Microbial transformations in phosphonate biosynthesis and catabolism, and their importance in nutrient cycling

Phosphorus cycling in the biosphere has traditionally been thought to involve almost exclusively transformations of the element in its pentavalent oxidation state. Recent evidence, however, suggests that a significant fraction of environmental phosphorus may exist in a more reduced form. Most abundant of these reduced phosphorus compounds are the phosphonates, with their direct carbon–phosphorus bonds, and striking progress has recently been made in elucidating the biochemistry of microbial phosphonate transformations. These advances are now presented in the context of their contribution to our understanding of phosphorus biogeochemistry and of such diverse fields as the productivity of the oceans, marine methanogenesis and the discovery of novel microbial antimetabolites.

Mr Micawber anticipates Feist: transformations of mental labour.

This paper traces transformations of mental labour and its distribution between human and machine from Mr Micawber's parody of arithmetical calculation (result happiness) in the mid-19th century to the late 20th century judgment of the Supreme Court of United States in Feist v. Rural (1991), concerned with copyright in databases.

The proof of Kontsevich's periodicity conjecture on noncommutative birational transformations

For an arbitrary associative unital ring RR, let J1J1 and J2J2 be the following noncommutative, birational, partly defined involutions on the set M3(R)M3(R) of 3×33×3 matrices over RR: J1(M)=M−1J1(M)=M−1 (the usual matrix inverse) and J2(M)jk=(Mkj)−1J2(M)jk=(Mkj)−1 (the transpose of the Hadamard inverse).

We prove the surprising conjecture by Kontsevich that (J2∘J1)3(J2∘J1)3 is the identity map modulo the DiagL×DiagRDiagL×DiagR action (D1,D2)(M)=D−11MD2(D1,D2)(M)=D1−1MD2 of pairs of invertible diagonal matrices. That is, we show that, for each MM in the domain where (J2∘J1)3(J2∘J1)3 is defined, there are invertible diagonal 3×33×3 matrices D1=D1(M)D1=D1(M) and D2=D2(M)D2=D2(M) such that (J2∘J1)3(M)=D−11MD2(J2∘J1)3(M)=D1−1MD2.