# Can we write the mass $M$, a Casimir invariant of the Galilean group, as a function of its generators?

According to Wikipedia, the mass $$M$$ is one of the Casimir invariants of the Galilean group. Casimir invariants of a group are made out of the generators, and they commute with all the generators of the group. For example, the Casimir invariant of the group $$SU(2)$$ is $$J^2$$ which is made out of $$J_1, J_2, J_3$$ as $$J^2:=J_1^2+J_2^2+J_3^2.\tag{1}$$ Another example is a Casimir invariant of the Poincare group $$P^\mu P_\mu$$ which is made out of $$P_0, P_1,P_2, P_2$$ as $$P^\mu P_\mu:=-P_0^2+P_1^2+P_2^2+P_3^2.\tag{2}$$

In the same manner, can we write $$M$$ as a function of the generators Galilean group?

• It is not qudratic. Jul 26, 2021 at 7:04
• @DanielC Can you elaborate? I am asking how to write M using the generators of the Galilean group. Jul 26, 2021 at 7:54

1. No, the mass operator $$M$$ is the central charge operator in the central extension [known as the Bargmann algebra (BA)] of the Galilean algebra (GA).
2. In other words, the mass operator $$M$$ is by definition not part of the GA. [The momenta and Galilean boosts commute by definition within the GA.]
3. If we define a Casimir invariant of a Lie algebra as a central element of its universal enveloping algebra (UEA), then the mass operator $$M$$ is a Casimir invariant for the BA but not for the GA.