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It's all in the question! What is an example of a sub-bialgebra of a Hopf algebra that is not a Hopf subalgebra?

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    $\begingroup$ Perhaps a group algebra $k[G]$ along with a choice of submonoid $M$ of $G$ gives you what you want? $\endgroup$
    – Jim
    Commented Jan 14, 2021 at 15:12
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    $\begingroup$ Indeed, if $H$ is the group algebra of the free group on two generators $x,y$, then the sub-algebra generated by $x,y$ is a subbialgebra but not sub-Hopf. $\endgroup$
    – Adrien
    Commented Jan 14, 2021 at 15:47

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The Hopf algebra $H = k[x^{\pm 1}]$ with $\Delta(x) = x \otimes x$ is the group algebra of $\mathbb{Z}$, the free group on one generator. Its subalgebra $k[x]$ is the "monoid algebra" for the submonoid $\mathbb{N} \subset \mathbb{Z}$, the free monoid on one generator. $k[x] \subset k[x^{\pm 1}]$ is not a Hopf subalgebra just because $\mathbb{N}$ is not a group.

(Note that $k[x]$ does have a natural Hopf structure with $\Delta(x) = x\otimes 1 + 1 \otimes x$, but it is not the one that extends along $k[x] \subset k[x^{\pm1}]$.)

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    $\begingroup$ Related to @Jim's comment. $\endgroup$
    – LSpice
    Commented Jan 14, 2021 at 17:23
  • $\begingroup$ @LSpice Yes, absolutely. Jim could have posted an answer, and I suspect it would have been accepted (although I cannot speak for OP). I'm a little tired of writing comments of the form "Dear so and so, your comment answers the question, please post it as an answer". Questions that are not "answered" get occasionally bumped to the top of the homepage by the MO daemon, and also writing an "answer" allows for a little more explanation/education. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 15, 2021 at 14:22
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    $\begingroup$ I wasn't complaining or criticising, just trying to reflect the answer and comment history. If you know a more polite way to post such informational comments, then I would be happy to do it. $\endgroup$
    – LSpice
    Commented Jan 15, 2021 at 16:30
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    $\begingroup$ Sorry, I mistook your tone of voice. I think you were being perfectly polite. I also have, I'm sure, left comments on MO that I had meant without complaint or criticism, but that could be read that way depending on the reader's mindset. I should have just written "yes, absolutely, thank you for pointing it out" or something along those lines. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 16, 2021 at 16:19

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