LECTURES ON RECONSTRUCTION ALGEBRAS I

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1 LETURES ON REONSTRUTION ALGEBRAS I MIHAEL WEMYSS. Inroducion Noncommuive lgebr (=quivers) cn be used o solve boh explici nd non-explici problems in lgebric geomery, nd hese lecures will ry o explin some of he feures of boh pproches. I wn o use hese noes o give genle (!) inroducion o he subjec, nd will ry nd mke hem s self-conined s possible. Since I wn o evenully end up doing non-oric geomery, hroughou I shll never dop he lnguge of oric geomery, even if he exmple I m considering is oric. Firs some moivion: From noncommuive perspecive we would like o ke singulriy X = Spec R nd produce N ring A from which we cn exrc resoluion(s) of X. We cn hen sk wheher he N ring hs some geomericl mening, nd if so wheher his gives informion bou A. We cn lso sk wh A sys bou X nd is resoluions. From more geomeric perspecive we my lredy hve some resoluion Y of X nd would like produce oher resoluions, for exmple by flopping cerin curves. We my lso wn o describe he derived cegory of Y. This cn someimes be done using noncommuive lgebr. In prcice however hings re no quie s simple s his, since mos of he ime specific problem will be mixure of he wo bove problems. Someimes i is esier o solve he problem using he geomery, someimes i is esier using quivers. Thus geomery cn give us resuls in noncommuive lgebr nd noncommuive lgebr cn give us resuls in geomery; i is he process of plying he wo sides off ech oher which gives us he sronges resuls. Tody I m going o define quivers nd ell you how o hink of hem, hen following King [King] I ll lk bou heir moduli spce(s) of finie dimensionl represenions. Time permiing I ll hen show how o clcule he moduli spces in some esy exmples. 2. Quivers nd Represenions Any lgebr wih finie number of generors nd finie number of relions (i.e. lmos ll lgebrs you cn hink off) cn be wrien s quiver wih relions. You wn o do his since he quiver gives you wy o visulize he lgebr, nd more impornly i gives you wy o visulize he finie dimensionl modules (see ler). Definiion 2.. A quiver Q is jus finie direced grph. A his sge loops, double rrows,... re ll llowed, nd he direced grph need no be conneced. For exmple Q = is n exmple of quiver. A smll echnicl poin: for every verex i we cully lso dd in rivil loop h verex nd denoe i by e i, bu we do no drw hese loops. In he bove exmple, he loops drwn re he non-rivil loops. Denoing he verices of Q by Q 0 nd he rrows by Q, you cn view he direced grph Q s simply piece of combinoril d (Q 0, Q, h, ) where h nd re mps Q Q 0. The mp h (he hed ) ssigns o n rrow is hed, nd he mp (he il ) ssigns o n rrow is il. This cnno be done in unique wy

2 2 MIHAEL WEMYSS Definiion 2.2. A non-rivil ph of lengh n in Q is jus sequence of rrows n in Q wih h( i ) = ( i+ ) for ll i n. We cll his ph cycle if h( n ) = ( ). We wn o dd more srucure o he combinoril d of quiver by producing n lgebr: Definiion 2.3. For given quiver Q, he ph lgebr kq is defined o be he k-lgebr wih bsis given by he phs, wih muliplicion { pq h(p) = (q) pq := 0 else for ny phs p nd q. e i p := { p (p) = i 0 else pe i := { p h(p) = i 0 else This is n lgebr, wih ideniy kq = i Q 0 e i. Noe we re using he convenion h pq mens p hen q; be wre h some svge brbrins 2 use he opposie convenion. Noe h by he definiion of muliplicion he ph lgebr is ofen noncommuive: for exmple if Q = b hen b b since b = 0. In fc in his exmple kq is esy o describe: he bsis of kq is e, e 2, e 3,, b, b. Is no hrd o convince yourself h kq = k k k 0 k k. 0 0 k Exercise 2.4. Le Q be quiver, hen kq is finie dimensionl if nd only if Q hs no non-rivil cycles. For quivers Q wihou cycles, he resuling ph lgebrs kq hve been used in geomery lhough heir use is generlly limied o projecive vrieies; since in hese lks we re going o be resolving singulriies we need o mke one more definiion: Definiion 2.5. For given quiver Q, relion is simply k-liner combinion of phs in Q. Given finie number of relions, we cn form heir wo sided idel R in he ph lgebr, nd we hus define he lgebr kq/r o be quiver wih relions. We cn ssume (by removing rrows if necessry) h he lengh of every ph in every relion is greer hn or equl o wo. Noe h wih relions i is possible h kq/r cn be finie dimensionl even when Q hs cycles, hough in hese lecures mos of he exmples will involve infinie dimensionl lgebrs. In prcice you should hink of he relion p q s sying going long ph p is he sme s going long ph q, since p = q in he quoien kq/r. Now s is sndrd in ring heory (nd geomery), we end o sudy ring by insed sudying is module cegory (=coheren sheves), since his is n belin cegory nd so we hve he mchinery of homologicl lgebr our disposl. Represenion heoriss would ell us h we re we re sudying he ring s represenions - I ll now mke his more precise. Definiion 2.6. Le kq/r be quiver wih relions. A finie dimensionl represenion of kq/r is he ssignmen o every verex i of Q finie dimensionl vecor spce V i, nd o every rrow liner mp f : V () V h(), such h he relions R beween he liner mps hold. Denoe α i = dimv i nd le α = (α i ) be he collecion of ll he α i. We cll α he dimension vecor of he represenion. 2 you know who you re

3 LETURES ON REONSTRUTION ALGEBRAS I 3 For exmple le kq/r be b subjec o bc = 0. Denoing c M := 4 ( 0 ) 2 N := ( 0 ) 3 hen M is represenion of dimension vecor (,, 2) wheres N is no represenion of dimension vecor (,, ). We lso hve he obvious noion of morphism beween wo represenions: Definiion 2.7. Le V = (V i, f ) nd W = (W i, g ) be finie dimensionl represenions of kq/r. A morphism ψ from V o W is given by specifying, for every verex i, liner mp ψ i : V i W i such h for every rrow Q, V () ψ () W () f g V h() ψ h() Wh() commues. Noe ψ is n isomorphism if nd only if ech ψ i is liner isomorphism. Also noe we hve he obvious noion of subrepresenion. I is firly cler h in his wy he finie dimensionl represenions form cegory, which we denoe by frep(kq, R) The whole poin o ll his is he following: Lemm 2.8. Le A = kq/r be quiver wih relions. Denoe by fdmoda he finie dimensionl modules of A. Then here is cegoricl equivlence frep(kq, R) fdmoda Proof. This is cully quie uologicl. Given represenion (V i, f ) hen i Q0 V i is he corresponding module. onversely given ny finie dimensionl module W, seing W i = e i W (where e i is he rivil ph verex i) gives us he corresponding represenion. Thus we now see he benefi of wriing n lgebr A s quiver wih relions, s by he bove lemm we hve wy o visulize he finie dimensionl modules of A. 3. Moduli nd GIT In his secion we consider quiver wih relions A = kq/r nd define vrious moduli spces of finie dimensionl represenions. In he process we hve o ke very fs deour hrough he world of geomeric invrin heory (GIT). For fixed dimension vecor α we my consider ll represenions of A = kq/r wih dimension vecor α: R := Rep(A, α) = {represenions of A of dimension α} This is n ffine vriey, so denoe he co-ordine ring by k[r]. The vriey (hence he co-ordine ring) crries nurl cion of G := i Q 0 GL(α i ) cing on n rrow s g = g () g h(). Acully is relly n cion of PGL since he digonl one-prmeer subgroup = {(λ,, λ) : λ k } cs rivilly, bu his won concern us much. Anywy, by liner lgebr he isomorphism clsses of represenions of A = kq/r re in nurl oneo-one correspondence wih he orbis of his cion. To undersnd his spce is normlly n impossible problem (e.g. wild quiver ype), so we wn o hrow wy some represenions nd ke wh is known s GIT quoien.

4 4 MIHAEL WEMYSS To mke GIT quoien we need o dd he exr d of chrcer χ of G. Now he chrcers χ of G = i Q 0 GL(α i ) re given by powers of he deerminns χ(g) = i Q 0 de(g i ) θi for some collecion of inegers θ i Z Q0. Since such χ deermines nd is deermined by he θ i, we usully denoe χ by χ θ. Now consider he mp θ : fdmoda Z M i Q 0 θ i dimm i This is ddiive on shor exc sequences, so relly is mp K 0 (fdmoda) Z. Now ssume h our chrcer sisfies χ θ ( ) = {} (his is need o use Mumford s numericl crierion [King, 2.5]). I no oo hrd o see h his condiion rnsles ino i Q 0 θ i α i = 0. Hence for hese χ θ, θ(m) = 0 if M hs dimension vecor α. We rrive he key definiion [King,.] Definiion 3.. Le A be n belin cegory, nd θ : K 0 (A ) Z n ddiive funcion. We cll θ chrcer of A. An objec M A is clled θ-semisble if θ(m) = 0 nd every subobjec M M sisfies θ(m ) 0. Such n objec M is clled θ-sble if he only subobjecs M wih θ(m ) = 0 re M nd 0. We cll θ generic if every M which is θ- semisble is cully θ-sble. For A = kq/r s before, we re ineresed in he bove definiion for he cse A = fdmoda. We shll see how his works in prcice in he nex secion. The reson King gve he bove definiion is h i is equivlen o he oher noion of sbiliy from GIT, which we now describe: R is n ffine vriey wih n cion of linerly reducive group G = i Q 0 GL(α i ). Since G is reducive, we hve quoien R R /G = Speck[R] G which is dul o he inclusion k[r] G k[r]. Is he reduciveness of he group which ensures h k[r] G is finiely genered k-lgebr, nd so Speck[R] G is relly vriey, no jus scheme. Virully by definiion he bove is cegoricl quoien (quie wek condiion); furher is cully good quoien (if you don know wh his mens, don worry) To mke GIT quoien we hve o dd o his picure he exr d of χ, some chrcer of G. Definiion 3.2. f k[r] is semi-invrin of weigh χ if f(g x) = χ(g)f(x) for ll g G nd ll x R. We wrie he se of such f s R G,χ. We define R / χ G := Proj n 0 k[r] G,χn Definiion 3.3. x R is clled χ-semisble (in he sense of GIT) if here exiss some semi-invrin f of weigh χ n wih n > 0 such h f(x) 0, oherwise x R is clled unsble. The se of semisble poins R ss forms n open subse of R; in fc we hve morphism q : R ss R / χ G which is good quoien. One more definiion: Definiion 3.4. x R is clled χ-sble (in he sense of GIT) if i is χ-semisble, he G orbi conining x is closed in R ss nd furher he sbilizer of x is finie.

5 LETURES ON REONSTRUTION ALGEBRAS I 5 In fc q is geomeric quoien on he sble locus R s, mening h R s / χ G relly is n orbi spce. The poin in he bove discussion is he following resul [King, 3.], which sys he wo noions re he sme Proposiion 3.5. Le M Rep(A, α) = R, choose θ s in Definiion 3.. Then M is θ-semisble (in he sense of Definiion 3.) if nd only if M is χ θ -semisble (in he sense of GIT). The sme holds replcing semisbiliy wih sbiliy. Thus we use he mchinery from he GIT side o define for quivers he following: Definiion 3.6. For A = kq/r choose dimension vecor α nd chrcer θ sisfying i Q 0 α i θ i = 0. Denoe Rep(A, α) = R nd G = GL(α). We define θ (A, α) := R / χθ G := Proj n 0 M ss k[r] G,χn nd cll i he moduli spce of θ-semisble represenions of dimension vecor α. This is by definiion projecive over he ordinry quoien R /G = Speck[R] G. We mke some remrks (i) If k[r] G = k hen M ss θ (A, α) is projecive vriey. (ii) In he resoluion of singulriies we idelly would like he zeroh piece Speck[R] G o be he singulriy since hen he moduli spce is projecive over i! However, even in cses where we use N rings o resolve singulriies, Speck[R] G migh no be he hing we wn; see Exmple 4.6 ler. (iii) Noe h M ss θ (A, α) my be empy. (iv) One wy o compue his spce is o compue semi-invrins, bu his in generl is quie hrd. One smll poin before we coninue: we cn jus cll M ss θ (A, α) moduli spce, we relly hve o jusify h i is moduli spce, i.e. why i prmeerizes cerin objecs. We shll describe his more precisely in fuure secion. For now hough we shll concern ourselves wih showing how o clcule he moduli spce in some exmples: 4. Exmples The ls secion ws quie bsrc, here we show how i works in prcice. For A = kq/r, we my wn o consruc spce X from A s moduli spce of θ-sble A-modules. Wh his mens [King2]: To specify such moduli spce we mus give dimension vecor α nd weigh vecor (or chrcer ) θ sisfying i Q 0 θ i α i = 0. The moduli spce of θ-sble A-modules of dimension vecor α is hen he prmeer spce for hose A-modules which hve no proper submodules wih ny dimension vecor β for which i Q 0 θ i β i 0. For compuionl ese I will only compue moduli wih dimension vecor (,..., ) in his secion; I will reurn nd do compuion of some oher dimension vecors in fuure secion. There re mny differen (nd beer) wys o view he following exmple, bu here I give he esies: Exmple 4.. onsider he quiver wih no relions. hoose α = (, ) nd θ = (, ). Wih hese choices, since θ i α i = 0 we cn form he moduli spce. Now represenion of dimension vecor α = (, ) is θ- semisble by definiion if θ(m ) 0 for ll subobjecs M. Bu he only possible subobjecs in his exmple re of dimension vecor (0, 0), (0, ) nd (, 0), nd θ is 0 on ll bu he ls (in fc is esy o see h θ is generic in his exmple). Thus represenion of dimension

6 6 MIHAEL WEMYSS vecor (, ) is θ-semisble if nd only if i hs no submodules of dimension vecor (, 0). Now ke n rbirry represenion M of dimension vecor (, ) M = Noice h M hs submodule of dimension vecor (, 0) if nd only if = b = 0, since he digrm = mus commue. Thus by our choice of sbiliy θ, M is θ-semisble M hs no submodule of dim vecor (, 0) 0 or b 0. nd so we see h he semisble objecs prmerize P vi he rio ( : b), so he moduli spce is jus P. Anoher wy o see his: we hve wo open ses, one corresponding o 0 nd he oher o b 0. Afer chnging bsis we cn se hem o be he ideniy, nd so we hve U 0 = { b : b } U = { : } Now he gluing is given by, whenever U 0 b 0 which is evidenly jus P. U 0 b = b b b. = b = b U This lecure series is devoed o resolving singulriies, so we wrm up by blowing up he origin in 2 : Exmple 4.2. onsider he quiver wih relions b b = b nd gin choose dimension vecor (, ) nd sbiliy θ 0 = (, ). Excly s bove if M = b hen M is θ-semisble M hs no submodule of dim vecor (, 0) 0 or b 0. For he firs open se in he moduli U 0 (when 0): fer chnging bsis so h = we see h he open se is prmeerized by he wo sclrs b nd subjec o he single relion (subsiuing = ino he quiver relions) b = b. Bu his lwys holds so i isn relly relion, hus he open se U 0 is jus 2 wih co-ordines b,. We wrie his s 2 b,. Similrly for he oher open se: b U 0 = 2 b, Now he gluing is given by, whenever b 0 U = 2,. U 0 (b, ) = b = b b = (b, b) U nd so we see h his is jus he blowup of he origin of 2. Exercise 4.3. Wh does he sbiliy θ = (, ) give us in he bove exmple? Exmple 4.4. onsider he group 3 (, ) := ( ε 3 ) 0 0 ε 3 where ε3 is primiive hird roo of uniy. This cs on 2 giving us quoien singulriy [x, y] 3 (,). onsider he quiver wih relions (he reconsrucion lgebr) c 2 c 2 = c 2 2 c = c 2 2 k c k = c 2 2 k c = 2 c 2

7 LETURES ON REONSTRUTION ALGEBRAS I 7 hoose dimension vecor (,). We re going o clcule he moduli spce for sbiliy θ 0 = (, ), hen clcule he moduli spce for sbiliy θ = (, ). (i) Tke θ 0 = (, ). As in he exmples bove, for module M = c 2 2 k o be semisble requires c 0 or c 2 0 nd so we hve wo open ses U 0 = (c 0) nd U = (c 2 0). hnging bsis so h hese re, by he relions we hve c 2 c 2 c 2 2 c c 2 k c k k Now he gluing is given by, whenever c 2 0 U 0 = 2 c 2, U = 2 c,k. U 0 (c 2, ) = c2 c 2 = c 2 2 c 2 c 2 c 2 2 c 3 2 = (c 2, c3 2 ) U since we red off he co-ordines in U in he c nd k posiions. Thus by inspecion we see h our spce is O P ( 3), he miniml resoluion. (ii) Tke θ = (, ). Is cler h we now hve 3 open ses U 0 = ( 0), U = ( 2 0), U 2 = (k 0). onsider firs U 0 : fer chnging bsis so h =, we hve c c 2 c 2 = c 2 2 c = c 2 2 c k = c 2 2 k c = 2 c 2 k which is prmeerized by he hree vribles c, 2, k subjec o he one relion c k = c 2 2 i.e. c (k 2 2) = 0. This is singulr in dimension! If we drw U 0, i looks somehing like I hs wo componens, nmely he c = 0 componen nd he k = 2 2 componen. The k = 2 2 componen is he one h we wn, since i ends up giving us (pr of) he miniml resoluion. From he bove exmple we see h moduli spce my no be smooh nd migh hve componens. Noe h in he bove exmple here is one componen which is priculrly nice, however he nex exmple shows h moduli spce my be boh irreducible nd singulr.

8 8 MIHAEL WEMYSS ( ) Exmple 4.5. onsider he group 5 (, 2, 3) := ε ε 2 0 : ε 5 = giving hree dimensionl quoien singulriy. The lgebr o consider 0 0 ε 3 is z 4 x 2 x 3 z 3 y y 3 y 2 x y 5 y 4 x 4 z 5 x y 2 = y x 3 x z 2 = z x 4 y z 3 = z y 4 x 2y 3 = y 2x 4 x 2z 3 = z 2x 5 y 2z 4 = z 2y 5 x 3y 4 = y 3x 5 x 3z 4 = z 3x y 3z 5 = z 3y x 4y 5 = y 4x x 4z 5 = z 4x 2 y 4z = z 4y 2 x 5y = y 5x 2 x 5z = z 5x 3 y 5z 2 = z 5y 3 x 5 z 2 z onsider α = (,,,, ) wih sbiliy θ = ( 4,,,, ). onsider he open se given by x 0, y 0, y 3 0 nd z 0. Afer chnging bsis so h hese re he ideniy we hve z 4 x 2 z 3 y 2 y 5 y 4 x 3 x 4 z 5 y2 = x 3 z 2 = x 4 z 3 = y 4 x 2 = y 2x 4 x 2z 3 = z 2x 5 y 2z 4 = z 2y 5 x 3y 4 = x 5 x 3z 4 = z 3 z 5 = z 3 x 4y 5 = y 4 x 4z 5 = z 4x 2 y 4 = z 4y 2 x 5 = y 5x 2 x 5 = z 5x 3 y 5z 2 = z 5 x 5 z 2 from which eliminion of vribles gives h his open se is prmeerized by = y 5, b = x 3, c = z 4 nd d = x 4 subjec o he one relion d = bc. This is singulr he origin nd so consequenly he moduli spce is singulr. In fc in his exmple i is lso irreducible. Exmple 4.6. onsider he group 3 (,, 0) giving he hree dimensionl singulriy [x, y, z] 3 (,,0) = [x, y] 3 (,) [z] i.e. relly jus surfce crossed wih. In his cse he lgebr o consider is he higherdimensionl reconsrucion lgebr z c c 2 z 2 2 k c 2 = c 2 2c = c 2 z c = c z 2 z 2 = z c k = c 2 2 k c = 2c 2 z c 2 = c 2z 2 z 2 2 = 2z z 2k = k z An esy clculion shows h for α = (, ) nd θ 0 = (, ) we resolve he singulriy; unsurprisingly is jus he miniml resoluion crossed wih. Agin he sme is rue for θ = (, ) bu gin we hve o pss o componens. The poin in his exmple is h lhough for θ 0 = (, ) he moduli spce is projecive over [x, y, z] 3 (,,0), he zeroh pr of he grded ring which we ke he Proj of (i.e. he invrins k[r] G ) is no [x, y, z] 3 (,,0), so lile cre should be ken. The reson for his is h boh z nd z 2 belong o k[r] G, nd here is no relion which ells us hey re he sme (hey re however he sme s soon s c 0 or c 2 0). Thus k[r] G hs n exr z. One of he dvnges of quivers is h hey llow you o resolve singulriies explicly in exmples you wouldn be ble o do oherwise, especilly in he cse of quoiens by non-belin group: we will illusre his principle in more compliced exmples in fuure lecure.