By Duchamp Gérard H. E.


2015-04-04 16:35:53 8 Comments

It is known from the general theory that, given a bialgebra (over a field $k$) \begin{equation} \mathcal{B}=(B,\mu,1_B,\Delta,\epsilon) \end{equation} the Sweedler's dual $\mathcal{B}^0$ (called also Hopf or restricted dual, i.e. the space of linear functionals $f\in B^*$ such that $f\circ\mu\in \mathcal{B}^*\otimes_k \mathcal{B}^*$) is a bialgebra (with the transposed elements). It seems that it can happen that $\mathcal{B}$ admit no antipode whereas $\mathcal{B}^0$ does (and then be a Hopf algebra). My evidence (however neither an example nor a counterexample) is the following :

Let $k$ be a field and $q\in k^\times$. On the usual algebra of polynomials, $(k[x],.,1)$, let us define a structure of bialgebra by $$ \Delta(x)=x\otimes 1+ 1\otimes x + q x\otimes x $$ one checks easily that $\mathcal{B}=(k[x],.,1,\Delta,\epsilon)$ ($\epsilon(P)=P(0)$ as usual) is a bialgebra and, as $g=(1+qx)$ (which is group-like) has no inverse, $\mathcal{B}$ is a bialgebra without antipode. However all the elements of $\mathcal{B}$ are dualizable (indeed, the dual of $\Delta$ is the infiltration product of Chen, Fox and Lyndon, Free differential calculus) within $k[x]$ and the bialgebra so obtained $\mathcal{B}^\vee=(k[x],\mu_\Delta,1,\Delta_{conc},\epsilon)$ admits an antipode.

Remarks i) (Foissy). The Hopf algebra $\mathcal{B}^\vee$ can be considered as a subbialgebra of $\mathcal{B}^0$ (the full Sweedler's dual) which is NOT a Hopf algebra. To see this, remark that the characters of $\mathcal{B}$ are in $\mathcal{B}^0$ and the values of a character $\chi$ are fixed by $\chi(x)$. Call $F_a$ be the character s.t. $F_a(x)=a$. Convolution of characters satisfies $F_a*F_b(x)=a+b+qab$. Now, for all $b\in k$, we have $F_{-1/q}*F_b=F_{-1/q}$ which proves that the set of characters (i.e. group-like elements of $\mathcal{B}^0$) is NOT a group under convolution.

ii) The comultiplication above was introduced, for $q=1$, by Chen, Fox, and Lyndon, Free differential calculus IV, Ann. Math. 1958. It can be defined on a noncommutative alphabet $X$ (i.e. to complete as a bialgebra a free algebra $A\langle X\rangle$) by $$ \Delta_{\uparrow}(x):=x\otimes 1+ 1\otimes x + q x\otimes x $$ forall $x\in X$. remark that it is compatible with all types of commutation between the letters.

The comultiplication of such an infiltration bialgebra $(A\langle X\rangle,conc,1_{X^*},\Delta_{\uparrow},\epsilon)$ has a pretty combinatorial expression. It reads, on a word $w\in X^*$, $$ \Delta_{\uparrow}(w)=\sum_{I\cup J=[1..n]}\,q^{|I\cap J|}w[I]\otimes w[J] $$ where $n=|w|$ is the length of the word.

My questions are the following (bwa=bialgebra without antipode, ha=Hopf algebra)

Q1) Are there significant families of examples of the situation ($\mathcal{B}$ bwa and $\mathcal{B}^0$ ha) known ? (combinatorial, easy to check examples are preferred)

.

Q2) Are there general statements ? ($\mathcal{B}$ bwa +some condition $\Longrightarrow$ $\mathcal{B}^0$ ha)

1 comments

@Ender Wiggins 2018-10-22 10:11:21

Actually your guess is true. Consider the left Hopf algebra $\widetilde{SL_q(2)}$ from Iyer, Taft, The dual of a certain left quantum group. This is constructed essentially as $SL_q(2)$, but imposing just half of the relations. Namely, if $SL_q(2)$ is given by $\Bbbk\langle X_{i,j}\mid i,j=1,2\rangle$ with relations \begin{gather} X_{21}X_{11}=qX_{11}X_{21}, \tag{$\star$}\\ X_{12}X_{11}=qX_{11}X_{12}, \\ X_{22}X_{12}=qX_{12}X_{22}, \tag{$*$}\\ X_{22}X_{21}=qX_{21}X_{22}, \\ X_{22}X_{11}=qX_{12}X_{21}+1, \tag{$\circ$} \\ X_{21}X_{12}=qX_{11}X_{22}-q\cdot 1, \tag{$\bullet$} \\ X_{21}X_{12}=X_{12}X_{21}, \\ \end{gather} then $\widetilde{SL_q(2)}$ is obtained by just requiring $(\star),(*),(\circ)$ and $(\bullet)$. It turns out that $\widetilde{SL_q(2)}$ admits a left convolution inverse of the identity which is not a right convolution inverse (hence it is not an antipode). However, its Sweedler dual coincides with $SL_q(2)^\circ$, whence it is Hopf.

@Duchamp Gérard H. E. 2018-10-22 21:14:50

Very interesting (+1), thanks. Let me some time to digest it.

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