I’m reading about affine group schemes by Waterhouse and in the proof of showing the (Jordan) decomposition of Abelian affine group scheme (equivalently cocommutative Hopf algebra), I came across the argument, saying:

“Since the Hopf algebra $A$ is cocommutative, the antipode $S$ is a coalgebra morphism and therefore sends $ A’ $ (a sub-bialgebra of $A$) into $A’$“

I know that the antipode is an algebra automorphism of $A$, and in this case a bialgebra automorphism of A, but I still don’t understand why does the restriction of $S$ to $A’$ would send $A’$ back to itself.

P/S: we are assuming that Hopf Algebras are commutative here.

  • $\begingroup$ It is not per se true that a sub-bialgebra of a commutative cocommutative Hopf algebra must be a sub-Hopf algebra. After all, submonoids of abelian groups aren't always subgroups. Maybe the context says something helpful here; perhaps finiteness conditions? $\endgroup$ – darij grinberg Sep 30 '17 at 15:48
  • $\begingroup$ Yes, I should mention that $A’$ is a directed union of finite dimensional subcoalgebra of $A$. It was later shown that by construction $A’$ preserves multiplication and contains the unit - which makes it a bi-subalgebra. $\endgroup$ – AffineSpace Sep 30 '17 at 15:53
  • 1
    $\begingroup$ That's not helpful. Try the "positive part" of the group ring of $\mathbb Z $; it satisfies all your conditions. $\endgroup$ – darij grinberg Sep 30 '17 at 16:00
  • $\begingroup$ There is a given coalgebra projection $p : A \rightarrow A’$ that was also shown to be an algebra projection (and thus a bialgebra projection too). The only reasoning that I can now give is that for $a’ \in A’$, we have $p(S(a’)) = S(p(a’)) = S(a’)$, which shows that $S(A’) \subseteq A’$. But I still don’t see the logical connection that the author was trying to claim in the line he wrote. $\endgroup$ – AffineSpace Sep 30 '17 at 16:26
  • $\begingroup$ but isn't that a valid proof, given this projection $p$ ? $\endgroup$ – darij grinberg Oct 1 '17 at 1:09

I do not have access to the book you are citing, but if i understand correctly, you are speaking about a commutative, cocommutative hopf algebra.
If your assumptions are complemented by: finite dimensional hopf algebra over an algebraically closed field of characteristic zero, then the author's claim:

Since the Hopf algebra $A$ is cocommutative, the antipode $S$ is a coalgebra morphism and therefore sends $ A’ $ (a sub-bialgebra of $A$) into $A’$

seems to be correct.
Here is my argument:

Under these assumptions (i.e.: f.d., commutative, cocommutative over alg. closed field $k$ with $chark=0$), it is relatively easy to show that the hopf algebra $H$ is actually isomorphic to the group algebra of the group $G(H)$ formed by its grouplike elements: $$ H\cong kG(H) $$ (This isomorphism can also be viewed as a direct consequence of the Cartier-Konstant-Milnor-Moore theorem applied in the finite dimensional setting.)
Thus, by the same argument, the sub-bialgebras $A'$ of $H$ -which are again f.d., commutative, cocommutative, over an algebr. closed field of $chark=0$- are group algebras $A'=kN$, for some subgroup $N$ of $G(H)$. But then, since for any grouplike $S(g)=g^{-1}$ we get that:
$$ S(\sum_{g\in N}a_g g)=\sum_{g\in N}a_g S(g)= \sum_{g\in N}a_g g^{-1}\in kN $$ thus $S(kN)\subseteq kN \ \ $ i.e.: $ \ S(A')\subseteq A'$.

  • $\begingroup$ Why are they group algebras $kN$? $\endgroup$ – darij grinberg Oct 1 '17 at 0:05
  • 1
    $\begingroup$ By the same argument as in the initial case of $H$: The sub-bialgebras are again f.d., commutative, cocommutative hopf algebras and thus they are group algebras of some (sub)group. $\endgroup$ – KonKan Oct 1 '17 at 0:08
  • $\begingroup$ Why are they Hopf? $\endgroup$ – darij grinberg Oct 1 '17 at 0:08
  • 1
    $\begingroup$ because $S(g)=g^{-1}$ for all grouplikes. $\endgroup$ – KonKan Oct 1 '17 at 0:11
  • $\begingroup$ @darij grinberg, i've edited hoping to become more clear. $\endgroup$ – KonKan Oct 1 '17 at 0:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.