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My naive picture of the local Langlands correspondence for $GL(2,\mathbf{C})$ is this. The Weil group of $\mathbf{C}$ is canonically $\mathbf{C}^\times$. On the Galois side then we're looking at 2-dimensional semisimple representations of $\mathbf{C}^\times$, that is, pairs of continuous group homomorphisms $(\chi_1,\chi_2)$ with $\chi_i:\mathbf{C}^\times\to\mathbf{C}^\times$, modulo $(\chi_1,\chi_2)\sim(\chi_2,\chi_1)$.

On the representation theory side we're looking at irreducible admissible complex representations of $GL(2,\mathbf{C})$ and it's a standard fact that we can build them all from principal series (in the sense described below). Given a pair $(\chi_1,\chi_2)$ as above we can build a 1-dimensional representation of the upper triangular matrices in $GL(2,\mathbf{C})$ and then induce up (normalised induction) to get a principal series representation $I(\chi_1,\chi_2)$ of $GL(2,\mathbf{C})$.

If all the principal series representations were irreducible, and $I(\chi_1,\chi_2)\cong I(\chi_2,\chi_1)$ life would be great: we match up $\chi_1\oplus\chi_2$ with $I(\chi_1,\chi_2)$ and there's the correspondence.

I don't think life is quite so easy though, because there are some reducible principal series. Now the standard trick seems to be that you order the $\chi_i$ by rate of growth of absolute value and then show $I(\chi_1,\chi_2)$ has a unique irreducible quotient $J(\chi_1,\chi_2)$, and match $\chi_1\oplus\chi_2$ with $J(\chi_1,\chi_2)$. I think that this is what the local Langlands correpondence is really supposed to be.

I don't get it. If $I(\chi_1,\chi_2)$ has, say, two Jordan-Hoelder factors (one finite-dimensional say) then we get two representations of $GL(2,\mathbf{C})$ "attached" to $(\chi_1,\chi_2)$ and they're probably not going to be isomorphic, so they had better correspond to two different representations of the Weil group. But all we have is $\chi_1$ and $\chi_2$ and they can't both correspond to $\chi_1\oplus\chi_2$. Is the idea that when this happens, one of them is $I(\chi_3,\chi_4)$ for some different pair of characters? Presumably this is easy to see on the representation theory side but I can't spot the intertwiner. Have I got the picture wrong?

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My knowledge of this specific instance is very limited, but if you induce a $1$-dimensional representation from the upper triangular matrices to $GL_2(\mathbb{C})$ then you get either $0$ or an irreducible representation (at least when you view these as algebraic groups. I don't know if this is some slightly different type of induction). –  Tobias Kildetoft Mar 21 '14 at 7:54
Yes I'm not talking about algebraic induction -- I'm talking about the kind of representations of GL(2,C) which the local Langlands correspondence applies to, so typically infinite-dimensional representations of GL(2,C) on Hilbert spaces modulo infinitesimal equivalence. The issue I'm confused about is that if the characters are algebraic then the algebraic representations you're talking about will show up as subspaces of the infinite dimensional representations I'm talking about. –  eric Mar 21 '14 at 19:00
I have chosen to delete my answer, because it seems not to address your question. As Marty points out $I(\chi_1,\chi_2)$ is isomorphic to $I(\chi_2,\chi_1)$ if both are irreducible. If they are reducible, they are not. What is an irreducible subquotient in one of them is an irreducible subrepresentation in the other. So that's why considering only unique subquotients hits all irr reps. –  Marc Palm Mar 22 '14 at 16:10
Also there are other references for SL(2,C). Knapps book on semisimple groups or Wallachs book on reductive groups. Wallach seems more informative concerning things like temperedness and unitarity. SL(2,C) and GL(2,C) are very similar, not like SL(2,R) and GL(2,R). –  Marc Palm Mar 22 '14 at 16:17
Marc: here is the issue I'm trying to address. If they are reducible, then we get more than one irreducible subquotient. But there's only one representation of the Weil group, namely chi1+chi2. And yet local Langlands is a bijection! So what are you going to do with more than one subquotient? That's the question. Marty's answer is the answer. –  eric Mar 23 '14 at 21:19

1 Answer 1

up vote 10 down vote accepted

This is a common point of confusion, and the OP is on exactly the right track.

A good reference for the representation theory is Chapter 1, Section 6, of Jacquet-Langlands book "Automorphic forms on GL(2)," especially Theorem 6.2 (which has a tiny typo in its statement). This is freely available online, from Langlands page at the IAS. Thanks to the IAS for putting so many publications online recently!

Here's how it goes, beginning with your characters $\chi_1$ and $\chi_2$: Let $\chi = \chi_1 \chi_2^{-1}$ be the resulting character of $C^\times$. The normalized principal series $I(\chi_1, \chi_2)$ is irreducible unless $\chi$ has the form $z \mapsto z^p \bar z^q$ with $p,q$ both positive integers or both negative integers.

Decomposing with respect to the compact subgroup $SU(2)$ yields a series of irreps of $SU(2)$ with multiplicity $1$. This is the really the key to seeing what's going on, IMHO.

In the case with $p,q$ positive, $I(\chi_1, \chi_2)$ has a finite-dimensional quotient, of dimension $$d = \# \{ n : n < p+q, n = p+q \text{ mod } 2 \}.$$

In the case with $p,q$ negative, $I(\chi_1, \chi_2)$ has a finite-dimensional sub, of dimension $$d = \# \{ n : \vert p-q \vert \leq n \leq p+q, n = p+q \text{ mod } 2 \}.$$

When $I(\chi_1, \chi_2)$ is reducible, Langlands defines $\pi(\chi_1, \chi_2)$ to be the (equivalence class of the) finite-dimensional Jordan-Holder constituent of $I(\chi_1, \chi_2)$. When $I(\chi_1, \chi_2)$ is irreducible, $\pi(\chi_1, \chi_2)$ is defined to be the equivalence class of $I(\chi_1, \chi_2)$. This gives the local Langlands correspondence.

What's so confusing about this is the followingy: what happened to those perfectly nice infinite-dimensional constituents of the reducible $I(\chi_1, \chi_2)$? As the OP suspects, they are equivalent to irreducible representations $I(\chi_3, \chi_4)$ for another pair of characters. This is explained in Theorem 6.2, (vi), of Jacquet-Langlands book.

Remember that the reducible $I(\chi_1, \chi_2)$ correspond to $p,q$ integers of the same sign. Well, the Theorem 6.2 (vi) cited above guarantees that there exist characters $\chi_3, \chi_4$ such that:

  1. $\chi_3 \chi_4^{-1} (z) = z^p \bar z^{-q}$, so in particular $I(\chi_3, \chi_4)$ is irreducible.

  2. $\chi_3 \chi_4 = \chi_1 \chi_2$, so the central characters coincide.

  3. The infinite-dimensional chunk of $I(\chi_1, \chi_2)$ coincides with the irrep $I(\chi_3, \chi_4)$.

Langlands demonstrates this coincidence by using a result of Harish-Chandra. Basically, one checks that the infinitesimal character of $I(\chi_1, \chi_2)$ coincides with that of $I(\chi_3, \chi_4)$; then if there are constituents with a single $SU(2)$-type in common, the constituents are isomorphic.

The intertwiner is not so easy to spot -- maybe it's called a Knapp-Stein intertwining operator in this setting (after their joint PNAS paper?). I don't know the full history and common terminology in the archimedean setting. The idea is to take functions $f$ in a principal series $I(\chi_1, \chi_2)$ and look at the function $F(x) = \int_{\bar U} f(\bar u x) d \bar u$, where $\bar U$ is the opposite unipotent radical. This gives a family (as $\chi_1, \chi_2$ vary) of intertwiners to a principal series for the opposite parabolic, which can be conjugated back to principal series for the original parabolic subgroup. Convergence is an issue, but the point is that the intertwiners kill finite-dimensional subrepresentations at reducibility points. Tracking through the modular characters, Weyl elements, etc., one could (and others have, I'm sure) get the Theorem 6.2 (vi) above.

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The intertwiner seems wrong. You need $\int\limits_{U} f(wux) du$ for $w$ the Weyl element, or not? –  Marc Palm Mar 22 '14 at 15:57
That one should work too, with U the (say) upper-triangular unipotent radical. It's equivalent, I think after substitution, to the one I wrote down with the lower-triangular unipotent radical. The important thing is to use whatever unipotent radical is opposite of the one used for parabolic induction in $I(\chi_1, \chi_2)$. –  Marty Mar 23 '14 at 1:01
Are you sure? I can't see the invariance. The integral transfer needs to map $I(\mu)$ to $I(\mu^w)$. –  Marc Palm Mar 23 '14 at 13:18
Thanks a lot Marty. Although I didn't put it in the question, I half-wanted to say "surely we can't have $\{\chi_3,\chi_4\}\not=\{\chi_1,\chi_2\}$ because one can look at the infinitesimal character". But there's a subtlety. Very briefly: the point is that the inf char is obtained by regarding $GL(2,C)$ as a real Lie group and then complexifying, and there are two ways to match up the two inf chars you get in this way; if $\chi_1,\chi_2$ correspond to one way then $\chi_3,\chi_4$ correspond to the other. This subtlety is what I'd missed. Many thanks! –  eric Mar 23 '14 at 21:25
Example: if $\chi_1=z^p\overline{z}^q$ and $\chi_2=1$ then $\chi_3=z^p$ and $\chi_4=\overline{z}^q$. In both cases the inf chars are (with obvious notation) $\{p,0\}$ and $\{q,0\}$ but the $q$ has switched from the character-with-$z^p$-in to the other one. –  eric Mar 23 '14 at 21:27

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