1.45. Tensor categories with finitely many simple objects. Frobenius- Perron dimensions. Let A be a Z + -ring with Z + -basis I. Definition

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1 Tensor categores wth fntely many smple objects. Frobenus- Perron dmensons. Let A be a Z + -rng wth Z + -bass I. Defnton We wll say that A s transtve f for any X, Z I there exst Y 1, Y 2 I such that XY 1 and Y 2 X nvolve Z wth a nonzero coeffcent. Proposton If C s a rng category wth rght duals then Gr(C) s a transtve untal Z + -rng. Proof. Recall from Theorem that the unt object 1 n C s smple. So Gr(C) s untal. Ths mples that for any smple objects X, Z of C, the object X X Z contans Z as a composton factor (as X X contans 1 as a composton factor), so one can fnd a smple object Y 1 occurrng n X Z such that Z occurs n X Y 1. Smlarly, the object Z X X contans Z as a composton factor, so one can fnd a smple object Y 2 occurrng n Z X such that Z occurs n Y 2 X. Thus Gr(C) s transtve. Let A be a transtve untal Z + -rng of fnte rank. Defne the group homomorphsm FPdm : A C as follows. For X I, let FPdm(X) be the maxmal nonnegatve egenvalue of the matrx of left multplcaton by X. It exsts by the Frobenus-Perron theorem, snce ths matrx has nonnegatve entres. Let us extend FPdm from the bass I to A by addtvty. Defnton The functon FPdm s called the Frobenus-Perron dmenson. In partcular, f C s a rng category wth rght duals and fntely many smple objects, then we can talk about Frobenus-Perron dmensons of objects of C. Proposton Let X I. (1) The number α = FPdm(X) s an algebrac nteger, and for any algebrac conjugate α of α we have α α. (2) FPdm(X) 1. Proof. (1) Note that α s an egenvalue of the nteger matrx N X of left multplcaton by X, hence α s an algebrac nteger. The number α s a root of the characterstc polynomal of N X, so t s also an egenvalue of N X. Thus by the Frobenus-Perron theorem α α. (2) Let r be the number of algebrac conjugates of α. Then α r N(α) where N(α) s the norm of α. Ths mples the statement snce N(α) 1.

2 Proposton (1) The functon FPdm : A C s a rng homomorphsm. (2) There exsts a unque, up to scalng, element R A C := A Z C such that XR = FPdm(X)R, for all X A. After an approprate normalzaton ths element has postve coeffcents, and satsfes FPdm(R) > 0 and RY = FPdm(Y )R, Y A. (3) FPdm s a unque nonzero character of A whch takes nonnegatve values on I. (4) If X A has nonnegatve coeffcents wth respect to the bass of A, then FPdm(X) s the largest nonnegatve egenvalue λ(n X ) of the matrx N X of multplcaton by X. Proof. Consder the matrx M of rght multplcaton by X I X n A n the bass I. By transtvty, ths matrx has strctly postve entres, so by the Frobenus-Perron theorem, part (2), t has a unque, up to scalng, egenvector R A C wth egenvalue λ(m) (the maxmal postve egenvalue of M). Furthermore, ths egenvector can be normalzed to have strctly postve entres. Snce R s unque, t satsfes the equaton XR = d(x)r for some functon d : A C. Indeed, XR s also an egenvector of M wth egenvalue λ(m), so t must be proportonal to R. Furthermore, t s clear that d s a character of A. Snce R has postve entres, d(x) = FPdm(X) for X I. Ths mples (1). We also see that FPdm(X) > 0 for X I (as R has strctly postve coeffcents), and hence FPdm(R) > 0. Now, by transtvty, R s the unque, up to scalng, soluton of the system of lnear equatons XR = FPdm(X)R (as the matrx N of left multplcaton by X I X also has postve entres). Hence, RY = d (Y )R for some character d. Applyng FPdm to both sdes and usng that FPdm(R) > 0, we fnd d = FPdm, provng (2). If χ s another character of A takng postve values on I, then the vector wth entres χ(y ), Y I s an egenvector of the matrx N of the left multplcaton by the element X. Because of transtvty of A the matrx N has postve entres. By the Frobenus-Perron theorem there exsts a postve number λ such that χ(y ) = λ FPdm(Y ). Snce χ s a character, λ = 1, whch completes the proof. Fnally, part (4) follows from part (2) and the Frobenus-Perron theorem (part (3)). Example Let C be the category of fnte dmensonal representatons of a quas-hopf algebra H, and A be ts Grothendeck rng. Then by Proposton , for any X, Y C X I dm Hom(X H, Y ) = dm Hom(H, X Y ) = dm(x) dm(y ), 87

3 88 where H s the regular representaton of H. Thus X H = dm(x)h, so FPdm(X) = dm(x) for all X, and R = H up to scalng. Ths example motvates the followng defnton. Defnton The element R wll be called a regular element of A. Proposton Let A be as above and : I I be a bjecton whch extends to an ant-automorphsm of A. Then FPdm s nvarant under. Proof. Let X I. Then the matrx of rght multplcaton by X s the transpose of the matrx of left multplcaton by X modfed by the permutaton. Thus the requred statement follows from Proposton (2). Corollary Let C be a rng category wth rght duals and fntely many smple objects, and let X be an object n C. If FPdm(X) = 1 then X s nvertble. Proof. By Exercse (d) t s suffcent to show that X X = 1. Ths follows from the facts that 1 s contaned n X X and FPdm(X X ) = FPdm(X) FPdm(X ) = 1. Proposton Let f : A 1 A 2 be a untal homomorphsm of transtve untal Z + -rngs of fnte rank, whose matrx n ther Z + -bases has non-negatve entres. Then (1) f preserves Frobenus-Perron dmensons. (2) Let I 1, I 2 be the Z + -bases of A 1, A 2, and suppose that for any Y n I 2 there exsts X I 1 such that the coeffcent of Y n f(x) s non-zero. If R s a regular element of A 1 then f(r) s a regular element of A 2. Proof. (1) The functon X FPdm(f(X)) s a nonzero character of A 1 wth nonnegatve values on the bass. By Proposton (3), FPdm(f(X)) = FPdm(X) for all X n I. (2) By part (1) we have (1.45.1) f( X)f(R 1 ) = FPdm(f( X))f(R 1 ). X I 1 X I 1 But f( X I 1 X) has strctly postve coeffcents n I 2, hence f(r 1 ) = βr 2 for some β > 0. Applyng FPdm to both sdes, we get the result. Corollary Let C and D be tensor categores wth fntely many classes of smple objects. If F : C D be a quas-tensor functor, then FPdm D (F (X)) = FPdm C (X) for any X n C.

4 Example (Tambara-Yamagam fuson rngs) Let G be a fnte group, and T Y G be an extenson of the untal based rng Z[G]: T Y G := Z[G] ZX, where X s a new bass vector wth gx = Xg = X, X 2 = g G g. Ths s a fuson rng, wth X = X. It s easy to see that FPdm(g) = 1, FPdm(X) = G 1/2. We wll see later that these rngs are categorfable f and only f G s abelan. Example (Verlnde rngs for sl 2 ). Let k be a nonnegatve nteger. Defne a untal Z + -rng Ver k = Ver k (sl 2 ) wth bass V, = 0,..., k (V 0 = 1), wth dualty gven by V = V and multplcaton gven by the truncated Clebsch-Gordan rule: mn(+j,2k (+j)) (1.45.2) V V j = V l. l= j,+j l 2Z It other words, one computes the product by the usual Clebsch-Gordan rule, and then deletes the terms that are not defned (V wth > k) and also ther mrror mages wth respect to pont k +1. We wll show later that ths rng admts categorfcatons comng from quantum groups at roots of unty. Note that Ver 0 = Z, Ver 1 = Z[Z 2 ], Ver 2 = T Y Z2. The latter s called the Isng fuson rng, as t arses n the Isng model of statstcal mechancs. Exercse Show that FPdm(V j ) = [j+1] q := qj+1 q j 1, where q q 1 π q = e k+2. Note that the Verlnde rng has a subrng Ver 0 k spanned by V j wth even j. If k = 3, ths rng has bass 1, X = V 2 wth X 2 = X + 1, X = X. Ths rng s called the Yang-Lee fuson rng. In the Yang-Lee rng, FPdm(X) s the golden rato Note that one can defne the generalzed Yang-Lee fuson rngs Y L n n Z +, wth bass 1, X, multplcaton X 2 = 1+nX and dualty X = X. It s, however, shown n [O2] that these rngs are not categorfable when n > 1. Proposton (Kronecker) Let B be a matrx wth nonnegatve nteger entres, such that λ(bb T ) = λ(b) 2. If λ(b) < 2 then λ(b) = 2 cos(π/n) for some nteger n 2. Proof. Let λ(b) = q + q 1. Then q s an algebrac nteger, and q = 1. Moreover, all conjugates of λ(b) 2 are nonnegatve (snce they are 89

5 90 egenvalues of the matrx BB T, whch s symmetrc and nonnegatve defnte), so all conjugates of λ(b) are real. Thus, f q s a conjugate of q then q +q 1 s real wth absolute value < 2 (by the Frobenus-Perron theorem), so q = 1. By a well known result n elementary algebrac number theory, ths mples that q s a root of unty: q = e 2πk/m, where k and m are coprme. By the Frobenus-Perron theorem, so k = ±1, and m s even (ndeed, f m = 2p + 1 s odd then q p + q p > q + q 1 ). So q = e π/n for some nteger n 2, and we are done. Corollary Let A be a fuson rng, and X A a bass element. Then f F P dm(x) < 2 then F P dm(x) = 2cos(π/n), for some nteger n 3. Proof. Ths follows from Proposton , snce F P dm(xx ) = F P dm(x) Delgne s tensor product of fnte abelan categores. Let C, D be two fnte abelan categores over a feld k. Defnton Delgne s tensor product C D s an abelan category whch s unversal for the functor assgnng to every k-lnear abelan category A the category of rght exact n both varables blnear bfunctors C D A. That s, there s a bfunctor : C D C D : (X, Y ) X Y whch s rght exact n both varables and s such that for any rght exact n both varables bfunctor F : C D A there exsts a unque rght exact functor F : C D A satsfyng F = F. Proposton (cf. [D, Proposton 5.13]) () The tensor product C D exsts and s a fnte abelan category. () It s unque up to a unque equvalence. () Let C, D be fnte dmensonal algebras and let C = C mod and D = D mod. Then C D = C D mod. (v) The bfunctor s exact n both varables and satsfes Hom C (X 1, Y 1 ) Hom D (X 2, Y 2 ) = Hom C D (X 1 X 2, Y 1 Y 2 ). (v) any blnear bfunctor F : C D A exact n each varable defnes an exact functor F : C D A. Proof. (sketch). () follows from the unversal property n the usual way. () As we know, a fnte abelan category s equvalent to the category of fnte dmensonal modules over an algebra. So there exst fnte dmensonal algebras C, D such that C = C mod, D = D mod. Then one can defne C D = C D mod, and t s easy to show that

6 t satsfes the requred condtons. Ths together wth () also mples (). (v),(v) are routne. A smlar result s vald for locally fnte categores. Delgne s tensor product can also be appled to functors. Namely, f F : C C and G : D D are addtve rght exact functors between fnte abelan categores then one can defne the functor F G : C D C D. Proposton If C, D are multtensor categores then the cate gory C D has a natural structure of a multtensor category. Proof. Let X 1 Y 1, X 2 Y 2 C D. Then we can set (X 1 Y 1 ) (X 2 Y 2 ) := (X 1 X 2 ) (Y 1 Y 2 ). and defne the assocatvty somorphsm n the obvous way. Ths defnes a structure of a monodal category on the subcategory of C D consstng of -decomposable objects of the form X Y. But any object of C D admts a resoluton by -decomposable njectve objects. Ths allows us to use a standard argument wth resolutons to extend the tensor product to the entre category C D. It s easy to see that f C, D are rgd, then so s C D, whch mples the statement Fnte (mult)tensor categores. In ths subsecton we wll study general propertes of fnte multtensor and tensor categores. Recall that n a fnte abelan category, every smple object X has a projectve cover P (X). The object P (X) s unque up to a non-unque somorphsm. For any Y n C one has (1.47.1) dm Hom(P (X), Y ) = [Y : X]. Let K 0 (C) denote the free abelan group generated by somorphsm classes of ndecomposable projectve objects of a fnte abelan category C. Elements of K 0 (C) Z C wll be called vrtual projectve objects. We have an obvous homomorphsm γ : K 0 (C) Gr(C). Although groups K 0 (C) and Gr(C) have the same rank, n general γ s nether surjectve nor njectve even after tensorng wth C. The matrx C of γ n the natural bass s called the Cartan matrx of C; ts entres are [P (X) : Y ], where X, Y are smple objects of C. Now let C be a fnte multtensor category, let I be the set of somorphsm classes of smple objects of C, and let, denote the rght and left duals to, respectvely. Let Gr(C) be the Grothendeck rng of C, spanned by somorphsm classes of the smple objects X, I. In ths rng, we have X X j = k N j k k X k, where N j are nonnegatve ntegers. Also, let P denote the projectve covers of X. 91

7 92 Proposton Let C be a fnte multtensor category. Then K 0 (C) s a Gr(C)-bmodule. Proof. Ths follows from the fact that the tensor product of a projectve object wth any object s projectve, Proposton Let us descrbe ths bmodule explctly. Proposton For any object Z of C, P = j,k N = j,k N Z kj [Z : X j ]P k, Z P jk[z : X j ]P k. Proof. Hom(P Z, X k ) = Hom(P, X k Z ), and the frst formula follows from Proposton The second formula s analogous. Proposton Let P be a projectve object n a multtensor category C. Then P s also projectve. Hence, any projectve object n a multtensor category s also njectve. Proof. We need to show that the functor Hom(P, ) s exact. Ths functor s somorphc to Hom(1, P ). The functor P s exact and moreover, by Proposton , any exact sequence splts after tensorng wth P, as an exact sequence consstng of projectve objects. The Proposton s proved. Proposton mples that an ndecomposable projectve object P has a unque smple subobject,.e. that the socle of P s smple. For any fnte tensor category C defne an element R C K 0 (C) Z C by (1.47.2) R C = FPdm(X )P. I Defnton The vrtual projectve object R C s called the regular object of C. Defnton Let C be a fnte tensor category. Then the Frobenus- Perron dmenson of C s defned by (1.47.3) FPdm(C) := FPdm(R C ) = FPdm(X ) FPdm(P ). Example Let H be a fnte dmensonal quas-hopf algebra. Then FPdm(Rep(H)) = dm(h). Proposton (1) Z R C = R C Z = FPdm(Z)R C for all Z Gr(C). (2) The mage of R C n Gr(C) Z C s a regular element. I

8 Proof. We have FPdm(X ) dm Hom(P, Z) = FPdm(Z) for any ob ject Z of C. Hence, FPdm(X ) dm Hom(P Z, Y ) = FPdm(X ) dm Hom(P, Y Z ) = FPdm(Y Z ) = FPdm(Y ) FPdm(Z ) = FPdm(Y ) FPdm(Z) = FPdm(Z) FPdm(X ) dm Hom(P, Y ). Now, P (X) Z are projectve objects by Proposton Hence, the formal sums FPdm(X )P Z = R C Z and FPdm(Z) FPdm(X )P = FPdm(Z)R C are lnear combnatons of P j, j I wth the same coeffcents. Remark We note the followng useful nequalty: (1.47.4) FPdm(C) N FPdm(P ), where N s the number of smple objects n C, and P s the projectve cover of the neutral object 1. Indeed, for any smple object V the projectve object P (V ) V has a nontrval homomorphsm to 1, and hence contans P. So FPdm(P (V )) FPdm(V ) FPdm(P ). Addng these nequaltes over all smple V, we get the result Integral tensor categores. Defnton A transtve untal Z + -rng A of fnte rank s sad to be ntegral f FPdm : A Z (.e. the Frobenus-Perron dmnensons of elements of C are ntegers). A tensor category C s ntegral f Gr(C) s ntegral. Proposton A fnte tensor category C s ntegral f and only f C s equvalent to the representaton category of a fnte dmensonal quas-hopf algebra. Proof. The f part s clear from Example To prove the only f part, t s enough to construct a quas-fber functor on C. Defne P = FPdm(X )P, where X are the smple objects of C, and P are ther projectve covers. Defne F = Hom(P, ). Obvously, F s exact and fathful, F (1) = 1, and dm F (X) = FPdm(X) for all X C. Usng Proposton , we contnue the functors F ( ) and F ( ) F ( ) to the functors C C Vec. Both of these functors are exact and take the same values on the smple objects of C C. Thus these functors are somorphc and we are done. 93

9 94 Corollary The assgnment H Rep(H) defnes a bjecton between ntegral fnte tensor categores C over k up to monodal equvalence, and fnte dmensonal quas-hopf algebras H over k, up to twst equvalence and somorphsm Surjectve quas-tensor functors. Let C, D be abelan categores. Let F : C D be an addtve functor. Defnton We wll say that F s surjectve f any object of D s a subquotent n F (X) for some X C. 13 Exercse Let A, B be coalgebras, and f : A B a homomorphsm. Let F = f : A comod B comod be the correspondng pushforward functor. Then F s surjectve f and only f f s surjectve. Now let C, D be fnte tensor categores. Theorem ([EO]) Let F : C D be a surjectve quas-tensor functor. Then F maps projectve objects to projectve ones. Proof. Let C be a fnte tensor category, and X C. Let us wrte X as a drect sum of ndecomposable objects (such a representaton s unque). Defne the projectvty defect p(x) of X to be the sum of Frobenus-Perron dmensons of all the non-projectve summands n ths sum (ths s well defned by the Krull-Schmdt theorem). It s clear that p(x Y ) = p(x) + p(y ). Also, t follows from Proposton that p(x Y ) p(x)p(y ). Let P be the ndecomposable projectve objects n C. Let P P j = k Bj k P k, and let B be the matrx wth entres Bj k. Also, let B = B. Obvously, B has strctly postve entres, and the Frobenus-Perron egenvalue of B s FPdm(P ). On the other hand, let F : C D be a surjectve quas-tensor functor between fnte tensor categores. Let p j = p(f (P j )), and p be the vector wth entres p j. Then we get p p j k Bj k p k, so ( p )p Bp. So, ether p are all zero, or they are all postve, and the norm of B wth respect to the norm x = p x s at most p. Snce p FPdm(P ), ths mples p = FPdm(P ) for all (as the largest egenvalue of B s FPdm(P )). Assume the second opton s the case. Then F (P ) do not contan nonzero projectve objects as drect summands, and hence for any projectve P C, F (P ) cannot contan a nonzero projectve object as a drect summand. However, let Q be a projectve object of D. Then, 13 Ths defnton does not concde wth a usual categorcal defnton of surjectvty of functors whch requres that every object of D be somorphc to some F (X) for an object X n C.

10 snce F s surjectve, there exsts an object X C such that Q s a subquotent of F (X). Snce any X s a quotent of a projectve object, and F s exact, we may assume that X = P s projectve. So Q occurs as a subquotent n F (P ). As Q s both projectve and njectve, t s actually a drect summand n F (P ). Contradcton. Thus, p = 0 and F (P ) are projectve. The theorem s proved Categorcal freeness. Let C, D be fnte tensor categores, and F : C D be a quas-tensor functor. Theorem One has FPdm(C) (1.50.1) F (R C ) = R D. FPdm(D) Proof. By Theorem , F (R C ) s a vrtually projectve object. Thus, F (R C ) must be proportonal to R D, snce both (when wrtten n the bass P ) are egenvectors of a matrx wth strctly postve entres wth ts Frobenus-Perron egenvalue. (For ths matrx we may take the matrx of multplcaton by F (X), where X s such that F (X) contans as composton factors all smple objects of D; such exsts by the surjectvty of F ). The coeffcent s obtaned by computng the Frobenus-Perron dmensons of both sdes. Corollary In the above stuaton, one has FPdm(C) FPdm(D), and FPdm(D) dvdes FPdm(C) n the rng of algebrac ntegers. In fact, FPdm(C) (1.50.2) FPdm(D) = FPdm(X ) dm Hom(F (P ), 1 D ), where X runs over smple objects of C. Proof. The statement s obtaned by computng the dmenson of Hom(, 1 D ) for both sdes of (1.50.1). Suppose now that C s ntegral,.e., by Proposton , t s the representaton category of a quas-hopf algebra H. In ths case, R C s an honest (not only vrtual) projectve object of C, namely the free rank 1 module over H. Theorefore, multples of R C are free H-modules of fnte rank, and vce versa. Then Theorem and the fact that F (R C ) s proportonal to R D mples the followng categorcal freeness result. Corollary If C s ntegral, and F : C D s a surjectve quas-tensor functor then D s also ntegral, and the object F (R C ) s free of rank FPdm(C)/ FPdm(D) (whch s an nteger). 95

11 96 Proof. The Frobenus-Perron dmensons of smple objects of D are coordnates of the unque egenvector of the postve nteger matrx of multplcaton by F (R C ) wth nteger egenvalue FPdm(C), normalzed so that the component of 1 s 1. Thus, all coordnates of ths vector are ratonal numbers, hence ntegers (because they are algebrac ntegers). Ths mples that the category D s ntegral. The second statement s clear from the above. Corollary ([Scha]; for the semsmple case see [ENO1]) A fnte dmensonal quas-hopf algebra s a free module over ts quas- Hopf subalgebra. Remark In the Hopf case Corollary s well known and much used; t s due to Nchols and Zoeller [NZ].

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