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Let $\mathfrak g = \mathfrak sl(2,\mathbb C)$ with standard basis $H = \begin{pmatrix} 1&0\\0&-1\end {pmatrix}, E= \begin{pmatrix} 0&1\\0&0\end {pmatrix}, F= \begin{pmatrix} 0&0\\1&0\end {pmatrix}$. Let $M(\lambda)$ denote the highest weight module of $\mathfrak g$ for $\lambda \in \mathbb C$. Formally $M(\lambda)$ is a complex vector with basis $v_k, k \in \mathbb N_0$, with action of $\mathfrak g$ given by $h v_k = (\lambda-2k) v_k$, $f v_k = v_{k+1}$ and $e v_k = (k(\lambda-k+1))v_{k-1}$. How does one proof that this $\mathfrak g$-module does not integrate to a (unitary?) representation of $G=SL(2,\mathbb C)$? In [Rossmann:Lie groups], p 230, Exercise 6, a hint is to consider the element $s=\begin{pmatrix} 0&-1\\1&0\end {pmatrix}$ and how it would act on $M(\lambda)$. So let us assume $(\pi,V)$ is a (unitary) representation of $G$ that differentiates to the $\mathfrak g$-module $M(\lambda)$. Now $s$ is the infitesimal generator of the compact subgroup $K = SO(2) = \{u_\theta = \begin{pmatrix} cos \theta&-sin \theta\\sin \theta&cos \theta\end {pmatrix}\}$ in $SL(2,\mathbb R) \subset SL(2,\mathbb C)$. If we assume $\pi$ to be unitary then since $K$ is abelian, $V$ decomposes into one-dimensional $K$-eigenspaces and $u_\theta$ acts on some $v \in V$ as the scalar $e^{i n \theta}$ for some integer $n$. Then it follows that $s$ sends $v$ to $i n v$ ($H' = i s$ sends $v$ to $n v$). In $SL(2,\mathbb C)$, the matrix $s$ is conjugate to $iH$ which is the infitesimal generator of the torus $\tilde K = SO(2) = \{g_\theta = \begin{pmatrix} e^{i \theta}&0\\0&e^{-i \theta}\end {pmatrix}\}$. The analogous argument as above for $K$ is true for $\tilde K$. Let $V_n$ denote the eigenspace of vectors $v$ with $\pi(g_\theta) v = e^{i n \theta} v$. Since $s$ conjugates $g_\theta$ to its inverse, $\pi(s)$ maps a vector $v \in V_n$ to $V_{-n}$. That is, the set of weights (w.r.to $\tilde K$) is closed under the action of $s$ (the generator of the Weyl group $\mathbb Z/2\mathbb Z$ of $SU(2)$ resp. $SL(2,\mathbb C)$), but the set of Lie algebra-weights of $M(\lambda)$ is not closed under $\mu \mapsto -\mu$.

1) Is this the argument to show that $M(\lambda)$ does not come from a unitary representation of $G=SL(2,\mathbb C)$? How to show it could not come from a non-unitary representation?

2) In the mentioned exercise in [Rossmann] one should show that the restriction of the $\mathfrak g$-module to a $\mathfrak sl(2,\mathbb R)$-module also does not integrate to a representation of $SL(2,\mathbb R)$. How to see this?

3) Also, doesnt the group $SL(2,\mathbb R)$ have contrary to $SL(2,\mathbb C)$ discrete series representations? Don't these representations differentiate to highest(/lowest) weight modules? How does this fit together with 2)?

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