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I came to know from the paper Left Hopf Algebras by Green, Nichols and Taft that one may consider a Hopf algebra whose antipode satisfies only the left (resp. right) antipode condition.

To be more precise, let $\Bbbk$ be a field and $(B,\mu,\eta,\Delta,\varepsilon)$ a $\Bbbk$-bialgebra. We say that $B$ is a left Hopf algebra if there exists a linear endomorphism $S:B\to B$ such that $$S(b_1)b_2=\varepsilon(b)1$$ for every $b\in B$ (i.e. $S$ is a left convolution inverse of the identity morphism).

In Section 3 of Left Hopf Algebras an "artificial" (in my opinion) example of such an object is provided. In a "recent" paper, A Left Quantum Group, Rodriguez-Romo and Taft provided a new example, which I still consider a bit "artificial". Is anybody aware of some more "concrete" or "natural" examples of this construction?

This question is strictly related to this other question on MSE. Maybe that was not the right place where to ask. The only answer it received concerns the quotient $B$ of the free non-commutative k-algebra $k\{X,Y\}$ by the ideal generated by the relation $XY-1$. This is, however, not a one-sided Hopf algebra because the condition $S(x)x=1$ together with $xy=1$ would imply $S(x)=y$ and hence would force $B$ to be a Hopf algebra ($x$ denotes the class of $X$ in the quotient and similarly for $y$).

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Did you try $B=k\{\dots,x_{-1},x_1,x_2,\dots, x_n,\dots \}/(x_nx_{n+1}-1: n\in\mathbb Z)$ with $\Delta x_n=x_n\otimes x_n$ for all $n$?

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    $\begingroup$ Will not this quotient out all $x_n-x_{n+2}$ too? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 25, 2019 at 14:34
  • $\begingroup$ ops, you are right. It's ok because cocommutativity implies $S^2=Id$... I was thinking in Takeuchi's construction, given $C$ a coalgebra, define $C_n=C$ if $n$ odd and $C_n=C^{op}$ if $n$ is even. Then take $T(\oplus_n C_n)$ and $S$ the extension of the "identity" map going from $C_n\to C_{n+1}$. After that, one should divide by 2 relations: left antipode formula and right antipode formula. Probably the smallest example should be with the smallest non cocommutative coalgebra. $\endgroup$
    – Marco Farinati
    Commented May 25, 2019 at 16:06
  • $\begingroup$ @MarcoFarinati. Two things: 1) the example you mention in the answer has the same problem of your example on MSE. $S(x_n)x_n=1$ and $x_nx_{n+1}=1$ implies $S(x_n)=x_{n+1}$ and hence the outcome is a Hopf algebra. 2) the example you mention in the comment is exactly the one from the paper "Left Hopf Algebras" referenced in the question: as I already mentioned, it seems to me pretty artificial. I am looking for some genuine and "natural" example, such as quantum groups, physical one-sided symmetries.. $\endgroup$
    – Ender Wiggins
    Commented May 25, 2019 at 16:39
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    $\begingroup$ Could I get one of you to explain what is going on in the "Left Hopf Algebras" paper? At the top of page 409, the author introduce a free algebra on noncommutative variables $X_{ij}^r$ for $1 \leq i,j \leq n$ and $r \geq 0$, modulo relations $(\ast)$ and $(\ast \ast)$, and claim that these relations don't imply $(\ast \ast \ast)$. If we write $Y_r$ for the matrix with entries $X_{ij}$, then $(\ast)$ is $Y_{r+1}^T Y_r = \text{Id}$ for $r \geq 0$, $(\ast \ast)$ is $Y_r Y_{r+1}^T = \text{Id}$ for $r \geq 1$ and $(\ast \ast \ast)$ is $Y_0 Y_1^T = \text{Id}$ (continued). $\endgroup$
    – David E Speyer
    Commented Oct 2, 2023 at 14:41
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    $\begingroup$ But, taking the transpose of $(\ast \ast)$, we have $Y_2 Y_1^T = \text{Id}$, and thus $Y_0 = Y_2 Y_1^T Y_0 = Y_2$ (second equality is $(\ast)$). So $Y_0 Y_1^T = Y_2 Y_1^T = \text{Id}$ (second equality is $(\ast \ast)$), so we have deduced $(\ast \ast \ast)$ from the relations $(\ast)$ and $(\ast \ast)$ after all. I assume I am missing something basic; what? $\endgroup$
    – David E Speyer
    Commented Oct 2, 2023 at 14:45

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