canonical extensions
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2019 ◽  
Vol Volume 3 ◽  
Author(s):  
Hélène Esnault ◽  
Michel Harris

We prove that the $\ell$-adic Chern classes of canonical extensions of automorphic vector bundles, over toroidal compactifications of Shimura varieties of Hodge type over $\bar{ \mathbb{Q}}_p$, descend to classes in the $\ell$-adic cohomology of the minimal compactifications. These are invariant under the Galois group of the $p$-adic field above which the variety and the bundle are defined. Comment: 28 pages


2018 ◽  
Vol 28 (03) ◽  
pp. 521-541 ◽  
Author(s):  
W. Morton ◽  
C. J. van Alten

It is known that a poset can be embedded into a distributive lattice if, and only if, it satisfies the prime filter separation property. We describe here a class of “prime filter completions” for posets with the prime filter separation property that are completely distributive lattices generated by the poset and preserve existing finite meets and joins. The free completely distributive lattice generated by a poset can be obtained through such a prime filter completion. We also show that every completely distributive completion of a poset with the prime filter separation property is representable as a canonical extension of the poset with respect to some set of filters and ideals. The connections between the prime filter completions and canonical extensions are described and yield the following corollary: the canonical extension of any distributive lattice is the free completely distributive lattice generated by the lattice. A construction that is a variant of the prime filter completion is given that can be used to obtain the free distributive lattice generated by a poset. In addition, it is shown that every distributive lattice extension of the poset can be represented by such a construction. Finally, we show that a poset with the prime filter separation property and the free distributive lattice generated by it generates the same free completely distributive lattice.


2018 ◽  
Vol 79 (1) ◽  
Author(s):  
Guram Bezhanishvili ◽  
Patrick J. Morandi ◽  
Bruce Olberding

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