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
    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Jan 17 '17 at 9:59
  • $\begingroup$ Ok. 1) and 2) are actually easily verified $\endgroup$ – tomak Jan 17 '17 at 10:18
  • $\begingroup$ What is true is that the set of grouplike elements is a group. Please correct your title, so that it makes sense. $\endgroup$ – Mariano Suárez-Álvarez Jan 18 '17 at 3:06

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 $1$.

  • $\begingroup$ Ok. 1) and 2) are actually easily verified $\endgroup$ – tomak Jan 17 '17 at 10:18
  • $\begingroup$ I don't understand the final conclusion. Since $S(g)$ is the inverse of g, it follows... $\endgroup$ – tomak Jan 17 '17 at 10:54
  • $\begingroup$ I'm going to edit it, it's not complete. $\endgroup$ – Mathematician 42 Jan 17 '17 at 11:04
  • $\begingroup$ 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. $\endgroup$ – tomak Jan 17 '17 at 11:34
  • 1
    $\begingroup$ 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. $\endgroup$ – Mathematician 42 Jan 17 '17 at 14:24

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.


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.