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.
 A: 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. 
A: First of all, the algebra appears in your question is in fact the central extension of the Galilean algebra.
The Galilean group of three dimensional Euclidean space, denoted by $\mathrm{Gal}_{3}$, takes the form $$(s,\vec{a},\vec{v},R)\rightarrow\begin{pmatrix}
R & \vec{v} & \vec{a} \\
0 & 1 & s \\
0 & 0 & 1
\end{pmatrix}$$
where $R\in O(3)$, $\vec{v}\in\mathbb{R}^{3}$ and $\vec{a}\in\mathbb{R}^{3}$. It can be viewed as a $\rho:\mathrm{Gal}(3)\rightarrow\mathrm{GL}(5,\mathbb{R})$ representation, where $\vec{v}$ and $\vec{a}$ are both column vectors.
Its Lie algebra, denoted by $\mathfrak{gal}_{3}$, takes the form $$(\epsilon,\vec{b},\vec{u},X)=\begin{pmatrix}
X & \vec{u} & \vec{b} \\
0 & 0 & \epsilon \\
0 & 0 & 0
\end{pmatrix}$$
The commutation relations of are
$$[H,J^{i}]=0\quad [H,P^{i}]=0\quad [H,K^{i}]=-P^{i} \tag{1.a}$$
$$[J^{i},J^{j}]=\sum_{k=1}^{3}\epsilon^{ijk}J^{k}\quad [J_{i},K_{j}]=\sum_{k=1}^{3}\epsilon^{ijk}K^{k}\quad[J^{i},P^{j}]=\sum_{k=1}^{3}\epsilon^{ijk}P^{k} \tag{1.b}$$
$$[K^{i},P^{j}]=0\quad[K^{i},K^{j}]=0 \tag{1.c}$$
$$[P^{i},P^{j}]=0, \tag{1.d}$$
where $H$ generates time translation, $\vec{P}$ generates spatial translation, $\vec{J}$ generates rotation, and $\vec{K}$ generates the Galilean boost. Since this is not a semi-simple Lie algebra, one cannot expect to find its Killing form, and there is no standard way to find its Casimir invariants.
Remember that for a given Lie algebra $\mathfrak{g}$, its Casimir invariant is defined as follows:

Definition: Let $\mathfrak{g}$ be a Lie algebra. The direct sum of all possible tensorial powers, $$\bigotimes(\mathfrak{g})\equiv\bigoplus_{p=0}^{\infty}\otimes^{p}\mathfrak{g},$$
known as the tensor algebra of $\mathfrak{g}$, contains a two-sided ideal generated by $$x\otimes y-y\otimes x-[x,y],$$
where $x$, $y\in\mathfrak{g}$. Then the quotient algebra $$\mathcal{U}(\mathfrak{g})\equiv\bigotimes(\mathfrak{g})/\langle x\otimes y-y\otimes x-[x,y]\rangle$$
is called the universal enveloping algebra of $\mathfrak{g}$. Then, elements in the center $Z(\mathcal{U}(\mathfrak{g}))$ are called the Casimir invariants of $\mathfrak{g}$.

In general, a Casimir operator $C$ of an $n$-dimensional Lie algebra $\mathfrak{g}$ is said to be of order $m$ if it can be expressed as a polynomial of degree $m$ in the basis of $\mathfrak{g}$. i.e $$C=\sum_{k_{1},\cdots,k_{n}\geq 0}^{k_{1}+\cdots+k_{n}\leq m}f_{k_{1}\cdots k_{n}}X_{1}^{k_{1}}\cdots X_{n}^{k_{n}}, \tag{$\star$}$$
where $\left\{X_{i}\right\}_{i=1,\cdots,n}$ is a basis of $\mathfrak{g}$.
For the Galilean algebra $\mathfrak{gal}_{3}$, one can seek for help from the quasi-regular representation of $\mathrm{Gal}_{3}$, which is given by $$\Gamma(s,\vec{a},\vec{v},R)f(t,\vec{x})=f(t-s,R^{-1}(\vec{x}-\vec{v}t-\vec{a}+s\vec{v})),$$
where $f$ is an square-integrable function. Then, it's easy to attain the following representation of the Galilean algebra:
$$H=\frac{\partial}{\partial t},\quad P_{i}=\frac{\partial}{\partial x^{i}},\quad J_{i}=\sum_{k=1}^{3}\epsilon_{ijk}x_{j}\frac{\partial}{\partial x^{k}},\quad K_{i}=t\frac{\partial}{\partial x^{i}}.$$
From the above differential operators, one can easily check that their commutation relations indeed satisfy equations (1.a) (1.b) (1.c) and (1.d). Now it should be clear that the Galilean algebra $\mathfrak{gal}_{3}$ has only one Casimir invariant, which is $$C=\|\vec{P}\|^{2}.$$
Also notice that in the Galilean algebra, $\vec{K}\times\vec{P}=0$, which is a trivial element in the center of its universal enveloping algebra.
Next, consider the central extension of the Galilean algebra.

Definition: Let $G$ be a group and $A$ be an Abelian group. A group $E$ is called a central extension of $G$ by $A$ if the following diagram is a short exact sequence: $$1\xrightarrow{}A\overset{\iota}{\hookrightarrow}E\overset{\pi}{\twoheadrightarrow}G\xrightarrow{}1, \tag{2.a}$$
where image of $\iota$ is contained in the center of $E$, i.e. $\text{im}(\iota)\subseteq Z(E)$. 
Definition: Let $\mathfrak{a}$ be an Abelian Lie algebra, and $\mathfrak{g}$ a Lie algebra, the Lie algebra $\mathfrak{e}$ is a central extension of $\mathfrak{g}$ by $\mathfrak{a}$ if the following Lie algebra homomorphisms are a short exact sequence: $$0\xrightarrow{}\mathfrak{a}\overset{\iota}{\hookrightarrow}\mathfrak{e}\overset{\pi}{\twoheadrightarrow}\mathfrak{g}\xrightarrow{}0, \tag{2.b}$$
so that $[\mathfrak{a},\mathfrak{e}]=0$, and $\mathfrak{g}\simeq\mathfrak{g}/\mathfrak{a}$.

It is known in group cohomology that central extensions (2.a) of $G$ by $A$ are classified by the second cohomology group $\mathrm{H}^{2}(G,A)$. Similarly, the second cohomology group $\mathrm{H}^{2}(\mathfrak{g},\mathfrak{a})$ classifies central extensions (2.b) of $\mathfrak{g}$ by $\mathfrak{a}$.
As for the case of the Galilean group and its Lie algebra, it is shown in Lie Groups Lie Algebras Cohomology and some Applications in Physics by Jose A. De Azcarraga that the only possible central extensions of $\mathfrak{gal}_{3}$ are as follows
$$[H,J_{i}]=0\quad [H,P_{i}]=0\quad [H,K_{i}]=-P_{i} \tag{3.a}$$
$$[J_{i},J_{j}]=\sum_{k=1}^{3}\epsilon_{ijk}J_{k}\quad [J_{i},K_{j}]=\sum_{k=1}^{3}\epsilon_{ijk}K_{k}\quad[J_{i},P_{j}]=\sum_{k=1}^{2}\epsilon_{ijk}P_{k} \tag{3.b}$$
$$[K_{i},P_{j}]=m\delta_{ij}\quad[K_{i},K_{j}]=0 \tag{3.c}$$
$$[P_{i},P_{j}]=0, \tag{3.d}$$
where $m\in\mathbb{R}$.
Comparing them with (1.a) (1.b) (1.c) and (1.d), one finds that there appears an anomaly which is $[K_{i},P_{j}]=m\delta_{ij}$. This paramter is known as the mass of the particle in Newtonian mechanics. It parameterizes central extensions of $\mathfrak{gal}_{3}$, which will be denoted as $\mathfrak{gal}_{3}(m)$ in the following. In terms of Lie algebra coholomogy, it suggests that $$\mathrm{H}^{2}(\mathfrak{gal}_{3},\mathbb{R})\simeq\mathbb{R}. \tag{4}$$
Also, notice that the anomalous commutator $[K_{i},P_{j}]=m\delta_{ij}$ indicates that there is no finite dimensional representation of $\mathfrak{gal}_{3}(m)$, because otherwise $\mathrm{tr}\left([K_{i},P_{j}]\right)=0=3m$ which is absurd when $m\neq 0$. This also suggests that, in general, $\vec{K}\times\vec{P}$ may not vanish in the extended algebra $\mathfrak{gal}_{3}(m)$.
To find the Casimir invariants of $\mathfrak{gal}_{3}(m)$, one can study the Hamiltonian mechanics of a classical Newtonian particle, i.e $$S[\vec{x}]=\int dt L(\vec{x},\dot{\vec{x}},t)=\frac{m}{2}\int dt|\dot{\vec{x}}(t)|^{2}. \tag{5}$$
It is a mathematical theorem that a non-vanishing second cohomology classes of $\mathrm{Gal}_{3}$ exhibits a topological obstruction to lifting its projective representation to a linear representation. The appearance of anomaly (4) suggests that the Lagrangian in (5) may not be totally invariant under a generic Galilean transformation. Details are explained here. One can easily check that under a Galilean boost, the Lagrangian in (5) changes by a total time derivative, which still makes the action invariant. By using Noether's first theorem, one obtains the following conserved charges associated with Galilean transformations:
$$\vec{p}=\frac{\partial L}{\partial\dot{\vec{x}}}=m\dot{\vec{x}} \tag{6.a}$$
$$E=\vec{p}\cdot\dot{\vec{x}}-L=\frac{m}{2}|\dot{\vec{x}}|^{2}=\frac{|\vec{p}|^{2}}{2m} \tag{6.b}$$
$$\vec{l}=m\vec{x}\times\dot{\vec{x}}=\vec{x}\times\vec{p} \tag{6.c}$$
$$\vec{g}=m(\vec{x}-t\dot{\vec{x}})=m\vec{x}-t\vec{p}, \tag{6.d}$$
where $\vec{p}$, $E$, $\vec{l}$ and $\vec{g}$ are conserved charges generated by spatial translation $\vec{P}$, time translation $H$, rotation $\vec{J}$, and Galilean boost $\vec{K}$ in (1.a), (1.b), (1.c), and (1.d), respectively.
In the Hamiltonian formalism, where on the phase space the Poisson bracket $\left\{x_{i},p_{j}\right\}_{PB}=\delta_{ij}$ is defined, one can easily check that the Poisson brackets of above Noether charges are a canonical realization of the Lie algebra $\mathfrak{gal}_{3}(m)$, i.e
$$\left\{E,l_{i}\right\}_{PB}=0\quad \left\{E,p_{i}\right\}_{PB}=0\quad \left\{E,g_{i}\right\}_{PB}=-p_{i} \tag{7.a}$$
$$\left\{l_{i},l_{j}\right\}_{PB}=\sum_{k=1}^{3}\epsilon_{ijk}l_{k}\quad \left\{l_{i},g_{j}\right\}_{PB}=\sum_{k=1}^{3}\epsilon_{ijk}g_{k}\quad\left\{l_{i},p_{j}\right\}_{PB}=\sum_{k=1}^{2}\epsilon_{ijk}p_{k} \tag{7.b}$$
$$\left\{g_{i},p_{j}\right\}_{PB}=m\delta_{ij}\quad\left\{g_{i},g_{j}\right\}_{PB}=0 \tag{7.c}$$
$$\left\{p_{i},p_{j}\right\}_{PB}=0. \tag{7.d}$$
Thus, in canonical formalism of classical Newtonian mechanics there appears a classical anomaly of the Galilean symmetry, eventhough the action is Galilean invariant. From physics perspective this is easy to understand. Simply by observing the trajectory of a classical free particle one cannot tell how massive the particle is. One can only measure its mass by measuring its momentum in different inertial reference frames related by Galilean boosts.
The advantage of using the commutation relations (7) instead of (3) is that one can easily finds that the Noether charges (6) satisfy the following two constraints: $$E-\frac{|\vec{p}|^{2}}{2m}=0,\quad\quad\vec{l}-\frac{1}{m}\vec{g}\times\vec{p}=0, \tag{8}$$
which are useful in the search of its Casimir invariants.
Now perform an analysis of the relative dimensions in the algebra (3). For convenience, the dimensions of $H$, $\vec{P}$, $\vec{J}$, and $\vec{K}$ are denoted as $h$, $p$, $j$ and $k$, respectively. From the commutation relations (3), one has $$hk=p,\quad j^{2}=j,\quad jk=k,\quad jp=p,\quad kp=m.$$
From the above dimensional analysis, one finds that $j$ is dimensionless, and $k=\frac{m}{p}$, and $h=\frac{p^{2}}{m}$. From equation $(\star)$, it is clear that the operators $H$, $\vec{P}$, $\vec{J}$, and $\vec{K}$ can only appear in the Casimir invariants with positive exponents. This suggests that one should really write the constraints of their relative dimensions as
$$\frac{kp}{m}=j=1$$
$$h-\frac{p^{2}}{m}=0.$$
Comparing them with equation (8), and noticing that only terms with the same relative dimension can be summed together in the polynomial $(\star)$ of Casimir invariants, one concludes that for the extended Lie algebra $\mathfrak{gal}_{3}(m)$, its Casimir invariants can only be monomials of $$\vec{A}=f_{0}\vec{J}+f_{1}\frac{1}{m}\vec{K}\cdot\vec{P}+f_{2}\frac{1}{m}\vec{K}\times\vec{P},\quad\mathrm{and}\quad B=g_{0}H+g_{1}\frac{1}{m}\|\vec{P}\|^{2},$$
where $f_{0}$, $f_{1}$, $f_{2}$, and $g_{0}$, $g_{1}$ are dimensionless constants to be determined.
Looking at equation (8), an easy guess would be considering the following combinations:
$$C_{1}\equiv M=mI$$
$$C_{2}\equiv U=H-\frac{1}{2m}\|\vec{P}\|^{2}$$
$$\vec{S}\equiv\vec{J}-\frac{1}{m}\vec{K}\times\vec{P}.$$
Obviously, $C_{1}$ and $C_{2}$ are Casimir invariants. Whether $\vec{S}$ has anything to do with Casimir invariants need some further work. In doing so, one can ask for help from the Poisson brackets (7) because they are totally isomorphic to the Lie brackets (3).
If an operator constraint among the Noether charges (6) (denoted as $\mathcal{Q}_{n}$ in the following discussion for convenience), say $\mathcal{O}(\vec{x},\vec{p})=0$ in the universal enveloping algebra of the charges (6), is not a Casimir invariant, then using the Leibniz rule of Poisson brackets, one has $$\left\{\mathcal{O}^{2},\mathcal{Q}_{n}\right\}_{PB}=\left\{\mathcal{O},\mathcal{Q}_{n}\right\}_{PB}\mathcal{O}+\mathcal{O}\left\{\mathcal{O},\mathcal{Q}_{n}\right\}_{PB}=0,$$
which implies that $\mathcal{O}^{2}$ must be a Casimir invariant. This suggests that $\|\vec{S}\|^{2}$ must be a Casimir invariant.
In conclusion, one finds three indepent Casimir invariants of the centrally extended Galilean algebra:
$$C_{1}\equiv M=mI$$
$$C_{2}\equiv U=H-\frac{1}{2m}\|\vec{P}\|^{2}$$
$$C_{3}\equiv\|\vec{S}\|^{2}=\|\vec{J}-\frac{1}{m}\vec{K}\times\vec{P}\|^{2}.$$
$C_{1}$ is known as the mass of the particle. $C_{2}$ is known as its internal energy, and $C_{3}$ is its spin.
