If $\mathfrak{g}$ is semisimple and complex, then we can identify a Cartan subalgebra $\mathfrak{h}$ and with choice of positive roots, a Borel subalgebra containing this Cartan.

We can then consider $U(\mathfrak{g})$ and with respect to our choice of Borel sugalgebra $\mathfrak{b}$ above, we can choose a $1$-dimensional $\mathfrak{b}$ representation. By PBW we can quickly understand the Verma modules of $U(\mathfrak{g})$ and thus by an equivalence of categories, consider the representations of $\mathfrak{g}$.

We can consider $U(\mathfrak{g})$ for any Lie algebra, not just those that are semisimple, but without semisimplicity we cannot find a Cartan subalgebra, and hence cannot find a Borel subalgebra to induce a representation from.

Is there an analogue to Verma modules, for reductive Lie algebras or general Lie algebras?

  • $\begingroup$ Which properties of the Vermas would you require? All simples to be the simple head of a unique one? $\endgroup$ – Tobias Kildetoft Sep 2 '17 at 10:42
  • $\begingroup$ @TobiasKildetoft A weakening of the universal property in some way: "If $V$ is any representation generated by a highest weight $\lambda$, there is a surjective $\mathfrak{g}$-homomorphism $M_\lambda \to V$." $\endgroup$ – Representation theory Sep 2 '17 at 10:50
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    $\begingroup$ But that requires you to define what a highest weight vector is. And once that is done, the universal property is really just the universal property of induced representations. $\endgroup$ – Tobias Kildetoft Sep 2 '17 at 10:59
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    $\begingroup$ Borel subalgebras are just maximal solvable subalgebras, and these always exist (although they may not always be conjugate). Once you have these you can define Verma modules as induced representations as usual. I don't know how useful this is beyond the reductive case, e.g. for solvable algebras nothing interesting happens. $\endgroup$ – Qiaochu Yuan Sep 2 '17 at 19:16

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