Local Langlands for $GL(2,\mathbf{C})$ and reducible principal series 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?
 A: 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:


*

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

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

*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.
