# Commutation relation under an arbitrary Lie algebra representation

This is an exercise in Woit's book, B9, Problem 2:

For the case of the Euclidean group $$E(2)$$, show that in any representation $$\pi'$$ of its Lie algebra, there is a Casimir operator $$|\vec{P}|^2 = \pi'(p_1)\pi'(p_1) +\pi'(p_2)\pi'(p_2)$$ that commutes with all the Lie algebra operators $$\pi'(p_1), \pi'(p_2), \pi'(l)$$.

I have a couple of doubts regarding this computation. Suppose we want to prove that it commutes with $$\pi'(p_1)$$. One has to compute $$[\pi'(p_1)\pi'(p_1) +\pi'(p_2)\pi'(p_2),\pi'(p_1)] = 0,$$

but since $$\pi'(\cdot)$$ is arbitrary, I would like to "use" the definition of the Lie algebra representation

A Lie algebra representation $$(\phi,V)$$ of a Lie algebra $$\mathbb{g}$$ on an $$n$$-dimensional complex vector space $$V$$ is given by a real-linear map $$\phi: X\in \mathbb{g}\rightarrow \phi(X) \in \mathbb{gl}(n,\mathbb{C})$$ satisfying $$\phi([X,Y]) = [\phi(X),\phi(Y)].$$

Of course this definition is for finite vector spaces (it is the one in my book), but still I can use it in my case (e.g. the Schrödinger representation is still unitary and fulfills the commutator relation above), right?

If this holds then $$\pi'([p_1^2+p_2^2,p_1]),$$ where the Lie bracket is now the Poisson bracket and thus $$\pi'(0)$$.

Is the process above valid? Is it $$\pi'(0)=0$$?

• I suspect the average reader will not understand your question, because there is precious little background and references on the π' "representation", nor what PBs have to do with anything here. Commented Nov 26, 2021 at 17:30
• I rewrote the question, adding more details. Commented Nov 29, 2021 at 9:07
• You need to be careful about one thing here. The way the Casimir element was defined using a representation makes sense ($\pi(p_1)^2+\pi(p_2)^2$), as the representations live in the space of linear operators on a vector space where such products exist. However, in your last expression you write things like $p_1^2$ which does not make sense in a Lie algebra as the only product available is the commutator. There are two ways to make sense of the above 1) Define the concept of a "Universal Enveloping Algebra". Which I guess the book has not done yet. Commented Nov 29, 2021 at 13:24
• 2) Use a simple identity, to relate the commutator you want to the Lie algebra commutator. For any associative algebra, the commutator satisfies (easy to show): $$[AB,C] = A[B,C] + [A,C]B$$ Try using this on $[\pi(p_1)^2+\pi(p_2)^2,\pi(\cdot)]$ and then use the fact that $\pi$ is a Lie algebra homomorphism. Commented Nov 29, 2021 at 13:24
• I am not sure I understand the comment about Poisson Brackets. I don't know the book and not sure if there is some special context here. Commented Nov 29, 2021 at 13:26

Is the process above valid? Is it $$\pi'(0)=0$$?
Specifically, you have the E(2) brackets, $$[\pi'(p_1),\pi'(p_2)]=0, \qquad [\pi'(l),\pi'(p_1)]=\pi'(p_2), \qquad [\pi'(l),\pi'(p_2)]=-\pi'(p_1),$$ where your Casimir works; but it is not in the Lie algebra.
In point of fact, your target bracket can also be computed by inspection as a plain commutator of the Schroedinger realization, $$\pi'(l)= -q_1\partial/\partial q_2 + q_2\partial/\partial q_1 ,\qquad \pi'(p_1)=-\partial/\partial q_1, \qquad \pi'(p_2)=-\partial/\partial q_2, \\ \pi^{'~ 2}(p_1)+ \pi^ {'~ 2}(p_2)= \frac{\partial^2 }{\partial q_1^2}+ \frac{\partial^2 }{\partial q_2^2}~~.$$ Note the last line is not in the Lie algebra, but it is still "represented" faithfully by this "crypto-quantization" realization.
• Q. The Schrödinger representation is "the only one that exists" on a Hilbert space for finite spatial dimensions due to the Stone-von Neumann theorem. I cannot think on another representation that might be useful from the Physics point of view. Do you know if exists another one, so that the "general representation" $\pi'$ is justified to use? Commented Nov 30, 2021 at 7:58