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Let $X,Y,H$ be the standard base for the Lie algebra $\mathrm{sl}_2({\mathbb{C}})$, i.e. $H=\begin{pmatrix} 1 & 0\\ 0 &-1\end{pmatrix}$, $X=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}$, $Y=\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}$. Let $V$ be some finite dimensional irreducible representation of $\mathrm{sl}_2$. I'm sorry if my question is for children but why the action of $H$ is diagonalizable?

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  • $\begingroup$ I don't think your definition of $Y$ is correct for the usual basis. See here. $\endgroup$ – Viktor Vaughn May 27 '14 at 22:02
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An elementary proof relies on the folllowing observation:

Let $\lambda\in\Bbb C$ be an eigenvalue of $H$ acting on $V$ (any finite dimensional representation of $\mathfrak{sl}_2(\Bbb C)$). Then the following sum of eigenspaces of $H$ $$W=\bigoplus_{n\in\Bbb Z}E_{\lambda+2n}(H)$$ is a subrepresentation of $V$.

Here, for every complex number $\alpha$, $E_{\alpha}(H)=\lbrace v\in V\mid H\cdot{}v=\alpha v\rbrace$ is either $0$ or an eigenspace of $H$ when $\alpha$ is an eigenvalue.

The fact that $\lambda$ is eigenvalue is actually irrelevant to the proof of the highlighted claim. Of course, if it isn't an eigenvalue of $H$, then $W$ might very well be $0$.


Let $\lambda\in\Bbb C$ be a complex number and $W$ defined as above. Let us show that $W$ is a subrepresentation of $V$, that is, let us show that it is stable under the action of $\mathfrak{sl}_2(\Bbb C)$. By construction it is stable under $H$. Let $n\in\Bbb Z$ be an integer, and $v\in E_{\lambda+2n}(H)$. Then

$$ \begin{array}{rvl} H\cdot\big(X\cdot v\big)&=&[H,X]\cdot v+X\cdot\big(H\cdot v\big)\\ &=& 2X\cdot v+X\cdot(\lambda+2n)v\\ &=&(\lambda+2(n+1))X\cdot v \end{array}$$ that is, $X\cdot v\in E_{\lambda+2(n+1)}(H)\subset W$. Similarly, $Y\cdot v\in E_{\lambda+2(n-1)}(H)\subset W$, and the claim follows. This shows that $W$ is stable under $X, H$ and $Y$, so that it is indeed a subrepresentation of $V$.


Now suppose $V$ is irreducible and $\lambda$ is an eigenvalue of the action of $H$ on $V$ (because $V$ is finite dimensional over $\Bbb C$, there exists at least one such eigenvalue). From what precedes we know that $W$ is a (nonzero) subrepresentation of of $V$ so that it is all of $V$. By definition, $H$ is diagonalizable on $W$, hence $H$ acts diagonally on $V$.

If $V$ is irreducible, then $W$ is all of $V$ and we further have that $$\mathrm{Spec}(H)\subset\lbrace\lambda+2n\mid n\in\Bbb Z\rbrace$$ Actually if $0\leq m,n$ are maximal such that $\lbrace\lambda+2k\mid -m\leq k\leq n\rbrace\subset\mathrm{Spec}(H)$ then $$\mathrm{Spec}(H)=\lbrace\lambda+2k\mid -m\leq k\leq n\rbrace)$$

The slightly stronger claim follows at once and is very easily established. It requires more work to show that all those subspaces $E_{\lambda+2k}$, for $-m\leq k\leq m$ are one dimensional.

I warmly recommend Problem 1.55 from this free and wonderful Introduction to representation theory.

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There are two ways to see this one is the fact that the Jordan decomposition is preserved under representations of semisimple Lie algebras. The second is Weyl's unitary trick which is based on the fact that all representations of compact groups are completely reducible. You can find more on both in Fulton and Harris: Representation Theory A First Course.

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    $\begingroup$ One note: Jordan decomposition is preserved under representations of semisimple Lie algebras, which applies in this case. It does not hold in general. For example, if you take the Lie algebra $\mathfrak{g}$ of strictly upper triangular $2 \times 2$ matrices (which has only nilpotent elements), we can find a one-dimensional representation on which each element of $\mathfrak{g}$ acts semisimply. $\endgroup$ – Siddharth Venkatesh May 27 '14 at 22:23
  • $\begingroup$ Thanks for pointing that out, I edited my answer. $\endgroup$ – Rene Schipperus May 27 '14 at 22:37

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