Video Lectures

Given a K3 surface XX over a number field KK, we prove that the set of primes of KK where the geometric Picard rank jumps is infinite, assuming that XX has everywhere potentially good reduction. This result is formulated in the general framework of...

A theorem of Bernstein identifies the center of the affine Hecke algebra of a reductive group GG with the Grothendieck ring of the tensor category of representations of the dual group G∨G∨. Gaitsgory constructed a functor which categrorifies this...

Some years ago, I proved with Shulman and Sørensen that precisely 12 of the 17 wallpaper groups are matricially stable in the operator norm. We did so as part of a general investigation of when group C∗C∗-algebras have the semiprojectivity and weak...

Expander graphs are graphs which simultaneously are both sparse and highly connected. The theory of expander graphs received a lot of attention in the past half a century, from both computer science and mathematics. In recent years, a new theory of...

Valiant (1980) showed that general arithmetic circuits with negation can be exponentially more powerful than monotone ones. We give the first qualitative improvement to this classical result: we construct a family of polynomials P-n in n variables...

One of the earliest fundamental applications of Lagrangian Floer theory is detecting the non-displaceablity of a Lagrangian submanifold. Many progress and generalisations have been made since then but little is known when the Lagrangian submanifold...

Continuation of the talk from Feb 4.

This is the second in a series of talks on "two realizations." We discuss equivariantization and de-equivariantization for a group GG, which relate categories with GG-actions and categories with RepG-actions, and how this relates to the notion of...

We show that the direct product of an infinite, finitely generated Kazhdan Property (T) group and a finitely presented, not residually finite amenable group admits no sofic approximation by expander graphs. Joint work with Andreas Thom.