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It is easy to show that every irreducible finite-dimensional real $\mathfrak{sl}_2(\mathbb C)$-representation is a weight module. The operator commonly denoted as $h$ in $\mathfrak{sl}_2$ has an eigenvalue since $\mathbb C$ is algebraically closed and from this one gets those Verma-modules of dimension $d + 1$ with basis $(v_{-d}, v_{-d + 2}, \dots, v_{d - 2}, v_d)$, where $v_\lambda$ is an eigenvector of $h$ with eigenvalue $\lambda$ and the action of the other two generators is defined accordingly.

This argument can not be used for the real case. So does there exist an irreducible finite-dimensional real $\mathfrak{sl}_2(\mathbb R)$-representation that is not a weight module and if not is there an easy proof?

HDB
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    The structure theory of $\mathfrak{sl}_2(k)$-modules is virtually identical for all characteristic $0$ fields $k$. The eigenvalues of $h (\in \mathfrak{sl}_2(\mathbb Z))$ are integers. Analogous results hold true for all semisimple Lie algebras over characteristic $0$ fields which are "split", i.e. contain a CSA consisting of $ad$-diagonalisable elements. – Torsten Schoeneberg Nov 18 '21 at 16:47
  • By the way, even in the non-split case (example: $\mathfrak g= \mathfrak{su}2$), one can use complexification and classify the irreducible complex representations of $\mathfrak g$ as exactly the irreducible complex representations of $\mathfrak g{\mathbb C}$, restricted to $\mathfrak g$. (Cf. https://math.stackexchange.com/q/2292267/96384, https://math.stackexchange.com/q/1408894/96384, https://math.stackexchange.com/a/3258221/96384.) So you still classify with highest weights. – Torsten Schoeneberg Nov 18 '21 at 16:56
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    However classifying the real irreducible reps of $\mathfrak{su}_2$ is a bit different. There's one irreducible in every odd dimension, which is absolutely irreducible. While there is no irreducible in dimension $4n+2$, and one in each dimension $4n\ge 4$, which is not absolutely irreducible. – YCor Nov 22 '21 at 00:06

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