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By Theorem 1.66 (page 62) of Knapp's Lie groups: beyond an introduction, 2nd ed:

For each positive integer $n$, if $\pi$ is an $(n+1)$-dimensional irreducible representation of $\mathfrak{sl}(2,\mathbb C)$, then there exists a basis $v_0,v_1,...,v_n$ such that \begin{align} &\pi(h)(v_i)=(n-2i)v_i, i=0,1,...,n;\\ &\pi(e)(v_i)=i(n-i+1)v_{i-1}, i=0,1,...,n;\\ &\pi(f)(v_i)=v_{i+1}, i=0,1,...,n~( \text{with the convention } v_{n+1}=0). \end{align} where $h = \begin{bmatrix} 1 & ~~0\\ 0 & -1 \end{bmatrix}, e = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$, and $f = \begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}$ is the so-called Cartan-Weyl basis.

When I looked through the proof of this, I noticed that the complex number field is used to show that $\pi(h)$ has an eigenvector with eigenvalue $\lambda$ (possibly a complex number at this point) and by the commutativity of $\pi(h)$ and $\pi(e)$ we have distinct eigenvalues $\lambda, \lambda+2, \lambda+4...$. By finite dimensionality of the representation, one finds a eigenvector $v_0$ (with eigenvalue $\lambda$ redefined) killed by $\pi(e)$. It turns out that the eigenvalue $\lambda=n$, a real number!

When I thought about this again, I realized that maybe we could start with the real representation of $\mathfrak{sl}(2,\mathbb R)$ instead. I mean we could still start with an eigenvalue $\lambda\in \mathbb C$ and conclude in the end that $\lambda=n \in \mathbb R$.

Is my reasoning correct? If so, then the structural theorem above for the irreducible reprensentation of $\mathfrak{sl}(2,\mathbb C)$ also applies to $\mathfrak{sl}(2,\mathbb R)$, right?

taylor
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1 Answers1

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It is certainly true that the representations of $\mathfrak{sl}(2,\mathbb{R})$ correspond to the representations of $\mathfrak{sl}(2,\mathbb{C})$ but we have to be a bit more careful about what we mean.

Specifically, if $\mathfrak{g}_\mathbb{R}$ is a real form of $\mathfrak{g}$ then every $\mathfrak{g}$-representation on a complex vector space $V$ restricts to a representation of $\mathfrak{g}_\mathbb{R}$ and conversely every $\mathfrak{g}_\mathbb{R}$-representation on complex $V$ extends to a representation of $\mathfrak{g}$ by simply extending complex-linearly.

However, this does not guarantee that $\mathfrak{g}_\mathbb{R}$ has a corresponding representation on a real vector space. For example $\mathfrak{su}(2)$, another real form of $\mathfrak{sl}(2,\mathbb{C})$ has no (non-trivial) representation on $\mathbb{R}^2$.

In the case of $\mathfrak{sl}(2,\mathbb{R})$ there is something special going on. It is not just any real form it is the split real form. The key property here is that we can choose a Cartan subalgebra comprising only diagonalisable elements (more generally, we are only guaranteed semisimple elements i.e. diagonalisable over $\mathbb{C}$). The fact that these are diagonalisable is why you are seeing real eigenvalues. Of course we should note $\mathfrak{sl}(2,\mathbb{R})$ has Cartan subalgebras which are not diagonalisable as well, unlike $\mathfrak{sl}(2,\mathbb{C})$.

Split forms in many respects (but not all) follow the behaviour of their complexifications so you could deduce an corresponding theory of their representations that looks like the complex story with the word complex replaced by real at each step, but be aware that this is something special about split forms.

Callum
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