# Marc van Leeuwen

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 Name Marc van Leeuwen Member for 1 year Seen 2 hours ago Website Location Poitiers, France Age 53
professor of mathematics at the Université de Poitiers (France)
 May7 answered quasi-minuscule representations Apr27 comment p-quotient of an integer partitionAn intuitive description of the core-quotient bijection, illustrations, and precise definitions of the notions involved, can be found in section 2 of my paper "Edge sequences, ribbon tableaux, and an action of affine permutations" Europ. J. Combinatorics 20 (1999). It is available from my home page www-math.univ-poitiers.fr/~maavl. Apr15 revised bijection between number of partitions of 2n satisfying certain conditions with number of partitions of nclarify formulation Apr15 revised bijection between number of partitions of 2n satisfying certain conditions with number of partitions of noops, inconsistent varible name! Apr11 revised bijection between number of partitions of 2n satisfying certain conditions with number of partitions of none more argument added Apr11 answered bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n Apr4 revised Simply connected algebraic groups and reductive subgroups of maximal rankcorrected index calculation Apr4 answered Simply connected algebraic groups and reductive subgroups of maximal rank Apr3 comment bijection between number of partitions of 2n satisfying certain conditions with number of partitions of nIt would be helpful to have a description of conormal $j$ nodes. Am I correct in assuming that $j$ is to be interpreted modulo $2$ here? Also, with a definition of conormal nodes I found here www.iazd.uni-hannover.de/~bessen/tensalt.ps (where removable nodes below it have to be paired with intermediate addable nodes), it would seem that adding the lowest such nodes would not result in an invertible operation (having become removable, they could pair up with another addable node above). Are you sure you did not mean adding the highest conormal $j$ nodes? Apr3 comment bijection between number of partitions of 2n satisfying certain conditions with number of partitions of nCan you explain the relation with that sequence? OP wants a bijection with all partition of $n$. Apr2 accepted Randomized algorithm? Mar30 comment bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n@Rekha Biswal: c0 is just the empty partition, or diagram. Mar26 comment Geometric proof of Robinson-Schensted-Knuth correspondence?@darij grinberg: For the Knuth correspondence, since the matrix can have entries larger than one, the "growth" of shapes can correspondingly be by more than one square; the natural condition is growth by horizontal strips. The original reference is S. Fomin, “Schur operators and Knuth correspondences”, J. Combin. Theory, Ser. A 72,(1995), pp. 277–292, though I might unmodestly suggest that my discussion in section 2.2 of "Spin-preserving Knuth correspondences for ribbon tableaux" combinatorics.org/Volume_12/Abstracts/… is maybe more accessible (you can skip section 1). Mar23 revised Geometric proof of Robinson-Schensted-Knuth correspondence?spelling Mar21 awarded ● Necromancer Mar16 comment Generalization’s of Greene’s Theorem for the Robinson-Schensted correspondenceIf nothing else, you will be able to say something about the shape of the words in that article. Mar16 comment The four squares theorem from the Gauss-Legendre three squares theoremThe page you linked to just proves the easy part of the three squares theorem. Mar11 comment When is an algebra of commuting matrices (contained in one) generated by a single matrix?And see [this answer](mathoverflow.net/questions/29087/…) to another question. Notably it is indeed a theorem (Gerstenhaber 1961) that an algebra generated by two commuting matrices cannot have dimension greater than n (but it can be of dimension n while not being single-generated), and it is an open question whether an algebra generated by 3 matrices can. – Marc van Leeuwen 0 secs ago Mar10 comment Do these matrix rings have non-zero elements that are neither units nor zero divisors?Actually, as far as I've seen definitions, $0$ is usually not called a zero divisor (so an integral domain is a non-trivial commutative ring without zero divisors). But in the current setting saying "zero divisor of zero" all the time would indeed by tiresome.