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Given a Dynkin diagram, we can construct a presentation of a semisimple Lie algebra $\mathfrak g$ by the Cartan elements $H_\alpha$, raising operators $E_i$ and lowering operators $F_i$ satisfying the Serre relations. These are analogous to the ladder operators encountered in physics. However, there is a prominent feature of ladder operators not present a priori: Ladder operators work on Hilbert spaces (let's restrict to finite dimensional spaces for simplicity), and there is a certain adjoint relation between the operators.

Take $\mathfrak{so}(3)$ for example. We can generate this algebra by $L_z, L_+$ and $L_-$. Here $L_z$ is hermitian, serving as the Cartan subalgebra; $L_+$ and $L_-$ are the ladder operators, satisfying $L_+^\dagger = L_-$. This prompts the question:

Is it possible to equip all the finite dimensional complex representations of semisimple Lie algebras with an inner product, so that $H_\alpha$ is (anti)hermitian, and $E_i^\dagger = F_i$?

Notice that this is different from unitary representations for Lie groups (or, after a differentiation, a "hermitian representation" for Lie algebras) since not everything is hermitian. It also probably depends on the specific choice of Cartan algebra.

Trebor
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  • I assume your $L_z, L_+, L_-$ are not actually elements of the real Lie algebra $\mathfrak{so}(3)$, but of its complexification. – Torsten Schoeneberg Mar 20 '25 at 15:04
  • Please have a look at my answers to https://math.stackexchange.com/q/3318978/96384 and https://math.stackexchange.com/q/3711173/96384, which I think are strongly related to your question. The upshot is that I think what you want can be done if you start with a compact real semisimple Lie algebra. In the non-compact case, those operators still exist in the complexified setting, but they will not be contained in your original Lie algebra anymore, i.e. that one has no reason to and in general will not act via (anti)hermitian operators. – Torsten Schoeneberg Mar 20 '25 at 19:22

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