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What is the relationship between the Galilean group and the Poincaré group? Are they siblings within the Lie group? Or does the Poincaré group contain the Galilean group as a subgroup?

I'm not so much interested in the Galilean group being the limit of the Poincaré group for c -> inf.

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    $\begingroup$ The Galilean group is the group contraction of the Poincaré, not a subgroup. Why do you declare you are not interested in the only meaningful answer to your question? $\endgroup$ Sep 10 at 16:20
  • $\begingroup$ Thanks. Because I already knew the c to infinity relationship. Would you call those groups siblings? $\endgroup$ Sep 10 at 19:34
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    $\begingroup$ What is "Are they siblings within the Lie group" supposed to mean? What is "the Lie group"? What does it mean to be siblings? $\endgroup$
    – ACuriousMind
    Sep 10 at 22:08
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    $\begingroup$ Thanks Cosmas. If we take the open segment (0, 1) and 1 as analogy, the answer seem to suggest that because 0.99999... = 1, 1 is the amputated remnant of (0, 1). While (0, 1) and 1 are clearly distinct. Neither could 1 be considered a child of (0, 1). Is this a fair assessment? $\endgroup$ Sep 11 at 13:08
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    $\begingroup$ Your "amputated remnant" paradigm is fair and sound. I don't know how to unify the two in a common group structure. How one names this structure in kinship terms is likely subjective. Mathematicians call it group contraction, and physicists Wigner- Inönü contraction. $\endgroup$ Sep 11 at 14:37
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It is assumed you have appreciated Inönü, E.; Wigner, E. P. (1953), "On the Contraction of Groups and Their Representations" Proc. Natl. Acad. Sci. 39 (6): 510–24, and the super-helpful Gilmore text in Group contraction.

Very crudely, the Poincaré Lie algebra, $$ [J_m,P_n] = i \epsilon_{mnk} P_k ~, \qquad [J_i, P_0] = 0 ~, \\ [K_i,P_k] = i \delta_{ik} P_0 ~, \\ [P_0,P_i]=0 \qquad [P_i,P_j]=0 \qquad [K_i, P_0] = -i P_i ~, \\ [J_m,J_n] = i \epsilon_{mnk} J_k ~, \qquad [J_m,K_n] = i \epsilon_{mnk} K_k ~, \\ [K_m,K_n] = -i \epsilon_{mnk} J_k ~, $$ given relabelings $E=-cP_0$ and $K_i=cC_i$ contracts upon $c\to \infty$ to the Galilean algebra, $$ [J_m,P_n] = i \epsilon_{mnk} P_k ~, \qquad [J_i,E]=0 \\ [C_i,P_j]= 0,~\\ [E,P_i]=0, \qquad [P_i,P_j]=0, \qquad [C_i,E]=i P_i \\ [J_m,J_n] = i \epsilon_{mnk} J_k ~, \qquad [J_m,C_n] = i \epsilon_{mnk} C_k ~, \\ [C_i,C_j]=0 . $$

There are a few subtleties and wrinkles, extensions, to be sure, which I gather you are not focussing on, but, crudely, the third and the last commutation relations trivialized/collapsed. (There is more, but I am oversimplifying...). This collapse/amputation is the Lie algebraic manifestation of a group contraction.

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    $\begingroup$ Thanks Cosmas. I will mark your answer as the solution. Could you take a look at my comment above? $\endgroup$ Sep 11 at 14:30

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