Are all cuspidals induced?

Question

This is a follow-up to this question by Marc Palm asked 7 years ago:

Let $K$ be a finite extension of $\mathbb{Q}_p$, and $G$ a reductive group over $K$. Is every irreducible cuspidal representation induced from an open, compact-mod-center subgroup?

My specific questions are:

Question 1: Have there been any new results on this question in the last 7 years?

(Meta remark: I don't know of a different possibility to draw attention to an MO question again, asking for new results, other than asking the question again.)

Question 2: Do people generally believe that this is true, or are they expecting that there are some weird counterexamples somewhere?

This is of course a very hazy question, but sometimes there is a common "folklore" belief in a community about some open questions. I don't belong to the community of experts on this, that's why I'm asking.

Question 3: I once heard the name Bernstein Conjecture for the conjecture that the answer is "yes". Is this a common name?

Answer 1

Question 1. Yes indeed.

a) There are new results for classical groups and their inner forms (works of Shaun Stevens, Daniel Skodlerack, ...). In particular Skodlerack proved that in the case of "quaternionic forms" of classical groups, in residue characteristic not $2$, any irreducible supercuspidal representation is induced.

Daniel Skodlerack, "Cuspidal irreducible representations of quaternionic forms of p-adic classical groups for odd p" arXiv:1907.02922 math.RT math.NT

b) Jessica Fintzen has improved a result of J.K. Yu by proving that for general reductive groups, J.K. Yu's construction gives all supercuspidal representations as induced representations when the residue characteristic does not divide the order of the Weyl group.

Jessica Fintzen, "Tame cuspidal representations in non-defining characteristics" arXiv:1905.06374

She has also corrected errors in Yu's work and has improved his construction (cf. her Arxiv papers).

c) Martin Weissman gave a very short and elegant proof that for any rank $1$ reductive group, irreducible supercuspidal are induced. This gave new instances of induced supercuspidal representations. His proof is based on a deep result of Schneider and Stuhler (cf. IHES paper).

Martin Weissman, "An induction theorem for groups acting on trees", arXiv:1808.08944 .

Question 2. I think the community of experts does indeed think that the conjecture should hold true.

Question 3. I have personnally never heard of that.

Answer 2

Quite late to the party, however, to add to Paul's answer, for Question 3, the Bernstein conjecture (as far as I am aware) relies on the idea that we can decompose the category of smooth representations of your group $R(G(F))$ into the direct sum of indecomposable subcategories. What this is saying is that if we were to take a random smooth representation $\pi$ of $G$ and write it as a direct sum of pieces, each piece is contained in one block. These blocks are called Bernstein Blocks. We can categorize these in terms of Levi subgroups $M$ (and a supercuspidal representation $\sigma$ of the Levi).

The Bernstein conjecture says that for each block $R_{\{M, \sigma\}}(G(F))$, there exists a compact open subgroup $K$ with representation $\kappa$ such that $(\pi, V)$ belongs to $R_{\{M, \sigma\}}(G(F))$ if and only if $\text{Res}_K \pi$ contains $\kappa.$