# Casimir Invariants of the Galilean group

I had studied a couple of things about Galilean and Poincare group. But in the Galilean group, there is not enough clarity on how to calculate generators for boosts ($B_i$), which if I do it seems I should be able to obtain Mass ($M$) as a Casimir Invariant. $$[B_i,P_j] = iM\delta_{ij}$$

One attempt at a (scalar) representation of boosts are : $$B_i = v_i\delta t\frac{\partial}{\partial x^i}$$ But with this, how am I supposed to arrive at the commutators like the one above and also, $$[B_i,L_j] = i\epsilon_{ij}^{\:\:k}B_k$$

I am also interested in understanding how to find the Casimir Invariants of a given Lie algebra in general.

• Related: physics.stackexchange.com/q/63394/2451 and links therein. – Qmechanic Mar 14 '14 at 10:17
• But does it have any reference to this particular Galilean group and obtaining the commutator with mass $M$. – user35952 Mar 14 '14 at 10:19
• If you are interested in specific groups, it is recommended that you change the title accordingly. – Qmechanic Mar 14 '14 at 10:23
• The correct definition of the boost is vector, not scalar, $B_i= t {\partial\over \partial x^i}$, and then it obviously must transform like a vector under rotations. $v_i$ is just the vector of group parameters dotted onto these vector generators and exponentiated to produce generic group elements. – Cosmas Zachos Mar 17 '16 at 2:28

Your answer is not clear about the lack of clarity of the Galilean group WP article: it provides you with the centrally extended Galilean group Lie algebra, the Bargmann algebra, and reassures you M is central, i.e. it is the eleventh generator introduced to extend the configuration space Galilean algebra of this WP article to this Bargmann algebra which you are referring to and you cited in the WP article.

So, given just this lie algebra and the invariants of this algebra, M, as postulated in it; the mass-shell invariant $ME- P^2/ 2$; and $\vec{W}\cdot\vec{W}$ where $\vec{W} \equiv M \vec{L} + \vec{P}\times\vec{C}$, you should be able to work the commutators of the two later invariants with all generators, to see they both commute with all 11 Lie algebra elements. It is a brute force calculation. You note the last invariant, unlike the penultimate, is not purely quadratic in the generators, at least superficially.

Because of the structure of the Galilean algebra, it is not obvious how to construct invariants of it, but trial and error has worked miracles here. All you need to do is check commutativity with all generators.

If you were asking how M entered the picture beyond the 10-dim matrix algebra in the WP article I'm quoting, there have been complete discussions of the real subtleties of how this is achieved in well-answered questions, such as 10442 or 12341 or this one, 104216.

As for your question of quick candidates for Casimirs, in general, recall that, for the vanilla classical Lie algebras, there are as many independent Casimirs, quadratic, cubic, etc..., as the dimension of their root systems, or, equivalently, their rank (the dimension of their Cartan subalgebra}, introduced in every good Lie group theory book or review.

• Hi C., do you know where I can find the Casimirs of the homogeneous Galileo Group? Thanks! – AccidentalFourierTransform Feb 24 '17 at 11:50
• Gilmore's book has everything. I don't want to mislead you, but, superficially, if you just keep the 3 Ls and the 3 Cs, the latter behave just like the Ps, algebraically (!); so then you have E(3) ~ ISO(3), and the Casimirs are L·L and C·C , no? – Cosmas Zachos Feb 24 '17 at 15:13
• That's (more or less) what I thought, but I wanted to be sure. I'll check Gilmore for further details. Again, thank you :-) – AccidentalFourierTransform Feb 24 '17 at 15:20