infinitesimal character of Langlands quotient for GL(n,R) Let $G = GL(n,\mathbb{R})$. Consider a Langlands data $(Q_F, \sigma, \lambda)$ with $F \subset \Delta$ (the set of simple roots), $Q_F$ the associated standard parabolic subgroup, $\sigma$ an irreducible tempered representation of $M_F$, and $\lambda$ an element of $(\mathfrak{a}_F)^*_\mathbb{C} = Hom(\mathfrak{a},\mathbb{C})$ satisfying $\langle Re \lambda, \alpha \rangle \ge 0 $ for $\alpha \in \Delta \setminus F$
Here $\mathfrak{a}_F$ is the Lie algbera of $A_F$ appearing in the Langlands decompostion $Q_F = M_F A_F N_F$. 
Then the Langlands classification of irreducible admissible $(\mathfrak{g},K)$-modules is essentially given as the module $J(Q_F, \sigma, \lambda)$ occuring in the parabolically induced module $Ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1)_K$ (the subscript $K$ denotes  $K-$finite).
Is it the case that the infinitesimal character of Langland's quotient same as that of the $Ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1)$ (with out $K$ subscript?) 
(I am a novice in this subject.)
 A: It is quite some years ago that I used to know this, but as far as I remember the ansewr is YES. 
$J(Q_F, \sigma, \lambda)$ is the unique irreducible quotient of $Ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1)$. In general, if $W$ is a quotient of $V$ then the set of infinitesimal characters of $W$ is a subset of that of $V$, and in this case we know (by irreducibility) that $W$ only has a single infinitesimal character.
What remains to be shown (but is implicit in your question) is that $Ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1)$ only has a single infinitesimal character. This is true and in fact this infinitesimal character can be computed from $\lambda$ and the infinitesimal character of $\sigma$ in a way that I believe is described here: https://dspace.library.uu.nl/handle/1874/1625  (Induced representations and the Langlands classification by Erik van den Ban, lecture notes from a summer school in Edinburgh in the '90s.)
With regards to the $K$-subscript. The space indicated by $K$-subscript is a) dense in the space without the $K$-subscript en b) closed under the action of the Lie-algebra (and so in particular the center of the universal enveloping algebra) on the bigger (no-$K$-subscript) space. These two facts should be enough to see that you can read off the infinitesimal character of the bigger space from the smaller space. 
