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Let $(\pi,V)$ be an irreducible representation of the Lie algebra $sl(2,\mathbb C)$ on a possibly infinite-dimensional complex vector space $V$. Further let $h,e^+, e^-$ be the usual standard basis where $h$ is the semisimple matrix $\begin{pmatrix} 1 &0\\0&-1\end{pmatrix}$. Now, if $V$ is finite-dimensional, the usual argument goes like this: the operator $\pi(h)$ has at least one eigenvector $v_0$ and from there one finds $h$-eigenvectors with distinct eigenvalues by applying $\pi(e^+)$ resp. $\pi(e^-)$ several times to $v_0$. (I am not very precise here) Since $V$ is finite-dimensional one deduces that (because $\pi(h)$ can have only finitely many eigenvalues) this way one gets finitely many nonzero vectors which are $h$-eigenvectors and their span is an invariant subspace of $V$ which equals $V$ since $V$ is irreducible. In particular one sees that $V$ is $h$-semisimple and $h$-multiplicity free (in other words: $V$ is the direct sum of the nonzero $h$-eigenspaces and every nonzero eigenspace is one-dimensional).

My question: 1) I am now interested in $V$ an infinite-dimensional and irreducible representation of $sl(2,\mathbb C)$. Is it then also true that $V$ is $h$-semisimple and $h$-multiplicity free and is there a fairly elementary way to argument? Actually I guess in general a module will not be $h$-semisimple (we cannot even garantue the existence of one $h$-eigenvalue), so I guess I should ask if $h$-semisimplicity implies that in the decomposition of $V$ into $h$-eigenspaces any eigenspace is of dimension $1$?

'Add'-it: 2) Just to make sure my understanding is correct (it is probably not), but is it right that in general we cannot assume that the operator $\pi(h)$ is diagonalizable on $V$? So whenever in a text I read 'let $V = ..$' be the eigenspace decomposition of $V$ into eigenspaces of $\pi(h)$, the text is probably only treating the finite-dimensional case.

3) Assuming we have given one $h$-eigenvector $v_0$ with $h$-eigenvalue $\lambda$ and $V$ is infinite-dimensional, we can again deduce that the submodule $V_0$ generated by $v_0$ is $h$-semisimple and if $V$ is additionally irreducible, then of course $V$ itself is $h$-semisimple. I have read that under the additional assumption that $v_0$ is also an eigenvector for the Casimir element $C$, then all eigenspaces in $V_0$ are one-dimensional, but I don't understand why. Also, when is it the case that $C$ acts as scalar multiple of the identity on $V$?

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    $\begingroup$ The keyword you want to look up is "category O." $\endgroup$ – Qiaochu Yuan Oct 25 '15 at 18:45
  • $\begingroup$ @QiaochuYuan Irreducible objects of $\mathcal{O}$ for $\mathfrak{sl}_2$ are indeed multiplicity free, so the subtle part of the question has to do with weight modules that are not in $\mathcal{O}$. $\endgroup$ – Stephen May 16 '17 at 14:46
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1) This is part of the classification problem of weight-$sl_2$-modules, whose solution is attributed to Gabriel and can be found in Dixmier's Algèbres enveloppantes (7.8.16). Britten and Lemire classified all simple multiplicity-free weight modules [Britten, D.J., Lemire, F.W. A classification of simple Lie modules having a 1-dimensional weight space, Trans.Am.Math.Soc. 299, 1987].

2) When a text writes "Let $V=\bigoplus_\lambda V_\lambda$" be the eigenspace decomposition of $V$, ..." then the text only and exactly assumes that $V$ has an eigenspace decomposition. By the way, $V$ is then called weight module.

If $V$ is irreducible, it is in particular indecomposable, which implies that all $\lambda$ lie in a single $\mathbb{Z}\alpha$-coset of $\mathfrak{h}^*$.

3) Yes, given one $h$-eigenvector $v_\lambda$, the universal enveloping algebra $\mathcal{U}(sl_2)$ generates a before mentioned weight module.

4) Let $\{ e_+, e_-, h'\}$ be the dual basis of $\{e^+, e^-, h\}$ of $sl_2$ with respect to the Cartan-Killing form $B(x, y) = \mathrm{trace}(\mathrm{ad}(x)\mathrm{ad}(y))$. Then

$$C = hh'+ e_-e^- + e_+e^+$$

commutes with any $g\in sl_2$ and is called Casimir element. Thus the Casimir element $C$ is in the center $\mathcal{Z}(\mathfrak{g})$ of $\mathcal{U}(\mathfrak{g})$ and acts by diagonal matrices $\pi(c)$.

Actually, for semisimple $\mathfrak{g}$ the center $\mathcal{Z}(\mathfrak{g})$ of $\mathcal{U}(\mathfrak{g})$ is isomorphic to the polynomial ring over $\mathbb{C}$ in $\mathrm{rank}\,\mathfrak{g}$ variables [cf. Carter, Lie algebras of finite and affine type, Thm 11.32]. In our case of $sl_2$ this is just $\mathbb{C}[C]$.

The stated additional condition of $C$ acting by a scalar, say $Cv_0=\xi\cdot v_0$ implies now that on the generated module $V_0 = \mathcal{U}(sl_2)v_0$, the Casimir element $C$ will also act by scalars: $c(e_\pm v_0)=e_\pm cv_0= \xi\cdot (e_\pm v_0)$.

Next, let's equip the module $V_0$ with the action $hv_0 = \lambda v_0$, for some $\lambda\notin\mathbb{Z}$, and $e_-v_n := fv_n = (n+\varepsilon)v_{n+1}$ for $\varepsilon\in (0,1)$.

Lemma. If $n > 0$, then $ (i)\; hv_n = (\lambda -2n)v_n\\ (ii)\, ev_n = (\lambda - n + 1)v_{n-1}\\ $

Proof. (i) By induction on $n$:

$h\left(fv_n\right) = fhv_n + \left[h,f\right]v_n = f(\lambda -2n) v_n - 2fv_n = \left(\lambda -2(n+1)\right)\left(fv_n\right)$

(ii)

$$\left(n+\varepsilon\right)ev_{n+1} = efv_{n} = fev_n+\left[e,f\right]v_{n}\\ =\left(\lambda-n+1\right)fv_{n-1}+hv_{n}\\ =\left(\lambda-n+1\right)\left(n-1+\varepsilon\right)v_{n}+\left(\lambda-2n\right)v_{n}\\ =\left(\lambda-n\right)\left(n+\varepsilon\right)v_{n}$$

for $n\ge 0$. Now divide by $\left(n+\varepsilon\right)$.

q.e.d.

Now, we can extend the action of $e=e_+$ for $n \le 0$. This is consistent with the action of $h$, as demonstrated by

$h(ev_n)=ehv_n + [h,e]v_n = e(\lambda - 2n) v_n + 2ev_n = (\lambda -2(n-1))(ev_n).$

Surely, these modules are infinite-dimensional for $0 < \varepsilon < 1$ and their weight-spaces are one-dimensional.

Since $V_0$ is generated by the vector $v_0$ with $h$-action $\lambda$, the Harish-Chandra homomorphism $\phi:\mathcal{U}(\mathfrak{g})_0 \to \mathcal{U}(\mathfrak{h})$ allows to determine the central character from the the h-action on $v_\lambda$: $\chi_\lambda: \mathcal{Z}(\mathfrak{g}) \to \mathbb{C}$ by $\chi_\lambda(z) = \lambda(\chi(z))$. This is due to the fact, that $\mathcal{U}(\mathfrak{h})$ is isomorphic to a polynomial ring, in the $sl_2$-case just $\mathbb C[h]$ where $\lambda: \mathbb C[h] \to \mathbb C$ is just the evaluation homomorphism.

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