Marc van Leeuwen
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Registered User
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professor of mathematics at the Université de Poitiers (France)
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May 7 |
answered | quasi-minuscule representations |
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Apr 27 |
comment |
p-quotient of an integer partition An 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. |
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Apr 15 |
revised |
bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n clarify formulation |
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Apr 15 |
revised |
bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n oops, inconsistent varible name! |
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Apr 11 |
revised |
bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n one more argument added |
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Apr 11 |
answered | bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n |
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Apr 4 |
revised |
Simply connected algebraic groups and reductive subgroups of maximal rank corrected index calculation |
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Apr 4 |
answered | Simply connected algebraic groups and reductive subgroups of maximal rank |
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Apr 3 |
comment |
bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n It 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? |
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Apr 3 |
comment |
bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n Can you explain the relation with that sequence? OP wants a bijection with all partition of $n$. |
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Apr 2 |
accepted | Randomized algorithm? |
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Mar 30 |
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. |
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Mar 26 |
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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). |
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Mar 23 |
revised |
Geometric proof of Robinson-Schensted-Knuth correspondence? spelling |
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Mar 21 |
awarded | ● Necromancer |
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Mar 16 |
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Generalization’s of Greene’s Theorem for the Robinson-Schensted correspondence If nothing else, you will be able to say something about the shape of the words in that article. |
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Mar 16 |
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The four squares theorem from the Gauss-Legendre three squares theorem The page you linked to just proves the easy part of the three squares theorem. |
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Mar 11 |
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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 |
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Mar 10 |
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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. |
