Lie brackets and commuting generators

Question

I'm showing that $$J^2 = \frac{1}{2}\bigg[J_{\alpha\beta} J^{\alpha\beta} + \frac{2}{M^2} P_\alpha J^{\alpha\gamma} P^\beta J_{\beta\gamma}\bigg]$$ is a Casimir of the Poincaré algebra. The brackets of this algebra are given by \begin{align} \{P^\mu, P^\nu\}&= 0 \\ \{P^\mu, J^{\rho \sigma}\}&= \eta^{\mu\sigma}P^\rho - \eta^{\mu\rho}P^\sigma \\ \{J^{\mu\nu}, J^{\rho\sigma}\} &= \eta^{\mu\rho} J^{\nu\sigma} + \eta^{\nu\sigma}J^{\mu\rho} - \eta^{\nu\rho}J^{\mu\sigma} - \eta^{\mu\sigma} J^{\nu\rho}. \end{align} In my calculation I arrive at an expression that $$2\{J^2, P^\mu\} =2J^{\mu \alpha}P_\alpha + 2P_{\alpha} J^{\mu\alpha} + 2P_\alpha J^{\alpha \mu} + \frac{2}{M^2}P^\mu P^\gamma P^\beta J_{\beta\gamma} +2 P_\alpha J^{\alpha\mu} + \frac{2}{M^2}P^\alpha J_{\alpha\gamma}P^\mu P^\gamma.$$

I am confused about the first term, which comes from computing $$\{J^2, P^\mu\} = J_{\alpha\beta}\{J^{\alpha\beta}, P^\mu\} + \{J^{\alpha\beta}, P^\mu\}J_{\alpha\beta}.$$ It seems to me that we'll always get a term like "$J\cdot P$", but we can't necessarily write this as "$P\cdot J$" don't commute?

In the end my question boils down to this: for general Lie algebras we can't write the Lie brackets as commutators of the generators. Then what am I doing incorrectly here? Or, alternatively, why is it okay in this case to view the Lie bracket as the commutator?

Answer 1

I don't have anything to say about the Poincaré Casimir operator: it seems you just go through the algebra and verify it, if you want.

To answer the question 'why is it okay to view the Lie bracket as the commutator': because it is the commutator.

Consider a Lie group $G$, its Lie algebra $\mathfrak{g}$ is defined as the vector space of the left-invariant vector fields which act on $G$, with a Lie bracket $[~,~]:\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$ such that the Jacobi identity holds.

The Lie bracket defined for left invariant vector fields is such that for $X,Y\in \mathfrak{g}$, \begin{align} [X,Y]f=X(Yf)-Y(Xf) \end{align} at every point $p\in G$. These left-invariant vector fields are each uniquely identified by their behaviour at any point $p\in G$ that's why we also identify the Lie algebra as the tangent space at identity of the Lie group. When you represent the Lie algebra on some vector space $V$, these generators either become represented by matrices or operators, and the Lie bracket becomes the common commutator in Linear Algebra.

In short, a vector in any tangent space of a Lie group $G$ is a linear object which eats a smooth function and spits out a smooth function: it can be proved that if you require Lebniz rule to apply to this object, then this object will become differentiation operator essentially. A vector field is just a smooth function which assigns, to every point on the Lie group, a vector. This might help

To clarify things up: OP is asking, why can we treat $P^\mu$ and $J^{\mu\nu}$ like matrices and understand the Lie bracket as commutators.

In my opinion the reason falls to the following:

1. Without specifying a rule of product, the notation $J_{\alpha\beta}J^{\alpha\beta}$ in the definition of $J^2$ is ill-defined and meaningless.
2. It is known that the Poincaré algebra has representations on vector space, in particular a vector representation on four dimensional Minkowski space which makes the given Casimir operator well-defined and correct.
3. Though there might be unusual realisations of the Poincaré algebra on which one may be able to define the product $J_{\alpha\beta}J^{\alpha\beta}$ in a meaningful way, and hence ask us to very the $J^2$ given is a Casimir operator. We still need to know the exact definition of the product to proceed, which is not given in the original question.

Answer 2

There are two ways to extend a Lie algebra, $L$.

• Treat the Lie bracket as a commutator $[u,v] = uv - vu$ and embed the algebra, itself, into its Universal Enveloping Algebra.
• Expand the Lie bracket to a Poisson bracket. This treats $L$ as linear functions over the dual $L^\star$, and then generalizes to all smooth functions $C^∞\left(L^\star\right)$. If $\left(Y_a: 0 ≤ a < n\right)$ is a basis of $L$, and $y_a$ is (linear) coordinate function over $L^\star$, then the Poisson bracket of $F,G ∈ C^∞\left(L^\star\right)$ is $$\{F,G\} = \sum_{a,b} \frac{∂(F,G)}{∂(y_a,y_b)} \left\{y_a,y_b\right\},$$ where the fundamental Poisson brackets match the corresponding Lie brackets: $$\left[Y_a,Y_b\right] = \sum_{0≤c<n} f^c_{ab} Y_c\quad→\quad \left\{Y_a,Y_b\right\} = \sum_{0≤c<n} f^c_{ab} y_c, $$ and where I'm using the short-hand $$\frac{∂(u,v)}{∂(w,x)} = \frac{∂u}{∂w}\frac{∂v}{∂x} - \frac{∂u}{∂x}\frac{∂v}{∂w}.$$

In the second extension, the $y$'s are commuting c-numbers, and the Casimirs correspond to functions $F(y)$ such that $\{F(y),y_b\} = 0$, for $0 ≤ b < n$.

On the other hand, in the first extension, within the universal enveloping algebra, the $Y$'s are non-commuting q-numbers, and the polynomials over the $Y$'s have to use a specific operator ordering - I think: the Weyl Ordering. So, monomials would be symmetrically ordered: $$ \widehat{ab} = \frac{\hat{a}\hat{b} + \hat{b}\hat{a}}2,\quad \widehat{abc} = \frac{\hat{a}\hat{b}\hat{c} + \hat{b}\hat{c}\hat{a} + \hat{c}\hat{a}\hat{b} + \hat{a}\hat{c}\hat{b} + \hat{b}\hat{a}\hat{c} + \hat{c}\hat{b}\hat{a}}6, $$ and so on. The Weyl ordering is consistent with the derivation rule: $$[uv,w] = uvw - wuv = (uw - wu)v + u(vw - wv) = [u,w]v + u[v,w],$$ so that $\left[u^2,v\right] = [u,v]u + u[u,v]$.