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a short visit inside algebraic combinatorics
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A short visit inside Algebraic Combinatorics Jean-Christophe Novelli - - PowerPoint PPT Presentation

Introduction Combinatorial Answers Conclusion A short visit inside Algebraic Combinatorics Jean-Christophe Novelli Universit Paris-Est Marne-la-Valle Paris, 2015, CombinatoireS J.-C. Novelli Introduction Combinatorial Answers


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Introduction Combinatorial Answers Conclusion

A short visit inside Algebraic Combinatorics

Jean-Christophe Novelli

Université Paris-Est Marne-la-Vallée

Paris, 2015, CombinatoireS

J.-C. Novelli

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Introduction Combinatorial Answers Conclusion

Combinatorics? Ok. But Algebraic Combinatorics?

J.-C. Novelli

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Introduction Combinatorial Answers Conclusion

Combinatorics? Ok. But Algebraic Combinatorics?

Strict subpart of combinatorics aiming to connect combinatorics and algebra.

J.-C. Novelli

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Introduction Combinatorial Answers Conclusion

Combinatorics? Ok. But Algebraic Combinatorics?

Strict subpart of combinatorics aiming to connect combinatorics and algebra. Provide combinatorial answers to algebraic problems.

J.-C. Novelli

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Introduction Combinatorial Answers Conclusion

Combinatorics? Ok. But Algebraic Combinatorics?

Strict subpart of combinatorics aiming to connect combinatorics and algebra. Provide combinatorial answers to algebraic problems. Also provide algebraic reasons for combinatorial workarounds (in French: brandouillages combinatoires).

J.-C. Novelli

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Introduction Combinatorial Answers Conclusion

Combinatorics? Ok. But Algebraic Combinatorics?

Strict subpart of combinatorics aiming to connect combinatorics and algebra. Provide combinatorial answers to algebraic problems. Also provide algebraic reasons for combinatorial workarounds (in French: brandouillages combinatoires). The Ultimate Goal: provide constructions or proofs requiring (almost) no mathematical knowledge but offering great insights in the theory at work.

J.-C. Novelli

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Introduction Combinatorial Answers Conclusion

Combinatorics? Ok. But Algebraic Combinatorics?

Strict subpart of combinatorics aiming to connect combinatorics and algebra. Provide combinatorial answers to algebraic problems. Also provide algebraic reasons for combinatorial workarounds (in French: brandouillages combinatoires). The Ultimate Goal: provide constructions or proofs requiring (almost) no mathematical knowledge but offering great insights in the theory at work. Our enemies: theories with no examples (algebraic nonsense) and the induction process.

J.-C. Novelli

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Introduction Combinatorial Answers Conclusion Classical examples The dendriform operators The dendriform relations Combinatorial facts about the dendriform operad

Examples

Representation theory! Integer partitions encoding the irreducible representations

  • f the symmetric group,

Standard Young tableaux giving the size of their irreducible representation, Hive models giving insight on Littlewood-Richardson coefficients, Domino tableaux, ...

J.-C. Novelli

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Examples

Representation theory! Integer partitions encoding the irreducible representations

  • f the symmetric group,

Standard Young tableaux giving the size of their irreducible representation, Hive models giving insight on Littlewood-Richardson coefficients, Domino tableaux, ... Today: Operads!

J.-C. Novelli

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Cut the deck

Start with a deck of card. Cut it in half and shuffle together both

  • subdecks. What happens?

With 52 cards and two decks of say 26 cards, we get 52

26

  • different possibilities.

Do it again. And again. And again... Is it "random" after 6 shuffles?

J.-C. Novelli

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Cut the deck

Start with a deck of card. Cut it in half and shuffle together both

  • subdecks. What happens?

With 52 cards and two decks of say 26 cards, we get 52

26

  • different possibilities.

Do it again. And again. And again... Is it "random" after 6 shuffles? Oh sorry! I’m doing algebraic combinatorics not asymptotics. Too bad, the question is so nice...

J.-C. Novelli

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Back to Algebraic combinatorics

Now cut the deck into three parts. Shuffling A and B first then with C brings other possibilities than shuffling B and C then with A?

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Back to Algebraic combinatorics

Now cut the deck into three parts. Shuffling A and B first then with C brings other possibilities than shuffling B and C then with A? Of course not!

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Back to Algebraic combinatorics

Now cut the deck into three parts. Shuffling A and B first then with C brings other possibilities than shuffling B and C then with A? Of course not! So this operation is commutative and associative!

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Cut the shuffle

Consider two words u = u1 . . . un v = v1 . . . vp Their shuffle u v is u v := (u1 . . . un−1 v).un + (u v1 . . . vp−1).vp.

J.-C. Novelli

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Cut the shuffle

Consider two words u = u1 . . . un v = v1 . . . vp Their shuffle u v is u v := (u1 . . . un−1 v).un + (u v1 . . . vp−1).vp. This equation is clearly a sum of two parts. Separate these parts. u < v := (u1 . . . un−1 v).un u > v := (u v1 . . . vp−1).vp

J.-C. Novelli

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Kill the commuter

With these rules, u < v = v > u and nothing interesting can be expected.

J.-C. Novelli

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Kill the commuter

With these rules, u < v = v > u and nothing interesting can be expected. So define < and > as the components of the shifted shuffle: let u[k] be (u1 + k, . . . , un + k) and define u ⋒ v = u v[|u|].

J.-C. Novelli

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Kill the commuter

With these rules, u < v = v > u and nothing interesting can be expected. So define < and > as the components of the shifted shuffle: let u[k] be (u1 + k, . . . , un + k) and define u ⋒ v = u v[|u|]. Now 1 < 1 = 21 and 1 > 1 = 12.

J.-C. Novelli

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Introduction Combinatorial Answers Conclusion Classical examples The dendriform operators The dendriform relations Combinatorial facts about the dendriform operad

Kill the commuter

With these rules, u < v = v > u and nothing interesting can be expected. So define < and > as the components of the shifted shuffle: let u[k] be (u1 + k, . . . , un + k) and define u ⋒ v = u v[|u|]. Now 1 < 1 = 21 and 1 > 1 = 12. Please welcome the dendriform operators!

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Left and right

The dendriform operators < and > are not associative: they cannot both be or their sum wouldn’t.

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Left and right

The dendriform operators < and > are not associative: they cannot both be or their sum wouldn’t. With three words, there are 8 expressions using < and >:        (u < v) < w u < (v < w) (u < v) > w u < (v > w) (u > v) < w u > (v < w) (u > v) > w u > (v > w) Do they have some relations?

J.-C. Novelli

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Left and right

The dendriform operators < and > are not associative: they cannot both be or their sum wouldn’t. With three words, there are 8 expressions using < and >:        (u < v) < w u < (v < w) (u < v) > w u < (v > w) (u > v) < w u > (v < w) (u > v) > w u > (v > w) Do they have some relations? Of course: their sum is associative so both sums of both columns are equal.

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Left and right

The dendriform operators < and > are not associative: they cannot both be or their sum wouldn’t. With three words, there are 8 expressions using < and >:        (u < v) < w u < (v < w) (u < v) > w u < (v > w) (u > v) < w u > (v < w) (u > v) > w u > (v > w) Do they have some relations? Of course: their sum is associative so both sums of both columns are equal. Do they have other relations?

J.-C. Novelli

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Right, right

In all expressions        (u < v) < w u < (v < w) (u < v) > w u < (v > w) (u > v) < w u > (v < w) (u > v) > w u > (v > w)

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Right, right

In all expressions the last letter comes from a given word:        (u < v) < w ⇐ u u < (v < w) ⇐ u (u < v) > w ⇐ w u < (v > w) ⇐ u (u > v) < w ⇐ v u > (v < w) ⇐ v (u > v) > w ⇐ w u > (v > w) ⇐ w

J.-C. Novelli

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Right, right

In all expressions the last letter comes from a given word:        (u < v) < w ⇐ u u < (v < w) ⇐ u (u < v) > w ⇐ w u < (v > w) ⇐ u (u > v) < w ⇐ v u > (v < w) ⇐ v (u > v) > w ⇐ w u > (v > w) ⇐ w Hence    (u < v) < w = u < (v < w) + u < (v > w) (u > v) < w = u > (v < w) (u > v) < w + (u > v) > w = u > (v > w)

J.-C. Novelli

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Right, right

In all expressions the last letter comes from a given word:        (u < v) < w ⇐ u u < (v < w) ⇐ u (u < v) > w ⇐ w u < (v > w) ⇐ u (u > v) < w ⇐ v u > (v < w) ⇐ v (u > v) > w ⇐ w u > (v > w) ⇐ w Hence    (u < v) < w = u < (v < w) + u < (v > w) (u > v) < w = u > (v < w) (u > v) < w + (u > v) > w = u > (v > w) And there cannot be other relations with 3 words.

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Right and Wrong

Are there relations with 4 words not coming from the previous

  • nes?

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Right and Wrong

Are there relations with 4 words not coming from the previous

  • nes?

Well, no. Is there a good explanation for this?

J.-C. Novelli

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Right and Wrong

Are there relations with 4 words not coming from the previous

  • nes?

Well, no. Is there a good explanation for this? First, write any dendriform expression as a binary tree: < > u v w = (u > v) < w.

J.-C. Novelli

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Forbidden rights

Our relations can be written as rewriting rules on trees:                                  < < → < < + < > > < → < > > > → > > + > <

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Left overs

How many non-rewritable trees are there? Split them according to their root: S< = xS(S − S<) S> = xS. so that S = 1 + 2xS + x2S2 = (1 + xS)2. And one easily finds that S is the g.s. of the Catalan numbers.

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What’s right and what’s left (to be done)

All dendriform expressions with n operands can be rewritten as Catalan different (non-rewritable) trees. So the dendriform

  • perad on 1 generator (all leaves of the trees equal to 1) has

graded dimension at most Catalan. Converse property?

J.-C. Novelli

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What’s right and what’s left (to be done)

All dendriform expressions with n operands can be rewritten as Catalan different (non-rewritable) trees. So the dendriform

  • perad on 1 generator (all leaves of the trees equal to 1) has

graded dimension at most Catalan. Converse property? With the help of combinatorics!

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Different rights

Consider the five different trees with n = 3: < > > < > > < < < > When applied to 1 on each leaf, one gets 132 + 312 213 123 321 231

J.-C. Novelli

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Different rights

Consider the five different trees with n = 3: < > > < > > < < < > When applied to 1 on each leaf, one gets 132 + 312 213 123 321 231 Note that these are disjoint sets!

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Characterize these sets

Loday proved that two permutations are in the same subset iff their inverses satisfy that their decreasing trees have same shape. Is that explicit (combinatorial) enough?

J.-C. Novelli

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Characterize these sets

Loday proved that two permutations are in the same subset iff their inverses satisfy that their decreasing trees have same shape. Is that explicit (combinatorial) enough? Yes!

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Characterize these sets

Loday proved that two permutations are in the same subset iff their inverses satisfy that their decreasing trees have same shape. Is that explicit (combinatorial) enough? Yes! And No...

J.-C. Novelli

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Decreasing trees

If σ = 25481376, its decreasing tree is

  • 8
  • 5
  • 7
  • 2
  • 4
  • 3
  • 6
  • 1

J.-C. Novelli

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Decreasing trees

If σ = 25481376, its decreasing tree is

  • 8
  • 5
  • 7
  • 2
  • 4
  • 3
  • 6
  • 1

σ−1 = 51632874.

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Decreasing trees

If σ = 25481376, its decreasing tree is

  • 8
  • 5
  • 7
  • 2
  • 4
  • 3
  • 6
  • 1
  • 4
  • 2
  • 7
  • 1
  • 3
  • 6
  • 8
  • 5

The tree on the right is the Binary Search tree of σ−1 = 51632874.

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Decreasing trees

If σ = 25481376, its decreasing tree is

  • 8
  • 5
  • 7
  • 2
  • 4
  • 3
  • 6
  • 1
  • 4
  • 2
  • 7
  • 1
  • 3
  • 6
  • 8
  • 5

The tree on the right is the Binary Search tree of σ−1 = 51632874. So Loday’s result is equivalent to: two permutations have the same image iff they have the same BST.

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From BSTs to combinatorics on words

Can someone guess (without computations) other permutations having this same tree as BST?

  • 4
  • 2
  • 7
  • 1
  • 3
  • 6
  • 8
  • 5

J.-C. Novelli

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From BSTs to combinatorics on words

Can someone guess (without computations) other permutations having this same tree as BST?

  • 4
  • 2
  • 7
  • 1
  • 3
  • 6
  • 8
  • 5

Hint: the extremal ones are 13256874 and 85673124.

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From BSTs to combinatorics on words

Can someone guess (without computations) other permutations having this same tree as BST?

  • 4
  • 2
  • 7
  • 1
  • 3
  • 6
  • 8
  • 5

Hint: the extremal ones are 13256874 and 85673124. Complete answer: they are the linear extensions of the tree and an interval of the weak order on permutations.

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A monoid on trees, a sylvester monoid

Given a permutation, finding all permutations with the same BST does not require building the BST itself! It amounts to compute the transitive closure of the following rewriting rules: ac . . . b ≡ ca . . . b for all a < b < c. This is the sylvester monoid.

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From monoids to operads

On these objects, a one-line proof shows that < and > of two sylvester classes is a union of sylvester classes. The converse is also easy to prove: any sylvester class can be

  • btained as a linear combination of the dendriform operad

generated by 1. Write the dendriform expression of their corresponding tree.

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From monoids to operads

On these objects, a one-line proof shows that < and > of two sylvester classes is a union of sylvester classes. The converse is also easy to prove: any sylvester class can be

  • btained as a linear combination of the dendriform operad

generated by 1. Write the dendriform expression of their corresponding tree. So the free object has dimension smaller than Catalan and one

  • f its (maybe nonfree) instance has dimension greater than

Catalan.

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From monoids to operads

On these objects, a one-line proof shows that < and > of two sylvester classes is a union of sylvester classes. The converse is also easy to prove: any sylvester class can be

  • btained as a linear combination of the dendriform operad

generated by 1. Write the dendriform expression of their corresponding tree. So the free object has dimension smaller than Catalan and one

  • f its (maybe nonfree) instance has dimension greater than

Catalan. So the free dendriform operad has dimension Catalan exactly. And so is our instance on permutations which is btw free too.

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Byproducts

Simple proofs using no mathematical knowledge,

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Byproducts

Simple proofs using no mathematical knowledge, Easier way of designing generalizations: all combinatorial

  • bjects have analogs of their own:

permutations: packed words: i ∈ w → i − 1 ∈ w, parking functions, signed permutations, . . . binary trees: Cayley trees, Cambrian trees, . . . BST and Decreasing trees: repeated letters, fixed number

  • f repeated letters, . . .

sylvester monoid: plactic, hyposylvester, metasylvester, . . .

J.-C. Novelli

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Byproducts

Simple proofs using no mathematical knowledge, Easier way of designing generalizations: all combinatorial

  • bjects have analogs of their own:

permutations: packed words: i ∈ w → i − 1 ∈ w, parking functions, signed permutations, . . . binary trees: Cayley trees, Cambrian trees, . . . BST and Decreasing trees: repeated letters, fixed number

  • f repeated letters, . . .

sylvester monoid: plactic, hyposylvester, metasylvester, . . .

Combinatorial proofs available and reasonable for very technical examples (quadrigebras),

J.-C. Novelli

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Byproducts

Simple proofs using no mathematical knowledge, Easier way of designing generalizations: all combinatorial

  • bjects have analogs of their own:

permutations: packed words: i ∈ w → i − 1 ∈ w, parking functions, signed permutations, . . . binary trees: Cayley trees, Cambrian trees, . . . BST and Decreasing trees: repeated letters, fixed number

  • f repeated letters, . . .

sylvester monoid: plactic, hyposylvester, metasylvester, . . .

Combinatorial proofs available and reasonable for very technical examples (quadrigebras), Hook formulas and (q, t)-hooks now available without efforts,

J.-C. Novelli

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Introduction Combinatorial Answers Conclusion

Byproducts

Simple proofs using no mathematical knowledge, Easier way of designing generalizations: all combinatorial

  • bjects have analogs of their own:

permutations: packed words: i ∈ w → i − 1 ∈ w, parking functions, signed permutations, . . . binary trees: Cayley trees, Cambrian trees, . . . BST and Decreasing trees: repeated letters, fixed number

  • f repeated letters, . . .

sylvester monoid: plactic, hyposylvester, metasylvester, . . .

Combinatorial proofs available and reasonable for very technical examples (quadrigebras), Hook formulas and (q, t)-hooks now available without efforts, Noncommutative setting where algebraic proofs come easily, multistatistics on permutations for free, . . .

J.-C. Novelli

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Open problems

Combinatorial questions: Study more examples, Fill in the blanks: describe combinatorially and enumerate the intervals of orders on permutations, packed words, parking functions, . . . Algebraic or geometrical questions: Provide a general setting where the combinatorial algebras are related to polytopes, Get a non semi-simple algebra whose representation theory rings encode the (commutative) Catalan algebra, Find a polytope encoding clearly the algebra on parking functions, . . .

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Bibliography

  • F. BERGERON, Combinatorics of r-Dyck paths, r-Parking

functions, and the r-Tamari lattices, arXiv:1202.6269.

  • F. HIVERT, J.-C. NOVELLI, and J.-Y. THIBON, The algebra
  • f binary search trees, Theoretical Computer Science 339

(2005), 129–165. J.-L. LODAY, Dialgebras, arXiv:0102.053. J.-L. LODAY and M. O. RONCO, Hopf algebra of the planar binary trees, Adv. Math. 139 (1998) n. 2, 293–309. J.-C. NOVELLI m-dendriform algebras, arXiv:1406.1616. J.-C. NOVELLI and J.-Y. THIBON, Hopf Algebras of m-permutations, (m + 1)-ary trees, and m-parking functions, arXiv:1403.5962.

J.-C. Novelli