Three variations on the linear independence of grouplikes in a coalgebra
aa r X i v : . [ m a t h . QA ] S e p Three variations on the linearindependence of grouplikes in acoalgebra
Gérard H. E. Duchamp ∗ , Darij Grinberg † , andVincel Hoang Ngoc Minh ‡ LIPN, University Paris 13, Sorbonne Paris City, 93430 Villetaneuse, France Drexel University, Korman Center, Room 291, 15 S 33rd Street,Philadelphia PA, 19104, USA University of Lille, 1 Place Déliot, 59024 Lille, France
Abstract.
The grouplike elements of a coalgebra over a field areknown to be linearly independent over said field. Here we prove threevariants of this result. One is a generalization to coalgebras over a com-mutative ring (in which case the linear independence has to be replacedby a weaker statement). Another is a stronger statement that holds (un-der stronger assumptions) in a commutative bialgebra. The last variantis a linear independence result for characters (as opposed to grouplikeelements) of a bialgebra. ∗ [email protected] † [email protected] ‡ [email protected] ariations on linear independence of grouplikes page 2 Contents
1. Introduction 22. Background 3
3. Grouplikes in a coalgebra 74. Grouplikes over id-unipotents in a bialgebra 11
5. Linear independence in the dual algebra 32
1. Introduction
A classical result in the theory of coalgebras ([Sweedl69, Proposition 3.2.1 b)],[Radfor12, Lemma 2.1.12], [Abe80, Theorem 2.1.2 (i)]) says that the grouplike el-ements of a coalgebra over a field are linearly independent over said field. Weshall prove three variants of this result. The first variant (Theorem 3.1, in Section 3)generalizes it to coalgebras over an arbitrary commutative ring (at the expense ofobtaining a subtler claim than literal linear independence). The second variant(Theorem 4.7, in Section 4) gives a stronger independence claim under a strongerassumption (viz., that the coalgebra is a commutative bialgebra, and that the grou-plike elements and their pairwise differences are regular). The third variant (Theo-rem 5.3 (b) , in Section 5) is a linear independence statement in the dual algebra of abialgebra; namely, it claims that (again under certain conditions) a set of charactersof a bialgebra (i.e., algebra homomorphisms from the bialgebra to the base ring)are linearly independent not just over the base ring, but over a certain subalgebraof the dual. We discuss the connection between grouplike elements and characters.
Acknowledgments
We thank Jeremy Rickard for Example 5.10.
DG thanks the Mathematisches Forschungsinstitut Oberwolfach for its hospital-ity at the time this paper was finished. This research was supported through theprogramme “Oberwolfach Leibniz Fellows” by the Mathematisches Forschungsin-stitut Oberwolfach in 2020.
2. Background
We shall study coalgebras and bialgebras. We refer to the literature on Hopf alge-bras and bialgebras – e.g., [GriRei20, Chapter 1] or [Bourba89, Section III.11] – forthese concepts. We let N denote the set {
0, 1, 2, . . . } .Rings are always associative and have unity (but are not always commutative).We fix a commutative ring k . All algebras, linear maps and tensor signs thatappear in the following are over k unless specified otherwise. The symbol Homshall always stand for k -module homomorphisms. Recall that • a k -algebra can be defined as a k -module A equipped with a k -linear map µ : A ⊗ A → A (called multiplication) and a k -linear map η : k → A (called unitmap) satisfying the associativity axiom (which says that µ ◦ ( µ ⊗ id ) = µ ◦ ( id ⊗ µ ) ) and the unitality axiom (which says that µ ◦ ( η ⊗ id ) and µ ◦ ( id ⊗ η ) are the canonical isomorphisms from k ⊗ A and A ⊗ k to A ). • a k -coalgebra is defined as a k -module C equipped with a k -linear map ∆ : C → C ⊗ C (called comultiplication) and a k -linear map ǫ : C → k (calledcounit map) satisfying the coassociativity axiom (which says that ( ∆ ⊗ id ) ◦ ∆ = ( id ⊗ ∆ ) ◦ ∆ ) and the counitality axiom (which says that ( ǫ ⊗ id ) ◦ ∆ and ( id ⊗ ǫ ) ◦ ∆ are the canonical isomorphisms from C to k ⊗ C and C ⊗ k ). • a k -bialgebra means a k -module B that is simultaneously a k -algebra and a k -coalgebra, with the property that the comultiplication ∆ and the counit ǫ are k -algebra homomorphisms (where the k -algebra structure on B ⊗ B is thestandard one, induced by the one on B ). We note that terminology is not entirely standardized across the literature. What we call a “coal-gebra”, for example, is called a “counital coassociative cogebra” in [Bourba89, Section III.11].What we call a “bialgebra” is called a “bigebra” in [Bourba89, Section III.11]. (Note that Lie algebras are not considered to be k -algebras.)The multiplication and the unit map of a k -algebra A will always be denoted by µ A and η A . Likewise, the comultiplication and the counit of a k -coalgebra C willalways be denoted by ∆ C and ǫ C . We will occasionally omit the subscripts whenit is clear what they should be (e.g., we will write ∆ instead of ∆ C when it is clearthat the only coalgebra we could possibly be referring to is C ). If C is a k -coalgebra, and if A is a k -algebra, then the k -module Hom ( C , A ) itselfbecomes a k -algebra using a multiplication operation known as convolution. Wedenote it by ⊛ , and recall how it is defined: For any two k -linear maps f , g ∈ Hom ( C , A ) , we have f ⊛ g = µ A ◦ ( f ⊗ g ) ◦ ∆ C : C → A .The map η A ◦ ǫ C : C → A is a neutral element for this operation ⊛ .(Note that the operation ⊛ is denoted by ⋆ in [GriRei20, Definition 1.4.1].)If f is a k -linear map from a coalgebra C to an algebra A , and if n ∈ N , then f ⊛ n denotes the n -th power of f with respect to convolution (i.e., the n -th power of f inthe algebra ( Hom ( C , A ) , ⊛ , η A ◦ ǫ C ) ).If M is any k -module, then the dual k -module Hom k ( M , k ) shall be denoted by M ∨ . Thus, if C is a k -coalgebra, then its dual k -module C ∨ = Hom ( C , k ) becomesa k -algebra via the convolution product ⊛ . The unity of this k -algebra C ∨ is exactlythe counit ǫ of C . One of the simplest classes of elements in a coalgebra are the grouplike elements:
Definition 2.1.
An element g of a k -coalgebra C is said to be grouplike if it satisfies ∆ ( g ) = g ⊗ g and ǫ ( g ) = ǫ ( g ) =
1” condition by the weaker requirement “ g = k is a field, but ill-behaved and useless when k is merely a commutative ring.The following examples illustrate the notion of grouplike elements in different k -bialgebras. Example 2.2.
Let q ∈ k . Consider the polynomial ring k [ x ] in one variable x over k . Define two k -algebra homomorphisms ∆ : k [ x ] → k [ x ] ⊗ k [ x ] and ǫ : k [ x ] → k by setting ∆ ( x ) = x ⊗ + ⊗ x and ǫ ( x ) = k -algebra k [ x ] , equipped with the comultiplication ∆ and the counit ǫ , is a k -bialgebra. This is the “standard” k -bialgebra structure on the polynomial ring k . If k is a reduced ring (i.e., a commutative ring with nononzero nilpotent elements), then it is not hard to show that 1 is the only grouplikeelement of this k -bialgebra. On the other hand, if u ∈ k satisfies u =
0, then theelement 1 + ux ∈ k [ x ] is also grouplike. Example 2.3.
Let q ∈ k . Consider the polynomial ring k [ x ] in one variable x over k .Define two k -algebra homomorphisms ∆ ↑ q : k [ x ] → k [ x ] ⊗ k [ x ] and ǫ : k [ x ] → k by setting ∆ ↑ q ( x ) = x ⊗ + ⊗ x + qx ⊗ x and ǫ ( x ) = k -algebra k [ x ] , equipped with the comultiplication ∆ ↑ q and the counit ǫ , is a k -bialgebra. This k -bialgebra (cid:16) k [ x ] , ∆ ↑ q , ǫ (cid:17) is known asthe univariate q -infiltration bialgebra . Note the following special cases:1. If q =
0, then the univariate q -infiltration bialgebra (cid:16) k [ x ] , ∆ ↑ q , ǫ (cid:17) is the “stan-dard” k -bialgebra ( k [ x ] , ∆ , ǫ ) from Example 2.2. (Indeed, ∆ ↑ q = ∆ when q = q is nilpotent, the univariate q -infiltration bialgebra (cid:16) k [ x ] , ∆ ↑ q , ǫ (cid:17) is aHopf algebra (see, e.g., [GriRei20, Definition 1.4.6] or [Radfor12, Definition7.1.1] for the definition of this concept). In this case, the antipode of this Hopfalgebra sends x to − x / ( + qx ) .If k is a Q -algebra, then we can say even more: Assume that k is a Q -algebraand q is nilpotent. Let y be the element ∞ ∑ k = k ! q k − x k of k [ x ] . (This sum isactually finite, since q is nilpotent. It can be viewed as the result of evaluatingthe formal power series exp ( tx ) − t ∈ k [ x ] [[ t ]] at t = q .) Consider theunique k -algebra homomorphism Φ : k [ x ] → k [ x ] that sends x to y . Then, Φ is an isomorphism from the k -bialgebra (cid:16) k [ x ] , ∆ ↑ q , ǫ (cid:17) to the k -bialgebra ( k [ x ] , ∆ , ǫ ) (defined in Example 2.2).No such isomorphism exists in general when k is not a Q -algebra. For ex-ample, if k is an F -algebra, and if q ∈ k satisfies q = q =
0, thenthe Hopf algebra ( k [ x ] , ∆ , ǫ ) has the property that its antipode is the iden-tity, but the Hopf algebra (cid:16) k [ x ] , ∆ ↑ q , ǫ (cid:17) does not have this property. Sinceany k -bialgebra isomorphism between Hopf algebras must respect their an-tipodes (see, e.g., [GriRei20, Corollary 1.4.27]), this precludes any k -bialgebraisomorphism. See [ChFoLy58, Ducham01, Ducham15] for the multivariate case.
Let us return to the general case (with k and q arbitrary). From the above defini-tions of ∆ ↑ q and ǫ , it is easy to see that ∆ ↑ q ( + qx ) = ∆ ↑ q ( ) | {z } = ⊗ + q ∆ ↑ q ( x ) | {z } = x ⊗ + ⊗ x + qx ⊗ x = ( + qx ) ⊗ ( + qx ) and ǫ ( + qx ) = ǫ ( ) | {z} = + q ǫ ( x ) | {z } = = + qx of the univariate q -infiltration bialgebra is grouplike. Theelement 1 is grouplike as well (in fact, 1 is grouplike in any k -bialgebra). Example 2.4.
Let p be a prime. Let k be a commutative F p -algebra. Let q ∈ k . Let B be the quotient ring of the polynomial ring k [ x ] by the ideal I generated by x p .For any f ∈ k [ x ] , we let f denote the projection f + I of f onto this quotient B .(Thus, x p = x p = k -algebra homomorphisms ∆ ↑ q : B → B ⊗ B and ǫ : B → k by setting ∆ ↑ q ( x ) = x ⊗ + ⊗ x + qx ⊗ x and ǫ ( x ) = F p -algebra homomorphism) showsthat ( x ⊗ + ⊗ x + qx ⊗ x ) p = ( x ⊗ ) p + ( ⊗ x ) p + q p ( x ⊗ x ) p = x p |{z} = ⊗ + ⊗ x p |{z} = + q p x p |{z} = ⊗ x p |{z} = = F p -algebra B ⊗ B . The k -algebra B , equipped with the comulti-plication ∆ ↑ q and the counit ǫ , is a k -bialgebra – actually a quotient of the univariate q -infiltration bialgebra defined in Example 2.3. Unlike the latter, it is always a Hopfalgebra (whether or not q is nilpotent). Examples of grouplike elements in B are 1,1 + qx and ( + qx ) − = − qx + q x − q x ± · · · + ( − q ) p − x p − . Example 2.5.
Consider the polynomial ring k [ g , x ] in two (commuting) variables g and x over k . Let B be the quotient ring of this ring by the ideal J generated by gx .For any f ∈ k [ g , x ] , we let f denote the projection f + J of f onto this quotient B .Define a k -algebra homomorphism ∆ : B → B ⊗ B by ∆ ( g ) = g ⊗ g and ∆ ( x ) = x ⊗ + ⊗ x ,and define a k -algebra homomorphism ǫ : B → k by ǫ ( g ) = ǫ ( x ) = ( g ⊗ g ) ( x ⊗ + ⊗ x ) = · = k -algebra B into a k -bialgebra. Its element g is grouplike, and so are all its powers g i . Example 2.6.
Let M be a monoid, and let ( k [ M ] , µ M , 1 M ) be the monoid algebra of M . As a k -module, k [ M ] is k ( M ) , the set of finitely supported functions M → k .As usual, we identify each w ∈ M with the indicator function δ w ∈ k ( M ) = k [ M ] that sends w to 1 k and all other elements of M to 0. Thus, M becomes a basisof k [ M ] . The multiplication on k [ M ] is given by k -bilinearly extending the multi-plication of M from this basis to the entire k -module k [ M ] . Thus, k [ M ] becomesa k -algebra, with unity equal to the neutral element of M (or, more precisely, itsindicator function).The k -module dual ( k [ M ]) ∨ of k [ M ] can be identified with k M , the set of allfunctions M → k , via the natural k -bilinear pairing between k M and k ( M ) given by h S | P i = ∑ w ∈ M S ( w ) P ( w ) for all S ∈ k M and P ∈ k [ M ] = k ( M ) . (1)The k -linear map ∆ ⊙ : k [ M ] → k [ M ] ⊗ k [ M ] that sends each w ∈ M to w ⊗ w isa k -algebra homomorphism; it is called the diagonal comultiplication. The k -linearmap ǫ : k [ M ] → k sending each w ∈ M to 1 k is a k -algebra homomorphism aswell. Equipping k [ M ] with these two maps ∆ ⊙ and ǫ , we obtain a k -bialgebra B = ( k [ M ] , µ M , 1 M , ∆ ⊙ , ǫ ) , in which every w ∈ M is grouplike. More generally, a k -linear combination ∑ w ∈ M a w w (with a w ∈ k ) is grouplike in B if and only if the a w are orthogonal idempotents (i.e., satisfy a w = a w for all w ∈ M , and a v a w = v , w ∈ M ). We call B the monoid bialgebra of M .If M is a group, then B is a Hopf algebra. The converse is also true when k isnontrivial.We remark that there is some overlap between monoid bialgebras and q -infiltrationbialgebras. Indeed, let M be the monoid { x n | n ∈ N } of all monomials in one vari-able x . Let q ∈ k , and let (cid:16) k [ x ] , ∆ ↑ q , ǫ (cid:17) be the univariate q -infiltration bialgebraas in Example 2.3. Then, the k -algebra homomorphism k [ x ] → k [ M ] that sends x to 1 + qx is a homomorphism of k -bialgebras from the univariate q -infiltrationbialgebra (cid:16) k [ x ] , ∆ ↑ q , ǫ (cid:17) to the monoid bialgebra B = ( k [ M ] , µ M , 1 M , ∆ ⊙ , ǫ ) . When q is invertible, this homomorphism is an isomorphism.
3. Grouplikes in a coalgebra
Let us first state a simple fact in commutative algebra:
Theorem 3.1.
Let g , g , . . . , g n be n elements of a commutative ring A. Let c , c , . . . , c n be n further elements of A. Assume that n ∑ i = c i g ki = for all k ∈ N . (2) Then, c i ∏ j = i (cid:0) g i − g j (cid:1) = for all i ∈ {
1, 2, . . . , n } . (3) Consider the ring A [[ t ]] of formal power series in one variable t over A . Inthis ring, we have n ∑ i = c i · − tg i | {z } = ∑ k ∈ N ( tg i ) k = n ∑ i = c i · ∑ k ∈ N ( tg i ) k | {z } = t k g ki = n ∑ i = c i · ∑ k ∈ N t k g ki = ∑ k ∈ N n ∑ i = c i g ki | {z } = t k = n ∏ j = (cid:0) − tg j (cid:1) , we obtain n ∑ i = c i ∏ j = i (cid:0) − tg j (cid:1) =
0. (4)This is an equality in the polynomial ring A [ t ] (which is a subring of A [[ t ]] ).Now consider the ring A (cid:2) t , t − (cid:3) of Laurent polynomials in t over A . There is an A -algebra homomorphism A [ t ] → A (cid:2) t , t − (cid:3) , t t − (by the universal propertyof A [ t ] ). Applying this homomorphism to both sides of the equality (4), we obtain n ∑ i = c i ∏ j = i (cid:16) − t − g j (cid:17) = n ∑ i = c i ∏ j = i (cid:0) t − g j (cid:1)| {z } = t ( − t − g j ) = n ∑ i = c i ∏ j = i (cid:16) t (cid:16) − t − g j (cid:17)(cid:17) = n ∑ i = c i t n − ∏ j = i (cid:16) − t − g j (cid:17) = t n − n ∑ i = c i ∏ j = i (cid:16) − t − g j (cid:17)| {z } = =
0. (5)This is an equality in the polynomial ring A [ t ] (which is a subring of A (cid:2) t , t − (cid:3) );hence, we can substitute arbitrary values for t in it.Now, fix h ∈ {
1, 2, . . . , n } . Substitute g h for t in the equality (5). The result is n ∑ i = c i ∏ j = i (cid:0) g h − g j (cid:1) = Hence,0 = n ∑ i = c i ∏ j = i (cid:0) g h − g j (cid:1) = c h ∏ j = h (cid:0) g h − g j (cid:1) + ∑ i ∈{ n } ; i = h c i ∏ j = i (cid:0) g h − g j (cid:1)| {z } = g h − g h = ( here, we have split off the addend for i = h from the sum )= c h ∏ j = h (cid:0) g h − g j (cid:1) + ∑ i ∈{ n } ; i = h c i | {z } = = c h ∏ j = h (cid:0) g h − g j (cid:1) .Now forget that we fixed h . Thus, we have shown that c h ∏ j = h (cid:0) g h − g j (cid:1) = h ∈ {
1, 2, . . . , n } . Renaming h as i , we obtain (3). Theorem 3.1 is proven.Our first main result is now the following: Theorem 3.2.
Let g , g , . . . , g n be n grouplike elements of a k -coalgebra C. Let c , c , . . . , c n ∈ k . Assume that n ∑ i = c i g i = . Then,c i ∏ j = i (cid:0) g i − g j (cid:1) = in Sym
C for all i ∈ {
1, 2, . . . , n } . Here,
Sym
C denotes the symmetric algebra of the k -module C. Our proof of this theorem will rely on a certain family of maps defined for any k -coalgebra: Definition 3.3.
Let C be a k -coalgebra. We then define a sequence of k -linear maps ∆ ( − ) , ∆ ( ) , ∆ ( ) , . . ., where ∆ ( k ) is a k -linear map from C to C ⊗ ( k + ) for any integer k ≥ −
1. Namely, we define them recursively by setting ∆ ( − ) = ǫ and ∆ ( ) = id C and ∆ ( k ) = (cid:16) id C ⊗ ∆ ( k − ) (cid:17) ⊗ ∆ for all k ≥
1. These maps ∆ ( k ) are known as theiterated comultiplications of C , and studied further, e.g., in [GriRei20, Exercise1.4.20]. Proof of Theorem 3.2.
Every grouplike element g ∈ C and every k ∈ N satisfy ∆ ( k − ) g = g ⊗ k (6)(where g ⊗ k means g ⊗ g ⊗ · · · ⊗ g | {z } k times ∈ C ⊗ k ). Indeed, this can be easily proven byinduction on k . The g i (for all i ∈ {
1, 2, . . . , n } ) are grouplike. Thus, (6) (applied to g = g i ) showsthat ∆ ( k − ) g i = g ⊗ ki for each k ∈ N and each i ∈ {
1, 2, . . . , n } . Hence, applying the k -linear map ∆ ( k − ) to both sides of the equality n ∑ i = c i g i =
0, we obtain n ∑ i = c i g ⊗ ki = k ∈ N .But recall that the symmetric algebra Sym C is defined as a quotient of the tensoralgebra T ( C ) . Hence, there is a canonical projection from T ( C ) to Sym C that sendseach tensor x ⊗ x ⊗ · · · ⊗ x m ∈ T ( C ) to x x · · · x m ∈ Sym C . In particular, thisprojection sends g ⊗ ki to g ki for each i ∈ {
1, 2, . . . , n } and each k ∈ N . Thus, applyingthis projection to both sides of (7), we obtain n ∑ i = c i g ki = C for all k ∈ N .Thus, Theorem 3.1 can be applied to A = Sym C . We conclude that c i ∏ j = i (cid:0) g i − g j (cid:1) = C for all i ∈ {
1, 2, . . . , n } .This proves Theorem 3.2.From Theorem 3.2, we obtain the following classical fact [Radfor12, Lemma2.1.12]: Corollary 3.4.
Assume that k is a field. Let g , g , . . . , g n be n distinct grouplike elementsof a k -coalgebra C. Then, g , g , . . . , g n are k -linearly independent.Proof of Corollary 3.4. The k -module C is free (since k is a field). Thus, the k -algebraSym C is isomorphic to a polynomial algebra over k , and thus is an integral do-main (again since k is a field). Its elements g i − g j for j = i are nonzero (since g , g , . . . , g n are distinct), and thus their products ∏ j = i (cid:0) g i − g j (cid:1) (for i ∈ {
1, 2, . . . , n } )are nonzero as well (since Sym C is an integral domain).Let c , c , . . . , c n ∈ k be such that n ∑ i = c i g i =
0. Then, Theorem 3.2 yields c i ∏ j = i (cid:0) g i − g j (cid:1) = C for all i ∈ {
1, 2, . . . , n } .We can cancel the ∏ j = i (cid:0) g i − g j (cid:1) factors from this equality (since these factors ∏ j = i (cid:0) g i − g j (cid:1) are nonzero, and since Sym C is an integral domain). Thus, we obtain the equal-ities c i = C for all i ∈ {
1, 2, . . . , n } . In other words, c i = k for all i ∈ {
1, 2, . . . , n } (since k embeds into Sym C ).Now, forget that we fixed c , c , . . . , c n . We thus have proven that if c , c , . . . , c n ∈ k are such that n ∑ i = c i g i =
0, then c i = i ∈ {
1, 2, . . . , n } . In other words, g , g , . . . , g n are k -linearly independent. This proves Corollary 3.4. . Example 2.3 illustrates why we required k to be a field in Corollary3.4. Indeed, if q ∈ k is a zero-divisor but nonzero, then the two grouplike elements1 and 1 + qx of the k -bialgebra in Example 2.3 fail to be k -linearly independent.Theorem 3.2, however, provides a weaker version of linear independence that isstill satisfied.
4. Grouplikes over id-unipotents in a bialgebra
In this section, we shall use the following concept:
Definition 4.1.
Let B be a k -bialgebra. An element b ∈ B is said to be id-unipotentif there exists some m ∈ Z such that every nonnegative integer n > m satisfies ( ηǫ − id ) ⊛ n ( b ) = B ). We shall refer to such an m as a degree-upper bound of b .Before we move on to studying id-unipotent elements, let us show several exam-ples: Example 4.2.
Let B be a connected graded k -bialgebra. (The word “graded” heremeans “ N -graded”, and it is assumed that all structure maps µ , η , ∆ , ǫ preserve thegrading. The word “connected” means that the 0-th homogeneous component B is isomorphic to k .) Then, each b ∈ B is id-unipotent, and if b ∈ B is homogeneousof degree k , then k is a degree-upper bound of b . Example 4.3.
Let q , k [ x ] , ∆ ↑ q and ǫ be as in Example 2.3. Assume that q is nilpo-tent, with q m = m ∈ N . It is easy to see (by induction on n ) that ( ηǫ − id ) ⊛ n ( x ) = − ( − q ) n − x n in k [ x ] for each n ≥
1. Thus, for each n > m , wehave ( ηǫ − id ) ⊛ n ( x ) = − ( − q ) n − x n = q m = ( − q ) n − = x is id-unipotent in k [ x ] , with m being a degree-upper boundof x . Combining this observation with Corollary 4.24 further below, it followseasily that every element of k [ x ] is id-unipotent (albeit m will not always be adegree-upper bound). Example 4.4.
Let p , k , q and B be as in Example 2.4. It is easy to see (by inductionon n ) that ( ηǫ − id ) ⊛ n ( x ) = − ( − q ) n − x n for each n ≥
1. Thus, for each n ≥ p , wehave ( ηǫ − id ) ⊛ n ( x ) = − ( − q ) n − x n = x p = x n = x is id-unipotent in B , with p − x . Remark . The notion of id-unipotence is connected with the classical notion of ∆ + -nilpotence, but it is a weaker notion. Let us briefly describe the latter notionand the connection. Consider a k -bialgebra B with comultiplication ∆ , counit ǫ and unit map η . Then,we have a canonical direct sum decomposition B = k B ⊕ B + of the k -module B ,where B + = ker ( ǫ ) (and where 1 B = ǫ ( k ) is the unity of B ). The correspondingprojections are ηǫ (projecting B onto k B ) and id − ηǫ (projecting B onto B + ). Thus,we set id = id − ηǫ : B → B (this projection, in a way, “eliminates the constantterm”). For each k ∈ N , we define a map ∆ ( k )+ = id ⊗ ( k + ) ◦ ∆ ( k ) : B → B ⊗ ( k + ) ,where id ⊗ ( k + ) = id ⊗ id ⊗ · · · ⊗ id (with k + ∆ ( k ) : B → B ⊗ ( k + ) is the iterated comultiplication from Definition 3.3.Now, an element x of B is said to be ∆ + -nilpotent if there exists some m ∈ Z such that every nonnegative integer k > m satisfies ∆ ( k )+ ( b ) =
0. (9)It is easy to see that every n ≥ ( ηǫ − id ) ⊛ n = ( − id ) ⊛ n = ( − ) n µ ( n − ) ◦ ∆ ( n − )+ ,where µ ( n − ) : B ⊗ n → B is the “iterated multiplication” map that sends each puretensor b ⊗ b ⊗ · · · ⊗ b n to b b · · · b n ∈ B . Thus, every ∆ + -nilpotent element of B is id-unipotent. The converse, however, is not true. For instance, the element x inExample 4.4 is id-unipotent but (in general) not ∆ + -nilpotent. (But the elements x in Example 4.3 and b in Example 4.2 are ∆ + -nilpotent.) We shall also use a slightly generalized notion of linear independence:
Definition 4.6.
Let g , g , . . . , g n be some elements of a k -algebra A . Let S be asubset of A . (a) We say that the elements g , g , . . . , g n are left S -linearly independent if every n -tuple ( s , s , . . . , s n ) ∈ S n of elements of S satisfying n ∑ i = s i g i = ( s i = i ) . Nevertheless, at least in one important case, the two concepts are closely intertwined. Namely, if k is a Q -algebra and B is cocommutative, then we have the following chain of equivalences: ( every b ∈ B is id-unipotent ) ⇐⇒ ( B = U ( Prim ( B ))) ⇐⇒ ( every b ∈ B is ∆ + -nilpotent ) ,where “ B = U ( Prim ( B )) ” means that B is the universal enveloping bialgebra (see [Bourba98,ch II §1] for the definition) of its primitive elements. The two equivalence signs follow easilyfrom [DMTCN14, Theorem 1] (noticing that a filtration as in [DMTCN14, Theorem 1 condition1] forces every b ∈ B to be ∆ + -nilpotent). This is a variant of the well-known Cartier–Quillen–Milnor–Moore theorem [Car07, MM65] that relies on (formal) convergence of infinite sums in-stead of primitive generation of the Hopf algebra. (b) The notion of “right S -linearly independent” is defined in the same wayas “left S -linearly independent”, except that the sum n ∑ i = s i g i is replaced by n ∑ i = g i s i . (c) If the k -algebra A is commutative, then these two notions are identical, andwe just call them “ S -linearly independent”.We also recall that an element a of a commutative ring A is said to be regular ifit is a non-zero-divisor – i.e., if it has the property that whenever b is an element of A satisfying ab =
0, then b = Theorem 4.7.
Let B be a commutative k -bialgebra. Let L denote the set of all id-unipotentelements of B.Let g , g , . . . , g n be n grouplike elements of B. Let us make two assumptions: • Assumption 1: For any ≤ i < j ≤ n, the element g i − g j of B is regular. • Assumption 2: For any ≤ i ≤ n, the element g i of B is regular.Then, g , g , . . . , g n are L-linearly independent. Before we prove this theorem, some remarks are in order.
Remark . The L in Theorem 4.7 is a k -submodule of B . (This is easily seendirectly: If u ∈ B and v ∈ B are id-unipotent with degree-upper bounds p and q ,respectively, then λ u + µ v is id-unipotent with degree-upper bound max { p , q } forany λ , µ ∈ k .)It can be shown that L is a k -subalgebra of B . (See Corollary 4.24 further be-low; we will not need this to prove Theorem 4.7.) We do not know if L is a k -subcoalgebra of B . Remark . Example 2.5 illustrates why we required Assumption 2 to hold in The-orem 4.7. Indeed, in this example, g is grouplike, and x is id-unipotent (withdegree-upper bound 1), so the equality xg = gx = g is not L -linearly independent.The necessity of Assumption 1 can be illustrated by Remark 3.5 again. Question . Is the requirement for B to be commutative necessary? (It is certainlyneeded for our proof.)To prove Theorem 4.7, we shall use a few lemmas. The first one is a classicalresult [GriRei20, Exercise 1.5.11(a)]: Lemma 4.10.
Let C be a k -bialgebra. Let A be a commutative k -algebra. Let p : C → Aand q : C → A be two k -algebra homomorphisms. Then, the map p ⊛ q : C → A is also a k -algebra homomorphism. 24-09-2020 00:47ariations on linear independence of grouplikes page 14Proof of Lemma 4.10. Let a ∈ C and b ∈ C . We shall prove that ( p ⊛ q ) ( ab ) =( p ⊛ q ) ( a ) · ( p ⊛ q ) ( b ) .Using Sweedler notation, write ∆ ( a ) = ∑ ( a ) a ( ) ⊗ a ( ) and ∆ ( b ) = ∑ ( b ) b ( ) ⊗ b ( ) .Then, the multiplicativity of ∆ yields ∆ ( ab ) = ∑ ( a ) ∑ ( b ) a ( ) b ( ) ⊗ a ( ) b ( ) . Thus, ( p ⊛ q ) ( ab ) = ∑ ( a ) ∑ ( b ) p (cid:16) a ( ) b ( ) (cid:17)| {z } = p ( a ( ) ) p ( b ( ) ) (since p is a k -algebrahomomorphism) q (cid:16) a ( ) b ( ) (cid:17)| {z } = q ( a ( ) ) q ( b ( ) ) (since q is a k -algebrahomomorphism) = ∑ ( a ) ∑ ( b ) p (cid:16) a ( ) (cid:17) p (cid:16) b ( ) (cid:17) q (cid:16) a ( ) (cid:17) q (cid:16) b ( ) (cid:17) = ∑ ( a ) p (cid:16) a ( ) (cid:17) q (cid:16) a ( ) (cid:17)| {z } =( p ⊛ q )( a ) · ∑ ( b ) p (cid:16) b ( ) (cid:17) q (cid:16) b ( ) (cid:17)| {z } =( p ⊛ q )( b ) ( since A is commutative )= ( p ⊛ q ) ( a ) · ( p ⊛ q ) ( b ) .Now, forget that we fixed a and b . We thus have proven that ( p ⊛ q ) ( ab ) =( p ⊛ q ) ( a ) · ( p ⊛ q ) ( b ) for each a , b ∈ C . An even simpler argument shows that ( p ⊛ q ) ( ) =
1. Combining these results, we conclude that p ⊛ q : C → A is a k -algebra homomorphism. This completes the proof of Lemma 4.10. Lemma 4.11.
Let B be a commutative k -bialgebra. Let k ∈ N . Then, id ⊛ k : B → B is a k -algebra homomorphism.Proof of Lemma 4.11. This follows by straightforward induction on k . Indeed, thebase case k = η B ǫ B is a k -algebra homomorphism,while the induction step uses Lemma 4.10 (applied to C = B and A = B ) and thefact that id B is a k -algebra homomorphism. Lemma 4.12.
Let B be a k -bialgebra. Let g ∈ B be any grouplike element. Let k ∈ N .Then, id ⊛ k ( g ) = g k . (Here, id means id B .)Proof of Lemma 4.12. This follows easily by induction on k . Lemma 4.13.
Let B be a k -bialgebra. Let b ∈ B be id-unipotent and nonzero. Let m be adegree-upper bound of b. Then, m ∈ N .Proof of Lemma 4.13. Assume the contrary. Thus, m is negative (since m is an in-teger). Therefore, every n ∈ N satisfies n > m . Hence, by the definition of a degree-upper bound, we must have ( ηǫ − id ) ⊛ n ( b ) = n ∈ N . Thus, inparticular, we have ( ηǫ − id ) ⊛ ( b ) = ( ηǫ − id ) ⊛ ( b ) =
0. Hence, ( ηǫ − id ) ⊛ ( b ) | {z } = − ( ηǫ − id ) ⊛ ( b ) | {z } = = ( ηǫ − id ) ⊛ | {z } = ηǫ ( b ) − ( ηǫ − id ) ⊛ | {z } = ηǫ − id ( b ) = ( ηǫ ) ( b ) − ( ηǫ − id ) ( b ) = id ( b ) = b = b is nonzero). This contradiction completes the proof of Lemma 4.13. Lemma 4.14.
Let B be a k -bialgebra. Let b ∈ B be id-unipotent. Let m be a degree-upperbound of b. Then, in the ring B [[ t ]] of power series , we have ( − t ) m + ∑ k ∈ N id ⊛ k ( b ) t k = m ∑ k = ( − ) k ( ηǫ − id ) ⊛ k ( b ) t k ( − t ) m − k . (Here, id means id B .)Proof of Lemma 4.14. We know that m is a degree-upper bound of b . In other words,we have ( ηǫ − id ) ⊛ n ( b ) = n > m (by (8)). In otherwords, we have ( ηǫ − id ) ⊛ k ( b ) = k > m . (10)Let ι denote the canonical inclusion B ֒ → B [[ t ]] . This is an injective k -algebrahomomorphism, and will be regarded as an inclusion.The convolution algebra ( Hom ( B , B ) , ⊛ ) embeds in the convolution algebra ( Hom ( B , B [[ t ]]) , ⊛ ) (indeed, there is an injective k -algebra homomorphism fromthe former to the latter, which sends each f ∈ Hom ( B , B ) to ι ◦ f ∈ Hom ( B , B [[ t ]]) ).Again, we shall regard this embedding as an inclusion (so we will identify each f ∈ Hom ( B , B ) with ι ◦ f ).The convolution algebra ( Hom ( B , B [[ t ]]) , ⊛ ) has unity ηǫ = η B [[ t ]] ǫ B . Thus, fromthe classical power-series identity ( − u ) − = ∑ k ∈ N u k , we obtain the equality ( ηǫ − t f ) ⊛ ( − ) = ∑ k ∈ N ( t f ) ⊛ k = ∑ k ∈ N t k f ⊛ k (11)for every f ∈ Hom ( B , B [[ t ]]) . This ring is defined in the usual way even if B is not commutative. Thus, t will commute withevery power series in B [[ t ]] . Now, ∑ k ∈ N id ⊛ k ( b ) t k = ∑ k ∈ N t k id ⊛ k !| {z } =( ηǫ − t id ) ⊛ ( − ) (by (11), applied to f = id) ( b )= ( ηǫ − t id ) ⊛ ( − ) ( b ) . (12)But ηǫ − t id = ηǫ ( − t ) − t ( id − ηǫ ) = (cid:18) ηǫ − t ( id − ηǫ ) − t (cid:19) ( − t ) and therefore ( ηǫ − t id ) ⊛ ( − ) = (cid:18)(cid:18) ηǫ − t ( id − ηǫ ) − t (cid:19) ( − t ) (cid:19) ⊛ ( − ) = ( − t ) − · (cid:18) ηǫ − t ( id − ηǫ ) − t (cid:19) ⊛ ( − ) = ( − t ) − · ∑ k ∈ N t k (cid:18) ( id − ηǫ ) − t (cid:19) ⊛ k (by (11), applied to f = ( id − ηǫ ) − t ). Hence, (12) becomes ∑ k ∈ N id ⊛ k ( b ) t k = ( ηǫ − t id ) ⊛ ( − ) | {z } =( − t ) − · ∑ k ∈ N t k ( id − ηǫ ) − t ! ⊛ k ( b )= ( − t ) − · ∑ k ∈ N t k (cid:18) ( id − ηǫ ) − t (cid:19) ⊛ k ( b ) | {z } =( id − ηǫ ) ⊛ k ( b ) − t ! k = ( − t ) − · ∑ k ∈ N id − ηǫ | {z } = − ( ηǫ − id ) ⊛ k ( b ) (cid:18) t − t (cid:19) k = ( − t ) − · ∑ k ∈ N ( − ( ηǫ − id )) ⊛ k | {z } =( − ) k ( ηǫ − id ) ⊛ k ( b ) (cid:18) t − t (cid:19) k = ( − t ) − · ∑ k ∈ N ( − ) k ( ηǫ − id ) ⊛ k ( b ) (cid:18) t − t (cid:19) k | {z } All addends of this sum for k > m arezero, due to (10). = ( − t ) − · m ∑ k = ( − ) k ( ηǫ − id ) ⊛ k ( b ) (cid:18) t − t (cid:19) k .Multiplying both sides of this equality by ( − t ) m + , we get ( − t ) m + ∑ k ∈ N id ⊛ k ( b ) t k = ( − t ) m · m ∑ k = ( − ) k ( ηǫ − id ) ⊛ k ( b ) (cid:18) t − t (cid:19) k = m ∑ k = ( − ) k ( ηǫ − id ) ⊛ k ( b ) t k ( − t ) m − k .This proves Lemma 4.14. Lemma 4.15.
Let A be a commutative ring. Let m ∈ N and u ∈ A. Let F ( t ) ∈ A [ t ] bea polynomial of degree ≤ m. SetG ( t ) = t m F (cid:16) ut (cid:17) ∈ A h t , t − i . Then, G ( t ) is actually a polynomial in A [ t ] and satisfiesG ( u ) = u m F ( ) . (13) We have G ( t ) = t m F (cid:16) ut (cid:17) ∈ A [ t ] because F ( t ) has degree ≤ m . It remains to show that G ( u ) = u m F ( ) . It is tempting to argue this bysubstituting u for t in the equality G ( t ) = t m F (cid:16) ut (cid:17) , but this is not completely justi-fied (we cannot arbitrarily substitute u for t in an equality of Laurent polynomialsunless we know that u is invertible ). But we can argue as follows instead:The polynomial F ( t ) has degree ≤ m . Hence, we can write it in the form F ( t ) = a + a t + a t + · · · + a m t m for some a , a , . . . , a m ∈ A . Consider these a , a , . . . , a m .Thus, F ( ) = a + a + a + · · · + a m m = a + a + a + · · · + a m and F (cid:16) ut (cid:17) = a + a ut + a (cid:16) ut (cid:17) + · · · + a m (cid:16) ut (cid:17) m . The definition of G ( t ) yields G ( t ) = t m F (cid:16) ut (cid:17)| {z } = a + a ut + a (cid:18) ut (cid:19) + ··· + a m (cid:18) ut (cid:19) m = t m (cid:18) a + a ut + a (cid:16) ut (cid:17) + · · · + a m (cid:16) ut (cid:17) m (cid:19) = a t m + a ut m − + a u t m − + · · · + a m u m t .Substituting u for t in this equality of polynomials, we obtain G ( u ) = a u m + a uu m − + a u u m − + · · · + a m u m u = a u m + a u m + a u m + · · · + a m u m = u m ( a + a + a + · · · + a m ) | {z } = F ( ) = u m F ( ) ,and thus Lemma 4.15 is proven. Proof of Theorem 4.7.
Let b , b , . . . , b n ∈ L be such that n ∑ i = b i g i =
0. We must showthat b i = i .Assume the contrary. Thus, there exists some i such that b i =
0. We WLOGassume that each i satisfies b i =
0, since otherwise we can just drop the violating b i and the corresponding g i and reduce the problem to a smaller value of n .For each i ∈ {
1, 2, . . . , n } , the element b i ∈ B is id-unipotent (since it belongs to L ) and thus has a degree-upper bound. We let m i be the smallest degree-upperbound of b i . This is well-defined, because Lemma 4.13 (applied to b = b i ) showsthat every degree-upper bound of b i must be ∈ N . Thus, each m i belongs to N .For each i ∈ {
1, 2, . . . , n } , Lemma 4.14 (applied to b = b i and m = m i ) shows thatin the ring B [[ t ]] , we have ( − t ) m i + ∑ k ∈ N id ⊛ k ( b i ) t k = m i ∑ k = ( − ) k ( ηǫ − id ) ⊛ k ( b i ) t k ( − t ) m i − k (14) since the universal property of the Laurent polynomial ring A (cid:2) t , t − (cid:3) is only stated for invertibleelements (since m i is a degree-upper bound of b i ). The right-hand side of this equality isa polynomial in t of degree ≤ m i . Denote this polynomial by Q i ( t ) . Hence, (14)becomes ( − t ) m i + ∑ k ∈ N id ⊛ k ( b i ) t k = Q i ( t ) ,so that ∑ k ∈ N id ⊛ k ( b i ) t k = Q i ( t )( − t ) m i + . (15)Now, let k ∈ N . Lemma 4.11 yields that the map id ⊛ k is a k -algebra homomor-phism. Thus,id ⊛ k n ∑ i = b i g i ! = n ∑ i = id ⊛ k ( b i ) id ⊛ k ( g i ) | {z } = g ki (by Lemma 4.12,since g i is grouplike) = n ∑ i = id ⊛ k ( b i ) g ki .Hence, n ∑ i = id ⊛ k ( b i ) g ki = id ⊛ k n ∑ i = b i g i | {z } = =
0. (16)Forget that we fixed k . We thus have proved (16) for each k ∈ N . Now, in thecommutative ring B [[ t ]] , we have n ∑ i = Q i ( tg i )( − tg i ) m i + | {z } = ∑ k ∈ N id ⊛ k ( b i )( tg i ) k (by substituting tg i for t in (15)) = n ∑ i = ∑ k ∈ N id ⊛ k ( b i ) ( tg i ) k | {z } = t k g ki = n ∑ i = ∑ k ∈ N id ⊛ k ( b i ) t k g ki = ∑ k ∈ N n ∑ i = id ⊛ k ( b i ) g ki | {z } = t k = n ∏ j = (cid:0) − tg j (cid:1) m j + , we obtain n ∑ i = Q i ( tg i ) ∏ j = i (cid:0) − tg j (cid:1) m j + = B [ t ] . Hence, we can apply the B -algebrahomomorphism B [ t ] → B h t , t − i , t t − (into the Laurent polynomial ring B (cid:2) t , t − (cid:3) ) to this equality. We obtain n ∑ i = Q i (cid:16) g i t (cid:17) ∏ j = i (cid:18) − g j t (cid:19) m j + = t m + m + ··· + m n + n − , we transform this equality into n ∑ i = t m i Q i (cid:16) g i t (cid:17) ∏ j = i (cid:0) t − g j (cid:1) m j + = B (cid:2) t , t − (cid:3) ).For each i ∈ {
1, 2, . . . , n } , we set R i ( t ) = t m i Q i (cid:16) g i t (cid:17) ∈ B h t , t − i .This R i ( t ) is actually a polynomial in B [ t ] (since Q i is a polynomial of degree ≤ m i ). Using the definition of R i ( t ) , the equality (17) rewrites as n ∑ i = R i ( t ) ∏ j = i (cid:0) t − g j (cid:1) m j + =
0. (18)Now fix h ∈ {
1, 2, . . . , n } . We have m h ∈ N (since each m i belongs to N ), so that m h + m h + =
0. In other words, ( g h − g h ) m h + =
0. The equality (18) is an equality in the polynomial ring B [ t ] (since each m i belongsto N , and since each R i ( t ) is a polynomial in B [ t ] ), so we can substitute g h for t init. We thus obtain n ∑ i = R i ( g h ) ∏ j = i (cid:0) g h − g j (cid:1) m j + =
0. (19)But all addends on the left hand side of this equality are 0 except for the addendwith i = h (since the product ∏ j = i (cid:0) g h − g j (cid:1) m j + contains the factor ( g h − g h ) m h + = i = h ). Thus, the equality (19) simplifies to R h ( g h ) ∏ j = h (cid:0) g h − g j (cid:1) m j + = g h − g j (with j = h ) are regular elements of B (by Assumption1), we can cancel ∏ j = h (cid:0) g h − g j (cid:1) m j + from this equality, and obtain R h ( g h ) = Q h ( t ) ∈ B [ t ] is a polynomial of degree ≤ m h , and we have R h ( t ) = t m h Q h (cid:16) g h t (cid:17) (by the definition of R h ). Hence, (13) (applied to A = B , m = m h , u = g h , F ( t ) = Q h ( t ) and G ( t ) = R h ( t ) ) yields R h ( g h ) = g m h h Q h ( ) . Hence, m h h Q h ( ) = R h ( g h ) =
0, so that Q h ( ) = g h ∈ B is regular).But the definition of Q h yields Q h ( t ) = m h ∑ k = ( − ) k ( ηǫ − id ) ⊛ k ( b h ) t k ( − t ) m h − k .Substituting 1 for t in this equality, we find Q h ( ) = m h ∑ k = ( − ) k ( ηǫ − id ) ⊛ k ( b h ) k |{z} = ( − ) m h − k | {z } = mh − k =
1, if k = m h ;0, if k = m h = m h ∑ k = ( − ) k ( ηǫ − id ) ⊛ k ( b h ) (
1, if k = m h ;0, if k = m h = ( − ) m h ( ηǫ − id ) ⊛ m h ( b h ) .Hence, ( − ) m h ( ηǫ − id ) ⊛ m h ( b h ) = Q h ( ) =
0. Therefore, ( ηǫ − id ) ⊛ m h ( b h ) = m h is the smallest degree upper-bound of b h (since this is how m h wasdefined). Thus, every nonnegative integer n > m h satisfies the equality ( ηǫ − id ) ⊛ n ( b h ) = n = m h also satisfies thisequality (because we just showed that ( ηǫ − id ) ⊛ m h ( b h ) = n > m h − ( ηǫ − id ) ⊛ n ( b h ) =
0. In otherwords, m h − b h (by the definition of a degree upper-bound). Thus, m h is not the smallest degree-upper bound of b h (since m h − m h . This contradiction completes ourproof of Theorem 4.7. In this subsection, we shall show that the id-unipotent elements in a commutative k -bialgebra form a k -subalgebra. The proof will rely on a sequence of lemmas. Webegin with two identities for binomial coefficients: Lemma 4.16.
Let N ∈ N . Let ( a , a , . . . , a N ) and ( b , b , . . . , b N ) be two ( N + ) -tuplesof rational numbers. Assume thatb n = n ∑ i = ( − ) i (cid:18) ni (cid:19) a i for each n ∈ {
0, 1, . . . , N } . Then, a n = n ∑ i = ( − ) i (cid:18) ni (cid:19) b i for each n ∈ {
0, 1, . . . , N } . Lemma 4.17.
Let a , b , m ∈ N . Then, (cid:18) ma (cid:19)(cid:18) mb (cid:19) = a + b ∑ i = a (cid:18) ia (cid:19)(cid:18) aa + b − i (cid:19)(cid:18) mi (cid:19) .Both of these lemmas are not hard to prove; they are also easily found in theliterature (e.g., Lemma 4.16 is [Grinbe17, Proposition 7.26], while Lemma 4.17 is[Grinbe17, Proposition 3.37]). Note that we have (cid:18) ni (cid:19) = n , i ∈ N satisfying i > n . Thus, the nonzero addends on the right handside in Lemma 4.17 don’t start until i = max { a , b } .The following modified version of Lemma 4.16 will be useful to us: Lemma 4.18.
Let A be an abelian group, written additively. Let ( a , a , a , . . . ) and ( b , b , b , . . . ) be two sequences of elements of A. Assume thatb n = n ∑ i = ( − ) i (cid:18) ni (cid:19) a i for each n ∈ N . Then, a n = n ∑ i = ( − ) i (cid:18) ni (cid:19) b i for each n ∈ N . First proof of Lemma 4.18.
The proof of Lemma 4.16 that was given in [Grinbe17,proof of Proposition 7.26] can be immediately reused as a proof of Lemma 4.18(after removing all inequalities of the form “ n ≤ N ”, and after replacing everyappearance of “ {
0, 1, . . . , N } ” by “ N ”). Second proof of Lemma 4.18 (sketched).
We will interpret binomial transforms in termsof sequence transformations (see [GoF04] for motivations and details). A row-finitematrix will mean a family ( M [ n , k ]) n , k ≥ with M [ n , k ] ∈ k (i.e., a matrix of type ( N , N ) with elements in k , in the terminology of [Bourba89,Chapter II, §10]) with the property that for every fixed row index n , the sequence ( M [ n , k ]) k ≥ has finite support. We let L ( k N ) be the algebra of row-finite matrices;its product is given by the usual formula ( M M ) [ i , j ] = ∑ k ∈ N M [ i , k ] M [ k , j ] . (21) An “abelian group, written additively” means an abelian group whose binary operation is de-noted by + and whose neutral element is denoted by 0. This is just the definition in [Bourba89, Chapter II, §10, (3)], extended a bit.
Any row-finite matrix M ∈ L ( k N ) canonically acts on A N for any k -module A :Namely, if a = ( a n ) n ≥ ∈ A N is a sequence of elements of A , then Ma is defined tobe the sequence ( b n ) n ≥ ∈ A N with b n = ∑ k ≥ M [ n , k ] a k for all n ∈ N . (22)This action gives rise to a k -algebra homomorphism Φ : L ( k N ) → End ( A N ) , (23)which is injective in the case when A = k (but neither injective nor surjective inthe general case). Applying this to k = Q and A = Q , we thus have an injective Q -algebra homomorphism Φ : L ( Q N ) → End ( Q N ) . We also have a Q -vectorspace isomorphism Q N ∼ = → Q [[ z ]] , ( a n ) n ∈ N ∑ n ≥ a n z n n !(where z is a formal variable), thus a Q -algebra isomorphism End ( Q N ) → End ( Q [[ z ]]) .Composing the latter with Φ , we get an injective Q -algebra homomorphism Φ : L ( Q N ) → End ( Q [[ z ]]) . Explicitly, for any power series f = f ( z ) = ∑ n ≥ a n z n n ! ∈ Q [[ z ]] and any row-finite matrix M ∈ L ( Q N ) , we have Φ ( M )( f ) = ∑ n ≥ b n z n n ! = ∑ n ≥ ∑ k ≥ M [ n , k ] a k z n n ! . (24)Now, let M ∈ L ( Z N ) ⊆ L ( Q N ) be the row-finite matrix whose entries are M [ n , i ] : = ( − ) i (cid:18) ni (cid:19) . (25)Then, the assumption in Lemma 4.18 is saying that b = M a , where a , b ∈ A N are given by a = ( a n ) n ≥ and b = ( b n ) n ≥ . Likewise, the claim of Lemma 4.18 issaying that a = M b . Hence, in order to prove Lemma 4.18, it suffices to showthat M a = a . Thus, it suffices to show that M = I L ( Z N ) (the identity matrix in L ( Z N ) ).But this can be done via the injective Z -algebra homomorphism Φ : L ( Q N ) → End ( Q [[ z ]]) . Indeed, a direct calculation shows that any f ∈ Q [[ z ]] satisfies Φ ( M )( f )( z ) = f ( − z ) e z ,and hence Φ ( M )( f )( z ) = Φ ( M )( f ( − z ) e z ) = ( f ( − ( − z )) e − z ) e z = f ( z ) . This shows that Φ ( M ) = id = Φ ( I L ( Q N ) ) . Since Φ is injective, this entails M = I L ( Q N ) = I L ( Z N ) and proves Lemma 4.18 (as every abelian group is a Z -module).Our next lemma is a classical fact about finite differences: Lemma 4.19.
Let A be an abelian group, written additively. Let ( a , a , a , . . . ) ∈ A N bea sequence of elements of A. Let m ≥ − be an integer. Then, the following two statementsare equivalent:1. We have n ∑ i = ( − ) i ( ni ) a i = for every integer n > m.2. There exist m + elements c , c , . . . , c m of A such that every n ∈ N satisfies a n = m ∑ i = ( ni ) c i . The sequences ( a , a , a , . . . ) satisfying these two equivalent statements can beregarded as a generalization of polynomial sequences (i.e., sequences whose n -thentry is given by evaluating a fixed integer-valued polynomial at n ). Proof of Lemma 4.19.
We must prove the equivalence of Statement 1 and Statement2. We shall do so by proving the implications 1 = ⇒ = ⇒ Proof of the implication 1 = ⇒ Let us first show that Statement 1 implies State-ment 2.Indeed, assume that Statement 1 holds. In other words, we have n ∑ i = ( − ) i (cid:18) ni (cid:19) a i = n > m .We shall now show that Statement 2 holds.Indeed, let us set b n = n ∑ i = ( − ) i (cid:18) ni (cid:19) a i (27)for each n ∈ N . Then, for every integer n > m , we have b n = n ∑ i = ( − ) i (cid:18) ni (cid:19) a i = n as i in this statement, we obtain the following: For everyinteger i > m , we have b i =
0. (28)
Furthermore, recall that b n = n ∑ i = ( − ) i ( ni ) a i for each n ∈ N . Thus, Lemma 4.16shows that we have a n = n ∑ i = ( − ) i (cid:18) ni (cid:19) b i (29)for each n ∈ N .Now, let n ∈ N . Let N = max { n , m } ; then, N ≥ n and N ≥ m . Thus, 0 ≤ n ≤ N and 0 ≤ m ≤ N . Now, comparing N ∑ i = (cid:18) ni (cid:19) ( − ) i b i = n ∑ i = (cid:18) ni (cid:19) ( − ) i | {z } =( − ) i ( ni ) b i + N ∑ i = n + (cid:18) ni (cid:19)| {z } = i ≥ n + > n ) ( − ) i b i ( here, we have split the sum at i = n , since 0 ≤ n ≤ N )= n ∑ i = ( − ) i (cid:18) ni (cid:19) b i + N ∑ i = n + ( − ) i b i | {z } = = n ∑ i = ( − ) i (cid:18) ni (cid:19) b i = a n ( by (29) ) with N ∑ i = (cid:18) ni (cid:19) ( − ) i b i = m ∑ i = (cid:18) ni (cid:19) ( − ) i b i + N ∑ i = m + (cid:18) ni (cid:19) ( − ) i b i |{z} = i ≥ m + > m ) ( here, we have split the sum at i = m , since 0 ≤ m ≤ N )= m ∑ i = (cid:18) ni (cid:19) ( − ) i b i + N ∑ i = m + (cid:18) ni (cid:19) ( − ) i | {z } = = m ∑ i = (cid:18) ni (cid:19) ( − ) i b i ,we obtain a n = m ∑ i = (cid:18) ni (cid:19) ( − ) i b i .Forget that we fixed n . We thus have proved that every n ∈ N satisfies a n = m ∑ i = ( ni ) ( − ) i b i . Thus, there exist m + c , c , . . . , c m of A such that every n ∈ N satisfies a n = m ∑ i = ( ni ) c i (namely, these m + c i =( − ) i b i ). In other words, Statement 2 holds.We thus have proved the implication 1 = ⇒ Proof of the implication 2 = ⇒ Let us now show that Statement 2 implies State-ment 1.
Indeed, assume that Statement 2 holds. That is, there exist m + c , c , . . . , c m of A such that every n ∈ N satisfies a n = m ∑ i = (cid:18) ni (cid:19) c i . (30)Consider these c , c , . . . , c m .We must prove that Statement 1 holds. In other words, we must prove that wehave n ∑ i = ( − ) i ( ni ) a i = n > m .Extend the ( m + ) -tuple ( c , c , . . . , c m ) ∈ A m + to an infinite sequence ( c , c , c , . . . ) ∈ A N by setting ( c i = i > m ) . (31)Furthermore, define a sequence ( d , d , d , . . . ) ∈ A N by setting (cid:16) d i = ( − ) i c i for all i ∈ N (cid:17) . (32)Then, each i ∈ N satisfies ( − ) i d i |{z} =( − ) i c i (by (32)) = ( − ) i ( − ) i | {z } =( − ) i = c i = c i . (33)Let n ∈ N . Let N = max { n , m } ; then, N ≥ n and N ≥ m . Thus, 0 ≤ n ≤ N and0 ≤ m ≤ N . Now, comparing N ∑ i = (cid:18) ni (cid:19) c i = n ∑ i = (cid:18) ni (cid:19) c i + N ∑ i = n + (cid:18) ni (cid:19)| {z } = i ≥ n + > n ) c i ( here, we have split the sum at i = n , since 0 ≤ n ≤ N )= n ∑ i = (cid:18) ni (cid:19) c i + N ∑ i = n + c i | {z } = = n ∑ i = (cid:18) ni (cid:19) c i with N ∑ i = (cid:18) ni (cid:19) c i = m ∑ i = (cid:18) ni (cid:19) c i + N ∑ i = m + (cid:18) ni (cid:19) c i |{z} = i ≥ m + > m ) ( here, we have split the sum at i = m , since 0 ≤ m ≤ N )= m ∑ i = (cid:18) ni (cid:19) c i + N ∑ i = m + (cid:18) ni (cid:19) | {z } = = m ∑ i = (cid:18) ni (cid:19) c i = a n ( by (30) ) , we obtain a n = n ∑ i = (cid:18) ni (cid:19) c i |{z} =( − ) i d i (by (33)) = n ∑ i = (cid:18) ni (cid:19) ( − ) i d i = n ∑ i = ( − ) i (cid:18) ni (cid:19) d i .Now, forget that we fixed n . We thus have shown that a n = n ∑ i = ( − ) i ( ni ) d i foreach n ∈ N . Thus, Lemma 4.18 (applied to d n and a n instead of a n and b n ) showsthat d n = n ∑ i = ( − ) i (cid:18) ni (cid:19) a i for each n ∈ N . (34)Now, let n > m be an integer. Then, (31) (applied to i = n ) yields c n =
0. But (32)(applied to i = n ) yields d n = ( − ) n c n |{z} = =
0. But (34) yields d n = n ∑ i = ( − ) i ( ni ) a i .Comparing the latter two equalities, we obtain n ∑ i = ( − ) i ( ni ) a i = n . We thus have shown that n ∑ i = ( − ) i ( ni ) a i = n > m . In other words, Statement 1 holds.We thus have proved the implication 2 = ⇒ = ⇒ = ⇒
1, we conclude thatStatement 1 and Statement 2 are equivalent. This proves Lemma 4.19.
Definition 4.20.
Let A be an abelian group, written additively. Let ( a , a , a , . . . ) ∈ A N be a sequence of elements of A . Let m ≥ − ( a , a , a , . . . ) is m -polynomial if the two equivalent statements 1 and 2 ofLemma 4.19 are satisfied. Lemma 4.21.
Let A be an abelian group, written additively. Let p ≥ − and q ≥ − betwo integers such that p + q ≥ − . Let ( a , a , a , . . . ) ∈ A N be a p-polynomial sequenceof elements of A. Let ( b , b , b , . . . ) ∈ A N be a q-polynomial sequence of elements of A.Then, ( a b , a b , a b , . . . ) ∈ A N is a ( p + q ) -polynomial sequence of elements of A.Proof of Lemma 4.21. We have assumed that the sequence ( a , a , a , . . . ) is p -polynomial.In other words, this sequence satisfies the two equivalent statements 1 and 2 ofLemma 4.19 for m = p (by the definition of “ p -polynomial”). Thus, in particular,it satisfies Statement 2 of Lemma 4.19 for m = p . In other words, there exist p + c , c , . . . , c p of A such that every n ∈ N satisfies a n = p ∑ i = ( ni ) c i . Considerthese c , c , . . . , c p , and denote them by u , u , . . . , u p . Thus, u , u , . . . , u p are p + A with the property that every n ∈ N satisfies a n = p ∑ i = (cid:18) ni (cid:19) u i . (35) The same argument (but applied to the sequence ( b , b , b , . . . ) and the integer q instead of the sequence ( a , a , a , . . . ) and the integer p ) helps us construct q + v , v , . . . , v q of A with the property that every n ∈ N satisfies b n = q ∑ i = (cid:18) ni (cid:19) v i . (36)Consider these q + v , v , . . . , v q .For each i ∈ N , we set w i = ∑ ( j , k ) ∈ N × N ; j ≤ p ; k ≤ q ; j ≤ i ≤ j + k (cid:18) ij (cid:19)(cid:18) jj + k − i (cid:19) u j v k . (37)Thus, if i ∈ N satisfies i > p + q , then w i =
0. (38)[
Proof of (38) : Let i ∈ N satisfy i > p + q . Then, there exists no ( j , k ) ∈ N × N satisfying j ≤ p and k ≤ q and j ≤ i ≤ j + k (since any such ( j , k ) would satisfy i ≤ j |{z} ≤ p + k |{z} ≤ q ≤ p + q , which would contradict i > p + q ). Hence, the sum onthe right hand side of (37) is empty, and thus equals 0. Therefore, (37) rewrites as w i =
0. This proves (38).]Let n ∈ N . Then, (35) yields a n = p ∑ i = (cid:18) ni (cid:19) u i = p ∑ j = (cid:18) nj (cid:19) u j .Meanwhile, (36) yields b n = q ∑ i = (cid:18) ni (cid:19) v i = q ∑ k = (cid:18) nk (cid:19) v k . Multiplying these two equalities, we obtain a n b n = p ∑ j = (cid:18) nj (cid:19) u j ! q ∑ k = (cid:18) nk (cid:19) v k ! = p ∑ j = q ∑ k = | {z } = ∑ ( j , k ) ∈ N × N ; j ≤ p ; k ≤ q (cid:18) nj (cid:19)(cid:18) nk (cid:19)| {z } = j + k ∑ i = j ( ij )( jj + k − i )( ni ) (by Lemma 4.17,applied to m = n , a = j and b = k ) u j v k = ∑ ( j , k ) ∈ N × N ; j ≤ p ; k ≤ q j + k ∑ i = j |{z} = ∑ i ∈ N ; j ≤ i ≤ j + k (cid:18) ij (cid:19)(cid:18) jj + k − i (cid:19)(cid:18) ni (cid:19) u j v k = ∑ ( j , k ) ∈ N × N ; j ≤ p ; k ≤ q ∑ i ∈ N ; j ≤ i ≤ j + k | {z } = ∑ i ∈ N ∑ ( j , k ) ∈ N × N ; j ≤ p ; k ≤ q ; j ≤ i ≤ j + k (cid:18) ij (cid:19)(cid:18) jj + k − i (cid:19)(cid:18) ni (cid:19) u j v k = ∑ i ∈ N ∑ ( j , k ) ∈ N × N ; j ≤ p ; k ≤ q ; j ≤ i ≤ j + k (cid:18) ij (cid:19)(cid:18) jj + k − i (cid:19)(cid:18) ni (cid:19) u j v k = ∑ i ∈ N (cid:18) ni (cid:19) ∑ ( j , k ) ∈ N × N ; j ≤ p ; k ≤ q ; j ≤ i ≤ j + k (cid:18) ij (cid:19)(cid:18) jj + k − i (cid:19) u j v k | {z } = w i (by (37)) = ∑ i ∈ N (cid:18) ni (cid:19) w i = ∑ i ∈ N ; i ≤ p + q | {z } = p + q ∑ i = (cid:18) ni (cid:19) w i + ∑ i ∈ N ; i > p + q (cid:18) ni (cid:19) w i |{z} = = p + q ∑ i = (cid:18) ni (cid:19) w i + ∑ i ∈ N ; i > p + q (cid:18) ni (cid:19) | {z } = = p + q ∑ i = (cid:18) ni (cid:19) w i .Forget that we fixed n . We thus have shown that every n ∈ N satisfies a n b n = p + q ∑ i = ( ni ) w i . Hence, there exist p + q + c , c , . . . , c p + q of A such that every ∈ N satisfies a n b n = p + q ∑ i = ( ni ) c i (namely, these p + q + c i = w i ). In other words, Statement 2 of Lemma 4.19 with m and ( a , a , a , . . . ) replaced by p + q and ( a b , a b , a b , . . . ) is satisfied. Thus, the two equivalentstatements 1 and 2 of Lemma 4.19 with m and ( a , a , a , . . . ) replaced by p + q and ( a b , a b , a b , . . . ) is satisfied. In other words, the sequence ( a b , a b , a b , . . . ) is ( p + q ) -polynomial (by the definition of “ ( p + q ) -polynomial”). This provesLemma 4.21. Lemma 4.22.
Let B be a k -bialgebra. Let b ∈ B. Let m ≥ − be an integer.Then, m is a degree-upper bound of b if and only if the sequence (cid:16) id ⊛ ( b ) , id ⊛ ( b ) , id ⊛ ( b ) , . . . (cid:17) is m-polynomial.Proof of Lemma 4.22. The elements ηǫ and id of the convolution algebra ( Hom ( B , B ) , ⊛ ) commute (since ηǫ is the unity of this algebra). Thus, the binomial formula showsthat ( ηǫ − id ) ⊛ n = n ∑ i = ( − ) i (cid:18) ni (cid:19) id ⊛ i for every n ∈ N (again because ηǫ is the unity of the convolution algebra). Hence,for every n ∈ N , we have ( ηǫ − id ) ⊛ n ( b ) = n ∑ i = ( − ) i (cid:18) ni (cid:19) id ⊛ i ( b ) . (39)Now, we have the following chain of equivalences: ( m is a degree-upper bound of b ) ⇐⇒ (cid:16) ( ηǫ − id ) ⊛ n ( b ) = n > m (cid:17) ( by the definition of a “degree-upper bound” ) ⇐⇒ n ∑ i = ( − ) i (cid:18) ni (cid:19) id ⊛ i ( b ) = n > m ! ( by (39) ) ⇐⇒ (cid:16) Statement 1 of Lemma 4.19 holds for A = B and a i = id ⊛ i ( b ) (cid:17) ⇐⇒ (cid:16) the sequence (cid:16) id ⊛ ( b ) , id ⊛ ( b ) , id ⊛ ( b ) , . . . (cid:17) is m -polynomial (cid:17) ( by the definition of “ m -polynomial” ) .This proves Lemma 4.22. Proposition 4.23.
Let B be a commutative k -bialgebra. Let b , c ∈ B. Let p , q ≥ − betwo integers with p + q ≥ − . Assume that p is a degree-upper bound of b. Assume thatq is a degree-upper bound of c. Then, p + q is a degree-upper bound of bc.Proof of Proposition 4.23. For each k ∈ N , we know that the map id ⊛ k : B → B is a k -algebra homomorphism (by Lemma 4.11), and thus satisfies id ⊛ k ( bc ) = id ⊛ k ( b ) · id ⊛ k ( c ) . Hence, (cid:16) id ⊛ ( bc ) , id ⊛ ( bc ) , id ⊛ ( bc ) , . . . (cid:17) = (cid:16) id ⊛ ( b ) id ⊛ ( c ) , id ⊛ ( b ) id ⊛ ( c ) , id ⊛ ( b ) id ⊛ ( c ) , . . . (cid:17) . (40)Lemma 4.22 (applied to m = p ) shows that p is a degree-upper bound of b ifand only if the sequence (cid:16) id ⊛ ( b ) , id ⊛ ( b ) , id ⊛ ( b ) , . . . (cid:17) is p -polynomial. Thus,the sequence (cid:16) id ⊛ ( b ) , id ⊛ ( b ) , id ⊛ ( b ) , . . . (cid:17) is p -polynomial (since p is a degree-upper bound of b ). The same argument (applied to c and q instead of b and p )shows that the sequence (cid:16) id ⊛ ( c ) , id ⊛ ( c ) , id ⊛ ( c ) , . . . (cid:17) is q -polynomial. Hence,Lemma 4.21 (applied to A = B , a i = id ⊛ i ( b ) and b i = id ⊛ i ( c ) ) shows that (cid:16) id ⊛ ( b ) id ⊛ ( c ) , id ⊛ ( b ) id ⊛ ( c ) , id ⊛ ( b ) id ⊛ ( c ) , . . . (cid:17) is a ( p + q ) -polynomial se-quence of elements of B . In view of (40), this rewrites as follows: (cid:16) id ⊛ ( bc ) , id ⊛ ( bc ) , id ⊛ ( bc ) , . . . (cid:17) is a ( p + q ) -polynomial sequence of elementsof B .But Lemma 4.22 (applied to p + q and bc instead of m and b ) shows that p + q is adegree-upper bound of bc if and only if the sequence (cid:16) id ⊛ ( bc ) , id ⊛ ( bc ) , id ⊛ ( bc ) , . . . (cid:17) is ( p + q ) -polynomial. Hence, p + q is a degree-upper bound of bc (since thesequence (cid:16) id ⊛ ( bc ) , id ⊛ ( bc ) , id ⊛ ( bc ) , . . . (cid:17) is ( p + q ) -polynomial). This provesProposition 4.23. Corollary 4.24.
Let B be a commutative k -bialgebra. Let L denote the set of all id-unipotentelements of B. Then, L is a k -subalgebra of B.Proof of Corollary 4.24. We have already seen in Remark 4.8 that L is a k -submoduleof B . Thus, it suffices to show that 1 ∈ L and that all b , c ∈ L satisfy bc ∈ L .It is easy to see that 1 ∈ L : Indeed, it is easy to see (by induction) that ( ηǫ − id ) ⊛ n ( ) = n . Thus, 0 is a degree-upper bound of 1. Hence, theelement 1 is id-unipotent, i.e., we have 1 ∈ L .It remains to show that all b , c ∈ L satisfy bc ∈ L . So let b , c ∈ L be arbitrary.Thus, b and c are two id-unipotent elements of B . Clearly, b ∈ B has a degree-upperbound (since b is id-unipotent); let us denote this bound by p . We WLOG assumethat p is nonnegative (since otherwise, we can replace p by 0). Likewise, we canfind a nonnegative degree-upper bound q of c . Now, Proposition 4.23 shows that + q is a degree-upper bound of bc . Hence, bc is id-unipotent. In other words, bc ∈ L .We thus have shown that all b , c ∈ L satisfy bc ∈ L . As we have seen, thisconcludes the proof of Corollary 4.24.
5. Linear independence in the dual algebra
So far we have considered grouplike elements in coalgebras and bialgebras. Arelated (and, occasionally, equivalent) concept are the characters of an algebra.
We recall that a character of a k -algebra A means a k -algebra homomorphism from A to k . Before we study linear independence questions for characters, let us brieflysurvey their relation to grouplike elements.The simplest way to connect characters with grouplike elements is the following(easily verified) fact: Proposition 5.1.
Let C be a k -coalgebra that is free as a k -module. Let c ∈ C. Considerthe k -linear map c ∨∨ : C ∨ → k that sends each f ∈ C ∨ to f ( c ) ∈ k . Then, c ∨∨ is acharacter of the k -algebra C ∨ if and only if c is grouplike in C.Proof of Proposition 5.1 (sketched). We shall prove the two equivalences ( ∆ ( c ) = c ⊗ c ) ⇐⇒ (cid:0) c ∨∨ ( f ⊛ g ) = c ∨∨ ( f ) · c ∨∨ ( g ) for all f , g ∈ C ∨ (cid:1) (41)and ( ǫ ( c ) = ) ⇐⇒ (cid:0) c ∨∨ ( ǫ ) = (cid:1) . (42)Indeed, the equivalence (42) follows from the fact that c ∨∨ ( ǫ ) = ǫ ( c ) (which is adirect consequence of the definition of c ∨∨ ). Let us now prove the equivalence (41).The k -module C is free. Hence, it is easy to prove the following fact: Fact 1:
Let x and y be two elements of C ⊗ C . Then, x = y if and only ifwe have ( f ⊗ g ) ( x ) = ( f ⊗ g ) ( y ) for all f , g ∈ C ∨ .Fact 1 is often stated as “the pure tensors in C ∨ ⊗ C ∨ separate C ⊗ C ”. It can beproved by picking a basis ( e i ) i ∈ I of C and its corresponding dual “basis” (cid:0) e ∗ i (cid:1) i ∈ I of C ∨ (with e ∗ i (cid:0) e j (cid:1) being the Kronecker delta δ i , j for all i , j ∈ I ), and arguing that (cid:16) e ∗ i ⊗ e ∗ j (cid:17) ( i , j ) ∈ I × I is the dual “basis” to the basis (cid:0) e i ⊗ e j (cid:1) ( i , j ) ∈ I × I of C ⊗ C . Applying Fact 1 to x = ∆ ( c ) and y = c ⊗ c , we obtain the following equivalence: ( ∆ ( c ) = c ⊗ c ) ⇐⇒ (cid:0) ( f ⊗ g ) ( ∆ ( c )) = ( f ⊗ g ) ( c ⊗ c ) for all f , g ∈ C ∨ (cid:1) . (43)But the multiplication map µ k : k ⊗ k → k of k is bijective; thus, for any f , g ∈ C ∨ ,we have the following chain of equivalences: (( f ⊗ g ) ( ∆ ( c )) = ( f ⊗ g ) ( c ⊗ c )) ⇐⇒ ( µ k (( f ⊗ g ) ( ∆ ( c ))) = µ k (( f ⊗ g ) ( c ⊗ c ))) ⇐⇒ (( f ⊛ g ) ( c ) = f ( c ) · g ( c )) since µ k (( f ⊗ g ) ( ∆ ( c ))) = ( µ k ◦ ( f ⊗ g ) ◦ ∆ ) | {z } = f ⊛ g (by the definition of ⊛ ) ( c ) = ( f ⊛ g ) ( c ) and µ k (( f ⊗ g ) ( c ⊗ c )) = µ k ( f ( c ) ⊗ g ( c )) = f ( c ) · g ( c ) ⇐⇒ (cid:0) c ∨∨ ( f ⊛ g ) = c ∨∨ ( f ) · c ∨∨ ( g ) (cid:1)(cid:18) since the definition of c ∨∨ yields c ∨∨ ( f ⊛ g ) = ( f ⊛ g ) ( c ) and c ∨∨ ( f ) = f ( c ) and c ∨∨ ( g ) = g ( c ) (cid:19) .Hence, the equivalence (43) is saying the same thing as the equivalence (41). Thus,the latter equivalence is proven.We have now proved both equivalences (41) and (42). But the definition of grou-plike elements yields the following chain of equivalences: ( c is grouplike in C ) ⇐⇒ ( ∆ ( c ) = c ⊗ c ) and ( ǫ ( c ) = ) ⇐⇒ (cid:0) c ∨∨ ( f ⊛ g ) = c ∨∨ ( f ) · c ∨∨ ( g ) for all f , g ∈ C ∨ (cid:1) and (cid:0) c ∨∨ ( ǫ ) = (cid:1) ( by (41) and (42) ) ⇐⇒ (cid:0) c ∨∨ is a k -algebra homomorphism from C ∨ to k (cid:1)(cid:0) since the map c ∨∨ is k -linear, and ǫ is the unity of the k -algebra C ∨ (cid:1) ⇐⇒ (cid:0) c ∨∨ is a character of the k -algebra C ∨ (cid:1) .This proves Proposition 5.1.Proposition 5.1 can be used to characterize the characters of some algebras asgrouplikes. To wit: If A is a k -algebra that is finite free as a k -module, then thedual k -module A ∨ canonically receives the structure of a k -coalgebra , whose dual ( A ∨ ) ∨ is canonically isomorphic to the k -algebra A (via the standard k -module This structure can be defined as follows: Its comultiplication ∆ A ∨ : A ∨ → A ∨ ⊗ A ∨ is the compo-sition of the map µ ∨ A : A ∨ → ( A ⊗ A ) ∨ (which is the dual of the multiplication µ A : A ⊗ A → A of A ) with the canonical isomorphism ( A ⊗ A ) ∨ → A ∨ ⊗ A ∨ (which exists because A is finitefree). The counit ǫ A ∨ : A ∨ → k of A ∨ is the dual of the unit map η A : k → A of A . isomorphism A → ( A ∨ ) ∨ ). Thus, Proposition 5.1 yields that the characters of a k -algebra A that is finite free as a k -module are precisely the grouplike elementsof its dual coalgebra A ∨ .There are some ways to extend this result to k -algebras A that are not finite free;the best-known case is when k is a field. In this case, every k -algebra A has a Hopfdual A o (also known as the zero dual or Sweedler dual), which is a k -coalgebrawhose grouplike elements are precisely the characters of A . (See [Sweedl69, Section6.0] for the definition and fundamental properties of this Hopf dual and [Duc97,DucTol09] for questions linked to rationality.) When k is not a field, this Hopf dualis not defined any more. Indeed, the canonical map M ∨ ⊗ N ∨ → ( M ⊗ N ) ∨ that isdefined for any two k -modules M and N is known to be injective when k is a field,but may fail to be injective even when k is an integral domain . Other concepts canbe used to salvage the relation between grouplikes and characters: bases in duality,dual laws constructed directly [Bui12, BDMKT16, Ducham01], pseudo-coproducts[PatReu02, Section 2]. We shall not delve on these things; but the upshot for usis that while the concepts of characters of an algebra and grouplike elements of acoalgebra are closely connected, neither concept subsumes the other.For this reason, our third main result will be stated directly in the language ofcharacters on a bialgebra (i.e., algebra morphisms from this bialgebra to the basering). We begin with a simple fact:
Proposition 5.2.
Let B be a k -bialgebra. Then, the set Ξ ( B ) of characters of B is a monoidfor the convolution product ⊛ .Proof. We know that B ∨ is a k -algebra under convolution, and thus a monoid.Hence, we just need to prove that Ξ ( B ) is a submonoid of this monoid B ∨ . But thisfollows from the following two observations: • If p , q ∈ Ξ ( B ) , then p ⊛ q ∈ Ξ ( B ) . (This is a consequence of Lemma 4.10,applied to C = B and A = k .) • The neutral element η k |{z} = id ǫ B = ǫ B of ⊛ belongs to Ξ ( B ) .Thus, Proposition 5.2 is proved.The convolution monoid Ξ ( B ) is not always a group. This will happen when M ∨ ⊗ N ∨ has torsion (see Subsection 5.3 and Proposition 5.9 in particular). which has been briefly announced at https://mathoverflow.net/questions/310354 Theorem 5.3.
Let B be a k -bialgebra. As usual, let ∆ = ∆ B and ǫ = ǫ B be its comultipli-cation and its counit.Let B + = ker ( ǫ ) . For each N ≥ , let B N + = B + · B + · · · · · B + | {z } N times , where B + = B . Notethat (cid:0) B + , B + , B + , . . . (cid:1) is called the standard decreasing filtration of B .For each N ≥ − , we define a k -submodule B ∨ N of B ∨ by B ∨ N = ( B N + + ) ⊥ = n f ∈ B ∨ | f (cid:16) B N + + (cid:17) = o . (44) Thus, (cid:0) B ∨− , B ∨ , B ∨ , . . . (cid:1) is an increasing filtration of B ∨ ∞ : = S N ≥− B ∨ N with B ∨− = .Then: (a) We have B ∨ p ⊛ B ∨ q ⊆ B ∨ p + q for any p , q ≥ − (where we set B ∨− = ). Hence, B ∨ ∞ is a subalgebra of the convolution algebra B ∨ . (b) Assume that k is an integral domain. Then, the set Ξ ( B ) × of invertible characters(i.e., of invertible elements of the monoid Ξ ( B ) from Proposition 5.2) is left B ∨ ∞ -linearly independent.Proof. The map ǫ : B → k is a k -algebra homomorphism; hence, its kernel B + isan ideal of B . Thus, B + = BB + , so that B N + = B N − + B + for each N ≥
1. (45)Let us define a left action ⊲ of the k -algebra B on B ∨ by setting h u ⊲ f | v i = h f | vu i for all f ∈ B ∨ and u , v ∈ B .Here, h g | b i means g ( b ) whenever g ∈ B ∨ and b ∈ B . Thus, B ∨ is a left B -module.For a given u ∈ B , we shall refer to the operator B ∨ → B ∨ , f u ⊲ f as shifting by u or the u -left shift operator; it generalizes Schützenberger’s right u − in automatatheory [BeRe88, Schütz61].In the following, we shall use a variant of Sweedler notation: Given an u ∈ B ,instead of writing ∑ ( u ) u ⊗ u for ∆ ( u ) , we will write ∑ ( u ) u ( ) ⊗ u ( ) for ∆ ( u ) − u ⊗ − ⊗ u . Thus, ∆ ( u ) = u ⊗ + ⊗ u + ∑ ( u ) u ( ) ⊗ u ( ) (46)for each u ∈ B . Moreover, if u ∈ B + , then all of the u ( ) and u ( ) can be chosen tobelong to B + themselves (because it is easy to check that ∆ ( u ) − u ⊗ − ⊗ u =(( id − ηǫ ) ⊗ ( id − ηǫ )) ( ∆ ( u )) ∈ (( id − ηǫ ) ⊗ ( id − ηǫ )) ( B ⊗ B ) = B + ⊗ B + , since ( id − ηǫ ) ( B ) = B + ). We shall understand, in the following, that we choose u ( ) and u ( ) from B + when u ∈ B + .The following two lemmas give simple properties of the left action ⊲ : Lemma 5.4.
Let f , f ∈ B ∨ and u ∈ B + = ker ( ǫ ) . Then,u ⊲ ( f ⊛ f ) = ( u ⊲ f ) ⊛ f + f ⊛ ( u ⊲ f ) + ∑ ( u ) ( u ( ) ⊲ f ) ⊛ ( u ( ) ⊲ f ) . (47) Proof.
Immediate by direct computation.
Lemma 5.5.
Let k ≥ , and let f ∈ B ∨ k and u ∈ B + . Then, u ⊲ f ∈ B ∨ k − .Proof. Easy using (45).Let us now proceed with the proof of Theorem 5.3. (a)
We shall first show that B ∨ p ⊛ B ∨ q ⊆ B ∨ p + q for any p , q ≥ − p + q . Let f ∈ B ∨ p and f ∈ B ∨ q . Weintend to show that f ⊛ f ∈ B ∨ p + q . This is trivial if p or q is −
1, since B ∨− = p , q ≥ B p + q + + = B p + q + B + by (45). Thus, the k -module B p + q + + is spanned by theproducts uv for u ∈ B p + q + and v ∈ B + . Thus, in order to prove that f ⊛ f ∈ B ∨ p + q ,it suffices to show that f ⊛ f is orthogonal to all these products.So let u ∈ B p + q + and v ∈ B + . Then, we must prove that h f ⊛ f | uv i =
0. But wehave h f ⊛ f | uv i = h v ⊲ ( f ⊛ f ) | u i . (48)Applying (47) to v instead of u , we get v ⊲ ( f ⊛ f ) = ( v ⊲ f ) ⊛ f + f ⊛ ( v ⊲ f ) + ∑ ( v ) ( v ( ) ⊲ f ) ⊛ ( v ( ) ⊲ f ) (49)with all v ( ) and v ( ) lying in B + . Thus, Lemma 5.5 yields v ⊲ f ∈ B ∨ p − and v ⊲ f ∈ B ∨ q − and v ( ) ⊲ f ∈ B ∨ p − and v ( ) ⊲ f ∈ B ∨ q − . Thus, (49) becomes v ⊲ ( f ⊛ f ) = ( v ⊲ f ) | {z } ∈B ∨ p − ⊛ f |{z} ∈B ∨ q + f |{z} ∈B ∨ p ⊛ ( v ⊲ f ) | {z } ∈B ∨ q − + ∑ ( v ) ( v ( ) ⊲ f ) | {z } ∈B ∨ p − ⊛ ( v ( ) ⊲ f ) | {z } ∈B ∨ q − ∈ B ∨ p − ⊛ B ∨ q + B ∨ p ⊛ B ∨ q − + B ∨ p − ⊛ B ∨ q − ⊆ B ∨ p + q − (by the induction hypothesis, applied three times). This entails h v ⊲ ( f ⊛ f ) | u i = B ∨ p + q − . Thus, (48) yields h f ⊛ f | uv i =
0. Thus, we haveproved that f ⊛ f ∈ B ∨ p + q . This completes the proof of B ∨ p ⊛ B ∨ q ⊆ B ∨ p + q .The fact that B ∨ ∞ is a subalgebra of the convolution algebra B ∨ follows from thepreceding, since we also know that ǫ ∈ B ∨ ⊆ B ∨ ∞ . (b) We define the degree of an element b ∈ B ∨ ∞ to be the least index d ≥ − b ∈ B ∨ d . We denote this index by deg ( b ) . (Note that deg ( ) = − b ∈ B ∨ d and u ∈ B + , we have u ⊲ b ∈ B ∨ d − (by Lemma 5.5). Thus,for b ∈ B ∨ ∞ and u ∈ B + , we have u ⊲ b ∈ B ∨ ∞ . In other words, B + ⊲ B ∨ ∞ ⊆ B ∨ ∞ . Let us consider non-trivial relations of the form ∑ g ∈ F p g ⊛ g = F of Ξ ( B ) × and with nonzero coefficients p g ∈ B ∨ ∞ . If thereare no such relations with F = ∅ , then we are done. Otherwise, we pick onesuch relation of type (50) with | F | 6 = ∑ g ∈ F deg ( p g ) is minimal. WLOG, we assume that ǫ ∈ F (otherwise, pick g ∈ F and multiply both sides of the equation (50) by g ⊛ − ).It is impossible that F = { ǫ } (as p ǫ = g ∈ F \ { ǫ } .Having chosen this g , we observe that g ( B + ) = g ( B + ) = g = ǫ because of g ∈ Ξ ( B ) ); in other words, there exists some u ∈ B + such that h g | u i 6 =
0. Choose such a u .It is easy to see (from the definition) that each character g ∈ Ξ ( B ) and each v ∈ B + satisfy v ⊲ g = h g | v i g . (51)Our plan is now to shift both sides of (50) by u , and rewrite the resulting equalityagain as an equality of the form (50) (with new values of p g ). To do so, we introducea few notations.Write ∆ ( u ) as in (46), with u ( ) , u ( ) ∈ B + . For each g ∈ F , let us set p ′ g : = u ⊲ p g + h g | u i p g + ∑ ( u ) h g | u ( ) i ( u ( ) ⊲ p g ) ∈ B ∨ .Then, we have u ⊲ ( p g ⊛ g ) = p ′ g ⊛ g (52)for each g ∈ F , because Lemma 5.4 yields u ⊲ ( p g ⊛ g )= ( u ⊲ p g ) ⊛ g + p g ⊛ ( u ⊲ g ) + ∑ ( u ) ( u ( ) ⊲ p g ) ⊛ ( u ( ) ⊲ g )= ( u ⊲ p g ) ⊛ g + p g ⊛ ( h g | u i g ) + ∑ ( u ) ( u ( ) ⊲ p g ) ⊛ (cid:16) h g | u ( ) i g (cid:17) ( by (51) )= (cid:16) u ⊲ p g + h g | u i p g + ∑ ( u ) h g | u ( ) i ( u ( ) ⊲ p g ) (cid:17)| {z } = p ′ g ⊛ g = p ′ g ⊛ g .Thus, if we shift both sides of (50) by u , we find u ⊲ ∑ g ∈ F p g ⊛ g ! = Hence, 0 = u ⊲ ∑ g ∈ F p g ⊛ g ! = ∑ g ∈ F u ⊲ ( p g ⊛ g ) = ∑ g ∈ F p ′ g ⊛ g (53)(by (52)).For each g ∈ F , we have p ′ g ∈ B ∨ ∞ (by the definition of p ′ g , since p g ∈ B ∨ ∞ and B + ⊲ B ∨ ∞ ⊆ B ∨ ∞ ).It is easy to see that each g ∈ F satisfiesdeg (cid:16) p ′ g (cid:17) ≤ deg (cid:0) p g (cid:1) . (54)(Indeed, Lemma 5.5 shows that any nonzero f ∈ B ∨ ∞ satisfies deg ( u ⊲ f ) < deg f ;for the same reason, deg (cid:16) u ( ) ⊲ f (cid:17) < deg f . Thus, the definition of p ′ g represents p ′ g as a k -linear combination of elements of B ∨ ∞ having the same degree as p g orsmaller degree. This proves (54).)But the definition of p ′ ǫ easily yields p ′ ǫ = u ⊲ p ǫ (55)(since u , u ( ) ∈ B + entail u ⊲ p ǫ = u ( ) ⊲ p ǫ = ( p ′ ǫ ) = deg ( u ⊲ p ǫ ) < deg ( p ǫ ) (since Lemma 5.5 shows that any nonzero f ∈ B ∨ ∞ satisfiesdeg ( u ⊲ f ) < deg f ).Now, recall the equality (53). This equality (once rewritten as ∑ g ∈ F p ′ g ⊛ g =
0) hasthe same structure as (50) (since p ′ g ∈ B ∨ ∞ for each g ∈ F ), and uses the same set F ,but it satisfies ∑ g ∈ F deg (cid:16) p ′ g (cid:17) < ∑ g ∈ F deg (cid:0) p g (cid:1) (because each g ∈ F satisfies (54), butthe specific character g = ǫ satisfies deg ( p ′ ǫ ) < deg ( p ǫ ) ). Thus, if the coefficients p ′ g in (53) are not all 0, then we obtain a contradiction to our choice of relation(which was to minimize ∑ g ∈ F deg (cid:0) p g (cid:1) ). Hence, all coefficients p ′ g must be 0. Inparticular, we must have p ′ g =
0. Therefore,0 = p ′ g = u ⊲ p g + h g | u i p g + ∑ ( u ) h g | u ( ) i ( u ( ) ⊲ p g ) (by the definition of p ′ g ), so that h g | u i p g = − u ⊲ p g − ∑ ( u ) h g | u ( ) i ( u ( ) ⊲ p g ) ∈ B ∨ deg ( p g ) − (by Lemma 5.5, since u , u ( ) ∈ B + and p g ∈ B ∨ deg ( p g ) ). In other words, h g | u i p g is orthogonal to B deg ( p g ) + . Since h g | u i 6 =
0, this entails that p g is orthogonal We recall that we have chosen u ∈ B + such that h g | u i 6 = to B deg ( p g ) + as well (since its target k is an integral domain). This means that p g ∈ B ∨ deg ( p g ) − , or, equivalently, deg (cid:0) p g (cid:1) ≤ deg (cid:0) p g (cid:1) −
1. But this is absurd.This is the contradiction we were looking for. Thus, Theorem 5.3 (b) is proved.
Remark . i) The invertible characters in Ξ ( B ) × are also right B ∨ ∞ -linearly in-dependent. This can be proven similarly, using right shifts in the proof.ii) The reader should beware of supposing that the standard decreasing filtrationis necessarily Hausdorff (i.e., satisfies T n ≥ B n + = { } ). A counterexamplecan be obtained by taking the universal enveloping bialgebra of any simpleLie algebra (or, more generally, of any perfect Lie algebra); it will satisfy T n ≥ B n + = B + .iii) The property (i.e., the linear independence) does not hold if we consider theset of all characters (that is, Ξ ( B ) ) instead of Ξ ( B ) × . For example, let B bethe univariate q -infiltration bialgebra (cid:16) k [ x ] , ∆ ↑ q , ǫ (cid:17) from Example 2.3, andassume that q ∈ k is invertible. Define two k -linear maps f : B → k and g : B → k by f ( x n ) = δ n ,1 and g ( x n ) = (cid:18) − q (cid:19) n for all n ∈ N (where the δ n ,1 is a Kronecker delta). Note that g is a character of B , while f ∈ B ∨ ⊆ B ∨ ∞ .We claim that f ⊛ g =
0. In fact, it is easy to see (by induction on m ) that ∆ ↑ q ( x m ) = ∑ ( i , j , k ) ∈ N ; i + j + k = m (cid:18) mi , j , k (cid:19) q j x i + j ⊗ x j + k (56)holds in B for every m ∈ N . Using this equality, it is straightforward to seethat ( f ⊛ g ) ( x m ) = m ∈ N ; thus, f ⊛ g =
0. Since f is nonzero, thisshows that the set Ξ ( B ) is not left B ∨ ∞ -linearly independent.Note that our above proof of Theorem 5.3 is somewhat similar to the standard(Artin) proof of the linear independence of characters in Galois theory ([Artin71,proof of Theorem 12]). Shifting by u in the former proof corresponds to replacing x by α x in the latter. Corollary 5.7.
We suppose that B is cocommutative, and k is an integral domain.Let ( g x ) x ∈ X be a family of elements of Ξ ( B ) × (the set of invertible characters of B ), and let ϕ X : k [ X ] → ( B ∨ , ⊛ , ǫ ) be the k -algebra morphism that sends each x ∈ X to g x . In orderfor the family ( g x ) x ∈ X (of elements of the commutative ring ( B ∨ , ⊛ , ǫ ) ) to be algebraically “Separated” in [Bourba72]. ( B ∨ ∞ , ⊛ , ǫ ) , it is necessary and sufficient that the monomialmap m : N ( X ) → ( B ∨ , ⊛ , ǫ ) , α ϕ X ( X α ) = ∏ x ∈ X g α x x (57) (where α x means the x-th entry of α ) be injective.Proof. Indeed, as B is cocommutative (and k commutative), ( B ∨ , ⊛ , ǫ ) is a commu-tative algebra (see, e.g., [GriRei20, Exercise 1.5.5] or [Bourba89, Chapter III, §11.2,Proposition 2]). Thus, the algebraic independence of the family ( g x ) x ∈ X is equiva-lent to the statement that the family (cid:18) ∏ x ∈ X g α x x (cid:19) α ∈ N ( X ) be B ∨ ∞ -linearly independent.In other words, it is equivalent to the claim that the family ( m ( α )) α ∈ N ( X ) (with m as in (57)) be B ∨ ∞ -linearly independent (since m ( α ) = ∏ x ∈ X g α x x for each α ∈ N ( X ) ).It suffices to remark that the elements m ( α ) are invertible characters (since theyare products of invertible characters). Therefore, in view of Theorem 5.3 (b) , the ( m ( α )) α ∈ N ( X ) are B ∨ ∞ -linearly independent if they are distinct; but this amounts tosaying that m is injective. Example 5.8. i) Let k be an integral domain, and let us consider the bialgebra B = ( k [ x ] , ∆ , ǫ ) from Example 2.2 (the standard univariate polynomial bial-gebra). As it is a particular case of the situation of the free algebra , we willlet ⊔⊔ denote the convolution ⊛ on its dual k -module B ∨ = k [ x ] ∨ ∼ = k [[ x ]] ;thus, ( k [[ x ]] , ⊔⊔ , 1 ) becomes a commutative k -algebra.For every α ∈ k , there exists only one character of k [ x ] sending x to α ; wewill denote this character by ( α . x ) ∗ ∈ k [[ x ]] (see [Eil74, DuMiNg19, DucTol09,DGM2-20] for motivations about this notation). Thus, Ξ ( B ) = { ( α . x ) ∗ | α ∈ k } .It is easy to check that ( α . x ) ∗ ⊔⊔ ( β . x ) ∗ = (cid:0) ( α + β ) . x (cid:1) ∗ for any α , β ∈ k . Thus,any c , c , . . . , c k ∈ k and any α , α , . . . , α k ∈ N satisfy (cid:0) ( c . x ) ∗ (cid:1) ⊔⊔ α ⊔⊔ (cid:0) ( c . x ) ∗ (cid:1) ⊔⊔ α ⊔⊔ · · · ⊔⊔ (cid:0) ( c k . x ) ∗ (cid:1) ⊔⊔ α k = (cid:0) ( α c + α c + · · · + α k c k ) . x (cid:1) ∗ . (58)The monoid Ξ ( B ) is thus isomorphic to the abelian group ( k , + , 0 ) ; in partic-ular, it is a group, so that Ξ ( B ) × = Ξ ( B ) .The decreasing filtration of B is given by B n + = k [ x ] ≥ n (the ideal of polyno-mials of degree ≥ n ). Hence, the reader may check easily that B ∨ n = k [ x ] ≤ n (the module of polynomials of degree ≤ n ), whence B ∨ ∞ = k [ x ] .Now, let (( c i . x ) ∗ ) i ∈ I be a family of elements of Ξ ( B ) × . Taking X = I and g i = c i . x for each i ∈ I , we can then apply Corollary 5.7, and we conclude that the See [GriRei20, Proposition 1.6.7] and [Reuten93, Section 1.4]. family (( c i . x ) ∗ ) i ∈ I of elements of the power series ring k [[ x ]] is algebraicallyindependent over the subring ( B ∨ ∞ , ⊛ , ǫ ) = k [ x ] if and only if the monomialmap (57) is injective.But (58) shows that the monomial map (57) is injective if and only if the family ( c i ) i ∈ I is Z -linearly independent in k .To illustrate this, take k = Q (the algebraic closure of Q ) and c n = √ p n ∈ N ,where p n is the n -th prime number. What precedes shows that the fam-ily of series (cid:0) ( √ p n x ) ∗ (cid:1) n ≥ is algebraically independent over the polynomials(i.e., over Q [ x ] ) within the commutative Q -algebra (cid:0) Q [[ x ]] , ⊔⊔ , 1 (cid:1) . This exam-ple can be double-checked using partial fractions decompositions as, in fact, ( √ p n x ) ∗ = − √ p n x (this time, the inverse is taken within the ordinaryproduct in k [[ x ]] ) and (cid:16) − √ p n x (cid:17) ⊔⊔ n = − n √ p n x .ii) The preceding example can be generalized as follows: Let k still be an integraldomain; let V be a k -module, and let B = (cid:16) T ( V ) , conc, 1 T ( V ) , ∆ ⊠ , ǫ (cid:17) be thestandard tensor conc-bialgebra For every linear form ϕ ∈ V ∨ , there is anunique character ϕ ∗ of (cid:16) T ( V ) , conc, 1 T ( V ) (cid:17) such that all u ∈ V satisfy h ϕ ∗ | u i = h ϕ | u i . (59)Again, it is easy to check that ( ϕ ) ∗ ⊔⊔ ( ϕ ) ∗ = (cid:0) ϕ + ϕ (cid:1) ∗ for any ϕ , ϕ ∈ V ∨ , because, from Lemma 4.10, both sides are characters of (cid:16) T ( V ) , conc, 1 T ( V ) (cid:17) so that the equality has only to be checked on V . Again, from this, any ϕ , ϕ , . . . , ϕ k ∈ V ∨ and any α , α , . . . , α k ∈ N satisfy (cid:0) ( ϕ ) ∗ (cid:1) ⊔⊔ α ⊔⊔ (cid:0) ( ϕ ) ∗ (cid:1) ⊔⊔ α ⊔⊔ · · · ⊔⊔ (cid:0) ( ϕ k ) ∗ (cid:1) ⊔⊔ α k = (cid:0) α ϕ + α ϕ + · · · + α k ϕ k (cid:1) ∗ . (60)The decreasing filtration of B is given by B n + = L k ≥ n T k ( V ) (the ideal oftensors of degree ≥ n ) and the reader may check easily that, in this case, B ∨ ∞ is the shuffle algebra of finitely supported linear forms – i.e., for each Φ ∈ B ∨ ,we have the equivalence Φ ∈ B ∨ ∞ ⇐⇒ ( ∃ N ∈ N )( ∀ k ≥ N )( Φ ( T k ( V )) = { } ) . The one defined by ∆ ⊠ ( ) = ⊗ ∆ ⊠ ( u ) = u ⊗ + ⊗ u ; ǫ ( u ) = u ∈ V . For this bialgebra ⊔⊔ stands for ⊛ on the space Hom ( B , k ) . Then, Corollary 5.7 shows that ( ϕ ∗ i ) i ∈ I are B ∨ ∞ -algebraically independent within ( T ( V ) ∨ , ⊔⊔ , ǫ ) if the corresponding monomial map is injective, and (60) showsthat it is so iff the family ( ϕ i ) i ∈ I of linear forms is Z -linearly independent in V ∨ . In Subsection 5.1, we have mentioned the difficulties of defining the dual coalgebraof a k -algebra in the general case when k is not necessarily a field. As we said,these difficulties stem from the fact that the canonical map M ∨ ⊗ N ∨ → ( M ⊗ N ) ∨ (for two k -modules M and N ) is not generally an isomorphism, and may fail to beinjective even if k is an integral domain. Even worse, the k -module M ∨ ⊗ N ∨ mayfail to be torsionfree (which automatically precludes any k -linear map M ∨ ⊗ N ∨ → ( M ⊗ N ) ∨ , not just the canonical one, from being injective). We shall soon give anexample where this happens (Example 5.10). First, let us prove a positive result: Proposition 5.9.
Let k be an integral domain. Let M and N be two k -modules. Considerthe canonical k -linear map Φ : M ∨ ⊗ k N ∨ −→ ( M ⊗ k N ) ∨ defined by ( Φ ( f ⊗ g )) ( u ⊗ v ) = f ( u ) g ( v ) for all f ∈ M ∨ , g ∈ N ∨ , u ∈ M and v ∈ N . Then, the following are equivalent:1. The k -module M ∨ ⊗ k N ∨ is torsionfree.2. The map Φ is injective.First proof of Proposition 5.9. = ⇒ k -module M ∨ ⊗ k N ∨ is torsionfree.We must prove that the map Φ is injective. Assume the contrary. Thus, ker Φ = t ∈ M ∨ ⊗ N ∨ such that Φ ( t ) =
0. Considerthis t . Consider all choices of nonzero m ∈ k and of elements u , u , . . . , u r ∈ M ∨ and v , v , . . . , v r ∈ N ∨ such that mt = r ∑ i = u i ⊗ v i . (61)Among all such choices, choose one for which r is minimum.From the fact that r is minimum, we conclude that the elements v , v , . . . , v r of N ∨ are k -linearly independent . But the map Φ is k -linear; hence, Φ ( mt ) = Here is the proof in detail: Assume the contrary. Thus, λ v + λ v + · · · + λ r v r = λ , λ , . . . , λ r ∈ k , not all of which are zero. Consider these scalars, and assume WLOG Φ ( t ) = Φ ( t ) = = Φ ( mt ) = r ∑ i = Φ ( u i ⊗ v i ) (by (61)). Hence, forall x ∈ M and y ∈ N , we have0 = r ∑ i = Φ ( u i ⊗ v i ) ! ( x ⊗ y ) = r ∑ i = u i ( x ) v i ( y ) (62)(by the definition of Φ ). This can be rewritten as0 = r ∑ i = u i ( x ) v i in N ∨ for all x ∈ M . (63)But this entails that u i ( x ) = ≤ i ≤ r (since v , v , . . . , v r are k -linearlyindependent). Since this holds for all x ∈ M , we thus find that u i = ≤ i ≤ r . Hence, (61) shows that mt =
0, so that t = M ∨ ⊗ N ∨ istorsionfree). This contradicts t =
0. This contradiction completes the proof of 1. = ⇒ = ⇒ k -module ( M ⊗ k N ) ∨ is torsionfree, since the dual of any k -module is torsionfree. Hence, if Φ is injective, then M ∨ ⊗ k N ∨ is torsionfree (sincea submodule of a torsionfree k -module is always torsionfree). Second proof of Proposition 5.9 (sketched). = ⇒ k -module M ∨ ⊗ k N ∨ is torsionfree.We must prove that the map Φ is injective.Let F be the fraction field of k . For any k -module P , we let P F denote the F -vectorspace F ⊗ P (which can also be defined as the localization of P with respect to themultiplicatively closed subset k \ { } of k ), and we let ι P denote the canonical k -linear map P → P F . We note that ι P is injective when P is torsionfree.If V is an F -vector space, then we shall write V ∗ for the dual space Hom F ( V , F ) of V . This should be distinguished from the k -module dual of V , which we denoteby V ∨ . that λ r =
0. Hence, λ r m = k is an integral domain) and λ r v r = − r − ∑ i = λ i v i (since λ v + λ v + · · · + λ r v r = λ r , we obtain λ r mt = λ r r ∑ i = u i ⊗ v i = r ∑ i = u i ⊗ λ r v i = r − ∑ i = u i ⊗ λ r v i | {z } = λ r u i ⊗ v i + u r ⊗ λ r v r |{z} = − r − ∑ i = λ i v i = r − ∑ i = λ r u i ⊗ v i + u r ⊗ − r − ∑ i = λ i v i ! = r − ∑ i = λ r u i ⊗ v i − r − ∑ i = λ i u r ⊗ v i = r − ∑ i = ( λ r u i − λ i u r ) ⊗ v i .This is an equality of the same shape as (61), but with r − r (since λ r m = r . This contradiction completes our proof. For any k -module P , the canonical F -linear map ρ P : ( P ∨ ) F → ( P F ) ∗ (which sendseach f ∈ P ∨ to the unique F -linear map P F → F that extends f ) is injective. (This iseasily checked by hand, using the fact that the map k → F is injective.) Hence, weobtain two injective F -linear maps ρ M : ( M ∨ ) F → ( M F ) ∗ and ρ N : ( N ∨ ) F → ( N F ) ∗ .Their tensor product is an injective F -linear map ρ M ⊗ F ρ N : ( M ∨ ) F ⊗ F ( N ∨ ) F → ( M F ) ∗ ⊗ F ( N F ) ∗ (since the tensor product of two injective F -linear maps over F isinjective, because F is a field).For any k -modules M and N , there is a canonical isomorphism ν M , N : ( M ⊗ N ) F → M F ⊗ F N F . (This is a general property of base change.) Hence, M F ⊗ F N F ∼ =( M ⊗ N ) F , so that ( M F ⊗ F N F ) ∗ ∼ = (( M ⊗ N ) F ) ∗ .Also, it is known from linear algebra that the canonical map κ V , W : V ∗ ⊗ F W ∗ → ( V ⊗ F W ) ∗ is injective whenever V and W are two F -vector spaces.We assumed that M ∨ ⊗ N ∨ is torsionfree. Thus, the map ι M ∨ ⊗ N ∨ is injective. Wehave the following commutative diagram of k -modules: M ∨ ⊗ N ∨ Φ ! ! ❉❉❉❉❉❉❉❉❉❉❉❉❉❉❉❉❉❉❉❉❉ hH ι M ∨⊗ N ∨ u u ❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦ ( M ∨ ⊗ N ∨ ) F ν M ∨ , N ∨ ∼ = (cid:15) (cid:15) ( M ∨ ) F ⊗ F ( N ∨ ) F _(cid:127) ρ M ⊗ F ρ N (cid:15) (cid:15) ( M ⊗ N ) ∨ ι ( M ⊗ N ) ∨ (cid:15) (cid:15) ( M F ) ∗ ⊗ F ( N F ) ∗ _(cid:127) κ M F , N F (cid:15) (cid:15) (cid:0) ( M ⊗ N ) ∨ (cid:1) F ρ M ⊗ N } } ④④④④④④④④④④④④④④④④④④④④④ ( M F ⊗ F N F ) ∗ ∼ = ) ) ❘❘❘❘❘❘❘❘❘❘❘❘❘❘ (( M ⊗ N ) F ) ∗ (64)All maps along the left boundary of this diagram are injective; thus, their compo-sition is injective as well. But this composition equals the composition of the mapsalong the right boundary of the diagram. Hence, the latter composition is injective.Thus, Φ (being the initial map in this composition) must be injective. This proves1. = ⇒ = ⇒ M ∨ ⊗ N ∨ fails to be torsionfree: Example 5.10.
Let k be a field, I and J infinite sets, and A the k -subalgebra of k ( t )[ x i , y j : i ∈ I , j ∈ J ] generated by { x i , y j , tx i , t − y j : i ∈ I and j ∈ J } .Then, A is an integral domain. In this example, all duals are taken with respect to A (not with respect to k ); that is, M ∨ means Hom A ( M , A ) .The free A -modules A ( I ) , A ( J ) and A ( I × J ) satisfy (cid:16) A ( I ) (cid:17) ∨ ∼ = A I , (cid:16) A ( J ) (cid:17) ∨ ∼ = A J , (cid:16) A ( I × J ) (cid:17) ∨ ∼ = A I × J , and A ( I × J ) ∼ = A ( I ) ⊗ A A ( J ) . Moreover, there is a natural A -linearmap Ψ : A I ⊗ A A J → A I × J , ( a i ) i ∈ I ⊗ A (cid:0) b j (cid:1) j ∈ J (cid:0) a i b j (cid:1) ( i , j ) ∈ I × J that forms a commutative diagram A I ⊗ A A J Ψ / / ∼ = (cid:15) (cid:15) A I × J ∼ = (cid:15) (cid:15) (cid:16) A ( I ) (cid:17) ∨ ⊗ A (cid:16) A ( J ) (cid:17) ∨ Φ / / (cid:16) A ( I × J ) (cid:17) ∨ (65)with the canonical map Φ : (cid:16) A ( I ) (cid:17) ∨ ⊗ A (cid:16) A ( J ) (cid:17) ∨ → (cid:16) A ( I × J ) (cid:17) ∨ defined as in Propo-sition 5.9.Now, we define an element ξ = ( tx i ) i ∈ I ⊗ A ( t − y j ) j ∈ J − ( x i ) i ∈ I ⊗ A ( y j ) j ∈ J ∈ A I ⊗ A A J .It is clear by computation that Ψ ( ξ ) =
0. We shall now show that ξ = ξ =
0. Hence, ξ can be “shown to be zerousing only finitely many elements of A ” – i.e., there exists a finitely generated k -subalgebra B of A such that ( tx i ) i ∈ I ⊗ B ( t − y j ) j ∈ J − ( x i ) i ∈ I ⊗ B ( y j ) j ∈ J = A I ⊗ B A J . (66)(Indeed, A I ⊗ A A J can be viewed as a quotient of A I ⊗ Z A J modulo the Z -submodulespanned by all the differences ua ⊗ Z v − u ⊗ Z av for u ∈ A I , v ∈ A J and a ∈ A .Thus, ξ = ( tx i ) i ∈ I ⊗ Z ( t − y j ) j ∈ J − ( x i ) i ∈ I ⊗ Z ( y j ) j ∈ J of A I ⊗ Z A J is a Z -linear combination of finitely many such differences. Now takethe (finitely many) a ’s involved in these differences, and define B to be the k -subalgebra of A generated by these finitely many a ’s. Then, (66) holds, as desired.)Consider this B . Choose r ∈ I and s ∈ J so that B ⊆ k ( t )[ x i , y j : i = r and j = s ] .(Such r and s exist, since B is finitely generated.) If m is a monomial in the variables { x i , y j : i ∈ I , j ∈ J } , then deg x m shall denotethe total degree of m in the variables x i , while deg y shall denote the total degreeof m in the variables y j . A good monomial shall mean a product of the form t l m ,where m is a monomial in the variables { x i , y j : i ∈ I , j ∈ J } and l ∈ Z satisfies − deg y m ≤ l ≤ deg x m . As a k -vector space, A has a basis consisting of all goodmonomials.A good monomial t l m will be called strict if m is just a power of x r or just a powerof y s . (In particular, the monomial 1 is strict.) The non-strict good monomials spana proper ideal of A . Let ¯ A be the corresponding quotient algebra of A ; then theimage ¯ B of B in ¯ A is just (a copy of) k (since every element of B is a k -linearcombination of non-strict good monomials and of the monomial 1). For any f ∈ A ,we let f denote the canonical projection of f on the quotient ring ¯ A .Now, we have a k -linear map obtained by composing A I ⊗ B A J → ¯ A I ⊗ ¯ B ¯ A J ∼ = → ¯ A I ⊗ k ¯ A J → ¯ A ⊗ k ¯ A ,where • the first arrow sends each tensor u ⊗ B v ∈ A I ⊗ B A J to the tensor u ⊗ ¯ B v ∈ ¯ A I ⊗ ¯ B ¯ A J (with u denoting the projection of u ∈ A I to ¯ A I , and with v definedsimilarly); • the second arrow is due to ¯ B being (a copy of) k ; • the third arrow is obtained by tensoringthe canonical projection ¯ A I → ¯ A , ( a i ) i ∈ I a r with the canonical projection ¯ A J → ¯ A , (cid:0) b j (cid:1) j ∈ J b s .Applying this k -linear map to both sides of the equation (66), we obtain tx r ⊗ k t − y s − x r ⊗ k y s = A ⊗ k ¯ A .But this contradicts the fact that the four basis elements tx r , t − y s , x r , y s of ¯ A are k -linearly independent.Hence, our assumption ( ξ =
0) was false. Thus, ξ =
0. In view of Ψ ( ξ ) =
0, this shows that Ψ is not injective. Due to the commutative diagram (65), thismeans that Φ is not injective. Hence, Proposition 5.9 shows that the A -module (cid:16) A ( I ) (cid:17) ∨ ⊗ A (cid:16) A ( J ) (cid:17) ∨ is not torsionfree. In view of (65), this means in turn that the A -module A I ⊗ A A J is not torsionfree (despite being the tensor product of the twotorsionfree, and even torsionless, A -modules A I and A J ).The presence of torsion in A I ⊗ A A J can also be seen directly using the element ξ from the above argument: For any s ∈ I , we have x s ξ = x s (cid:16) ( tx i ) i ∈ I ⊗ A ( t − y j ) j ∈ J − ( x i ) i ∈ I ⊗ A ( y j ) j ∈ J (cid:17) = since x s (cid:16) ( tx i ) i ∈ I ⊗ A ( t − y j ) j ∈ J (cid:17) = ( tx s x i ) i ∈ I ⊗ A ( t − y j ) j ∈ J = ( x i ) i ∈ I ⊗ A ( x s y j ) j ∈ J = x s (cid:0) ( x i ) i ∈ I ⊗ A ( y j ) j ∈ J (cid:1) .Another example of a non-injective canonical map A I ⊗ A A J → A I × J (and thus,of a tensor product of the form M ∨ ⊗ A N ∨ having torsion) can be found in the lasttwo paragraphs of [Goodea72].Note that such examples can only exist when A is not Noetherian. Indeed, it hasbeen shown in [AbGoWi99, Proposition 1.2] that if A is a Noetherian ring, then thecanonical map A I ⊗ A A J → A I × J for any two sets I and J is injective. References [Abe80] E iichi A be , Hopf algebras , Cambridge Tracts in Mathematics .Cambridge University Press, Cambridge-New York, 1980.[AbGoWi99] J awad Y. A buhlail , J osé G ómez -T orrecillas , R obert W isbauer , Dual coalgebras of algebras over commutative rings , Journal of Pureand Applied Algebra (2000), pp. 107–120.[Artin71] E mil A rtin , Galois Theory , Notre Dame Mathemati-cal Lectures , University of Notre Dame Press, 1971. https://projecteuclid.org/euclid.ndml/1175197041 [BeRe88] J. B erstel , C. R eutenauer , Rational series and their languages ,Springer-Verlag, 1988.[GoF04] G. D uchamp , K.A. P enson , A.I. S olomon , A. H orzela and
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