subcaptionfont+=small,labelformat=parens,labelsep=space,skip=6pt,list=0,hypcap=0 subcaption ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, 2/16/2016

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1 subcaptiofot+=small,labelformat=pares,labelsep=space,skip=6pt,list=0,hypcap=0 subcaptio ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/06. Self-cojugate Partitios Recall that, give a partitio λ, we may speak of the cojugate of λ, deoted λ. This is most easily described i terms of Youg diagrams; the diagram of λ is the traspose of the diagram of λ. We may further discuss the cocept of self-cojugate partitios; that is, partitios λ such that λ = λ. Propositio. Let be a positive iteger. The we have the followig equality: p( self-cojugate) = p( distict odd parts) Proof. Cosider the sets correspodig to the coditioed partitio fuctio above; we aim to costruct a bijectio betwee the two. To see how such a bijectio may be costructed, cosider the partitio λ = (5, 5, 3,, ). This has a Youg diagram as see i Figure a. We ow defie a ew term, hook, i a Youg diagram. A hook is a L-shaped part of a diagram reachig from the far right edge to the bottom edge; Figure b highlights the each hook i a differet color. We refer to the umber of cells i a hook as its legth. For example, the red hook has legth 9, the gree hook has legth 6, ad the blue hook has legth. a. Youg Diagram b. Hooks Figure. Istace of Propositio Observe that all hook legths are odd if the partitio is self-cojugate, ad oe ca recostruct a self-cojugate partitio from its hook legths. Furthermore, the hook legths Prepared by Adrew Norto

2 ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/06 of a partitio themselves form a partitio! If the partitio is self-cojugate, the all parts of this hook legth partitio are distict ad odd. So, the bijectio we costruct performs the followig mappig: λ partitio cosistig of hook legths of λ. Euler s Petagoal Number Idetity.. Triagular Numbers. Before delvig ito the petagoal umbers ad their relatioship to the partitio fuctio, let us first cosider the triagle umbers which are more likely to be familiar to the reader. Cosider Figure, which displays four triagles formed of dots. The triagle associated with the iteger j has j dots at a side. Notice how the jth triagle is formed by addig a row of j dots to the (j )st triagle. This gives rise to a defiitio usig a commo summatio formula. Defiitio. For o-egative itegers j, the jth triagle umber, deoted j, is defied as follows: j(j + ) j = j = a. j = b. j = c. j = 3 d. j = 4 Figure. Triagle Numbers.. Petagoal Numbers. As mathematicias ted to do, we may exted the idea of a triagular umber to a petagoal umber. Just like the triagle umbers, the jth petagoal umber, which we will deote π j, is the umber of dots required to form a petago with j dots to a side. Diagrams associated with the first four petagoal umbers are give i Figure 3. These dot diagrams are less ituitive, sice the easiest way to draw them is ot as a regular petago. However, this particular arragemet is illumiatig whe tryig to determie a formula for π j. Notice that each diagram cosists of a triagle with a j (j ) dot rectagle joied to the bottom. Thus, the jth petagoal umber may be determied by the followig equatio: π j = j + j(j ) = j(3j ).

3 ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/06 3 a. j = b. j = c. j = 3 Figure 3. Petagoal Numbers d. j = 4.3. Euler s Idetity. To brig our discussio back from the world of triagular ad petagoal umbers, we cosider the followig idetity discovered by Euler. Theorem 3 (Euler s Petagoal Number Idetity (PNI) ). For a positive iteger, we have: p ( eve # distict parts) = p ( odd # distict parts) + e() where the error term, e() {, 0, }, is give by: e() = { ( ) j 0 else. = j(3j±) As is ofte the case, there are multiple ways oe may take to prove this idetity. We ofte have a choice of provig theorems by costructio of a bijectio betwee two sets or by usig the method of geeratig fuctios. Both ways have their ow advatages ad may provide uique isight ito the problem at had. We will start by provig this by costructio of a bijectio; later i the lecture, we will use geeratig fuctios to develop further idetities. Proof. To costruct this bijectio, we eed some way to chage the parity of l(λ). To do this, we will either move the smallest part to the diagoal or move the diagoal to form a ew smallest part. At most oe of the two choices is allowed for a give Youg diagram. Neither choice is valid exactly whe λ is petagoal. Hece we do ot have a perfect bijectio; istead, we will be off by ± depedig o whether λ is petagoal. The process of costructig this map is described below by correspodig etries i Figure 4 ad Figure 5 for the cocrete case j = 3 = l(λ). (That is, subfigure (A) i Figure 4 maps to subfigure (A) i Figure 5, ad so o.) Notice that, you caot move either the diagoal or the smallest part for Youg diagrams (c) ad (d). Also ote that the size of (c) is j(3j+) while the size of (d) is j(3j ) these are, of course, petagoal umbers.

4 4 ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/06 a b c d e Figure 4. Iitial Youg diagrams a Noe c b Noe d Figure 5. Fial Youg diagrams e 3. Geeratig Fuctios 3.. Geeral Methods. Give a umber sequece a 0, a,..., we represet it with the geeratig fuctio (a fuctio i a variable, q): A(q) = a k q k. k=0 Two sequeces of special ote are the sequece a i =, i 0 ad the sequece a 0 =,..., a N =, a N+ = 0, a N+ = 0,.... These have the followig geeratig fuctios,

5 respectively: ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/06 5 k=0 N k=0 q k = q, q k = qn+ q = [N + ]. Note that we do t cocer ourselves with the otio of covergece for these series. Some geeratig fuctios have clear partitio fuctio iterpretatio: q k = q k m=0 p (m parts i {k}) q k. Now, let us cosider what happes whe we geeralize beyod just parts i the sigleto set {k}. Let S Z be a set; for istace {,..., r }. (It is importat to ote that we do t eed S to be fiite; however, it is easier to form a ituitio if we cosider S to be fiite first.) Propositio 4. Let S Z. We have s S( + q s ) = 0 p ( distict parts i S). Proof. Let us cosider the cocrete case for S = {, }. The, we have: ( + q )( + q ) = + q + q + q }{{} +. }{{} oe part each two parts The proof i geeral is similar to the above special case. However, we may eve further geeralize by o loger requirig distict parts i each partitio. Propositio 5. Agai takig S as defied above, we have: q = p ( parts i S). s 0 s S Proof. As before we may cosider the cocrete case r = ; this traslates to the followig: q q = q k +k. k,k 0 (Note that there is o restrictio o the multiplicity of parts of a particular size.) The geeral case is similar. A special case that is of particular ote is whe we take S = Z. I this case, we have the followig: (3.) p()q = s= ( qs ). =0

6 6 ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/ Reformulatig Euler s Theorem. From the above, we kow that the followig two equatios hold: () p ( odd parts) q = ( q)( q 3 )( q 5 ) () p ( distict parts) q = ( + q)( + q )( + q 3 ) Theorem 6 (Euler s Theorem, Reformulated). The followig idetity holds: ( q)( q 3 )( q 5 ) = ( + q)( + q )( + q 3 ). Proof. I this direct proof, the key is to fill i the missig terms i the deomiator; the, by factorig the differeces of two squares, we may arrive immediately at the result: ( q)( q 3 )( q 5 ) = ( q )( q 4 ) ( q)( q )( q 3 )( q 4 )( q 5 ) = ( q)( + q) ( q )( + q ) ( q 3 )( + q 3 ) ( q) ( q ) ( q 3 ) = ( + q)( + q ) Remark 7. A similar reformulatio method occurs i oe of the homework problems Revisitg the PNI. Now, let us apply our kowledge of geeratig fuctios to the petagoal umber idetity. Cosider the followig expressios, which will be used throughout the sectio: (3.) p ( eve # distict parts) q (3.3) p ( odd # distict parts) q. We wat to fid (3.) (3.3), which should give us the geeratig fuctio for the error term sequece {e()} as described i Theorem 3. As a aside, first ote that the easier sum (3.) + (3.3), which is just Propositio 4. As before, we cosider the fiite case for r = (where r is the size of S as above). The we have the followig: ( q )( q ) = ( + q + ) ( q q ). By extedig this example, we ca see that the differece is as below: (3.4) (3.) (3.3) = ( q)( q ) Now, we cosider the error terms. We ca represet this as the sum of two geeratig fuctios as below: e()q = ( ) j q j(3j ) + ( ) j q j(3j+) j=0 j=

7 ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/06 7 Note that the sums are disjoit (ad that is why the latter starts at j = istead of j = 0). By a chage of variable j j i the secod sum, we have: (3.5) e()q = ( ) j q j(3j ). j= Combiig (3.4) ad (3.5), we may reformulate the petagoal umber idetity as the followig idetity: (3.6) q m=( m ) = ( ) j q j(3j. j Z The idetity (3.6) ca be rewritte further as ( (3.7) =0 p()q ) ( j Z ( ) j q j(3j If we fix a power of q, this the produces a recurrece relatioship for the th petagoal umber as below: ) =. p() p( ) p( ) + p( 5) + p( 7) = 0 This is sigificat because it eables the th petagoal umber to be computed efficietly i a recursive way.