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Let $(H, m, u, \Delta, \epsilon, S)$ be a $K$-Hopf Algebra. We call an element $x\in H$ grouplike if $\Delta(x) = x \otimes x$ and $\epsilon(x) = 1_K$. The set of all grouplike elements is a group (denoted $G(H)$).

How do we show this?

I thought I might have the group $(G(H), \mu, e, S')$ with $\ \mu : G(H) \times G(H) \rightarrow G(H) : (x,y) \rightarrow m(x\otimes y)\\ e: \{*\} \rightarrow G(H):* \rightarrow u(1_K) \\ S':G(H) \rightarrow G(H):x \rightarrow S(x)$

I can show that this structure gives us a group. But what I can't show is that the images of $\mu$ and $S'$ are indeed in $G(H)$.

We have to show that

1) $∆(m(x\otimes y)) = m(x\otimes y) \otimes m(x\otimes y)$ (EDIT: Solved)

2) $\epsilon(m(x\otimes y)) = 1_K$ (EDIT: Solved)

3) $∆(S(x)) = S(x) \otimes S(x)$

4) $\epsilon(S(x)) = 1_K$

Any hints would be appreciated

  • 1
    For 1) is is probably more obvious what is going on if you write $m(x\otimes y) = xy$ and remember that $\Delta$ is a homomorphism of algebras. – 2017-01-17
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    Ok. 1) and 2) are actually easily verified – 2017-01-17
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    What is true is that **the set** of grouplike elements is a group. Please correct your title, so that it makes sense. – 2017-01-18

2 Answers 2

3

Let $g$ and $h$ be grouplike elements. Then $\Delta(gh)=\Delta(g)\Delta(h)=(g\otimes g)(h\otimes h)=(gh\otimes gh)$. Indeed the multiplication on $H\otimes H$ uses the standard flip map $\tau$, thus $m_{H\otimes H}=(m\otimes m)\circ (1\otimes \tau\otimes 1)$. Thus the multiplication of grouplike elements is inner.

Now since $H$ is a Hopf algebra, the antipode axiom tells us that $S(g)g=\varepsilon(g)1$. Since $\varepsilon(g)=1$ (because $g$ is grouplike), this becomes $S(g)g=1$ and thus $S(g)$ is a left inverse for $g$. Similarly, $S(g)$ is a right inverse for $g$.

Now we show that $S(g)\in G(H)$. By applying $\Delta$ to $S(g)g=1$, we get that $$\Delta(S(g))(g\otimes g)=1\otimes 1.$$ Hence $\Delta(S(g))$ is the inverse of $g\otimes g$ in the algebra $H\otimes H$. By uniqueness of inverses, $\Delta(S(g))=S(g)\otimes S(g)$ as desired.

Questions 2 and 4 follows immediately from the fact that the counit applied to grouplike elements is $1$.

  • 0
    Ok. 1) and 2) are actually easily verified – 2017-01-17
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    I don't understand the final conclusion. Since $S(g)$ is the inverse of g, it follows... – 2017-01-17
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    I'm going to edit it, it's not complete. – 2017-01-17
  • 0
    Ok. So we get $∆(S(g)) = ((S\otimes S) \circ ∆)(g) = (S\otimes S)(g\otimes g) = S(g) \otimes S(g)$. And for 4) $(\epsilon \circ S)(g) = \epsilon(g) = 1$ (since S is a anti-algebra morphism. PS: by anti-algebra morphism you mean coalgebra morphism? PPS: I didn't know that S was a coalgebra morphism. I'll try to show that. – 2017-01-17
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    I edited it again, this is an easy argument. You can also use that $S$ is a bialgebra morphism from $H$ to $H^{\text{op cop}}$, but I believe the above argument is more streamlined. – 2017-01-17
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    Also, 4) follows since $\varepsilon(x)=1$ whenever $x$ is grouplike. Hence $\varepsilon(S(g))=1$ since by 3) $S(g)$ is also grouplike. – 2017-01-17
  • 0
    I like that more. I have not yet seen anything like $H^{op\ cop}$ yet so this new argument works better. However, I don't agree with your argument for 4) since 4) is a property required for "grouplikeness" so we don't know that S(g) is grouplike until we have proven 4). – 2017-01-17
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    Ok I got it forget my comment about 4) Everything works :-) Thanks – 2017-01-17
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    @Mathematician 42: you are handling the case of the antipode (i.e. $\Delta(S(g)) = S(g) \otimes S(g)$) without using that $S$ is a coalgebra antimorphism. Nice manoeuvre! +1 – 2017-01-18
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    @KonKan: Yeah, I know that most textbooks (Like Kassel for example) do use that $S$ is a coalgebra antimorphism. But this seems more natural to me. – 2017-01-18
2

First, note that the set of grouplikes $G(H)$ is closed wrt the multiplication of $Η$. This is a consequence of the fact that the comultiplication is an algebra homomorphism: $$ \Delta(gh) = \Delta(g) \Delta(h) = (g \otimes g)(h \otimes h) = gh \otimes gh $$ i.e. $g,h \in G(H)$ $\Rightarrow$ $gh \in G(H)$.

On the other hand, $1_{Η} \in G(H)$, since in any Hopf algebra we have: $\Delta(1_{Η}) = 1_{Η} \otimes 1_{Η}$.

Next, using that $S$ is a coalgebra antimorphism, i.e. if $\Delta(h)=\sum h_{(1)}\otimes h_{(2)}$ then $\Delta(S(h))=\sum S(h_{(2)})\otimes S(h_{(1)})$, implies that (for any $g \in G(Η)$): $$ \Delta(S(g)) = S(g) \otimes S(g) $$ thus: $g \in G(H)$ $\Rightarrow$ $S(g) \in G(H)$.

Finally, by the definition of the antipode, we get (for all $g \in G(Η)$): $$ S(g)g = gS(g) = 1_{H} $$ thus: for all $g \in G(Η)$ we have $S(g) = g^{-1}$ completing the proof.

P.S.1: If $Η$ is only a bialgebra then the above imply that $G(H)$ is a monoid. The presence of the antipode differentiates the situation, and makes $G(H)$ a group.

P.S.2: The above proof of the relation $\Delta(S(g)) = S(g) \otimes S(g)$, actually implements user's Mathematician_42 comment (posted in his answer above) that $S$ is a bialgebra morphism from $H$ to $H^{op\ cop}$ or equivalently an algebra antihomomorphism and a coalgebra antimorphism.