Let $(H,\Delta,\epsilon,S)$ be a Hopf algebra. Can there exist an algebra map $\phi:H \to H$ such that $$ \epsilon(\phi(g)) \neq \epsilon(g), ~~~~~ \textrm{ for some } g \in H? $$ Does the anti-pode really have anything to do with this, or is it truly a question about bialgebras?
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asked May 10, 2023 at 14:04
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$\begingroup$ I am confused, the antipode $S$ does not even occur in your equation. In any case, as stated the answers seems to be "yes they exist" for trivial reasons. E.g. let $G$ be a finite group, $H=k^G$ the set of function $f:G\to k$. Then $\epsilon(f)=f(1)$. Now it is not hard to find examples for $G$ (and the resulting $H$) and $\phi$ that have $\epsilon(\phi(g))\neq\epsilon(g)$. $\endgroup$– Max HornCommented Sep 18, 2023 at 15:55
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2 Answers
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Such a map can certainly exist. For instance, take the $k$-algebra $G = k \times k$, with $$ \begin{aligned} &\Delta(1,0) = (1,0) \otimes (1,0) + (0,1) \otimes (0,1), \\ &\Delta(0,1) = (1,0) \otimes (0,1) + (0,1) \otimes (1,0), \\ &S = \operatorname{id}, \\ &\epsilon(1,0) = 1, \\ &\epsilon(0,1) = 0. \end{aligned} $$ (This is the Hopf algebra associated with the finite group of order $2$.)
Then the map $\phi \colon G \to G \colon (a,b) \mapsto (b,a)$ is an algebra morphism, but it does not preserve the counit.
answered May 10, 2023 at 14:35
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$\begingroup$ Sorry but I don't see how this is isomorphic to the group Hopf algebra of $\mathbb{Z}_2$. $\endgroup$ Commented May 10, 2023 at 17:03
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$\begingroup$ @LorenzoDelVecchiopontopolos It's not isomorphic to the group Hopf algebra of $\mathbb{Z}_2$. It's the Hopf algebra of functions from $\mathbb{Z}_2$ to $k$. (This is the second example on en.wikipedia.org/wiki/Hopf_algebra#Examples.) $\endgroup$ Commented May 10, 2023 at 18:24
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$\begingroup$ Thanks for the reference. But I thought that for prime order, all finite-dimensional Hopf algebras were isomorphic . . . $\endgroup$ Commented May 10, 2023 at 20:17
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1$\begingroup$ @LorenzoDelVecchiopontopolos That depends on the field. For instance, the specific example I gave is indeed isomorphic to the group algebra $k\mathbb{Z}_2$ if $\operatorname{char}(k) \neq 2$, but not if $\operatorname{char}(k) = 2$. In general, the group algebra $k\mathbb{Z}_n$ is isomorphic to the Hopf algebra of functions from $\mathbb{Z}_n$ to $k$ if and only if the polynomial $f(x) = x^n - 1$ has $n$ different roots in $k$ (in other words, if it splits over $k$ and has no multiple roots). $\endgroup$ Commented May 11, 2023 at 7:33
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Let $H$ be the Hopf algebra of functions on an algebraic group $G$. The map $\phi$ defines a map of algebraic varieties $\hat{\phi}:G\rightarrow G$. The counit condition you try to impose is equivalent to $\hat{\phi} (1_G)=1_G$. There is no reason for arbitrary $\hat{\phi}$ to take $1$ to $1$.
answered Sep 18, 2023 at 10:16