In this answer on Phys.SE about reasons for the "discreteness" in quantum mechanics, the author mentioned
The discrete spectrum for Lie algebra generators of a compact Lie group, e.g. angular momentum operators.
I have taken a course in Lie groups and Lie algebras (though with applications to physics), but I cannot remember proving this fact. Since we're dealing with quantum mechanics, we're probably only concerned with unitary (projective?) representations of Lie groups.
Please could someone provide a proof of this statement? Bonus points if you can provide some connection to operators in quantum mechanics!
Attempt at a precise formulation of the question
Consider a compact Lie group $G$ and a Lie algebra $\mathcal{L}$ which generates $G$ via the usual exponential map. We can choose a basis $\{T_{i}\}_{i=1}^{n}$ of $\mathcal{L}$, which we refer to as the generators of $\mathcal{L}$. Consider a representation $\rho : \mathcal{L} \rightarrow \mathrm{End}(\mathcal{H})$, where $\mathcal{H}$ is an infinite-dimensional Hilbert space.
Question: Is the spectrum of the operator $\rho(T_{i})$ discrete?
Bonus question: Suppose $\mathcal{L}$ is equipped with a nondegenerate bilinear form (take the Killing form for concreteness), so given a basis $\{T_{i}\}_{i=1}^{n}$ of $\mathcal{L}$ we can canonically define a basis $\{T^{i}\}_{i=1}^{n}$ of the dual space $\mathcal{L}^{*}$. Define the quadratic Casimir $\Omega = \sum_{i=1}^{n} T^{i} T_{i}$. Given the representation $\rho$ defined above, we define the Casimir invariant $\rho(\Omega) = \sum_{i=1}^{n} \rho(T^{i}) \rho(T_{i})$. If $G$ is a compact Lie group, is the spectrum of the operator $\rho(\Omega)$ also discrete?