute, but if this is what I have to resort to...

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# I Commutator between Casimirs and generators for Lorentz group

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