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Left Hopf Algebras JOURNAL OF ALGEBRA 65, 399-411 (1980) Left Hopf Algebras J. A. GREEN Mathematics Institute, University of Warwick, Coventry CV4 7AL, England WARREN D. NICHOLS Department of Mathematics, The Pennsylvania State University, University Park, Pennsylvania 16802; and Department of Mathematics, Florida State University, Tallahassee, Florida 32306 AND EARL J. TAFT Department of Mathematics, Rutgers University, New Brunswick, New Jersey 08903; and School of Mathematics, Institute for Advanced Study, Princeton, New Jersey 08540 Communicated by N. Jacobson Received March 5, 1979 I. INTRODUCTION Let k be a field. If (C, A, ~) is a k-coalgebra with comultiplication A and co- unit ~, and (A, m,/L) is an algebra with multiplication m and unit/z: k-+ A, then Homk(C, A) is an algebra under the convolution productf • g = m(f (~)g)A. The unit element of this algebra is/zE. A bialgebra B is simultaneously a coalgebra and an algebra such that A and E are algebra homomorphisms. Thus Homk(B, B) is an algebra under convolution. B is called a Hopf algebra if the identify map Id of B is invertible in Homk(B, B), i.e., there is an S in Homk(B, B) such that S * Id =/zE = Id • S. Such an S is called the antipode of the Hopf algebra B. Using the notation Ax = ~ x 1 @ x~ for x ~ B (see [8]), the antipode condition is that Y'. S(xa) x 2 = E(x)l = • XlS(X~) for all x E B. A bialgebra B is called a left Hopf algebra if it has a left antipode S, i.e., S ~ Homk(B, B) and S • Id =/,E. We can similarly define a right Hopf algebra by the condition that B have a right antipode, i:e., Id is right invertible in Homk(B, B). In Section 3 we give an example of a left Hopf algebra which is not a Hopf algebra (Example 21). In Section 2 we give several conditions, each of which is sufficient for a left Hopf algebra to be a Hopf algebra (Theorem 3). We recall briefly some ideas concerning bialgebras. A group-like element a in a 399 0021-8693/80/080399-13 $02.00/0 Copyright © 1980 by Academic Press, Inc. All rights of reproduction in any form reserved 400 GREEN, NICHOLS, AND TAFT coalgebra is an element a # 0 with Aa = a @ a. In a bialgebra, the group-like elements form a monoid G(B). In a Hopf algebra, G(B) is a group. A coalgebra homomorphism f: C -~ D satisfies (f @f) A c = ADf and eDf = Ec . A co- algebra antihomomorphism g: C -+ D satisfies (g @ g) A c = (tw) ADg and cog = ~c, where tw is the twist map on D @ D. A bialgebra homomorphism is both an algebra and a coalgebra homomorphism. If H 1 and H 2 are left Hopf algebras with left antipodes S 1 and 8 3 , respectively, then a homomorphism of (H1, $1) to (H~, 83) is a bialgebra homomorphismf of H 1 to H 2 such thatfS 1 = SJ. If C is a coalgebra, then C* is the dual convolution algebra Homk(C, k). C is a left C*-module via f" c = ~f(q) q, where Ac = ~ q @ c 2 for c e C. Finally, we remark on notation. In general, we assume familiarity with the notation of [8], which can also serve as a general reference for the bialgebra ideas used here. Id will always denote the identity mapping of the appropriate object. M~(k) will denote the k-algebra of n-by-n matrices over k. Since M,~(k) is finite dimensional over k, its dual space Mn(k)* is a k-coalgebra. The basis (Xis I l ~ i, j ~ n} of M,~(k)*, which is dual to the basis of M,~(k) consisting of the usual matrix units, satisfies AXis = Y'.~=I Xik @ Xkt and e(Xit ) = ~is . All maps and tensor products are over k unless stated otherwise. We assume, unless otherwise stated, that k is a field. However, in Section 2 we give some modifications of the main Theorem 3, where k is a commutative ring with 1 (Proposition 6 through Corollary 10). 2. THE MAIN THEOREM We start with a short proof of the fundamental theorem of coalgebras whicl~ is relevant to the proof of part (2) of the main Theorem 3. LEMMA 1. Let C be a coalgebra. Let p: M ~ C @ M define a left C-comoduh structure on M. Let m ~ M. Then p(m) ~ F @ M for some finite-dimensional righl coidealF on C. Proof. Let {m¢} be a basis for M, with p(m)=Zci@m, and p(mi) = ~2t dit @ ms. Let F be the k-span of {ci}. Then (A @ i) p(m) = (I @ p) p(m implies Ac t = ~i ci @ dis ~F @ C, so F is a finite-dimensional right eoideal. THEOREM 2. (The Fundamental Theorem of Coalgebras). Let C be , coalgebra, c E C. Then c lies in a finite-dimensional subcoalgebra E of C. Proof. Let F be a finite-dimensional right coideal of C with dc ~F @ C Letfl .... ,f,, be a basis for F, Aft = ~ifi @ cis • Then coassociativity on the.] shows that Acit = ~k~ln ci~ @ %s , and the counit condition shows that e(cit) = LEFT HOPF ALGEBRAS 401 3ij. Thus E ~ ~i~,j=l kcij is a subcoalgebra, and the counit condition gives c • F C E. Note that E is a homomorphic image of M~(k)*. An n-by-n matrix {cij } with entries from a coalgebra C is called invariant if Acij = ~kn=l Cik @ Ckj and ~(%) = ~iJ . Then the above argument shows that every element of a coalgebra lies in the coefficient space of an invariant matrix, i.e., the span of the entries of an invariant matrix. Invariant matrices appear in Schur's dissertation, Berlin, 1901 (cf. [6]), where C is the coalgebra of polynomial functions on M~(k), k the complex numbers. THEOREM 3. Let B be a left Hopf algebra over a field k. Then B is a Hopf algebra over k if any one of the following conditions is satisfied: (1) B is finite dimensional over k; (2) B is a commutative algebra; (3) B is a pointed coalgebra; (4) The coradical B o of B is a cocommutative coalgebra. Proof. Let S denote a left antipode for B. (1) The convolution algebra Hom(B, B) is finite dimensional over k and thus has a faithful matrix representation. Thus S • Id =/zE implies Id • S =/z~. (2) Let b E B. By Theorem 2, b is in the coefficient space of an invariant matrix C = (cij). Let S(C)= (S(%)). Then (S. Id)(%)= tzE(ci~) says that S(C)C = I,~. Since B is commutative, this implies CS(C) = I,~, which means that (Id * S)(cij) = tze(ci~). Thus (Id * S(b) =/xe(b), since b is in the coefficient space of C. (3) Let G(B) denote the monoid of group-like elements of B. We claim that it is sufficient to show that G(B) is a group. For, since the restriction of Id to the coradical B 0 of B is then invertible in Hom(B0, B), it follows by Lemma 14 of [10] that Id is invertible in Hom(B, B). Another way to see this claim, which is in the spirit of the above proof of (2), is to note that when B is pointed as coalgebra, then any invariant matrix over B is equivalent to a triangular matrix C = (%). The elements cii• G(B), so if they are invertible, it follows easily by induction that C is invertible, As in (2), this implies that B is a Hopf algebra. To show G(B) is a group, let g • G. Then S(g)g = 1 (but we don't known yet that S(g) • G(B)). Set I = {b • B ] bg = e(b)l}. [ is the kernel of P: B --~ B given by P(b) = bg -- E(b)l. Let b • I. (P @ P) Ab = ~P(bl) @ P(be) = 2 (big -- e(bl)l) @ (beg -- e(be)l) = ~2 (big @ beg -- e(bl)l @ beg -- blg @ e(be)l -k e(bl)e(be)l ~) 1) = A(bg) -- 1 @ be -- bg @ 1 q- e(b)(1 @ 1) = e(b)(1 @ 1) -- e(b)(1 @ 1) -- e(b)(1 @ 1) + e(b)(1 @ 1) = 0. SoAICB @I+ I @ B. Now S(g) • I and since S(g)g = 1, we have e(S(g)) = l, so e(I) va 0. By Lemma 4 of [5], G(B) n I =/= ¢. Thus every element of G(B) has a left inverse in G(B), so G(B) is a group. Another proof of this is given in Proposition 4. 402 GREEN~ NICHOLS~ AND TAFT (4) Let ~ denote the algebraic closure of k. Letting (B,,} be the coradical filtration of B, it follows from Proposition 11.1.1 of [8] that (k @ B)0 _C/~ @ B0, since {/~ @ B~} is a filtration of/~ @ B. Thus/~ =/~ @ B is a pointed bialgebra over/~ by Lemma 8.0.1 of [8]. ~q = Id @ S is a left antipode for/~, so is an antipode by (3). It follows that S is an antipode for B. This completes the proof of Theorem 3. We give an alternate proof of the key point of part (3) of Theorem 3. PROPOSITION 4. Let B be a left Hopf algebra which is pointed as a coalgebra. Then G(B) is a group. Proof. By the splitting theorem of [7], there is a coalgebra projection ~r of/~ onto B 0 .
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