Crystal Interpretation of Kerov Kirillov Reshetikhin Bijection II

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1 arxv:math/6697v [math.qa] Jun 7 Crysta Interpretaton of Kerov Krov Reshethn Bjecton II Proof for sn Case Reho Saamoto Department of Physcs, Graduate Schoo of Scence, Unversty of Toyo, Hongo, Bunyo-u, Toyo, 3-33, Japan reho@monet.phys.s.u-toyo.ac.jp Abstract In provng the Fermonc formuae, combnatora bjecton caed the Kerov Krov Reshethn (KKR) bjecton pays the centra roe. It s a bjecton between the set of hghest paths and the set of rgged confguratons. In ths paper, we gve a proof of crysta theoretc reformuaton of the KKR bjecton. It s the man cam of Part I wrtten by A. Kunba, M. Oado, T. Taag, Y. Yamada, and the author. The proof s gven by ntroducng a structure of affne combnatora R matrces on rgged confguratons.

2 Introducton In ths paper, we treat the reatonshp between the Fermonc formuae and the we-nown soton ceuar automata box-ba system. The Fermonc formuae are certan combnatora denttes, and a typca exampe can be found n the context of sovabe attce modes. The bass of these formuae s a combnatora bjecton caed the Kerov Krov Reshethn (KKR) bjecton [,, 3], whch gves one-to-one correspondences between the two combnatora objects caed rgged confguratons and hghest paths. Precse descrpton of the bjecton s gven n Secton.. From the physca pont of vew, rgged confguratons gve an ndex set for egenvectors and egenvaues of the Hamtonan that appears when we use the Bethe ansatz under the strng hypothess (see, e.g., [4] for an ntroductory account of t), and hghest paths gve an ndex set that appears when we use the corner transfer matrx method (see, e.g., [5]). Therefore the KKR bjecton means that athough nether the Bethe ansatz nor the corner transfer matrx method s a rgorous mathematca theory, two ndex sets have one-to-one correspondence. Eventuay, t becomes cearer that the KKR bjecton tsef possesses a rch structure, especay wth respect to the representaton theory of crysta bases [6]. For exampe, an extenson of the rgged confguraton caed unrestrcted rgged confguraton s recenty ntroduced [7, 8], and ts crysta structure,.e., actons of the Kashwara operators on them s expcty determned. It gves a natura generazaton of the KKR bjecton whch covers nonhghest weght eements. (See, e.g., [9,, ] for other nformaton). On the other hand, the box-ba system has entrey dfferent bacground. Ths mode s a typca exampe of soton ceuar automata ntroduced by Taahash Satsuma [, 3]. It s an ntegrabe dscrete dynamca system and has a drect connecton wth the dscrete anaogue of the Lota Voterra equaton [4] (see aso [5]). Though the tme evouton of the system s descrbed by a smpe combnatora procedure, t beautfuy exhbts a soton dynamcs. Recenty, a remarabe correspondence between the box-ba systems and the crysta bases theory was dscovered, and t caused a ot of nterests (see, e.g., [6, 7, 8, 9, ] for reated topcs). In Part I [] of ths par of papers, a unfed treatment of both the Fermonc formua (or the KKR bjecton) and the box-ba systems was presented. It can be vewed as the nverse scatterng formasm (or Gefand Levtan formasm) for the box-ba systems. In Part I, generazatons to arbtrary nonexceptona affne Le agebras (the Oado Schng Shmozono bjecton []) are aso dscussed. In ths paper, we gve a proof of the resut announced n Part I for the genera s n case (see Secton.6 Man theorem of []). The precse statement of the resut s formuated n Theorem 3.3 of Secton 3 beow. Accordng to our resut, the KKR bjecton s nterpreted n terms of combnatora R matrces and energy functons of

3 the crystas (see Secton 3. for defntons). Orgnay the KKR bjecton s defned n a purey combnatora way, and t has no representaton theoretc nterpretaton for a ong tme. Therefore t s expected that our agebrac reformuaton w gve some new nsghts nto the theory of crystas for fnte-dmensona representatons of quantum affne Le agebras [3, 4, 5]. Recenty, as an appcaton of our Theorem 3.3, expct pecewse near formua of the KKR bjecton s derved [6]. Ths formua nvoves the so-caed tau functons whch orgnate from the theory of sotons [7]. Interestngy, these tau functons have drect connecton wth the Fermonc formua tsef. These resuts revea unexpected n between the Fermonc formuae and the soton theory and, at the same tme, aso gve rse to genera souton to the box-ba systems. Let us descrbe some more detas of our resuts. As we have descrbed before, man combnatora objects concernng the KKR bjecton are rgged confguratons and hghest paths. Rgged confguratons are the foowng set of data ( ) RC = (µ () ), (µ (),r () ),,(µ (n ),r (n ) ), () where µ (a) Z > and r (a) Z for a n and (a) ( (a) Z ). They obey certan seecton rue, whch w be gven n Defnton.. On the other hand, hghest paths are the hghest weght eements of B B B N, where B s the crysta of th symmetrc power of the vector (or natura) representaton of the quantum enveopng agebra U q (s n ). We regard eements of B as row-type sem-standard Young tabeaux fed n wth etters from to n. In ths paper, we ony treat a map from rgged confguratons to hghest paths. In order to reformuate the KKR bjecton agebracay, we notce that the nested structure arsng on rgged confguraton Eq.() s mportant. More precsey, we ntroduce the foowng famy of subsets of RC for a n ; RC (a) = ( (µ (a) ), (µ (a+),r (a+) ),,(µ (n ),r (n ) ) ). () On ths RC (a), we can aso appy the KKR bjecton. Then we obtan a path whose tensor factors are represented by tabeaux fed n wth etters from to n a. However, for our constructon, t s convenent to add a to each etter contaned n the path. Thus, we assume that the path obtaned from RC (a) contans etters a+ to n. Let us tentatvey denote the resutng path p (a). Then we can defne the foowng maps: p (a) Φ (a) C (a) p (a ). (3) We postpone a presce defnton of these maps Φ (a) C (a) unt Secton 3., but t shoud be stressed that the defnton uses ony conbnatora R matrces and energy functons. Note that the KKR bjecton on RC (n ) trvay yeds a path of the form p (n ) = (n ) = n µ(n ), where n µ s a tabeaux representaton of 3

4 crystas. Therefore, by successve appcatons of Φ (a) C (a) onto p (n ), we obtan the constructon p = Φ () C () Φ () C () Φ (n ) (n ) C (n ) n µ(n ), (4) where p s the path correspondng to the orgna rgged confguraton RC (Eq.()). = The pan of ths paper s as foows. In Secton, we revew defntons of rgged confguratons and the KKR bjecton. In Secton 3, we revew combnatora R matrces and energy functon foowng the graphca rue n terms of wndng and unwndng pars ntroduced n [8]. We then defne scatterng data n Eqs.(3) and (34) and defne the operators C (a) and Φ (a). Our man resut s formuated n Theorem 3.3. The rest of the paper s devoted to a proof of ths theorem. In Secton 4, we reca the Krov Schng Shmozono s resut (Theorem 4.). Ths theorem descrbes the dependence of a resutng path wth respect to orderngs of µ () of RC. We then ntroduce an mportant modfcaton of rgged confguratons. More precsey, we repace µ (a) of RC (a) by µ (a) µ (a+) ( L ), where the nteger L w be determned by Proposton 5.. We then appy Theorem 4. to ths modfed rgged confguraton and obtan the somorphsm of Proposton 4.4. Ths reduces our remanng tas to gvng nterpretaton of modes d (Eq.(34)) n terms of the KKR bjecton. Exampe of these arguments s gven n Exampe 4.6. In Secton 5, we connect modes d wth rgged confguraton n Proposton 5.. By usng ths proposton, we ntroduce a structure reated wth the energy functon n Secton 6. Ths s descrbed n Theorem 6. (see aso Exampes 6. and 6.3 as to the meanngs of ths theorem). In Secton 7, we gve a proof of Theorem 6. and hence compete a proof of Theorem 3.3. We do ths by drecty connectng the graphca rue of energy functon gven n Secton 3. wth rgged confguraton. In fact, we expcty construct a structure of unwndng pars on the rgged confguratons n Proposton 7.3. Premnares. Rgged confguratons In ths secton, we brefy revew the Kerov Krov Reshethn (KKR) bjecton. The KKR bjecton gves one-to-one correspondences between the set of rgged confguratons and the set of hghest weght eements n tensor products of crystas of symmetrc powers of the vector (or natura) representaton of U q (s n ), whch we ca paths. Let us defne the rgged confguratons. Consder the foowng coecton of data: µ (a) = ( µ (a),µ(a) ),,µ(a), ( a n, (a) Z (a), µ (a) Z > ). (5) 4

5 We use usua Young dagrammatc expresson for these nteger sequences µ (a), athough our µ (a) are not necessary monotoncay decreasng sequences. Defnton. () For a gven dagram µ, we ntroduce coordnates (row, coumn) of each boxes just e matrx entres. For a box α of µ, co(α) s coumn coordnate of α. Then we defne the foowng subsets: µ j := {α α µ,co(α) j}, (6) µ >j := {α α µ,co(α) > j}. (7) () For a sequence of dagrams (µ (),µ (),,µ (n ) ), we defne Q (a) j Q (a) j := (a) = by mn(j,µ (a) ), (8).e., the number of boxes n µ (a) j. Then the vacancy number p (a) j for rows of µ (a) s defned by p (a) j := Q (a ) j Q (a) j +Q (a+) j, (9) where j s the wdth of the correspondng row. Defnton. Consder the foowng set of data: ( RC := (µ () ), (µ (),r () ),,(µ (n ),r (n ) ) () If a vacancy numbers for (µ (),µ (),,µ (n ) ) are nonnegatve, p (a) µ (a) then RC s caed a confguraton. () If an nteger r (a) satsfes the condton ). (), ( a n, (a) ), () r (a) p (a) µ (a), () then r (a) s caed a rggng assocated wth row µ (a). For the rows of equa wdths,.e., µ (a) = µ (a) +, we assume that r(a) r (a) +. (3) If RC s a confguraton and f a ntegers r (a) are rggngs assocated wth row µ (a), then RC s caed s n rgged confguraton. In the rgged confguraton, µ () s sometmes caed a quantum space whch determnes the shape of the correspondng path, as we w see n the next subsecton. In the defnton of the KKR bjecton, the foowng noton s mportant. and corre- Defnton.3 For a gven rgged confguraton, consder a row µ (a) spondng rggng r (a). If they satsfy the condton then the row µ (a) s caed snguar. r (a) = p (a) 5 µ (a), (3)

6 . The KKR bjecton In ths subsecton, we defne the KKR bjecton. In what foows, we treat a bjecton φ to obtan a hghest path p from a gven rgged confguraton RC, where RC = φ : RC p B N B B, (4) ( (µ () ), (µ (),r () ),,(µ (n ) ),r (n ) ), (5) s the rgged confguraton defned n the ast subsecton, and N(= () ) s the ength of the partton µ (). B s the crysta of the th symmetrc power of the vector (or natura) representaton of U q (s n ). As a set, t s equa to B = {(x,x,,x n ) Z n x +x + +x n = }. (6) We usuay dentfy eements of B as the sem-standard Young tabeaux (x,x,,x n ) = x x x n {}}{{}}{{}}{ n n, (7).e., the number of etters contaned n a tabeau s x. Defnton.4 For a gven RC, the mage (or path) p of the KKR bjecton φ s obtaned by the foowng procedure. Step : For each row of the quantum space µ (), we re-assgn the ndces from to N arbtrary and reorder t as the composton ( ) µ () =. (8) µ () N,,µ(),µ() Taetherowµ (). Recathatµ() snotnecessarymonotoncaydecreasngnteger sequence. Step : We denote each box of the row µ () as foows: µ () = α () α () α (). (9) Correspondng to the row µ (), we tae p as the foowng array of empty boxes: p =. () Startng from the box α (), we recursvey tae α () µ () by the foowng Rue. Rue : Assume that we have aready chosen α ( ) µ ( ). Let g () be the set of a rows of µ () whose wdths w satsfy w co(α ( ) ). () 6

7 Let g s () ( g () ) be the set of a snguar rows (.e., ts rggng s equa to the vacancy number of the correspondng row) n a set g (). If g s (), then choose one of the shortest rows of g s () and denote by α () ts rghtmost box. If g s () =, then we tae α () = = α (n ) =. Step 3: From RC remove the boxes α (), α (),, α (j ) chosen above, where j s defned by j = max. () n, α () After remova, the new RC s obtaned by the foowng Rue. Rue : Cacuate agan a the vacancy numbers p (a) = Q (a ) Q (a) accordng to the removed RC. For a row whch s not removed, Q (a+) + tae the rggng equa to the correspondng rggng before remova. For a row whch s removed, tae the rggng equa to the new vacancy number of the correspondng row. Put the etter j nto the eftmost empty box of p : p = j. (3) Step 4: Repeat Step and Step 3 for the rest of boxes α (), α() 3,, α() n ths order. Put the etters j,j 3,,j nto empty boxes of p from eft to rght. Step 5: Repeat Step to Step 4 for the rest of rows µ (), µ () 3,, µ () N n ths order. Then we obtan p from µ (), whch we dentfy wth the eement of B. Then we µ () obtan p = p N p p (4) as an mage of φ. Note that the resutng mage p s a functon of the orderng of µ () whch we choose n Step. Its dependence s descrbed n Theorem 4. beow. The above procedure s summarzed n the foowng dagram. Step : Reorder rows of µ (), tae row µ () Step : Choose α () µ () Step 3: Remove a α () and mae new RC Step 4: Remove a boxes of row µ () Step 5: Remove a rows of µ () 7

8 Exampe.5 We gve one smpe but nontrva exampe. Consder the foowng s 3 rgged confguraton: µ () µ () µ () We wrte the vacancy number on the eft and rggngs on the rght of the Young dagrams. We reorder µ () as (,,,); thus we remove the foowng boxes : µ () µ () µ () We obtan p =. Note that, n ths step, we cannot remove snguar row of µ (), snce t s shorter than. After removng two boxes, cacuate agan the vacancy numbers and mae the row of µ () (whch s removed) snguar. Then we obtan the foowng confguraton: µ () µ () µ () Next, we remove the box from the above confguraton. We cannot remove µ (), snce a snguar rows are shorter than. Thus, we obtan p =, and the new rgged confguraton s the foowng: µ () µ () µ () Ths tme, we can remove µ () and µ () and obtan p = 3. Then we obtan the foowng confguraton: µ () µ () µ () From ths confguraton we remove the boxes and obtan p 3 =, and the new confguraton becomes the foowng: µ () µ () µ () 8

9 Fnay we obtan p 4 =. To summarze, we obtan as an mage of the KKR bjecton. p = 3, (5) 3 Crysta base theory and the KKR bjecton 3. Combnatora R matrx and energy functons In ths secton, we formuate the statement of our man resut. Frst of a, et us summarze the basc objects from the crysta bases theory, namey, the combnatora R matrx and assocated energy functon. FortwocrystasB andb ofu q (s n ), onecandefnethetensorproductb B = {b b b B,b B }. Then wehaveaunque somorphsmr : B B B B,.e. a unque map whch commutes wth actons of the Kashwara operators. We ca ths map combnatora R matrx and usuay wrte the map R smpy by. Foowng Rue 3. of [8], we ntroduce a graphca rue to cacuate the combnatora R matrx for s n and the energy functon. Gven the two eements x = (x,x,,x n ) B, y = (y,y,,y n ) B, we draw the foowng dagram to represent the tensor product x y: x {}} { x {}} { x {}} n { y {}}{ y {}}{ y n {}}{ The combnatora R matrx and energy functon H for B B (wth ) are cacuated by the foowng rue.. Pc any dot, say a, n the rght coumn and connect t wth a dot a n the eft coumn by a ne. The partner a s chosen from the dots whch are n the owest row among a dots whose postons are hgher than that of a. If there s no such a dot, we return to the bottom, and the partner a s chosen from the dots n the owest row among a dots. In the former case, we ca such a par unwndng, and, n the atter case, we ca t wndng.. Repeat procedure () for the remanng unconnected dots ( ) tmes. 9

10 3. Acton of the combnatora R matrx s obtaned by movng a unpared dots n the eft coumn to the rght horzontay. We do not touch the pared dots durng ths move. 4. The energy functon H s gven by the number of wndng pars. The number of wndng (or unwndng) pars s sometmes caed the wndng (or unwndng, respectvey) number of tensor product. It s nown that the resutng combnatora R matrx and the energy functons are not affected by the order of mang pars ([8], Propostons 3.5 and 3.7). For more propertes, ncudng that the above defnton ndeed satsfes the axom, see [8]. Exampe 3. The dagram for s By movng the unpared dot (etter 4) n the eft coumn to the rght, we obtan Snce ( we have one ) wndng par and two unwndng pars, the energy functon s H =. Bythedefnton, thewndngnumbersforx y andỹ xarethesamefx y ỹ x by the combnatora R matrx. 3. Formuaton of the man resut From now on, we reformuate the orgna KKR bjecton n terms of the combnatora R and energy functon. Consder the s n rgged confguraton as foows: ( ) RC = (µ () ), (µ (),r () ),,(µ (n ),r (n ) ). (6) By appyng the KKR bjecton, we obtan a path s (). In order to obtan a path s () by agebrac procedure, we have to ntroduce a nested structure on the rgged confguraton. More precsey, we consder the foowng subsets of gven confguraton (6) for a n : RC (a) := ( (µ (a) ), (µ (a+),r (a+) ),,(µ (n ),r (n ) ) ). (7)

11 RC (a) s a s n a rgged confguraton, and RC () s nothng but the orgna RC. Therefore we can perform the KKR bjecton on RC (a) and obtan a path s (a) wth etters,,,n a. However, for our constructon, t s convenent to add a to a etters n a path. Thus we assume that a path s (a) contans etters a+,,n. As n the orgna path s (), we shoud consder s (a) as hghest weght eements of tensor products of crystas as foows: s (a) = b b N B B N, ( = µ (a), N = (a) ). (8) The meanng of crystas B here s as foows. B s crysta of the th symmetrc power representaton of the vector (or natura) representaton of U q (s n a ). As a set, t s equa to B = {(x a+,x a+,,x n ) Z n a x a+ +x a+ + +x n = }. (9) We can dentfy eements of B as sem-standard Young tabeaux contanng etters a+,,n. Aso, we can naturay extend the graphca rue for the combnatora R matrx and energy functon (see Secton 3.) to ths case. The hghest weght eement of B taes the form (a+) = (a+) (a+) B. (3) Ths corresponds to the so-caed ower dagona embeddng of s n a nto s n. From now on, et us construct an eement of affne crysta s (a) from s (a) combned wth nformaton of rggngs r (a), s (a) := b [d ] b N [d N ] aff(b ) aff(b N ). (3) Here aff(b) s the affnzaton of a crysta B. As a set, t s equa to aff(b) = {b[d] d Z,b B}, (3) where ntegers d of b[d] are often caed modes. We can extend the combnatora R: B B B B to the affne case aff(b) aff(b ) aff(b ) aff(b) by the reaton b[d] b [d ] b [d H(b b )] b[d+h(b b )], (33) where b b b b s the somorphsm of combnatora R matrx for cassca crystas whch was defned n Secton 3.. Now we defne the eement s (a) of Eq.(3) from a path s (a) and rggngs r (a). Mode d of b [d ] of s (a) s defned by the formua where r (a) d := r (a) + < ( H b b (+) s the rggng correspondng to a row µ (a) eement b of s (a). The eements b (+) ), b := (a+) max, (34) of RC () whch yeded the ( < ) are defned by sendng b successvey

12 to the rght of b under the somorphsm of combnatora R matrces: b b b + b b b b b b + b b ( ) b b b b (+) b 3 b b. (35) Ths defnton of d s compatbe wth the foowng commutaton reaton of affne combnatora R matrx: b [d ] b + [d + ] b + [d + H] b [d +H], (36) where b b + b + b s an somorphsm by cassca combnatora R matrx (see Theorem 4. beow) and H = H(b b + ). We ca an eement of affne crysta s (a) a scatterng data. For a scatterng data s (a) = b [d ] b N [d N ] obtaned from the quantum space µ (a), we defne the norma orderng as foows. Defnton 3. For a gven scatterng data s (a), we defne the sequence of subsets S S S N S N+ (37) as foows. S N+ s the set of a permutatons whch are obtaned by ŝ n a combnatora R matrces actng on each tensor product n s (a). S s the subset of S + consstng of a the eements of S + whose th mode from the eft end are maxma n S +. Then the eements of S are caed the norma ordered form of s (a). Athough the above norma orderng s not unque, we choose any one of the norma ordered scatterng data whch s obtaned from the path s (a) and denote t byc (a) ( s (a) ). SeeRemar6.5foraternatvecharacterzatonofthenormaorderng. For C (a) ( s (a) ) = b [d ] b N [d N ] (b B ), we defne the foowng eement of s n a+ crysta wth etters a,,n: c = a d b a (d d ) b a (d N d N ) b N. (38) In the foowng, we need the map C (n ). To defne t, we use combnatora R of ŝ crysta defned as foows: n d n d n d H n d +H, (39) where H s now H = mn(,), and we have denoted b [d ] as b d. Ths s a speca case of the combnatora R matrx and energy functon defned n Secton 3., and ŝ corresponds to the s subagebra generated by e and f. We ntroduce another operator Φ (a), Φ (a) : aff(b ) aff(b N ) B B N (4)

13 where we denote = µ (a ) and N = (a ). Φ (a) s defned by the foowng somorphsm of s n a+ combnatora R: Φ (a)( C (a) ( s (a) ) ) ( N ) ( N ) a a d N c a, (4) = where c s defned n Eq.(38). Then our man resut s the foowng: Theorem 3.3 For the rgged confguraton RC (a) (see Eq.(7)), we consder the KKR bjecton wth etters from a+ to n. Then ts mage s gven by Φ (a+) C (a+) Φ (a+) C (a+) Φ (n ) (n ) C (n ) n µ(n ). (4) In partcuar, the KKR mage p of rgged confguraton (6) satsfes p = Φ () C () Φ () C () Φ (n ) (n ) C (n ) n µ(n ). (43) The mage of ths map s ndependent of the choce of maps C (a). In practca cacuaton of ths procedure, t s convenent to ntroduce the foowng dagrams. Frst, we express the somorphsm of the combnatora R matrx by the foowng vertex dagram: = = = a b b a (44) a b b a. If we appy combnatora R successvey as a b c b a c b c a, (45) then we express ths by jonng two vertces as foows: a b a c a. b c Aso, t s sometmes convenent to use the notaton a H b f we have H = H(a b). Exampe 3.4 We gve an exampe of Theorem 3.3 aong wth the same rgged confguraton we have consdered n Exampe.5. 3

14 µ () µ () µ () Frst we cacuate a path s (), whch s an mage of the foowng rgged confguraton (t contans the quantum space ony): µ () The KKR bjecton trvay yeds ts mage as s () = 3. We defne the mode of 3 usng Eq.(34). We put b = 3 and b = 3 (= s () ). Snce we have 3 3 and r () =, the mode s + =. Therefore we have ( ) C () 3 = 3. Note that 3 s trvay norma ordered. Next we cacuate Φ (). Let us tae the numberng of rows of µ () as (µ (),µ() ) = (,),.e., the resutng path s an eement of B () µ B () µ = B B. From 3, we create an eement 3 (see Eq.(38)) and consder the foowng tensor product (see the rght-hand sde of Eq.(4)): ( ) 3. We move 3 to the rght of and next we move to the rght, as n the foowng dagram: We have omtted framngs of tabeaux n the above dagram. Therefore we have ( ) Φ () 3 = 3. Note that the resut depend on the choce of the shape of path (B B ). 4

15 Let us cacuate C (). Frst, we determne the modes d, d of d 3 d. For d, we put b =, and the correspondng vaue of an energy functon s 3, and the rggng s r () = ; hence we have d = + =. For d, we need the foowng vaues of energy functons; 3 R 3, and the rggng s r () =. Hence we have d = ++ =. In order to determne the norma orderng of 3 ( R 3 ), foowng Defnton 3., we construct the set S 3 as { } S 3 = 3, 3. Therefore the norma ordered form s ( ) C () 3 = 3. Fnay, we cacuate Φ (). We assume that the resutng path s an eement of B B B B. From 3 we construct an eement 3. We consder the tensor product ( ) 3 (46) and appy combnatora R matrces successvey as foows: (47) Hence we obtan a path 3, whch reconstructs a cacuaton of Exampe.5. Remar 3.5 In the above cacuaton of Φ (), we have assumed the shape of path as B B. Then we cacuated modes and obtaned 3. Now suppose the path of the form B B on the contrary. In ths case, cacuaton proceeds as foows: 5

16 From the vaues of energy functons 3 and 3 R 3 andtherggngsr () = r () = weobtananeement 3. Comparng both resuts, we have 3 R 3. Ths s a genera consequence of the defnton of mode (Eq.(34)) and Theorem 4. beow. The rest of ths paper s devoted to a proof of Theorem Norma orderng from the KKR bjecton In the rest of ths paper, we adopt the foowng numberng for factors of the scatterng data: b N [d N ] b [d ] b [d ] aff(b N ) aff(b ) aff(b ), (48) snce ths s more convenent when we are dscussng about the reaton between the scatterng data and KKR bjecton. It s nown that the KKR bjecton on rgged confguraton RC admts a structure of the combnatora R matrces. Ths s descrbed by the foowng powerfu theorem proved by Krov, Schng and Shmozono (Lemma 8.5 of [3]), whch pays an mportant roe n the subsequent dscusson. Theorem 4. Pc out any two rows fromthe quantum space µ () anddenote these by µ a and µ b. When we remove µ a at frst and next µ b by the KKR bjecton, then we obtan tabeaux µ a and µ b wth etters,,n, whch we denote by A and B, respectvey. Next, on the contrary, we frst remove µ b and second µ a (eepng the order of other remova nvarant) and we get B and A. Then we have under the somorphsm of s n combnatora R matrx. B A A B, (49) Our frst tas s to nterpret the norma orderng whch appear n Defnton 3. n terms of purey KKR anguage. We can acheve ths transaton f we mae some trcy modfcaton on the rgged confguraton. Consder the rgged confguraton ( RC (a ) = (µ (a ) ), (µ (a),r (a) 6 ),,(µ (n ),r (n ) ) ). (5)

17 Then modfy ts quantum space µ (a ) as µ (a ) + := µ (a ) µ (a) ( L ), (5) where L s some suffcenty arge nteger to be determned beow. For the tme beng, we tae L arge enough so that confguraton µ (a) never becomes snguar whe we are removng µ (a ) part from quantum space µ (a ) µ (a) ( L ) under the KKR procedure. Then we obtan the modfed rgged confguraton RC (a ) + := ( (µ (a ) + ), (µ (a),r (a) ),,(µ (n ),r (n ) ) ), (5) where µ (a ) + s the th row of the quantum space µ (a ) +. In subsequent dscussons, we aways assume ths modfed form of the quantum space uness otherwse stated. For the KKR bjecton on rgged confguraton RC (a ) +, we have two dfferent ways to remove rows of quantum space µ (a ) +. We descrbe these two cases respectvey. Case. Remove µ (a) and ( L ) from µ (a ) +. Then the rgged confguraton RC (a ) + reduces to the orgna rgged confguraton RC (a ). Let us wrte the KKR mage of RC (a ) by p, then the KKR mage of modfed rgged confguraton RC (a ) + n ths case s (a) p = a µ(a) a L. (53) Case. Remove µ (a ) from µ (a ) + n RC (a ) +, then quantum space becomes µ (a) ( L ). Next, we remove the boxes of ( L ) one by one unt some rows n µ (a) become snguar. At ths tme, we choose any one of the snguar rows n µ (a) and ca t µ (a). We aso defne an nteger d ( L) such that ( L ) part of the quantum space s now reduced to ( d ). Then we have the foowng: Lemma 4. In the above settng, we remove row µ (a) n µ (a ) + by the KKR procedure (wth etters a,,n) and obtan a tabeau b B. On the other hand, consder the KKR procedure (wth etters a +,, n) on rgged confguraton RC (a), and remove row µ (a) as a frst step of the procedure. Then we obtan the same tabeau b. Proof. Consder the rgged confguraton RC (a ) + after removng µ (a ) ( L d ) from µ (a ) +. When we begn to remove row µ (a) n the quantum space µ (a ) +, we frst remove the rghtmost box of the row µ (a), ca box x. Then, by the defnton of d, the row µ (a) n the next confguraton µ (a) s snguar so that we can remove the rghtmost box of the row µ (a) µ (a). After removng x, the remanng row µ (a) \{x} µ (a) s made to be snguar agan. In the next step, we remove the box x µ (a ) + whch s on the eft of the box x. Then we can remove the correspondng box x µ (a). Contnung n ths way, we remove both rows µ (a) n quantum space µ (a ) + and µ (a) smutaneousy. We see 7

18 that ths box removng operatons on µ (a), µ (a+),, µ (n ) of RC (a ) + concdes wth the one that we have when we remove µ (a) of the quantum space of RC (a). Let us return to the descrpton of Case procedure, where we have just removed both µ (a) from quantum space µ (a ) + and µ (a). Agan, we remove boxes of ( d ) part of the quantum space one by one unt some snguar rows appear n partton µ (a) and choose any one of the snguar rows, whch we ca µ (a). At ths moment, the part ( d ) s reduced to ( d ). We then remove both µ (a) n quantum space and µ (a) just as n the above emma and obtan a tabeau b. We do ths procedure recursvey unt a boxes of the quantum space are removed. Therefore the KKR mage n ths Case s (a ) a d N b N a (d d 3 ) b a (d d ) b = a µ(a ), (54) where we wrte N = (a) and substtute L n µ (a ) + by d. Ths competes a descrpton of Case procedure. Notethat, nthsexpresson, theetterasseparated fromtheettersa+,,n contaned n b. By vrtue of ths property, we ntroduce the foowng: Defnton 4.3 In the above Case procedure, we obtan b and the assocated ntegers d by the KKR bjecton. From ths data we construct the eement C (a) := b N [d N ] b [d ] aff(b N ) aff(b ) (55) and ca ths a KKR norma ordered product. To obtan a KKR norma orderng, we have to refer the actua KKR procedure. AthoughtheKKRnormaorderng C (a) hasnotbeendentfed wththeonedefned n Defnton 3., these two procedures provde the nterpretaton of Φ (a) operator. More precsey, for each product c := a d N b N a (d d 3 ) b a (d d ) b (56) constructed from C (a), we have the foowng somorphsm. Proposton 4.4 For the rgged confguraton RC (a ) +, we have (a) (a ) p a µ(a) a d c a µ(a ), (57) = where the somorphsm s gven by the s n a+ combnatora R matrx wth etters a,,n, and p s a path obtaned by the KKR bjecton on the orgna rgged confguraton RC (a ). = 8

19 Proof. From the above constructon we see that a dfference between Case and Case s just the dfference of order of removng rows of µ (a ) + n RC (a ) +. Hence we can appy Theorem 4. to cam that both expressons are mutuay somorphc. Ths s just the Φ (a) part of Theorem 3.3. We contnue to study further propertes of ths KKR norma ordered product C (a). Let us perform the above Case procedure on RC (a ) + and obtan the KKR norma ordered product b N [d N ] b [d ] b [d ], (58) where each tabeau b comes from a row µ (a). However, there s an ambguty n the choce of snguar rows n Case. Assume that we obtan another KKR norma ordered product b j N [d j N ] b j [d j ] b j [d j ] (59) from the same confguraton RC (a ) +. We assume that each tabeau b j comes from a row µ (a) j. Then these two products have the foowng property. Lemma 4.5 In ths settng, we have by s n a combnatora R matrces. b N b b b j N b j b j (6) Proof. By usng the argument of the proof of Lemma 4., we regard the eft-hand sde of Eq.(6) as the path obtaned from RC (a) by removng rows µ (a),,µ (a) N of the quantum space n ths order. Smary, the rght-hand sde of Eq.(6) s the path obtaned by removng rows µ (a) j,,µ (a) j N n ths order. Therefore we can appy Theorem 4. to obtan the somorphsm. Exampe 4.6 We gve an exampe of genera argument gven n ths secton aong wth the foowng rgged confguraton RC: (µ () ) = ( 3 ), (µ (),r () ) = ( (4,),(3,),(,4) ), (µ (),r () ) = ( (,),(,) ), (µ (3),r (3) ) = ( (,) ) ; n the dagrammatc expresson, t s 3 µ () 9 4 µ () µ () µ (3) 9

20 For each Young dagram, we assgn the vacancy numbers (on the eft) and rggngs (on the rght) of the correspondng rows (for exampe, the vacancy numbers of µ () are,,9, and the correspondng rggngs are,,4, respectvey). By the usua KKR bjecton, we obtan the foowng mage (path) p: p = In the next secton, we w obtan a formua whch determnes the mode d (Proposton 5.). Usng the formua, we cacuate d as foows: d = max{q () Q () +r (), Q () 3 Q () 3 +r () 3, Q () 4 Q () 4 +r () 4 } = max{3 +4, 7 3+, 8 3+} = max{5,5,5} = 5. (6) Thus, n the modfed rgged confguraton RC () + (Eq.(5)), we have to tae a quantum space as foows: µ () + = µ () ( 3 ) ( 5 ) = {4,3,, 8 }. (6) The modfed rgged confguraton RC () + n ths case taes the foowng shape: µ () µ () µ () µ (3) 8 We remove boxes accordng to Case procedure gven above. In ths procedure, we remove the µ () ( 5 ) part from the quantum space µ () +. Then the remanng confguraton s exacty equa to the orgna one whose quantum space s µ (). Thus, n ths case, we obtan p 5 (63) as an mage of the KKR bjecton. Next, we appy Case procedure to the same modfed rgged confguraton. Frst, we remove the µ () = ( 3 ) part from the quantum space µ () +. Then we obtan 3 as a part of the mage, and the remanng rgged confguraton s µ () µ () µ () µ (3) 5

21 At ths tme, we recognze an mpcaton of the cacuaton n Eq.(6). From the above dagram we see that the quantum space now s µ () + = µ () ( d ), and a three rows of µ () become smutaneousy snguar. Ths s mped n the foowng term n Eq.(6): d = max{5, 5, 5}. (64) (Insde the max term, a factors are 5, and ths mpes that a three rows n µ () woud smutaneousy become snguar when quantum space becomes µ () + = µ () ( 5 ).) We further proceed aong the Case procedure. As we have sad above, we have three possbtes to remove a row of µ () µ () +. Let us remove the row {} from µ () µ () +. Then we have 4 as a part of the mage, and the remanng rgged confguraton s µ () + µ () µ () µ (3) 5 Agan, we encounter two possbtes to remove a row from µ () µ () + wth now We can nfer ths by appyng Proposton 5.: µ () + = µ () ( d ) = µ () ( 5 ). (65) d = max{q () 3 Q () 3 +r () 3, Q () 4 Q () 4 +r () 4 } = max{6 +, 7 +} = max{5, 5} = 5. (66) Let us remove the row {4} from µ () µ () + and remove the box {} from ( d ) µ () +. Then we have 33 as a part of the KKR mage, and the remanng rgged confguraton s µ () + µ () µ () µ (3) At ths tme, the quantum space µ () + becomes µ () ( d 3 ), where we can determne d 3 by Proposton 5. as foows: d 3 = max{q () 3 Q () 3 +r () 3 } = max{3 +} = 4. (67)

22 As a fna step of the KKR procedure, we remove {3} from µ () µ () +, and obtan 4 as the rest part of the KKR mage. Both Case and Case procedures above dffer from each other ony n the order of remova n the quantum space µ () + ; thus we can appy Krov Schng Shmozono s theorem (Theorem 4.) to get the foowng somorphsm: ( p ). (68) In order to cacuate the above somorphsm drecty, we use the dagrammatc method as n (47). That s, we comapare the rght-hand sde of Eq.(68) and Eq.(46). Then we can wrte down a smar vertex dagramas n (47) and obtan the eft-hand sde of (68) as an output. Beow we gve a st of a the scatterng data obtaned from the rgged confguraton here: s s what we have consdered n deta. s = s = s 3 = s 4 = Remar 4.7 By puttng etters on both ends of the above path p, we dentfy ths path as a state of the box-ba systems. For the sae of smpcty, we tentatvey omt frames of tabeaux and tensor products,.e., we wrte the path of the above exampe as p = 343. Then ts tme evouton s gven by t = : 334 t = : 334 t = 3: 334 t = 4: 343 t = 5: 343 t = 6: 343 t = 7: 343 t = 8: 343 t = 5 corresponds to the orgna path p. We see that there are three sotons of ength4, 3, and. Comparethswththeengthsofrowsofµ () above. Furthermore, compare the scatterng data s 3 at the end of the above exampe and t = path. Then we see that these three sotons concde wth three tabeaux of s 3. Ths s the orgn of the term scatterng data. In ths settng, norma orderng s the way to obtan physcay correct scatterng data.

23 5 Mode formua and coson states In the prevous secton, we ntroduce the KKR norma ordered product b N [d N ] b [d ]. (69) In order to determne the orderng of sequence b,,b N and assocated ntegers d, d N, we have to refer the actua KKR procedure. In ths and subsequent sectons, we determne these remanng pont by purey crysta bases theoretc scheme. In the KKR norma ordered product b N [d N ] b [d ], the mode d s determned by the foowng smpe formua. Ths formua aso determnes the correspondng row µ (a) from whch tabeau b arses. Proposton 5. For the KKR norma ordered product b N [d N ] b [d ] we obtan from the rgged confguraton RC (a ) + (Eq.(5)), the mode d has the foowng expresson: { } d = max Q (a) Q (a+) +r (a) (µ (a) ) µ (a) µ (a). (7) Proof. Consder the KKRbjecton onrggedconfguraton RC (a ) +. Wehave taen µ (a ) + = µ (a ) µ (a) ( d ), (7) assumng that, whe removng µ (a ), the confguraton µ (a) never becomes snguar. Remove µ (a ) from the quantum space µ (a ) +. Then we choose d so that, just after removng µ (a ), some snguar rows appear n µ (a) for the frst tme. We now determne ths d. To do ths, we tae arbtrary row µ (a) n the confguraton µ (a). Then the condton that ths row becomes snguar when we have just removed µ (a ) from µ (a ) + s ( ) d +Q (a) Q (a) +Q (a+) = r (a) µ (a) µ (a) µ (a), (7).e., the vacancy number of ths row s equa to the correspondng rggng at that tme. Thus, we have d = Q (a) Q (a+) +r (a) µ (a) µ (a). (73) These d s have dfferent vaues for dfferent rows µ (a). Snce we defne d so that the correspondng row s the frst to become snguar, we have to tae the maxmum of these d s. Ths competes the proof of the proposton. As a consequence of ths formua, we can derve the foowng near dependence of modes d on correspondng rggng r (a). Lemma 5. Suppose that we have the foowng KKR norma ordered product from rgged confguraton RC (a ) + : b N [d N ] b [d ] b [d ], (74) 3

24 where the tabeau b orgnates from the row µ (a), and the correspondng rggng s r (a). Now we change the rggng r(a) to r (a) + and construct a KKR norma ordered product. If we can tae the orderng of ths product to be b N,,b,,b agan, then the KKR norma orderng s b N [d N ] b [d +] b [d ], (75).e., d j (j ) reman the same, and d becomes d +. Proof. Wthout oss of generaty, we can tae =. From the assumpton that we have b [d ] at the rght end of Eq.(75), we see that the mode d does not change after we change the rggng r (a) (see Proposton 5.). We do a KKR procedure n the way descrbed n Case of the prevous secton. We frst remove the row µ (a), and next remove the ( d ) part of the quantum space µ (a ) + unt some snguar rows appear n the confguraton µ (a). At ths tme, we can appy Proposton 5. agan to ths removed rgged confguraton and obtan the next mode. Snce n formua of Proposton 5. we tae the maxmum of terms, the term correspondng to the row µ (a) s the maxmum one before we change the rggng r (a). Hence t st contrbutes as the maxma eement even f the rggng becomes r (a) +. From ths we deduce that the next mode s d +. After removng the row µ (a) and one more box from the quantum space by the KKR procedure, then the rest of the rgged confguraton goes bac to the orgna stuaton so that other terms n the KKR norma ordered product s not dfferent from the orgna one. To determne modes d s, t s convenent to consder the foowng state. Defnton 5.3 Consder the KKR bjecton on RC (a ) +. Remove rows of the quantum space µ (a ) + by Case procedure n the prevous secton,.e., we frst remove µ (a ) from µ (a ) +. Whe removng the ( d ) part of the quantum space, f more than one row of the confguraton µ (a) become smutaneousy snguar, then we defne that these rows are n coson state. We choose one of the KKR norma ordered products and fx t. Suppose that the rghtmost eements of t s B A. Then we have the foowng: Lemma 5.4 We can aways mae B and A n coson state by changng a rggng r B attached to row B. Proof. Let A be the wdth of a tabeau A. We can appy the above Lemma 5. to mae, wthout changng the other part of the KKR norma ordered product, d = Q (a) A Q(a+) A +r (a) A = Q(a) B Q(a+) B +r (a) B, (76) so that A and B are n coson state. Exampe 5.5 Consder s and s 3 n Exampe 4.6. In s, 33 and 4 are n the coson state, and, n s 3, 33 and 4 are n the coson state. 4

25 6 Energy functons and the KKR bjecton In the prevous sectons, we gve crysta nterpretaton for severa propertes of the KKR bjecton, especay wth respect to combnatora R matrces. Now t s a pont to determne a modes d n scatterng data by use of the H functon or the energy functon of a product B A. We consder the rgged confguraton RC (a ) + (Eq.(5)); that s, ts quantum space s µ (a ) + = µ (a ) µ (a) ( d ). (77) In the foowng dscusson, we tae a = wthout oss of generaty and remove µ () as a frst step. To descrbe the man resut, we prepare some conventons and notaton. For the KKR norma ordered product, we denote the rghtmost part as B[d ] A[d ], wheretheengthsoftabeauxare A = Land B = M. TabeauxAandB orgnate fromrows ofµ (), whch weaso denoteasrowaandrowb forthesae ofsmpcty. The dfference of Q () j s before and after the remova of row A s Q() j,.e., Q () j := (Q () j just before remova of A) (Q () j just after remova of A). (78) Then we have the foowng theorem. Theorem 6. If A and B n the KKR norma ordered product are successve (.e. B[d ] A[d ]), then we have Q () M = the unwndng number of B A. (79) Proof w be gven n the next secton. We gve two exampes of ths theorem. Exampe 6. Consder the foowng s 5 rgged confguraton: µ () + = {,, 3,3,3,,3,,4} 4 µ () µ () < > 3 µ (3) µ (4) Wehaveassumedthatwehadareadyremovedtheµ () partfromµ () + (the expresson of µ () + s reordered form, see Step of Defnton.4). Then, by the KKR procedure, we remove rows of µ () + from rght to eft n the above orderng and obtan the foowng KKR norma ordered product: (8) 5

26 The rghtmost part of the product satsfes the unwndng number of =. (8) In the above dagram, boxes wth cross n µ () mean that when we obtan 335 these boxes are removed by the KKR procedure. Snce the wdth of 44 s 3, we have Q () 3 =, whch agrees wth Theorem 6.. Exampe 6.3 Let us consder one more eaborated exampe n s 6. µ () + = { 4,4, 3,4, 3,6,,6, 4,8,,8} µ () µ () 4 < 8 > 3 µ (3) µ (4) µ (5) We have suppressed the vacancy numbers for the sae of smpcty. By the KKR procedure, we have the foowng KKR norma ordered product: (8) The rghtmost part of ths product satsfes the unwndng number of = 6. (83) Snce the wdth of 3345 s 8, ths means that Q () 8 = 6, and ths agrees wth the number of n µ () 8 of the above dagram. Impcaton of ths theorem s as foows. Wthout oss of generaty, we tae A and B as a coson state. We are choosng a normazaton for the H functon as H := the wndng number of B A. (84) By the above defnton, Q () M s equa to the number of boxes whch are removed from µ () M when we remove the row A by the KKR procedure. Thus, f we remove row A µ (), then Q () M s (reca that M := B and L := A ) Q () M = mn{m,l}. (85) From the above theorem we have Q () M Q() M = mn{m,l} the unwndng number of B A = the wndng number of B A = H(B A). (86) 6

27 On the other hand, snce A and B are n the coson state, from Proposton 5. we have d = Q () A Q() A +r A = Q () B Q() B +r B (87) just before we remove the row A. After removng the row A, we agan appy Proposton 5. to the rgged confguraton whch has been modfed by remova of the row A under the KKR procedure. We then have after removng A. By the defnton of Q () j, we have so that, combnng the above arguments, we obtan as a consequence of the above theorem. d = Q () B Q() B +r B (88) d d = Q () M Q() M, (89) d d = H, (9) Remar 6.4 From the pont of vew of the box-ba systems, Theorem 6. thus asserts that mnma separaton between two successve sotons s equa to the energy functon of the both sotons. Proof of Theorem 3.3. WhatwehavetodostodentfytheKKRnormaordered product (Defnton 4.3) wth norma orderng (Defnton 3.) whch s defned n termsofthecrysta basetheory. Agan, we consder therggedconfguratonrc (a ) + (Eq.(5)) wth a =. Frst, we gve an nterpretaton of the above arguments about the coson states n terms of the norma orderng of Defnton 3.. Consder the foowng norma ordered product n the sense of Defnton 3.: b N [d N ] b [d ] b [d ]. (9) Concentrate on a partcuar successve par b + [d + ] b [d ] wthn ths scatterng data. The somorphsm of the affne combnatora R then gves b + [d + ] b [d ] b [d H] b + [d + +H], (9) where H s a vaue of H functon on ths product, and b + b b b + s the correspondng somorphsm under the cassca combnatora R matrx. Snce the modes d depend neary on the correspondng rggng r (see Eq.(34)), we can adjust r to mae that both b + [d + ] b [d ] and b [d H] b + [d + +H] (93) are smutaneousy norma ordered, where the abbrevated parts n the above expresson are unchanged. From Defnton 3. we see that a norma ordered products 7

28 possess the common set of modes {d }. Thus, f ths adjustment s aready taen nto account, then the modes d + and d satsfy d d + = H, (94) whch s the same reaton as what we have seen nthe case of KKRnorma orderng. To summarze, both the KKR norma orderng and norma orderng share the foowng common propertes:. Each b s a tabeau whch s obtaned as a KKR mage of the rgged confguraton RC (a) (Eq.(7)) wth a =. They commute wth each other under the somorphsm of s n combnatora R matrces wth etters from to n.. Consder a norma ordered product. If we can change some rggngs r wthout changng the order of eements n norma orderng, then each mode d depends neary on the correspondng rggng r. 3. Concentrate on a partcuar product b b j nsde a scatterng data, then we can adjust correspondng rggng r to mae that both b [d ] b j [d j ] and b j[d j] b [d ], R where b b j b j b, are smutaneousy norma ordered. If we have aready adjusted the rggng r n such a way, then the dfference of the successve modes d and d j s equa to d j d = H, (95).e., the vaue of the H functon on ths product. From these observatons we see that the both modes d defned by Proposton 5. and Eq.(34) are essentay dentca. Thus the KKR norma ordered products are norma ordered products n the sense of Defnton 3.. On the contrary, we can say that a the norma ordered products are, n fact, KKR norma ordered. To see ths, tae one of the norma ordered products b N [d N ] b [d ] b [d ] S, (96) where S s defned n Defnton 3.. From ths scatterng data we construct the eement d N b N (d N d N ) b (d d ) b. (97) Then, n vew of the somorphsm of affne combnatora R matrces, each power d d s arger than the correspondng H functon (because f t s not the case, then we can permute b [d ] b [d ] to mae the ( )th mode as d + H,.e., arger than the orgna d n contradcton to the defnton of the norma orderng). From Theorem 6. and the arguments foowng t we see that t s a suffcent condton to be a KKR norma ordered product (by a sutabe choce of 8

29 rggngs r () ; snce other nformaton,.e., RC () can be determned from b N,,b n the gven scatterng data). Thus we can appy the nverse of the KKR bjecton and obtan the correspondng rgged confguraton. In the earer arguments, we have nterpreted the Φ (a) operator n terms of the KKR bjecton (Proposton 4.4). Now we nterpret the C (a) operators or, n other words, the norma orderng n terms of the KKR bjecton. Hence we compete the proof of Theorem 3.3. Remar 6.5 In the above arguments, we see that the norma ordered scatterng datacanbedentfedwththepathsobtanedfromtherggedconfguratonrc (a ) +. In partcuar, f the eement satsfes the two condtons. b N b b s a path of RC (a), s = b N [d N ] b [d ] b [d ] (98). every dfference of modes d satsfes the condton d d + H(b + b ), then s can be reazed as an mage of RC (a ) +. Therefore we obtan the foowng characterzaton of norma orderngs. Let S N+ be the set defned n Defnton 3.. Consder an eement s = b N [d N ] b [d ] b [d ] S N+. Then s S f and ony f the modes d of s satsfy the condton d d + H(b + b ) for a N. 7 Proof of Theorem 6. Proof of the theorem s dvded nto 6 steps. Step : Let us ntroduce some notaton used throughout the proof. Consder the rgged confguraton RC (a ) + (Eq.(5)) wth a =. Let the rghtmost eements of a KKR norma ordered product be B A. When we remove the th box from the rght end of the row A µ (), we remove the box α (j) from the confguraton µ (j) by the KKR bjecton. That s, when we remove the th box of a row A, the boxes α (), α(3),, α(n ) (99) are aso removed. In some cases, we have α (j ), α (j) =, α (j+) =,, () for some j n. The box adjacent to the eft of the box α (j) s the box α (j). We sometmes express a row by ts rghtmost box. Then we have the foowng: Lemma 7. For fxed j, co(α (j) ) monotonousy decrease wth respect to,.e., co(α (j) ) > co(α(j) + ). 9

30 Proof. Frst, we consder co(α () ). When we remove α(),.e., the th box from rght end of a row A µ (), then we remove the box α () from µ () and contnue as far as possbe. In the next step, we remove the box α () + from the row A whch satsfes co(α () + ) = co(α() ). () After the box α () +, we remove a box α() +, whch has the foowng two possbtes: () α () + and α() are on the same row, or () α () + and α() are on dfferent rows. In case (), we have co(α () + ) = co(α() ). In case (), we have co(α() + ) < co(α () ), snce f co(α() + ) = co(α() ), then we can choose α() + from the same row wth α (). In both cases, co(α() ) monotonousy decreases wth respect to. In the same way, we assume that unt some j, co(α (j) ) monotonousy decreases wth respect to. Then, from the reaton co(α (j) + ) co(α(j) ), () we can show that co(α (j+) ) aso monotonousy decreases wth respect to. By nducton, ths gves a proof of the emma. Step : When we remove boxes α (), α(), α(3),, the vacancy numbers of the rgged confguraton change n a specfc way. In ths step, we pursue ths characterstc pattern before and after the remova. Frst, consder the case α (+),.e., remove the boxes α (), α(),, α(+),. If co(α () ) < co(α(+) ), then the vacancy numbers attached to the rows α () ( α () ) of the confguraton µ() wthn the regon co(α () ) co(α() ) < co(α (+) ), (3) ncrease by (see Fg. ). To see ths, et us tentatvey wrte co(α () ) =. Reca that the vacancy number p () for ths row s After removng boxes α ( ) ; on the other hand, Q (+) p () = Q ( ) Q () +Q (+). (4), α (), α(+), we see that Q ( ) and Q () decrease by do not change (because of Eq.(3) combned wth co(α ( ) ) co(α () )). Summng up these contrbutons, the vacancy numbers p() ncrease by. It aso mpes that the coquantum numbers (.e., the vacancy numbers mnus rggngs for the correspondng rows) aso ncrease by. Smary, f we have the condton co(α () ) = co(α(+) ), then the vacancy numbers for rows α () of µ () n the regon co(α () ) co(α() ) (5) 3

31 µ ( ) µ () µ (+) α (+) α ( ) + α () Fgure : Schematc dagram of Eq.(3). To mae the stuaton transparent, three Young dagrams are taen to be the same. The shaded regon of µ () s Eq.(3) whose coquantum numbers are ncreased by. Imagne that we are removng boxes β (), β(), β(3), from the eft confguraton to the rght one accordng to the KKR procedure (see Step 3). We can thn of t as some nd of a partce traveng from eft to rght unt stopped. Then, when we remove a row B, the curved thc ne n the fgure oos e a potenta wa whch prevents the partce from gong rghtwards. do not change, snce Q (+) aso decrease by n ths case. Next, consder the case α () and α (+) = wth n. Then the vacancy numbers for rows α () ( α () ) of µ() n the regon co(α () ) co(α() ) (6) ncrease by, snce wthn the vacancy number p () = Q ( ) Q () +Q (+) for a row α () (wth wdth ), Q ( ) and Q () decrease by ; on the other hand, Q (+) do not change, snce now α (+) = (see Fg.). Therefore ts coquantum number aso ncrease by. The above arguments n Step are summarzed n (I), (II), and (III) of Lemma 7. beow. In the rest of ths Step, we show that once regons Eq.(3) or Eq.(6) of µ () become nonsnguar n the way descrbed above, then they never become snguar even when we are removng the rest of a row A. To begn wth, consder the effect of α ( ) +, α() +, and α(+) + α (+) =. We see that f α (+), then α (+). In what foows, we frst treat α(+) becomes snguar, snce we have removed a box α (+) satsfes α (+) because of the reaton, and then +. Ths s because: () the row α (+), and () the wdth of the row co(α () + ) co(α(+) ) = co(α (+) ) (7) co(α () + ) < co(α() ) co(α(+) ) (8) 3