Let $G$ be an unramified $p$-adic reductive group, $K$ a hyperspecial maximal compact subgroup, and $K/K^{+}\cong G(\mathbb{F}_q)$. Restricting an irreducible representation $V$ of $G(F)$ to $K$ gives a $G(\mathbb{F}_q)$-representation $V^{K^{+}}$. Prasad asked which $V$ have $V^{K^{+}}$ containing the Steinberg representation (or more generally any $\pi_{\lambda}$ attached to a Weyl group representation $\lambda$). This talk answers the first question completely. We also show that the irreducible components of $K^{+}$-fixed subspace lies in the Harish-Chandra principal series. The method also extends to the general case, providing a computable framework for the further study.

The talk will be streamed through "Tencent": https://meeting.tencent.com/dm/X6fMLc1Q60b3. meeting code: 148-207-215 To join via a web browser: go to https://voovmeeting.com/; click "Log in" in the upper-right corner and login either by a Google/Apple account, or by a verification code received on phone or email; click "Join now" in the upper-right corner and type the number of the meeting 148-207-215 in the "Enter meeting ID" field; when prompted, select "Join from browser".

Let $\pi$ be some set of prime numbers. A finite group is called a $\pi$-group if all the prime divisors of its order belong to $\pi$. Following Wielandt, it is said that the $\pi$-Sylow theorem is true for a finite group $G$ if in $G$ all maximal $\pi$-subgroups are conjugate; if the $\pi$-Sylow theorem is true for every subgroup of $G$, then it is said that the strong $\pi$-Sylow theorem is true for $G$. It is known that the strong $\pi$-Sylow theorem is true for a group if and only if it is true for every non-Abelian compositional factor of this group. The question of which finite simple non-Abelian groups the strong $\pi$-Sylow theorem is true was posed by Wielandt in 1979. By now, the answer is known for sporadic and alternating groups and Lie type groups of rank 1. The report will discuss new ideas for solving this problem for classical Lie type groups.

The talk will be streamed through “Kontur Talk”: