## Minkowski Problems

Abstract:

Abstract:

Abstract:

When the complex evolution of multi-physics systems is treated as a monolithic object, the time step selection is governed by the most rapidly varying component. However, the appropriate analysis can often reveal a splitting that allows rapid, efficient, and accurate simulation of the full system, by carefully coordinating the uncoupled computation of each subsystem.

Abstract:

The Hitchin system is an integrable system depending on a choice of a smooth complex Riemann surface and a positive integer. It was introduced by Nigel Hitchin in 1987. It received a huge amount of attention, partly because many classically known integrable systems can be embedded into the Hitchin system, partly because the system is related to many areas of mathematics such as algebraic geometry, Langlands program, and mathematical physics.

There are still completely open fundamental questions about one-variable polynomials. One example is Hilbert’s 13th Problem, which concerns formulas for the roots of a polynomial in terms of its coefficients.

Abstract:

In abstract Dirichlet spaces, we present a theory of Besov spaces which is based on the heat semigroup. This approach offers a new perspective on the class of bounded variation functions in settings including Riemannian manifolds, sub-Riemannian manifolds. In rough spaces like fractals it offers totally new research directions. The key assumption on the underlying space is a weak Bakry-Emery type curvature assumption.

The talk is based on joint works with Patricia Alonso-Ruiz, Li Chen, Luke Rogers, Nageswari Shanmugalingam and Alexander Teplyaev.

We regret to inform that the colloquium scheduled for April 5th have been cancelled. Dr. Qiang Du recently underwent knee surgery and was unable to travel. We apologize if this has caused any inconvenience.

If $A$ and $B$ are sets in ${\mathbb R}^n$, then

$$

A+B=\{a+b: a\in A, b\in B\}

$$

is the *Minkowski sum* of $A$ and $B$. It is not hard to see that if $A$ and $B$ are two convex sets then $A+B$ is also convex and $A+A=2A$, for a convex set $A$. Things become much less trivial if we would not assume that set $A$ is convex. Indeed, in such a case $A+A$ is not necessary equal to $2A$. But there is a strong feeling that $2A$ become "more" convex than $A$.