The Faltings height is a useful invariant for addressing questions in arithmetic geometry. In his celebrated proof of the Mordell and Shafarevich conjectures, Faltings shows the Faltings height satisfies a certain Northcott property, which allows him to deduce his finiteness statements. In this work we prove a new Northcott property for the Faltings height. Namely we show, assuming the Colmez Conjecture and the Artin Conjecture, that there are finitely many CM abelian varieties of a fixed dimension which have bounded Faltings height. The technique developed uses new tools from integral p-adic Hodge theory to study the variation of Faltings height within an isogeny class of CM abelian varieties. In special cases, we are able to use these techniques to moreover develop new Colmez-type formulas for the Faltings height.[-]

I will explain some new connections between the $abc$ conjecture and modular forms. In particular, I will outline a proof of a new unconditional estimate for the $abc$ conjecture, which lies beyond the existing techniques in this context. The proof involves a number of tools such as Shimura curves, CM points, analytic number theory, and Arakelov geometry. It also requires some intermediate results of independent interest, such as bounds for the Manin constant beyond the semi-stable case. If time permits, I will also explain some results towards Szpiro's conjecture over totally real number fields which are compatible with the discriminant term appearing in Vojta's conjecture for algebraic points of bounded degree.[-]

It is classical that, for example, there is a simple abelian variety of dimension $4$ which is not the jacobian of any curve of genus $4$, and it is not hard to see that there is one defined over the field of all algebraic numbers $\overline{\bf Q}$. In $2012$ Chai and Oort asked if there is a simple abelian fourfold, defined over $\overline{\bf Q}$, which is not even isogenous to any jacobian. In the same year Tsimerman answered ''yes''. Recently Zannier and I have done this over the rationals $\bf Q$, and with ''yes, almost all''. In my talk I will explain ''almost all'' the concepts involved.[-]