I want to classify all unitary representations of $GL(2,\mathbb{C})$ from the representation theory of $SL(2, \mathbb{C})$. Is this possible? Knapp claims that one obtains all irreducible representation by pasting a character on $\mathbb{C}^\times$, which coincides on $-1$ onto $SL(2, \mathbb{C})$ and that one exhausts in this fashion all irreducible representation, but I worry about the well definedness of the square root $g \mapsto \sqrt{\det g}$. How does this work?

Alternatively, a reference which list all irreducible representations is also enough for my current purpose.

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    $\begingroup$ Dear late_learner, It's slightly more subtle than Knapp suggests, although the general principle is correct. Have you looked in the {\em Corvalis} volumes? Best wishes, $\endgroup$ – Matt E Apr 14 '12 at 18:16
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    $\begingroup$ @Matt: You can get italics by enclosing text in asterisks like this: *Corvalis* produces Corvalis. $\endgroup$ – joriki Apr 14 '12 at 21:54
  • $\begingroup$ @Joriki: Dear Joriki, Thanks for the reminder. Whenever I've been on stackexchange for a while, my LaTeX code ends up full of *'s, and when I've been typing papers for so long, I end up typing {\em } commands in MathSE posts. Ah well ... . Best wishes, $\endgroup$ – Matt E Apr 14 '12 at 23:43
  • $\begingroup$ @Matt E: Yes, precisely, there I got the statement from on pag2 91, first volume. Where is the nitpick? I know how to inflate representations and how to twist by a character, but not how to paste a character. What does he mean, Obviously $SL_2(\mathbb{C})$ is not a quotient of $GL(2, \mathbb{C}$, but only $PSL(2, \mathbb{C}) \cong PGL_2(\mathbb{C})$ after fixing a branch for the square root. $\endgroup$ – Marc Palm Apr 15 '12 at 9:34

The group $GL_2(\mathbb C)$ is the product of its centre $\mathbb C^{\times}$ and its subgroup $SL_2(\mathbb C)$. This product is not direct; the two subgroups intersect in $\{\pm 1\}$ (the centre of $SL_2(\mathbb C)$).

Suppose $\pi$ is an irred. unitary rep. of $SL_2(\mathbb C)$, and let $\varepsilon$ be the character through which $\{\pm 1\}$ acts on $\pi$. Choose an extension of $\epsilon$ to a unitary character of $\mathbb C^{\times}$. Then there is a unique extension of $\pi$ to a unitary rep. of $GL_2(\mathbb C)$, on which $\mathbb C^{\times}$ acts via the character $\chi$.

Conversely, any irred. unitary rep. of $GL_2(\mathbb C)$ arises from one of $SL_2(\mathbb C)$ by such a construction.

[The case of $GL_2(\mathbb R)$ is a little more subtle, and this is what I was thinking of when I wrote my comment above. The reason is that in this case $GL_2(\mathbb R)$ is not the product of its centre and $SL_2(\mathbb R)$, because the squaring map $\mathbb R^{\times} \to \mathbb R^{\times}$ --- which is what arises from restricting the determinant map to the centre of $GL_2$ --- is not surjective, unlike in the case of $GL_2(\mathbb C)$.]

  • $\begingroup$ In the mean-while, I have used Jacquet-Langlands for $GL(2, \mathbb{C})$ and then proven for unitary central character that unitarizability, temperedness and square integrability hold if only if it holds for the restriction to $SL_2(\mathbb{C})$. This strategies works for every local field, but assumes that we know the rough classification already and the fine for $SL(2, \mathbb{C})$ as given in Wallach. But your answer was really helpful. Thanks. $\endgroup$ – Marc Palm Apr 15 '12 at 16:02
  • $\begingroup$ I'm sorry but what do you mean exactly by "the character through which $\{\pm 1 \}$ acts on $\pi$"? $\endgroup$ – Henrique Tyrrell Jan 28 '17 at 13:33

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