5
$\begingroup$

I've been working through Federico Ardila's online Hopf algebra lectures and hoped to check my understanding so far by constructing the Hopf algebra of a very small group ring from scratch. But I've failed, which has left me in a state of perplexity.

I chose $G=\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{k}=\mathbb{Z}/3\mathbb{Z}$. So my group ring $\mathbb{k}G$ is the set of all formal $\mathbb{k}$-linear combinations of the two elements of $G$ (I'll call them $e$ and $a$ to avoid confusion with the coefficients, which I'll write as 0, 1, 2).

On top of this structure we have:

  • Multiplication as defined here.
  • Comultiplication as $r\mapsto r\otimes r$, for each $r\in \mathbb{k}G$
  • The unit as $g\mapsto 1_\mathbb{k}$ for $g\in G$, extended linearly (so, for example, it maps $2e + a\mapsto 2 + 1 = 3 = 0 \mod 3$)
  • The counit as $\lambda \mapsto \lambda e$ for all $\lambda\in \mathbb{k}$
  • The antipode as $g\mapsto g^{-1}$, extended linearly. But in this case, that's just the identity.

With these definitions, I hoped this diagram would commute:

Hopf algebra diagram

but it blinking doesn't. Example:

  • Along the top path: $(e + a)\to (e + a)\otimes (e + a)\to (e + a)\otimes (e + a)\to (2e + 2a)$
  • Through the middle: $(e + a)\to 2\to (2e)$.

And it's obvious why: products in the group ring don't always annihilate the $a$ element, but the counit-unit sequence does. So this could never work.

Hopefully I've explained what I've done in enough detail that whatever fundamental thing I've misunderstood isn't elided in the process. I'm hoping someone will be kind and patient enough to unpick it and point to the thing I've stupidly misunderstood!

$\endgroup$
  • 2
    $\begingroup$ I don't believe your definition of comultiplication is correct. It's $r\mapsto r\otimes r$ for $r\in G$, not $r\in \mathbb{k}G$. $\endgroup$ – Matt Samuel Sep 7 '15 at 17:26
  • $\begingroup$ ...I see now that's in the posted answer. $\endgroup$ – Matt Samuel Sep 7 '15 at 17:27
  • $\begingroup$ Thank you -- and yes, that solves the problem. I feel slightly daft now, of course... $\endgroup$ – helveticat Sep 7 '15 at 17:41
6
$\begingroup$

The definition of comultiplication you have isn't quite right. $\Delta(g)=g\otimes g$ is true only for group elements, not every element of the algebra (that wouldn't even be a linear map!). Thus $\Delta (e+a) = \Delta(e) + \Delta(a) = e\otimes e + a \otimes a$, this goes to $e\otimes e +a\otimes a$ again, then to $e + a^2 = 2e$ agreeing with the path across the middle.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! I knew it would be something simple... I especially appreciate the quickness of your reply, this has been bothering me on and off all day and now I'll sleep soundly... $\endgroup$ – helveticat Sep 7 '15 at 17:40
  • 1
    $\begingroup$ You're welcome. The definition of $\Delta$ is a fine advert for the use of the Bourbaki dangerous bend symbol $\endgroup$ – Matthew Towers Sep 7 '15 at 17:42

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.