# Du al Singularities Igor Burban In tro duction consider quotien singularities G where is nite subgroup PDF document - DocSlides

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Since ha 57356nite ring extension x x the Krull dimension of x is 2 Note that x xy x In order to get kind of bijection et een 57356nite subgroups and quotien singularities need the follo wing de57356nition De57356nition 11 et GL 57356nit ID: 22262

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## Presentations text content in Du al Singularities Igor Burban In tro duction consider quotien singularities G where is nite subgroup

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Du al Singularities Igor Burban In tro duction consider quotien singularities =G where is nite subgroup. Since ha nite ring extension [[ x; ]] [[ x; ]], the Krull dimension of [[ x; ]] is 2. Note that [[ x; ]] [[( xy )]] [[ x; ]]. In order to get kind of bijection et een nite subgroups and quotien singularities need the follo wing denition. Denition 1.1 et GL nite sub gr oup. element is al le pseudo-reection if is onjugate to diag (1 wher gr oup is al le small if it ontains no pseudo-r ee ctions. Prop osition 1.2 1. et GL nite sub gr oup. Then [[ ]] [[ ]] wher the gr oup is ertain smal sub gr oup of GL 2. et 00 GL two smal sub gr oups. Then [[ ]] [[ ]] 00 if and only if and 00 to ar onjugate d. 3. et GL smal nite sub gr oup. Then [[ ]] is always Cohen-Mac aulay. 4. et GL smal nite sub gr oup. Then [[ ]] is Gor enstein i Remark 1.3 Note that every sub gr oup of is smal l. No an to answ er the follo wing question: what are nite subgroups of mo dulo conjugation?

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Finite subgroups of Lemma 2.1 Every nite sub gr oup of GL is onjugate to sub- gr oup of ). Pro of. Let hermitian inner pro duct on Dene u; := u; (Note that in case it holds )). Then is new hermitian inner pro duct on 1. u; 0. 2. u; implies 0. 3. u; It holds moreo er hu; hv hu; hv u; Hence is unitary with resp ect to Moreo er, has an orthonor- mal basis and let )) map sending ectors of the ho osen orthonormal basis of the left space to ectors of the canonical basis of the righ one. It holds u; u; ). kno that u; u; for all u; Hence get u; u; ), or, setting and instead of and u; u; for all and u; No an to describ all nite subgroups of (2). Recall that (2) jj So, from the top ological oin of view (2) Theorem 2.2 Ther is an exact se quenc of gr oup homomorphisms (2) (3) Mor pr cisely, er f op olo gic al ly is the map 2:1 RP

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rom this theorem follo ws that there is connection et een nite subgroups of (2) and (3). The classication of nite isometry groups of is classical result of F.Klein (actually of Platon). There are the follo wing nite subgroups of (3 ): 1. cyclic subgroup generated, for instance, cos sin( sin( cos 2. Dihedral group It is the automorphism group of prisma. It is generated rotation and reection whic satisfy the relations e; e; ab Note that the last relation can ho osen as ba or bab 3. Group of automorphisms of regular tetrahedron Note that 12. 4. Group of automorphisms of regular ctahedron 24. 5. Group of automorphisms of regular icosahedron 60. Observ that all non-cyclic subgroups of (3) ha ev en order. Lemma 2.3 is the only element of (2) of de gr 2. Pro of The pro of is simple computation. implies or Hence x; or ix; Moreo er since otherwise ust hold j 1. In case ha hence and 1. If ix is purely imaginary then j Let (2) nite subgroup, (2) (3) the surjec- tion. Consider cases. 1. is dd. Then (no elemen ts of order in ). So er and is an isomorphism. Hence is cyclic. 2. is ev en. The due to the Sylo w's theorem con tains subgroup of order and hence con tains an elemen of order 2. But there is exactly one elemen of the second order in (2) (see the lemma ab e). Hence er and )). So, in this case is the preimage of nite subgroup of (3).

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rom what as said get the full classication of nite subgroups of mo dulo conjugation. 1. cyclic subgroup Let its generator. implies where is some primitiv ro ot of of -th order. 2. Binary dihedral group nd the generators of ha to kno the explicit form of the map (2) (3). Skipping all details just write do wn the answ er. a; with relations bab concrete, exp 3. Binary tetrahedral group 24. where exp 4. Binary ctahedral group 48. This group is generated as in the case of and 5. Finally ha the binary icosahedral subgroup 120. where exp Remark 2.4 The pr oblem of classic ation of nite sub gr oups of GL is much mor omplic ate d. Inde d, every nite sub gr oup GL an emb dde into via the gr oup monomorphism GL 7! det Finite sub gr oups of GL give main series of nite sub gr oups of No can compute the corresp onding in arian subrings.

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Description of Du al singularities 1. cyclic subgroup 7! "x; 7! exp ). It is not dicult to see that and xy generate the whole ring of in arian ts. [[ x; ]] [[ ]] [[ x; ]] It is an equation of -singularit 2. Binary dihedral group 7! "x 7! 7! 7! where exp ). The set of in arian monomials is xy They satisfy the relation 2( +1) The standart computations sho that [[ x; ]] [[ x; ]] +1 It is an equation of +2 -singularit 3. It can hec ed that for the groups get singularities (a) 0, (b) 0, (c) 0. get the follo wing theorem: Theorem 3.1 Du al singularities ar pr cisely simple hyp ersurfac singular- ities an no to answ er our next question: what are minimal resolutions and dual graphs of Du al singularities? -singularit Consider the germ of -singularit Consider the blo w-up of this singularit (( x; )) xv u; xw u;

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ak the hart 0, i.e. 1. get xv xw What is nf Consider rst (it means that are lo oking for nf )). or In order to get should allo to arbitrary In this hart is cylinder (1 What is (0)? Ob viously it is the in tersection of with the exceptional plane ((0 0) )). In this hart just ha to set in addition to the equation of the surface (0) see that (0) is rational and since all harts of are symmetric, conclude that is smo oth, so No ha to compute the seln tersection um er do it using the follo wing tric k. Let the minimal resolution. It induces an isomorphism of elds of rational functions ). Let rational function. Then it holds: :E deg )) deg In particular it holds for :E Consider the function In the hart get xu What is the anishing set of 1. is an equation of 2. implies or Hence and ha the follo wing picture:

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So, :E :E :E 0. rom it follo ws 2. -singularit In this section an to compute minimal resolution and dual graph of singularit Let minimal resolution, (0) the exceptional divisor. In order to compute seln tersection um ers ha to consider the map ). Let then equalities :E will imply the seln tersection um ers of First step Let Consider the blo w-up of (( x; )) xv u; xw u; ak rst the hart (i.e. 1). get equations get the equation of the strict transform of assume that and or In this hart is smo oth: Jacobi criterium implies It is easy to see that this system has no solutions. In the hart the strict transform is again smo oth. Consider nally the hart 1.

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The strict transform is or Jacobi criterium implies that this surface has unique singular oin 0, or in the global co ordinates ((0 0) (0 1)). see that this oin indeed lies only in one of three ane harts of no need an equation of exceptional bre. The exceptional bre is the denition the in tersection ((0 0) )) get its lo cal equation in the hart just ha to set in the equation of implies 0. Hence get ((0 0) (0 1)) Going to the other harts sho ws that Finally the function in this hart gets the form Agreemen Since the um er of indices dep ends exp onen tially on the um er of blo wing-ups, shall denote the lo cal co ordinates of all harts of all blo wing-ups the letters x; ). Second step ha the follo wing situation: surface function xz exceptional divisor Consider again the blo wing-up of this surface. It is easy to see that the only in teresting hart is get the strict transform or Again is the only singularit of the blo wn-up surface. The exceptional bre of this blo wing-up has irreducible comp onen ts: implies iv (w call this comp onen ts and 00 ). What is the preimage (under preimage mean its strict transform) of implies 0. The function xz gets in this hart the form uv

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Third step ha the follo wing situation: surface function xy exceptional divisor exceptional divisor iz '' Let us consider the next blo wing-up. The strict transform is or It is an equation of -singularit (and it means that are almost done). The exceptional bre consists again of irreducible comp onen ts 00 They lo cal equations are iv 0. The function is uy It is easy to see that the preimage of is 0. What ab out the preimage of Our surface lies in the ane hart em- edded in to via the map u; 7! (( u; )). But then the condition ould imply that the preimage of lies in the exceptional plane ((0 0)( )). But it can not true! The solution of this parado is that the preimage of of lies in another co ordinate hart. Consider xu xv

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The strict transform is xv or xu The equations of the exceptional bre in this hart are 0, what implies The preimage of iz is giv en ixv xu hence arbitrary In the picture it lo oks lik e: E' E" E' E" (u=0, v=−i, x=0) (u=0,v=i, x=0) It is easy to see that all in tersections are transv ersal. ourth step ha the follo wing situation: there are co ordinate harts surface function xy exceptional divisor exceptional divisor iz surface xy function exceptional divisor exceptional divisor i: 10

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'' '' Our next step is the blo wing-up at the oin (0 0) in the rst co ordinate hart. Again, in order to get equations of the preimages of and ha to consider co ordinate harts. The strict transform is cylinder The preimage of is giv en equations 0, the exceptional bre is giv en 0, our function uy In another hart ha xu xv The strict transform is giv en uv or uv The exceptional bre is giv en uv 0, the preimages of and 00 are giv en Hence our exceptional bre is giv en the follo wing conguration of pro jectiv lines: 11

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E' E' E" E" The dual graph of this conguration is Fifth step ha to tak in to accoun three co ordinate harts of minimal resolution. xy xy i: No ha to compute the divisor ). Let normal surface, closed curv e, rational function. Supp ose that is the prime ideal corresp onding to Then is discrete aluation ring and ult al 1. Consider the rst hart. Let xy 0. implies or 1. 0, is an equation of is the strict transform of the curv in What is the ultiplicit of The generator of the maximal ideal of the ring x; )) is and Therefore ult 3. giv es an equation of It is easy to see that ult 6. Note that the curv has transv ersal in tersection with at the oin 1. 2. Consider the second hart. In this hart holds is imp ossible, cut out the equation of and equations of and 00 The same computation as ab sho ws that ult (what is not surprise and mak es us sure that did not mak mistak in computations) and ult ult 00 4. 12

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3. In the same obtain that ult ult 00 Therefore obtain: 4( 00 2( 00 ha :E 1, all other in tersection um ers of with irreducible comp onen ts of are zero. In tersection um ers of irreducible comp onen ts are co ded in the dual graph (whic is see the picture ab e). The whole job as done in order to compute self-in tersections. :E In the same conclude that the other seln tersection um ers are 2. Remark 5.1 et normal surfac singularity, its minimal esolution, =1 (0) the exc eptional divisor. Supp ose that is go esolution, al and Then is simple hyp ersurfac singularity. Indeed kno that the in tersection matrix :E i;j =1 is negativ ely de- nite. Let the dual graph of Then the quadratic form giv en in tersec- tion matrix coinside with the Tits form of the dual graph: 2( =1 1= i ij where ij is the um er of arro ws connecting ertices and rom the theorem of Gabriel kno that is negativ ely denite (and quiv er is represen tation nite) if and only if Since our singularit is rational, it is taut and uniquely determined its dual graph. 2-dimensional McKa corresp ondence Recall that dened Du al singularities as quotien singularities [[ x; ]] where (2) is some nite subgroup. natural question is: are there an connections et een the represen tation theory of and geometry of the minimal resolution of singularit y? Let us recall some standart facts ab out represen tations of nite groups. Theorem 6.1 (Mashk e) et nite gr oup. Then the ate gory of mo dules is semi-simple. This theorem means that an exact sequence of ]-mo dules splits. In partic- ular, ev ery nite-dimensional ]-mo dule is injectiv and pro jectiv e. But an 13

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indecomp osable pro jectiv mo dule theorem of Krull-Sc hmidt is isomorphic to direct summand of the regular mo dule. Let =1 direct sum decomp osition of ]. Then the whole list of indecomp osable ]-mo dules. Lemma 6.2 et =1 de omp osition of the gular mo dule into dir ct sum of inde omp osable ones. Then it holds dim ( In articular, the fol lowing identity is true: =1 Denition 6.3 et gr oup, ( End( its epr esentation. The char acter of epr esentation is the function dene by the rule (( )) Remark 6.4 1. It is asy to se that the char acter do es not dep end on the choic of epr esentative fr om the isomorphism class of epr esentation: (( )) ( 2. It holds: In other wor ds, denes rings homomorphism fr om the Gr othendie ck ring of to 3. It holds: (( )) (( ( )( )) (( )) It me ans that is cen tral function i.e. function which is is onstant on onjugacy classes of Theorem 6.5 nite dimensional epr esentation of nite gr oup is uniquely determine by its char acter. 14

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Idea of the pro of Let '; cen tral functions on Set '; := It denes an hermitian inner pro duct on the space of all cen tral functions on The theorem follo ws from the fact that is an orthonormal basis of this ector space. Indeed, let an nite-dimensional represen tation of Then kno that Then it ob viously holds Corollary 6.6 The numb er of inde omp osable epr esentations of nite gr oup is qual to the numb er of its onjugacy classes. Denition 6.7 (McKa quiv er) et (2) nite sub gr oup, al inde omp osable epr esentations of et the trivial epr esentation, nat the natur al epr esentation (i.e. the epr esentation given by the inclusion (2) ). Dene the McKay gr aph of as the fol lowing: 1. ertic es ar indexe by (we skip ). 2. et nat =0 ij (or, the same nat =0 ij Then we onne ct vertic es and by ij vertic es. It is easy to see that nat nat Remark 6.8 It holds ij Indeed, ij nat nat nat (here use that and hence ( id. rom this follo ws ( diag( and ( diag( diag( Since nat is the natural represen tation, all nat (2) G: Let (2). If diag( a; then diag( b; ab 1.) Therefore ha nat nat ). Then can con tin ue our equalit y: nat nat nat 15

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Example 6.9 Let binary dihedral group. As already kno w, 12. The group has generators a; whic satisfy the follo wing relations: aba The group has 1-dimensional represen tations 1; 1; and The natural represen tation is also kno wn: it is just where exp i: There is also another one irreducible 2-dimensional represen tation: cos sin sin cos ha found all indecomp osable represen tations of 12 can sum up the obtained information in to the haracter table. dim trivial natural rom this table can deriv the whole structure of the Grothendiec ring of nat can only In the same nat get the McKa graph of Observ that obtained the dual graph of the -singularit Note that the fundamen tal cycle of the -singularit is fund But the co etien ts of this decomp osition are the same as the dimensions of the represen tations corresp onding to the ertices of the McKa quiv er. 16

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Theorem 6.10 (McKa observ ation) et (2) nite sub gr oup, [[ x; ]] the orr esp onding invariant subring. Then the McKay quiver of oinside with the dual gr aph of [[ x; ]] dimensions of the epr esentation or- esp onding to vertex of McKay quiver is qual to the multiplicity of the orr e- sp onding omp onent of the exc eptional br in the fundamental cycle. 17