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I want to find decompose $\mathfrak{sl}_2$ as a direct sum of irreducible representations of $U(\mathfrak{sl}_2)$ which is the universal enveloping algebra of $\mathfrak{sl}_2$. I have a theorem that basically does the job for me by classifying all irreducible representations of $U(\mathfrak{sl}_2)$ saying for each $n\in \mathbb N^{*}$ there is a unique (up to iso) irreducible representation of dimension $n$ and the action of $U(\mathfrak{sl}_2)$ is given by

So to my understanding if it decomposes as $V_1\oplus V_2$ for $\dim V_1=1$ it means that there is a non zero element of $\mathfrak{sl}_2$ stable under the matrices (0),(0),(0) ($f,e,h$ in dimension 1), so we can take any element, say $e$.

Now for $V_2$ let's take $\operatorname{span}\{f,h\}$, we directly see that we have a problem since $[e,h]=-2e\notin V_2$, and I think we will get similar problems for other combinations or even if we don't chose $e$ so I'm misunderstanding something.

Here how can we decompose $\mathfrak{sl}_2$ into irreducible representations of $U(\mathfrak{sl}_2)$ ?

raisinsec
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    The irreducible reps of $U(\mathfrak{g})$ are the same as those of $\mathfrak{g}$ and the adjoint rep here is irreducible so you cannot decompose further – Callum May 09 '23 at 22:11
  • @Callum Ah ok so I guess the fact that we get absurdities if we suppose we have an irreducible subrepresentation of dimension 1 "proves" that the adjoint representation is irreducible – raisinsec May 10 '23 at 08:36
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    In effect, yes, although we need to already know representations decompose into irreducibles (only true for $\mathfrak{g}$ semisimple). Altenratively, the fact that we can get from any given line $\langle ae+bh+ cf\rangle$ to $\langle e\rangle$ by repeatedly applying $\langle e \rangle$ and from there we can get to $\langle h \rangle$ and $\langle f \rangle$ pretty much seals the deal. – Callum May 10 '23 at 12:07

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