On Kaplansky's Fifth Conjecture · On Kaplansky’s Fifth Conjecture Yorck Sommerhäuser* Universität M ünchen, Mathematisches Institut, Theresienstr. 39, D-80333 Munchen, Germany¨

journal of algebra 204, 202–224 (1998)article no. JA977337

On Kaplansky’s Fifth Conjecture

Yorck Sommerhauser*

Universitat Munchen, Mathematisches Institut, Theresienstr. 39,D-80333 Munchen, Germany

Communicated by Susan Montgomery

Received February 26, 1997

We prove that the antipode of a semisimple Hopf algebra is an involution if thecharacteristic of the base field is very large. © 1998 Academic Press

1. INTRODUCTION

In his Chicago lecture notes [5], I. Kaplansky set up a series of ten con-jectures on Hopf algebras that he considered as important problems of thistheory. Nearly all of these conjectures turned out to be puzzling as well asfundamental, and therefore have stimulated a lot of research in the area.Recently, important progress has been made on the first (cf. [17]), the sixth(cf. [18, 14]), the eighth (cf. [4, 35, 11]), and the tenth (cf. [31]) of theseconjectures. The reader is referred to [16, 30] for more precise informationon the status of these conjectures. Closely related to the eighth conjectureis the classification problem for semisimple Hopf algebras, where A. Ma-suoka has contributed important results (cf. [12] and the references there).

Kaplansky’s fifth conjecture states that the antipode of a semisimple Hopfalgebra is an involution. This was proved by R. Larson and D. Radford intwo closely related papers (cf. [7, 8]) in the case of a base field of charac-teristic zero. Their proof is carried out in two steps, the first one being toshow that the Hopf algebra under consideration is also cosemisimple; thesecond one being to prove that the antipode of a semisimple and cosemisim-ple Hopf algebra is an involution. Their methods used for the second stepwere also powerful enough to prove Kaplansky’s seventh conjecture. In thefirst step, their proof rests on the observation that a complex number times

*E-mail: [email protected].

202

0021-8693/98 $25.00Copyright © 1998 by Academic PressAll rights of reproduction in any form reserved.

on kaplansky’s fifth conjecture 203

its conjugate yields a nonnegative real number, and therefore does not eas-ily generalize to fields of positive characteristic. In the present paper, weimprove on this step and give a proof of the conjecture in the case of a fi-nite dimensional Hopf algebra over a field of large positive characteristic.More precisely, we prove that the antipode of a semisimple Hopf algebrais an involution if the characteristic p of the base field satisfies the inequal-ity p > mm−4 where m = 2 �dimH�2. Our techniques rely on the analysisof the structure of the character ring of H, as do the techniques used byG. I. Kac, Y. Zhu and M. Lorenz to prove Kaplansky’s eighth conjecturein characteristic zero and the techniques used by W. D. Nichols and M. B.Richmond to prove results on Kaplansky’s sixth conjecture.

The article is organized as follows: In Section 2, we discuss a techniqueto adjoin a grouplike element in such a way that the square of the antipodebecomes the conjugation with the adjoined grouplike element. In Section 3,we study the character of the adjoint representation in order to prove thatthe character ring of a semisimple Hopf algebra is itself semisimple if thecharacteristic is sufficiently large. The results of both sections are combinedin the final section to prove the stated result on Kaplansky’s fifth conjecture.

All vector spaces are defined over a base field that is denoted by K. Weassume familiarity with the basic notions of Hopf algebra theory that canbe found for example in [13, 21, 30, 33].

2. INNER AUTOMORPHISMS AND THE SQUARE OFTHE ANTIPODE

2.1. In this section, H denotes a finite dimensional Hopf algebra.We denote the coproduct, counit, and antipode of H by 1H , εH , and SH ,respectively. We shall use the following variant of the Heyneman–Sweedlersigma notation for the coproduct:

1H�h� = h1 ⊗ h2:

The square of the antipode of H is a Hopf algebra automorphism of H.It is known that in general this is not an inner automorphism (cf. [25; 28,p. 598), although it is an inner automorphism if H is semisimple (cf. [20,Folgerung 3.3.2, p. 13] and Subsection 3.2). Here an inner automorphismis understood to be the conjugation by an invertible element. It would beanother step to the proof of the general case of Kaplansky’s fifth conjec-ture if it could be shown in the semisimple case that the square of theantipode is given by conjugation with a grouplike element. In this section,we prove that every finite dimensional Hopf algebra H can be embeddedinto another finite dimensional Hopf algebra, denoted by E�H�, in whichthe square of the antipode is the conjugation with a grouplike element. We

204 yorck sommerhauser

do not consider the obvious generalization to an arbitrary Hopf algebraautomorphism because it will not be needed in the sequel.

2.2. Define n x= 2 dimH and denote by G = �/n� the cyclic groupof order n. We denote the generator 1 of G by g. We know from [26, Propo-sition 6, p. 347] that the fourth power of the antipode is the compositionof the conjugation with a grouplike element, namely the modular elementof H, and the coconjugation with a character of H, namely the modularfunction. Both mappings commute. Since the order of a grouplike elementobviously divides the order of the group G�H� of all grouplike elements,and this order in turn divides the dimension of H by the Nichols–Zoellertheorem (cf. [17, Theorem 7, p. 384], see also [13, Theorem 3.1.5, p. 30]),we conclude that S2n

H = idH . Therefore it is possible to turn H into a leftmodule over the group ring K�G� by specifying the action of the generatoras

g→ h x= S2H�h�:

We also turn H into a left K�G�-comodule via the trivial coaction:

δH :H → K�G� ⊗H; h 7→ 1⊗ h:In this way, H becomes a left Yetter–Drinfel’d module. (See [13, Defini-tion 10.6.10, p. 213] for the definition of Yetter–Drinfel’d modules, whichwere introduced in [34].) It is easy to see that H is even a Hopf algebra in-side the category of Yetter–Drinfel’d modules (cf. [13, Sect. 10.5]). We cantherefore form the Radford biproduct (cf. [27; 13, Theorem 10.6.5]):

Definition. Define the Hopf algebra E�H� to be the Radford biprod-uct of the group ring K�G� and the Hopf algebra H:

E�H� x= H ⊗K�G�:It is a Hopf algebra with multiplication

�h⊗ gk��h′ ⊗ gl� = hS2kH �h′� ⊗ gk+l;

tensor product comultiplication

1E�H��h⊗ gk� = �h1 ⊗ gk� ⊗ �h2 ⊗ gk�;unit 1E�H� = 1H ⊗ 1K�G�, counit εE�H� = εH ⊗ εK�G� and antipodeSE�H��h⊗ gk� = �1H ⊗ g−k��SH�h� ⊗ 1K�G��.

It is obvious that the dimension of E�H� is dimE�H� = 2 �dimH�2.

2.3. Inside E�H�, the square of the antipode is the conjugationwith a grouplike element:


Proposition. (1) Define gE x= 1H ⊗ g. Then gE is a grouplike elementof E�H� that satisfies S2

E�H��e� = gEeg−1E for all e ∈ E�H�.

(2) E�H� is semisimple if and only if H is semisimple and the charac-teristic p of the base field K does not divide n = 2 dimH.

(3) E�H� is cosemisimple if and only if H is cosemisimple.

Proof. It is obvious that gE is a grouplike element. Now, on the onehand we have

S2E�H��h⊗ gk� = S2

E�H��h⊗ 1K�G�� S2E�H��1H ⊗ gk� = S2

H�h� ⊗ gk

and on the other hand we have

gE�h⊗ gk� = S2H�h� ⊗ gk+1 = �S2

H�h� ⊗ gk�gE:This implies S2

E�H��h⊗ gk� = gE�h⊗ gk�g−1E :

To prove the second statement, observe that it follows easily from [27,Proposition 3, p. 333] and the Larson–Sweedler–Maschke theorem (cf. [9,Proposition 3, p. 84; 13, Theorem 2.2.1, p. 20],) that E�H� is semisimpleif and only if H and K�G� are both semisimple. Since K�G� is semisim-ple if and only if p¦n, the assertion follows. The third assertion followssimilarly from [27, Proposition 4, p. 335] and the fact that K�G� is alwayscosemisimple (cf. [13, Examples 2.1.2, p. 17; 13, Theorem 2.2.1, p. 20]).

3. CHARACTERS AND ORDERS

3.1. In this section, H denotes a semisimple Hopf algebra. H istherefore finite dimensional (cf. [32, Corollary 2.7, p. 330] or [33, Chap. V,Ex. 4, p. 108]). We shall assume throughout the whole section that the basefield K is algebraically closed. By Wedderburn’s theorem, H is thereforeisomorphic to a finite product of full matrix rings. We denote the simplecomponents of H by I1; : : : ; Ik:

H =k⊕i=1

Ii:

Choose a system V1; : : : ; Vk of irreducible modules of H such that Ii isisomorphic to End�Vi�, and denote the corresponding representation by

ρi:H → End�Vi�; i = 1; : : : ; k:

The dimension of Vi as a K-vector space will be denoted by dim Vi = ni.For every i = 1; : : : ; k, we introduce the character χi of Vi as the functionon H defined by

χi:H → K; h 7→ Tr�ρi�h��:


We can assume that V1 = K, the base field, regarded as a trivial H-modulevia εH , which implies that χ1 = εH . The subspace of H∗ generated by thecharacters χ1; : : : ; χk is called the character ring of H and is denoted byCh�H�. It is easy to see that it really is a subalgebra of H∗ which consistsprecisely of the cocommutative elements of H∗.

3.2. We summarize some known properties of the character ringthat will be needed in the sequel. For every module Vi, the dual vectorspace V ∗i is again an irreducible module, and therefore is isomorphic to oneof these modules, which is denoted by Vi. We know that separable algebrasare symmetric Frobenius algebras (see Subsections 3.4 and 3.8 for defi-nitions and references). Therefore, we can conclude from [20, Folgerung3.3.2, p. 13] that the square of the antipode is an inner automorphism. Thisimplies that Vi ∼= V ∗∗i which means that the map i 7→ i is an involution.Since we have SH∗�χi� = χi, where SH∗ is the transpose of SH , we see thatthe transpose of the antipode restricts to an involution of the characterring.

It is important to note that the character ring is in fact defined over �.This �-form is called the Grothendieck ring and will be denoted by G0�H�.It is also called the representation ring in K-theory or the fusion ring inconformal field theory. It is defined as follows: The tensor product of twoirreducible modules decomposes into a direct sum of irreducible modules:

Vi ⊗ Vj ∼=k⊕l=1

VNlij

l :

The number Nlij is called the multiplicity of Vl in Vi ⊗ Vj . Now define the

Grothendieck ring G0�H� to be the free �-module with basis χ1; : : : ; χkand the multiplication which is defined on the basis elements by

χiχj =k∑l=1

Nlijχl:

This turns G0�H� into a ring with unit χ1. The map χi 7→ χi extends to aring antihomomorphism

¯:G0�H� → G0�H�; χ 7→ ¯χ:

Similarly, we have a ring homomorphism

εD:G0�H� → �; χi 7→ ni

which will be called the dimension character. We shall refer to the elementsof the Grothendieck ring as virtual characters.


The character ring can be obtained from the Grothendieck ring by changeof scalars: Namely, the map

K ⊗� G0�H� → Ch�H�; 1⊗ χi 7→ χi

is a ring isomorphism. Under this isomorphism, the above antihomomor-phism corresponds to the transpose of the antipode of H, whereas thedimension character corresponds to the evaluation at 1H , i.e., the counit ofH∗ restricted to Ch�H�.

3.3. Two characters will play an exceptional role in the sequel: Thecharacter of the regular representation and the character of the adjointrepresentation. Define the left regular representation to be

rgH :H → End�H�; h 7→ �h′ 7→ hh′�:The character of the left regular representation will be denoted by χR:χR�h� = Tr�rgH�h��.

On the other hand, we have the left adjoint representation, which isdefined as

adH :H → End�H�; h 7→ �h′ 7→ h1h′SH�h2��:

We shall denote the character of the adjoint representation by χA: χA�h� =Tr�adH�h��.

We now want to express the characters of the regular representation andthe adjoint representation in terms of the irreducible characters. First ofall it is clear that the two-sided ideals I1; : : : ; Ik are invariant subspacesfor both representations. The restriction of the regular representation tothe ideal Ii corresponds via ρi after a choice of a basis in Vi to the leftmultiplication with a matrix inside a matrix ring MK�ni × ni�. Here, thespace of matrices with arbitrary entries in some column and zero entries inall other columns forms again an invariant subspace, and the whole matrixring is the direct sum of ni invariant subspaces of this form. Stated in termsof ideals, we have

Ii ∼= V niiif Ii is considered as a submodule of the left regular representation. Thisimplies

χR =k∑i=1

niχi:

On the other hand, if Ii is regarded as submodule of the adjoint represen-tation, then it is easy to see that the map

Ii → End�Vi� → Vi ⊗ V ∗i


which is the composition of ρi and the inverse of the canonical isomorphismVi ⊗ V ∗i → End�Vi�, v ⊗ φ 7→ �v′ 7→ φ�v′�v� is the composition of two H-linear isomorphisms and therefore itself an H-linear isomorphism. Thisimplies

χA =k∑i=1

χiχi:

These calculations provide the motivation for defining the following twoelements in the Grothendieck ring

χR x=k∑i=1

niχi; χA x=k∑i=1

χiχi:

The elements χR and χA will be called the virtual characters of the regularand of the adjoint representation, respectively. Note that both elementshave been studied before, mostly in their dual form. For example, in [7, 8],the analogue of χR is denoted by x, whereas in [19] the analogue of χAis denoted by z. The basic properties of these two elements are stated inthe next proposition. The first statement and its proof are taken from [30,Lemma 3.12, p. 35]. The dual of the second statement also appears in [19,Remark 21].

Proposition. (1) For all virtual characters χ ∈ G0�H�, we have χχR =χRχ = εD�χ�χR.

(2) For all virtual characters χ ∈ G0�H�, we have χχA = χAχ.

Proof. We shall denote H by Hrg if it is viewed as an H-module via theleft regular representation, and by Had if viewed as an H-module via theleft adjoint representation. If V is an arbitrary H-module, we denote byVεH the H-module that is obtained by regarding the vector space V as atrivial H-module via εH . Consider the map

f :Hrg ⊗ VεH → Hrg ⊗ V; h⊗ v 7→ h1 ⊗ �h2 → v�;where the arrow denotes the module action. This map is an H-linear iso-morphism with inverse

f−1:Hrg ⊗ V → Hrg ⊗ VεH ; h⊗ v 7→ h1 ⊗ �SH�h2� → v�:Therefore, the virtual characters of both modules are equal, which impliesχRχ = εD�χ�χR, which proves the second equality. Since ni = dim Vi =dim V ∗i = ni, we see that χR is invariant under the antihomomorphism.Applying the antihomomorphism to the second equality yields the firstequality.


To prove the second assertion, look at the map

g:Had ⊗ V → V ⊗Had; h⊗ v 7→ �h1 → v� ⊗ h2:

We prove that g is H-linear:

g(h→ �h′ ⊗ v�) = g(h1h

′SH�h2� ⊗ �h3 → v�)= (h1h

′1SH�h4�h5 → v

)⊗ h2h′2SH�h3�

= h→ g�h′ ⊗ v�:Obviously, g is invertible with inverse

g−1:V ⊗Had → Had ⊗ V; v⊗ h 7→ h2 ⊗(S−1H �h1� → v

):

Therefore, the virtual characters of the modules Had ⊗ V and V ⊗Had areequal, which proves the second assertion.

We remark that the module Had ⊗ V contains the module V with at leastmultiplicity 1 since the map

Had ⊗ V → V; h⊗ v 7→ �h→ v�is an H-linear surjection. This observation will be improved in Subsec-tion 3.7.

3.4. Some properties of the character ring can be better understoodby looking at it as a symmetric Frobenius algebra. Recall that a Frobeniusalgebra is a finite-dimensional algebra A which admits a nondegeneratebilinear form �·; ·� x A⊗A→ K which is associative in the sense that wehave �aa′, a′′� = �a, a′a′′� for all a, a′ and a′′ ∈ A. Such a form obviouslycan be written as

�a; a′� = f �aa′�for some linear form f :A→ K which is determined by the bilinear formvia f �a� = �a, 1� = �1, a�. This linear form is called the Frobenius homo-morphism. (This notion is not related to the same notion used in Galoistheory.) A Frobenius algebra is called symmetric if the bilinear form aboveis symmetric.

The following proposition was explained to me by H.-J. Schneider (cf.also [26, Proposition 3, p. 340]):

Proposition. Suppose that A is a Frobenius algebra which is augmentedby ε:A→ K. Then the space of left integrals �x ∈ A � ∀a ∈ A: ax = ε�a�x�and also the space of right integrals �x ∈ A � ∀a ∈ A:xa = ε�a�x� is one-dimensional.


Proof. Consider the left coregular action of A on A∗: An element a ofA acts on an element g of A∗ yielding the element a → g of A∗ whichis defined as �a → g��a′� = g�a′a�. One can restate the definition of aFrobenius algebra by saying that the mapping

A→ A∗; a 7→ �a→ f �is bijective, that is, A∗ is a free cyclic A-module generated by the Frobeniushomomorphism f . Our assertion will be proved if we can show that x is aleft integral if and only if x→ f is a multiple of ε. But observe that

∃λ ∈ K:x→ f = λε⇔ ∃λ ∈ K ∀a ∈ A: f �ax� = λε�a�⇔ ∀a ∈ A: f �ax� = f �x�ε�a�⇔ ∀a; a′ ∈ A: f �aa′x� = f �x�ε�aa′� = f �x�ε�a�ε�a′�⇔ ∀a; a′ ∈ A: f �aa′x� = f �aε�a′�x�⇔ ∀a′ ∈ A: a′x = ε�a′�x:

The assertion on right integrals follows by considering the opposite algebraAop instead of A.

Since the form �·; ·� is nondegenerate, we can choose dual basesx1; : : : ; xn and y1; : : : ; yn satisfying �yi; xj� = δij . From linear algebra weknow that we have a = ∑n

i=1�a; xi�yi =∑ni=1�yi; a�xi for all a ∈ A. This

implies that we haven∑i=1

axi ⊗ yi =n∑i=1

xi ⊗ yia:

The element∑ni=1 xi ⊗ yi is therefore called the Casimir element of A. It

does not depend on the choice of the dual bases, but of course it doesdepend on the bilinear form. It is clear that

∑ni=1 xiyi is a central element

of A. This element is sometimes called the Casimir element, too (cf. [1,Sect. 5]).

3.5. Now, it turns out that the character ring is a symmetric Frobe-nius algebra. This fact is also observed in [11], and similar statements canbe found in several references in conformal field theory (cf., for example,[1, Sect. 5.8, p. 13]). Essentially, this is equivalent to the well-known or-thogonality relations for the characters, which in turn follow easily fromSchur’s Lemma. Note first that it is clear that the character χi vanishes onIj if i 6= j. In particular, since I1 is precisely the one-dimensional subspaceof integrals, we have for some integral 3H that χi�3H� = 0 if i 6= 1, that is,χi 6= εH . If we assume that εH�3H� = 1, then, since the component of thetrivial representation inside the H-module HomK�Vi; Vj� ∼= Vj ⊗ V ∗i is the


space of H-linear maps HomH�Vi; Vj� which has dimension one or zero bySchur’s Lemma, we have

�χjχi��3H� = δij:These are the orthogonality relations for the characters which appear in adualized form for arbitrary Hopf algebras in [6].

Proposition. Ch�H� is a symmetric Frobenius algebra with respect to theFrobenius homomorphism

tC : Ch�H� → K; χ 7→ χ�3H�:The corresponding Casimir element is

∑ki=1 χi ⊗ χi: The space of integrals for

the character

εC : Ch�H� → K; χ 7→ χ�1H�is spanned by χR.

Proof. We have to prove that the bilinear form

�·; ·�: Ch�H� × Ch�H� → K; �χ;χ′� 7→ tC�χχ′�is nondegenerate. But this is obvious since we have already found dualbases with respect to this form, since we have �χi; χj� = δij . Since thisexpression is symmetric in i and j, the form is also symmetric. The formof the Casimir element follows from the definition and the form of theintegrals from Proposition 3.3. Note that χR 6= 0 since χR�3H� = 1.

This also gives us another proof for the fact that χA is central inCh�H� which we observed in Proposition 3.3, since we now see thatχA =

∑ki=1 χiχi comes from a Casimir element via multiplication of the

tensorands. We note that all the structures considered above are alreadydefined over �, that is, on the level of the Grothendieck ring, since if wedefine

tG:G0�H� → �; χi 7→ δi1;

then by the same proof we have tG�χiχj� = δij , and εC and χR are alsodefined on the level of the Grothendieck ring.

3.6. The question that we study next is the question under whichcircumstances the element χA ∈ Ch�H� is invertible, because this will implythat the character ring is separable. For this purpose, it is useful to intro-duce some more notation. First of all, we shall use a modification of thebilinear form arising from the Frobenius homomorphism tG. Define

�·; ·�∗:G0�H� ×G0�H� → �; �χ; χ′� 7→ �χ; χ′�∗ x= tG�χχ′�:


We shall use the same notation for the bilinear form on G0�H� ⊗� � whichis obtained by extension of scalars. This modified bilinear form has theadvantage that it puts the orthogonality relations for the characters intothe more symmetric form �χi; χj�∗ = δij . This equation shows in particularthat this bilinear form is symmetric and positive definite. We shall also usethe convention to denote the left regular representation of the characterring or the Grothendieck ring by Lχ resp. Lχ, that is, we define

Lχ: Ch�H� → Ch�H�; χ′ 7→ Lχ�χ′� x= χχ′:The adjoint of Lχ with respect to our scalar product then is L ¯χ, that is,we have �Lχ�χ′�; χ′′�∗ = �χ′; L ¯χ�χ′′��∗. This holds because of the simplecalculation, which of course is well known in the representation theory ofgroups and is carried out similarly in [1, Sect. 5; 19, Theorem 8],

�χχ′; χ′′�∗ = tG�χχ′χ′′� = tG�χ′χ′′χ� = tG�χ′χχ′′� = �χ′; ¯χχ′′�∗:Besides these, we have a third bilinear form on G0�H�, namely the traceform

G0�H� ×G0�H� → �; �χ; χ′� 7→ Tr�Lχχ′ �:These three bilinear forms are linked as follows (cf. [1, Sect. 5, Eq. (5.8)]):

Proposition. For all χ; χ′ ∈ G0�H�, we have

Tr�Lχχ′ � = tG�χAχχ′� = �χAχ; χ′�∗:Proof. The second equality follows directly from the definitions. For the

first equality, it obviously suffices to prove Tr�Lχ� = tG�χAχ�. Now, sinceχ1; : : : ; χk is an orthonormal basis with respect to �·; ·�∗, we know fromlinear algebra that Lχ�χj� =

∑ki=1 �Lχ�χj�; χi�∗χi. This implies

Tr�Lχ� =k∑j=1

�χχj; χj�∗ =k∑j=1

tG�χχjχj� = tG�χχA�

which proves the assertion since χA is central.

3.7. We now introduce a major object of our investigation, namelythe matrix representation of the left multiplication by χA:

Definition. Define M = �mij�i;j=1;:::;k to be the matrix representationof the left multiplication by χA with respect to the basis χ1; : : : ; χk, thatis, we have

χAχj =k∑i=1

mijχi:


Note that since χ1; : : : ; χk is an orthonormal basis with respect to �·; ·�∗,we have χAχj =

∑ki=1�χAχj; χi�∗χi. The matrix elements mij therefore can

be expressed as

mij = �χAχj; χi�∗ = tG�χAχjχi� = Tr�Lχjχi�:We now summarize the basic properties of the matrix M in the followingtheorem:

Theorem. The matrix M has the following properties:

(1) The diagonal elements of M satisfy dim Ch�H� ≤ mii ≤ dimH,where m11 = dim Ch�H�.

(2) M is symmetric and positive definite.(3) The eigenvalues of M are positive real algebraic integers.(4) dimH is the greatest eigenvalue of M .(5) dimH divides detM .

Proof. Although most of the statements are obvious, we proceed insteps.

(1) We have that tG�χA� = Tr�Lχ1� = dim Ch�H� = k. This means

that we have for some nonnegative integers q2; : : : ; qk,

χA = kχ1 +k∑i=2

qiχi

and this implies χAχj = kχj +∑ki=2 qiχiχj and therefore mjj ≥ k. On

the other hand, it is clear from dimension considerations that the moduleHad ⊗ Vj cannot contain the module Vj with a multiplicity which is greaterthan dimH. This proves the first statement.

(2) To prove the symmetry of M , we calculate

mij = �χAχj; χi�∗ = �χj; ¯χAχi�∗ = �χj; χAχi�∗ = �χAχi; χj�∗ = mji:

To prove definiteness, we observe that M is the fundamental matrix of thebilinear form �χ; χ′� 7→ �χAχ; χ′�∗ on G0�H� with respect to the basisχ1; : : : ; χk. The definiteness of M therefore follows from the definitenessof this bilinear form which in turn follows from the definiteness of �·; ·�∗:

�χAχ; χ�∗ =k∑i=1

�χiχiχ; χ�∗ =k∑i=1

�χiχ; χiχ�∗ ≥ 0

and �χAχ, χ�∗ = 0 implies χiχ = 0 for all i = 1; : : : ; k, which for i = 1implies χ = 0.


(3) The eigenvalues of a symmetric positive definite matrix are realand positive. On the other hand, since M has integer entries, the character-istic polynomial of M is a monic integral polynomial. Since the eigenvaluessatisfy the characteristic equation, they are algebraic integers.

(4) First of all, we observe that dimH really is an eigenvalue of M ,since we have by Proposition 3.3 that χAχR = dimHχR. We now proceedto prove that this is the greatest eigenvalue of M . Since M is symmet-ric, we can achieve by changing the enumeration of the virtual charactersχ1; : : : ; χk that M attains a block form

M =

M1

M2 0: : :

0 Ml−1Ml

;

where M1; : : : ;Ml are indecomposable matrices (in the sense of [3, Def-inition 2, p. 395]) and the entries outside these blocks are zero. Now weknow from the Perron–Frobenius theorem (cf. [3, Sect. 13.2, p. 398]) thateach Mi has a unique greatest eigenvalue λi which is strictly positive, whosealgebraic multiplicity is one, i.e., which is a simple root of the characteris-tic polynomial of Mi, and whose corresponding eigenvector of Mi can bechosen with strictly positive coordinates. For every Mi, we can enlarge thiseigenvector to an eigenvector of M by filling up with zeros. In this waywe see that, for every i = 1; : : : ; l, M has a strictly positive eigenvalue λisuch that the corresponding eigenvector xi = �xij�j=1; : : : ;k has nonnegativecoordinates. Define

χ =k∑j=1

xijχj:

The equation Mxi = λixi then yields χAχ = λiχ. Applying the dimensioncharacter to this equation we get

dimHεD�χ� = εD�χA�εD�χ� = λiεD�χ�:

Since εD�χ� =∑kj=1 xijnj 6= 0, we conclude that λi = dimH. Obviously,

the greatest eigenvalue of M is the greatest eigenvalue of some Mi, andtherefore is equal to dimH. Note that we have proved in addition that thealgebraic multiplicity of the eigenvalue dimH of M equals the geometricmultiplicity, i.e., the multiplicity of dimH as a root of the characteristicpolynomial equals the dimension of the eigenspace belonging to dimH,and this equals l, the number of indecomposable blocks of M .


(5) Denote the eigenvalues of M by µ1; : : : ; µk, where µk = dimH.Then we see that

detMdimH

=k−1∏i=1

µi

is on the one hand a rational number and on the other hand an algebraicinteger. Since � is integrally closed, it must be an integer.

3.8. Recall that a finite dimensional algebra A is called separableif there is an element

∑i xi ⊗ yi ∈ A⊗A, called the separability element,

such that∑i xiyi = 1 and

∑i axi ⊗ yi =

∑i xi ⊗ yia for all a ∈ A. In

this case, A is semisimple. A separable algebra is a symmetric Frobeniusalgebra (cf. [2, Chap. X, Theorem (71.6), p. 482]). This is particularly easyto prove in the case where the base field is algebraically closed, becausein this case the ordinary trace function yields a Frobenius homomorphismfor every simple component. With our preparations, we can now prove thatthe character ring is a separable algebra if the characteristic of the basefield is large enough. In characteristic zero, this is of course well known,see for example, [35; 1; 19, Theorem 9], where in all cases the proof isbased on the fact that the positive definiteness of the bilinear form �·; ·�∗contradicts the existence of a nilpotent ideal. We exclude from the firstassertion in the following theorem the following three cases: dimH = 2and charK = 2, dimH = 3 and charK = 3, dimH = 4 and charK = 2. Inthe first case we have the counterexample H = K��2�∗, in the second casewe have the counterexample H = K��3�∗, and in the third case we havethe counterexamples H = K��4�∗ and H = K��2 ×�2�∗. Note that K��2�∗,K��3�∗, K��4�∗ and K��2 × �2�∗ are the only semisimple Hopf algebrasof dimension ≤ 4 over an algebraically closed field, since by dimensionconsiderations H must be commutative, and therefore [13, Theorem 2.3.1,p. 22] applies.

Theorem. (1) Suppose that we do not have: dimH = 2 and charK= 2, dimH = 3 and charK = 3, dimH = 4 and charK = 2. If the charac-teristic p of K is zero or greater than �dimH��dimH−4�, then the characteristicof K does not divide the determinant of M .

(2) The character χA ∈ Ch�H� is invertible in Ch�H� if and only ifcharK does not divide detM .

(3) If χA is invertible, then Ch�H� is a separable algebra with separa-bility element

∑ki=1 χi ⊗ χ−1

A χi.(4) If the characteristic of K does divide the dimension of H, then χA

is not invertible and Ch�H� is not semisimple.

Proof. First observe that the second statement is obvious, because χAis invertible if and only if the left multiplication LχA with χA is invertible,


and the matrix representation of LχA is the matrix M reduced modulo p.So LχA is invertible if and only if its determinant is nonzero, that is, charKdoes not divide detM .

To prove the first statement, we first rule out the trivial case that H iscommutative. In this case, the adjoint representation is the trivial repre-sentation with multiplicity dimH. Therefore, M is dimH times the identitymatrix, and the determinant of M is �dimH�k. Since under our assumptionscharK does not divide dimH, it does also not divide detM .

We now turn to the more interesting case where H is not commutative.In this case, one of the simple components I1; : : : ; Ik has dimension atleast 4; therefore we have k ≤ dimH − 3. We assume on the contrary thatp divides detM . If dimH = µ1 ≥ µ2 ≥ : : : ≥ µk are the eigenvalues of M ,then detM is the product of the two integers dimH and

∏ki=2 µi. Since p

does not divide dimH, it must divide∏ki=2 µi. But this is not possible since

every eigenvalue µi is smaller than or equal to dimH, and therefore wesee that:

k∏i=2

µi ≤ �dimH�k−1 ≤ �dimH��dimH−4� < p:

This implies that LχA , and therefore χA, is invertible. Now the fact thatthe element

∑ki=1 χi ⊗ χ−1

A χi is a separability element follows from Propo-sition 3.5.

The third assertion follows directly from Proposition 3.5.To prove the fourth statement, we observe that χAχ

−1A = εH implies

dimHχ−1A �1� = χA�1�χ−1

A �1� = 1

and therefore p¦ dimH. On the other hand, assume that Ch�H� issemisimple. Then we have that ker εC = �χ ∈ Ch�H� �χ�1H� = 0� is atwo-sided ideal of codimension one which is therefore complemented bya one-dimensional ideal spanned by some nonzero element χC . Now, ifχ ∈ Ch�H� is arbitrary, then we have χ− εC�χ�εH ∈ ker εC and therefore�χ− εC�χ�εH�χC = 0. This implies

χχC = εC�χ�χC;i.e., χC is a nonzero left integral with respect to the character εC . Sincewe already know from Proposition 3.5 that χR is also a left integral withrespect to εC , and the space of left integrals is one-dimensional by Proposi-tion 3.4, we can assume that χC = χR. But this means that χR /∈ ker εC , andtherefore dimH = εC�χR� 6= 0 ∈ K, which means that the characteristic pdoes not divide dimH.

3.9. The Grothendieck ring of a semisimple Hopf algebra hasstrong similarities with the ring of integers in an algebraic number field:


Both are orders in the sense of [29]. In this subsection, we comment brieflyon the interrelation of the properties of orders and the properties of thematrix M considered above. A more detailed analysis of the interrelationof the theory of orders and the representation theory of Hopf algebras willbe carried out in the complete version of the author’s Dissertation.

Suppose that R is a Dedekind ring and denote by L its field of fractions.Define A x= G0�H� ⊗� L. If the characteristic of L is zero, it follows fromthe preceding discussion that A is a separable algebra. (As we have alreadypointed out, similar statements also appear in [35, 30, 10, 19, 11].) DefineB x= G0�H� ⊗� R ⊂ A. It is obvious that B is an R-order in A, that is, asubring which is finitely generated as an R-module such that LB = A (cf.[29, Sect. 8, p. 108]). We shall use the following notation for the images ofthe virtual characters in B:

xi x= χi ⊗ 1; xH x= χH ⊗ 1; xA x= χA ⊗ 1:

The following proposition should be compared with [29, Theorem (41.1),p. 379]:

Proposition. (1) If C is another R-order in A that contains B, then wehave xAC ⊂ B.

(2) If detM is invertible as an element of R, then B is a maximal R-order.

(3) If B is a maximal R-order in A, then dimH is invertible in R if itis invertible in L.

Proof. To prove the first statement, suppose we are given x ∈ C. Byelementary properties of integral ring extensions (cf. [24, Theorem 8.5,p. 104]), the element xxi ∈ C is integral over R. Therefore, the eigen-values of the left multiplication Lxxi also satisfy an integral equation, andtheir sum, the trace of Lxxi , is an integral element of L which is thereforecontained in R. But by Proposition 3.6, we have that Tr�Lxxi� = �xAx; xi�∗,where we have extended the bilinear form �·; ·�∗ to an L-bilinear form onA. With respect to this form, x1; : : : ; xk form an orthonormal basis, andtherefore we have

xAx =k∑i=1

�xAx; xi�∗xi ∈ B:

To prove the second statement, observe that the matrix represention of theleft multiplication with xA with respect to the basis x1; : : : ; xk is of courseM , regarded as a matrix with entries in R. The adjoint of M thereforealso has entries in R. If detM is invertible in R, then M−1 has entries in R,which means that LxA is invertible as an endomorphism of B, which implies


that xA is invertible in B. It therefore follows from the first statement thatfor every order C that contains B we have C = x−1

A xAC ⊂ x−1A B ⊂ B.

To prove the third statement, observe that e x= xH/ dimH is a centralidempotent in A since we have x2

H = dimHxH by Proposition 3.3. It isclear that Be and B�1− e� are R-orders in Ae resp. A�1− e�, and thereforeBe+B�1− e� is an R-order in A. But B ⊂ Be+B�1− e�, and therefore weconclude from the maximality of B that B = Be+ B�1− e�, which impliese ∈ B. But since

e = 1dimH

x1 +k∑i=2

nidimH

xi;

we can conclude that 1/ dimH ∈ R.

In particular, if p is a prime number that does not divide detM , thenwe see that G0�H� ⊗� ��p� is a maximal ��p�-order in G0�H� ⊗� �, where��p� denotes the localization of � at the prime ideal �p�.

3.10. We proceed to interrelate the matrix M and the element xAwith the discriminant and the different of B. Differents and discriminantsare defined for arbitrary orders, cf. [29, Sects. 10 and 25]. There, differentsand discriminants are defined via the reduced trace map (cf. [29, Sect. 9]).We shall adopt here a different version of these notions via the unreducedtrace map already considered in Subsection 3.6. For the sake of clarity, wemake this explicit in the following definition:

Definition. (1) The unreduced discriminant D�B� of B over R is theideal of R which is generated by the elements det�Tr�Lyiyj ��i; j=1;:::;k for allpossible k-tuples y1; : : : ; yk ∈ B.

(2) The unreduced inverse different $′�B� of B over R is defined as

$′�B� = {x ∈ A � ∀y ∈ B: Tr�Lxy� ∈ R}:

(3) The unreduced different $�B� of B over R is defined as

$�B� = {x ∈ A � ∀y; z ∈ $′�B�: yxz ∈ $′�B�}:The unreduced discriminant and the unreduced different can be ex-

pressed as follows:

Theorem. Suppose that the characteristic of L is zero or does not dividedetM .

(1) The unreduced discriminant is the principal ideal of R generated bydetM: D�B� = �detM�

(2) The unreduced different is the principal ideal of B generated by xA:$�B� = �xA� x= xAB


(3) $�B� is a two-sided ideal of B. If L is an algebraic number fieldand R is its ring of integers, then the order of the quotient ring is finite andgiven explicitly as card�B/$�B�� = �detM��Lx��.

Proof. As in [29, Theorem (10.2)] it can be shown that, since B is a freeR-module with basis x1; : : : ; xk, D�B� is the principal ideal of R generatedby det�Tr�Lxixj ��. But this determinant is equal to det�Tr�Lxixj ��, up to asign which is equal to the the determinant of the permutation matrix thatdescribes the change of basis from x1; : : : ; xk to x1; : : : ; xk. Now we knowfrom Subsection 3.7 that this is precisely the determinant of the transposeof M .

To prove the second statement, we first calculate the unreduced inversedifferent. We have x ∈ $′�B� if and only if Tr�Lxy� ∈ R for all y ∈ B. SinceB is a free R-module with basis xi; i = 1; : : : ; k, this will happen if andonly if we have Tr�Lxxi� ∈ R for all i = 1; : : : ; k. But we have Tr�Lxxi�= �xAx, xi�∗ by Proposition 3.6, and since x1; : : : ; xk is an orthonormalbasis with respect to this form, we have: xAx =

∑ki=1 �xAx, xi�∗xi. This

implies that x ∈ $′�B� if and only if xAx ∈ B.Now, if charL is zero or does not divide detM , the same argument as in

Subsection 3.8 proves that xA ∈ A is invertible. We therefore have $′�B� =x−1A B. By definition, we see that x ∈ $�B� if and only if x−1

A yxx−1A z ∈ x−1

A Bfor all y; z ∈ B, which is, since xA is central, equivalent to yxz ∈ xAB.Therefore, we see that $�B� = xAB.

We now prove the third statement. The fact that $�B� is a two-sidedideal follows from the fact that xA is central. Consider the exact sequence

0→ G0�H�LχA−→G0�H� → G0�H�/�χA� → 0:

By the Weierstrass elementary divisors theorem, there exist �-basesy1; : : : ; yk and z1; : : : ; zk of G0�H� such that we have

LχA�yi� = dizi; i = 1; : : : ; k;

where d1�d2� : : : �dk. This implies that

G0�H�/�χA� ∼=k∏i=1

�/di�

and therefore we have that cardG0�H�/�χA� = d1 · : : : · dk. But now, if Qdenotes the matrix of the base change from y1; : : : ; yk to χ1; : : : ; χk, andP the matrix of the base change from z1; : : : ; zk to χ1; : : : ; χk, we have forD = diag�d1; : : : ; dk�,

M = PDQ−1:


Since P , Q ∈ GL�k;��, their determinants are units in �, and therefore wehave detM = ±d1 · : : : · dk. Since M is positive definite, detM is positive,and therefore the positive sign in the last equation is correct.

Now consider the commutative diagram

G0�H� ⊗� RLχA⊗ id−→ G0�H� ⊗� R → G0�H�/�χA� ⊗� R

↓ ↓ ↓B −→ B → B/$�B�

LxA

From the diagram we conclude that B/$�B� ∼= G0�H�/�χA� ⊗� R. SinceR is a free �-module of rank �L x �� (cf. [15, Satz (2.10), p. 13]), wehave that G0�H�/�χA� ⊗� R ∼= �G0�H�/�χA��Lx��, which implies thatcardB/$�B� = �detM��Lx��.

The proof shows that without the assumption on the characteristic of L,the first statement still holds, while from the proof of the second statementwe get that $′�B� = �x ∈ A � xAx ∈ B�. It is easy to see that this implies atleast that xAB ⊂ $�B�. The third statement allows the following analogywith algebraic number theory: By a theorem of Dedekind, the norm ofthe different is the discriminant (cf. [29, Theorem (25.2), p. 218]), and inaddition the norm of a principal ideal is the norm of the generating element(cf. [15, Kap. I, Sect. 6, p. 37]), i. e. xA, which is precisely �detM��Lx��. Weshall call the ring G0�H�/�χA� the adjunction quotient ring of H.

3.11. To conclude this section, we change our viewpoint and inves-tigate what can be said about the eigenvalues of M if the conclusions wewant to derive are satisfied. We shall see that over fields of characteristiczero, the eigenvalues are actually integers. In order to formulate the result,we introduce some notation.

In this subsection, we shall assume that the antipode of H is an involutionand that the character ring Ch�H� is semisimple. Suppose that U1; : : : ; Ul isa system of irreducible Ch�H�-modules of dimensions m1; : : : ;ml such thatevery irreducible Ch�H�-module is isomorphic to precisely one of these.We denote the representation and the character corresponding to Uj by σjresp. ξj:

ξj�χ� = Tr�σj�χ��:Every H∗-module can be restricted to a Ch�H�-module. In particular, theleft regular representation of H∗ restricts to a Ch�H�-module, and we geta decomposition

H∗ ∼=l⊕j=1

Uqjj :


We refer to the integer qj as the multiplicity of the module Uj in the(restricted) left regular representation of H∗.

As in Subsection 3.9, we denote by ��p� the localization of � at the primeideal �p�, where p = charK. In particular, we have ��0� = �. We denoteby �p x= �/p� the field that contains p elements (and not the ring ofp-adic integers), for p = 0 we define �0 = �. By the universal propertyof localizations (cf. [24, Theorem 7.8, p. 81]), we have a canonical map��p� → �p, which is the identity in the case p = 0. If we apply this mapto all entries of M , we get a matrix Mp which we call the reduction of Mmodulo p. Now it turns out that often the eigenvalues of Mp are containedin the prime field �p.

Theorem. Suppose that SH is an involution and that Ch�H� is semisim-ple. Suppose furthermore that, if p > 0, the dimensions m1; : : : ;ml of the irre-ducible Ch�H�-modules are not divible by p. Then the multiplicities q1; : : : ; qlare also not divisible by p. The eigenvalues of Mp are contained in the primefield �p, they are given explicitly as the images of the rational numbers

dimHm1

q1; : : : ; dimH

ml

ql

under the canonical map ��p� → �p, occurring with multiplicitiesm21; : : : ;m

2l .

Proof. We have seen that χA is a central element of Ch�H�. If f1; : : : ; flare the primitive central idempotents of Ch�H�, we can write

χA =l∑j=1

βjfj:

Since Mp is the matrix representation of the left multiplication by χA, itis obvious that β1; : : : ; βl are the eigenvalues of Mp, occurring with multi-plicities m2

1; : : : ;m2l .

We know from [8, Theorem 4.4, p. 279] that for ϕ ∈ H∗, we have χ∗R�ϕ� =dimH ϕ�3H�, where 3H is a (two-sided) integral of H satisfying εH�3H�= 1 and χ∗R denotes the character of the left regular representation of H∗.If ϕ = χ ∈ Ch�H�, this equality reads

dimH tC�χ� =l∑j=1

qjξj�χ�:

Inserting χ = χAfi = βifi, we can conclude from Proposition 3.6 that

dimHm2i = dimH Tr�Lfi� = dimH tC�χAfi� = βiqiξi�fi� = βiqimi:

Since we have assumed that Ch�H� is semisimple, we can conclude fromTheorem 3.8 that dimH is not divisible by p. Therefore the left hand side inthe last equality is nonzero in �p, which means that βi and qi are nonzeroin K. This implies the assertion.


If the characteristic of the base field is zero, the above theorem assertsthat the eigenvalues of M are rational numbers. Since we have alreadyseen in Theorem 3.7 that they are algebraic integers, they must be naturalintegers. In his very interesting recent article [11], M. Lorenz derives amethod which can be used to give a rather different proof of the abovetheorem for fields of characteristic zero. His method, which is only slightlymore complicated than the one above, yields the refined result that alreadythe number dimH/qj is an integer, which is the main assertion in the so-called class equation for Hopf algebras first proved by G. I. Kac and Y. Zhu(cf. [4, 35]).

4. KAPLANSKY’S FIFTH CONJECTURE

4.1. We now combine the results of the two preceding sections toobtain a proof of Kaplansky’s fifth conjecture over fields of large positivecharacteristic. In this section, H continues to denote a semisimple Hopfalgebra, but we do no longer assume that the base field is algebraicallyclosed, since the more general case does not offer any additional difficulty.We summarize the technical work in the following proposition. Note that wehave already seen in Subsection 3.8 that the antipode of a semisimple Hopfalgebra of dimension ≤ 4 is an involution, even if H is not cosemisimple.

Proposition. Suppose that dimH ≥ 5. Suppose that the characteristic pof K is zero or satisfies p > nn−4 where n = dimH. Suppose that H containsa grouplike element g that induces the square of the antipode:

S2H�h� = ghg−1:

Then H is cosemisimple and the antipode is an involution.

Proof. Because H is separable, we can assume that K is algebraicallyclosed. Pick right integrals 0H ∈ H and ρH ∈ H∗ satisfying ρH�0H� = 1. Weknow from Theorem 3.8 that χA is invertible: χAχ

−1A = εH . This implies

χA�g�χ−1A �g� = 1 and therefore

Tr�S2H� = Tr�adH�g�� = χA�g� 6= 0:

But by [8, Theorem 2.5, p. 274] we know that Tr�S2H� = εH�0H�ρH�1H�

and therefore we have ρH�1H� 6= 0, that is, H is cosemisimple. If dimH≥ 6, we now conclude from [7, Theorem 3, p. 194] that the antipode is aninvolution. If dimH = 5, we conclude from the Nichols–Zoeller theorem(cf. [17, Theorem 7, p. 384], see also [13, Theorem 3.1.5, p. 30]) that theset of grouplike elements G�H� has order 1 or order 5. In the second case,we have H ∼= K��5� and therefore S2

H = id; in the first case we have g = 1and therefore S2

H = id. In both cases, we have εH�0H�ρH�1H� = Tr�S2H� =

5 6= 0 and therefore H is cosemisimple.


4.2. The proof of the main theorem is now trivial:

Theorem. Suppose that H is a semisimple Hopf algebra. Suppose thatthe characteristic p of K is zero or satisfies p > mm−4 where m = 2�dimH�2.Then H is cosemisimple and the antipode of H is an involution.

Proof. We can assume that dimH ≥ 2. We then have that dimE�H� =2�dimH�2 ≥ 8 and that p > 2�dimH�. Therefore, E�H� is semisimple byProposition 2.3. Now Proposition 4.1 implies that E�H� is cosemisimpleand that SE�H� is an involution. This implies by Proposition 2.3 and theform of SE�H� that H is cosemisimple and that SH is an involution.

ACKNOWLEDGMENTS

This work is part of the author’s Dissertation. The author thanks theMathematics Department of the University of Munich for the permis-sion to publish these results. He also thanks D. Husemoller, A. Masuoka,B. Pareigis, P. Schauenburg, H.-J. Schneider, and M. Takeuchi for in-teresting discussions. H.-J. Schneider also kindly pointed out Ref. [19].A. Masuoka gave detailed comments on an earlier draft of the manuscript.His important remarks led to an improvement of the results and thepresentation.

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