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Let $G$ be an algebraic group defined over $\mathbb{Q}$ with maximal unipotent radical $N$. Let $\pi$ be an admissible representation of $G(\mathbb{Q}_p)$, we say that this representation is supercuspidal if $\pi/\langle\pi(n)v-v\rangle = 0$. This condition is equivalent to the fact that their matrix coefficients have compact support modulo $Z(\mathbb{Q}_p)$, the centre of $G(\mathbb{Q}_p)$.

Let $K$ be a maximal compact subgroup of $G(\mathbb{Q}_p)$, we say that $\pi$ is spherical or unramified if $\pi^K$, the space of $K$-fixed vectors of $\pi$ has dimension bigger than $0$.

We say that $\pi$ is a discrete series representation if the matrix coefficients of $\pi$ are $2$-integrable modulo $Z(\mathbb{Q}_p)$.

Usually I have found that people divide the admissible representations into three disjoint sets: Supercuspidals, non supercuspidals and discrete series and spherical. Is this decomposition true? Can a supercuspidal representation be spherical? (If not, why not?) Can a discrete series be spherical? (If not, why not?) Why the classical principal series representation (induction of characters of the Torus) are not discrete series?

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    $\begingroup$ Your definition of supercuspidal is not right; one requires that the Jacquet module down to any parabolic, not just to a minimal parabolic (i.e., a Levi of a parabolic with maximal unipotent radical), is trivial. $\endgroup$
    – LSpice
    Commented May 12, 2022 at 0:48

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As it happens, (admissible) supercuspidals cannot be spherical, because (by Borel–Casselman–Matsumoto) admissible repns with Iwahori-fixed vectors have non-trivial maps to and from unramified principal series. The Jacquet-module vanishing condition for supercuspidals is exactly that (via Frobenius Reciprocity, etc.) they admit no (non-zero) homomorphisms to principal series.

Many people would count supercuspidals as discrete series.

For $p$-adic groups, I myself do not know much about discrete series that are not supercuspidals.

And, again, if a discrete series repn were spherical, it would be a sub and quotient of principal series, which is not possible (EDIT … (thanks @Amitay for comments) for hyperspecial maximal compacts. For $\operatorname{GL}_n$, all maximal compacts are conjugate, and are hyperspecial. For $\operatorname{SL}_n$, they are all hyperspecial, but there are $n$ conjugacy classes. At least every (EDIT: thanks @LSpice) reductive group that splits over an unramified extension has at least one (conjugacy class of) hyperspecial maximal compact, but/and some (EDIT: split groups!) do have non-hyperspecials: for example, the affine apartments of $\operatorname{Sp}_4$ have two different types of vertices, one with fewer edges touching it. The latter is not hyperspecial.

Principal series are generically irreducible. For classical groups, we can explicitly compute the $L^2$ norm (mod center) of the (essentially unique) spherical vector, (EDIT for hyperspecial maximal compact) and it's not finite….

Yes, there are some square-integrable subrepns of principal series, at some special values of the parameters. The archimedean case has the well-known examples of holomorphic discrete series subreps of principal series far away from the "unitary range".

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    $\begingroup$ More simply (?) than computing, if $(\sigma, M)$ is smooth with central character $\omega$, $K'$ (any compact open subgroup) admits an Iwahori decomposition with respect to opposite parabolic subgroups $P^\pm$ with common Levi $M$, and $\sigma$ contains a ($K' \cap M$)-fixed vector $v$, then, for every $v^* \ne 0$, the extension by $0$ to $G$ of $k'\cdot m\cdot n^+ \mapsto \langle v^*, \sigma(m)v\rangle$ is a matrix coefficient of $\operatorname{Ind}_P^G \sigma$ that transforms by $\omega$ under the central torus of $M$, hence is not square integrable modulo $Z(G)(F)$ unless $M$ equals $G$. $\endgroup$
    – LSpice
    Commented May 11, 2022 at 20:40
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    $\begingroup$ @LSpice, ah, better, I think. :) $\endgroup$
    – paul garrett
    Commented May 11, 2022 at 20:46
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    $\begingroup$ Many people would count supercuspidals as discrete series - Does anyone not? $\endgroup$
    – Kimball
    Commented May 11, 2022 at 21:55
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    $\begingroup$ @Kimball, I don't really know of any instances, and it would not be harmonious to me, but I might be able to imagine that some people are interested in non-cuspidal discrete series, and at least locally/temporarily want to make "discrete series" exclude supercuspidal. Stranger things have happened. :) $\endgroup$
    – paul garrett
    Commented May 11, 2022 at 21:58
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    $\begingroup$ Two short remarks: 1. There are representations that are square integrable (i.e., discrete series) with Iwahori fixed vector. One is the so-called "Steinberg representation". Those that come from characters of the Iwahori-Hecke algebra are classified in Borel's work (link.springer.com/article/10.1007/BF01390139). 2. Sometimes there are spherical representations that are square integrable. This happens if the maximal compact is not "hyperspecial". See e.g. mathoverflow.net/questions/407630/… $\endgroup$
    – Amitay
    Commented May 12, 2022 at 7:12

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