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I apologize in advance if the answer to this question is well-known to experts.

So let $F$ be a $p$-adic field, and $G$ a reductive group over $F$. Let $P$ be an $F$-parabolic subgroup of $G$ and $M$ the Levi of $P$. (For convenience, I just use the same notations for an algebraic group and a group of rational points.) We know that roughly, irreducible admissible (smooth) representations of $G$ come from (normalised) parabolic induction: if $\pi$ is a non-cuspidal irreducible representation of $G$, then $\pi$ can be embedded into some $I_Q^G(\sigma)$ for some representation $\sigma$ of Levi of some proper parabolic $Q$.

Meanwhile, we know what is local Langlands correspondence. Let $WD(F):=W(F)\times\operatorname{SL}_2(\mathbb{C})$ be the Weil-Deligne group, then an $L$-parameter of $G$ is a conjugacy class of homomorphism $$\phi:WD(F)\rightarrow {^LG}.$$ Suppose now we have an irreducible representation $\sigma$ of $M$ (setting as above), and then we consider $I_P^G(\sigma)$, and take some irreducible composition factor $\pi$, which is an irreducible representation of $G$. Suppose $\phi_\sigma:WD(F)\rightarrow {^LM}$ is the $L$-parameter of $\sigma$. We know that $M\subset G$ provides us with ${^LM}\subset {^LG}$, so we get a parameter of $G$: $WD(F)\rightarrow {^LM}\subset {^LG}$.

Question: some expert told me that this new parameter is just the parameter for the Langlands quotient of $I_P^G(\sigma)$. This is related to the Langlands classification of $p$-adic groups (for which I'm a total beginner). I want to ask where can I find the proof of this fact in the literature? And I also want to ask: in general, is there any systematic relation between the parameter of $\sigma$ and the parameter of $\pi$ (a composition factor of the parabolic induction)? How to characterise tempered/discrete series/cuspidal representations via parameters? (I know such facts about $\operatorname{GL}_n$, but I'm curious about whether there's a general result for general reductive groups, which so far I didn't find in any paper, but seems to be folklore to experts...)

Again, I want to apologize if this question is standard for veterans/experts. Thanks a lot in advance for any explanation!

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I think what you say is a part of the local Langlands Conjecture. See Conjecture 4.1 (7)(8)(10)in Kaletha and Taibi's Lecture notes on LLC for IHES 2022. The local Langlands conjectures for different groups should be compatible with parabolic inductions. As a special case, you can consider spherical representations and calculate their Satake parameters.

For your other problem, here is my understanding (these can also be found in the lecture mentioned above):

  • A Langlands parameter is tempered if its image under the projection to $\widehat{G}(\mathbb{C})$ is bounded (Thanks to Kenta's comment, here "the image of the Weil group is bounded" should be more precise). By the LLC for quasi-split groups, there should be a surjective map from the set of isomorphism classes of tempered irreducible admissible representations to the set of equivalence classes of tempered parameters.
  • We should have: the image of a tempered Langlands parameter is not contained in any proper parabolic subgroup of the L-group (equivalently, its centralizer in $\widehat{G}(\mathbb{C})$ is finite modulo $Z(\widehat{G})^{G_{F}}$) if and only if one element (equivalently, every element) in its L-packet is essentially square-integrable. We call this kind of parameter essentially discrete.
  • Here we just assume the coefficient field $C=\mathbb{C}$ or $\mathbb{Q}_{l},l\neq p$ so that we can view cuspidal and supercuspidal as the same thing. We should have: a Langlands parameter is essentially discrete and trivial on Deligne's $\mathrm{SL}_{2}(\mathbb{C})$ if and only if every member in its L-packet is supercuspidal. However, even for $\mathrm{GL}_{n}$ it can happen that for an essentially discrete Langlands parameter nontrivial on $\mathrm{SL}_{2}(\mathbb{C})$ its L-packet contains a supercuspidal representation. Of course you can try to describe the correspondence in terms of enhanced Langlands parameters.
  • For discrete series, it should correspond to some enhanced Langlands parameter consisting of an essentially discrete Langlands parameter and a character on its component group satisfying some extra condition. Notice that here we need the refined LLC.

I am also not so familiar with the representation theory of $p$-adic groups so my answer may contain some errors and comments are welcome!

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  • $\begingroup$ Thanks a lot! I think the problem is that I didn't even know a precise statement of LLC, and a main ingredient is its compatibility with parabolic induction. What made me puzzled before was that to state such compatibility, since the Levi can be some other reductive groups, now you need LLC for another reductive group (I just thought you should formalise it for each single $G$). But now it seems that this is so. Thank you very much for the recommendation of IHES video series! $\endgroup$
    – youknowwho
    Commented Feb 8, 2023 at 12:47
  • $\begingroup$ By the way, is the refined version due to Vogan? Can you recommend any reference for the refined LLC? $\endgroup$
    – youknowwho
    Commented Feb 8, 2023 at 12:49
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    $\begingroup$ I think they also have lecture notes but I am not sure they have already posted them on their homepages. And yes the refined version is due to Vogan, this version is also discussed by Taibi. From my point of view, this version considers the component group in order to distinguish different representations in the same L-packet. I remember Aubert also wrote a paper on this and its name should be Langlands conjecture and Springer correspondence. If you are only interested in classical groups, this refined LLC is stated in detail in the famous GGP astérisque paper. $\endgroup$
    – Erica
    Commented Feb 8, 2023 at 13:15
  • $\begingroup$ Thank you very much! $\endgroup$
    – youknowwho
    Commented Feb 8, 2023 at 13:34
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    $\begingroup$ @KentaSuzuki It means the image of the Weil group $W_{F}$ is bounded, which is equivalent to that the image of the Frobenius lies in a compact subgroup of $\widehat{G}(\mathbb{C})$. $\endgroup$
    – Erica
    Commented Apr 4, 2023 at 8:17

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