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?

  • $\begingroup$ @DietrichBurde So it is not generally true that the spectrum of a generator of a compact Lie group is discrete? Are there any conditions under which this is true? $\endgroup$
    – anon1802
    Jan 31, 2018 at 18:04
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    $\begingroup$ If the Hilbert space is finite dimensional… how could the spectra fail to be discrete? $\endgroup$ Jan 31, 2018 at 20:45
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    $\begingroup$ If the group $G$ is $S^1$, then the Lie algebra $L$ is just $\mathbb R$, and a represenation $\rho:L\to End(H)$ (notice it is End and not GL…) can have image whatever operator you want, with spectrum as ugly as you will) $\endgroup$ Jan 31, 2018 at 20:46
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    $\begingroup$ On the other hand, if the represenation of the Lie algebra arises from a representation of the group by differentiation, and the latter is unitary (which is probably what you want in your context) then this follows from the Peter—Weyl theorem that asserts that $H$ is a direct sum of finite dimensional reps. $\endgroup$ Jan 31, 2018 at 20:52
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    $\begingroup$ The Casimir operator preserves the decomposition of $H$ as a (completed) direct sum of finite dimensional representations in that case, so its spectrum is just as discrete as everything else. This applies to all elements of the enveloping algebra of L, in fact. $\endgroup$ Jan 31, 2018 at 20:59


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