We prove the BMV (Bessis, Moussa, Villani, [1]) conjecture, which states that the function $${t \mapsto \mathop{\rm Tr}\exp(A-tB)}$$ , $${t \geqslant 0}$$ , is the Laplace transform of a positive measure on [0,∞) if$A$and$B$are $${n \times n}$$ Hermitian matrices and$B$is positive semidefinite. A semi-explicit representation for this measure is given.

Using the Nakayama functor, we construct an equivalence from a Serre quotient category of a category of finitely generated modules to a category of finite-dimensional modules. We then apply this result to the categories FI_G and VI_q, and answer positively an open question of Nagpal on representation stability theory.

We give bounds for various homological invariants (including Castelnuovo-Mumford regularity, degrees of local cohomology, and injective dimension) of finitely generated VI-modules in the non-describing characteristic case. It turns out that the formulas of these bounds for VI-modules are the same as the formulas of corresponding bounds for FI-modules.

Let $\Lambda$ be a finite dimensional algebra and $G$ be a finite group whose elements act on $\Lambda$ as algebra automorphisms. Under the assumption that $\Lambda$ has a complete set $E$ of primitive orthogonal idempotents, closed under the action of a Sylow $p$-subgroup $S \leqslant G$, we show that the skew group algebra $\Lambda G$ and $\Lambda$ have the same finitistic dimension and the same strong global dimension if the action of $S$ on $E$ is free. Using a homological characterization of piecewise hereditary algebras proved by Happel and Zacharia, we deduce that $\Lambda G$ is piecewise hereditary if and only if $S$ acts freely on $E$ and $\Lambda$ is piecewise hereditary as well.

For any unital separable simple infinite-dimensional nuclear$C$^{∗}-algebra with finitely many extremal traces, we prove that $$ \mathcal{Z} $$ -absorption, strict comparison and property (SI) are equivalent. We also show that any unital separable simple nuclear$C$^{∗}-algebra with tracial rank zero is approximately divisible, and hence is $$ \mathcal{Z} $$ -absorbing.