# HW 2 #1

Does anyone know how to figure out what $\Delta(1)$ or $\Delta(xg)$ equals?  I’ve tried playing around with the map $(\text{id}\otimes\epsilon)\circ\Delta\circ\mu(g\otimes g)=g\otimes g$.  This tells me that $\Delta(1)$ should equal something that maps to $g\otimes g$ under $\text{id}\otimes\epsilon$, (or $\epsilon\otimes\text{id}$), but I can’t think of such an element, since $\epsilon$ sends $g$ to 1.  Any advice would be appreciated.  Thanks.

1. $\Delta$ is supposed to be an algebra homomorphism, and algebra homomorphisms are supposed to map the unity to the unity.

2. Also, as a homomorphism it should preserve multiplication so $\delta (xg)$ should equal $\delta x \delta g = (g\otimes g)(x\otimes 1+g\otimes x) = gx\otimes g + 1\otimes gx$.

3. $\delta$ is not assumed to be a alg. morphism. It is given that it is linear, but not an alg. morphism. In the definition of bialgebra, we need to show either $\delta and \epsilon$ are Algebra morphisms or ${m} and {u}$ are colagebra morphisms. So for you to assume that we already have algebra morphisms, then we’d be done from the get go. I am stuck in the same place that Zach is in. Either I assume we have Alg. morphisms to define where we map 1 and gx, or we need to show where 1 and gx get mapped to and then show the maps are Algebra morphisms. FML

4. Well, the problem isn’t posed really precisely. If $\Delta$ and $\latex \epsilon$ were not supposed to be algebra morphisms, then they wouldn’t be uniquely defined. If they are (and this is what I believe), then the problem is not to prove that it’s a bialgebra (this is obvious), but to prove that $\Delta$ and $\latex \epsilon$ are well-defined at all (I mean, if you would replace $\Delta = g\otimes g$ by $\Delta = g\otimes g + 1\otimes 1$, then there wouldn’t exist such an algebra morphism $\Delta$).

• Yea Im having some trouble with what exactly the question wants me to do with the given information. I feel that we need to show that $\Delta , \epsilon$ are Alg. Morphisms by showing multiplication and identities are preserved for all elements of ${H_{4}}$, but in order to do that I need to know where 1 and gx are mapped to. Am I completely off?

5. I was somewhat wrong: the bialgebra is not obvious even after the existence of $\Delta$ and $\epsilon$ is ensured. In fact, you have to prove two things:

1) that there exist algebra morphisms $\Delta$ and $\epsilon$ satisfying $\Delta (g) = g \otimes g$, $\Delta (x) = x \otimes 1 + g \otimes x$, $\epsilon (g) = 1$ and $\epsilon (x) = 0$;

2) that, endowed with these morphisms, $H_4$ becomes a bialgebra. (The only thing you have to check here are the coalgebra axioms.)

6. Ok everybody calm down. I just assume we have a morphism (homomorphism). The kind we all like and let
$\Delta(gx) = \Delta(g)\Delta(x)$ and $\Delta(ag + a x) = \Delta(ag) + \Delta(ax)$
Then just work things out from there and you will see that we do indeed have a Coalgebra. Ask if you would like a better explanation. Crista 🙂

7. Brian Cruz

Hey all,

I might be rewording what some of you have already said, but I was having some problems with this at first, so here’s what I was thinking.

Definitely if $\Delta(gx)\neq\Delta (g) \Delta (x)$ then there is no way this could be a bialgebra. So, since we’re the ones defining $\Delta$ we should define it in a way to make it a homomorphism, thus setting $\Delta(gx)=\Delta (g) \Delta (x)$ and $\Delta(1)=1$. We still do need to check to see that it is a homomorphism by checking with two arbitrary elements from $H_4$ (a+bg+cx+dgx, etc.). That is to say, the values given for $\Delta g$ and $\Delta x$ may not be compatible with the notion of $\Delta$ being a homomorphism when we use them to define $\Delta gx$ and $\Delta 1$ and then extend linearly. Heck, I found it helpful to see that $\Delta (xx)=\Delta x \Delta x$ is actually true given this definition.

It’s kinda like when you’re looking for a linear transformation it’s helpful to define it on a basis, which is what we’re doing when we’re extending a function linearly. Now that we’re looking for a multiplication-preserving map, we only need to initially define our values on a few of the basis elements, since the rest (of the values on the other basis elements) can come from our supposed homomorphism. Unlike with plain linear transformations, however, we cannot choose just any values (a trait of vector spaces that makes them so nice), so instead we need to check that the values with which we start actually define values on the rest of the basis elements that in turn extend linearly to a bonafide homomorphism.

So really, I just read that part of the problem as, “does there exist some homomorphism such that $\Delta g=g\otimes g$ and $\Delta x=...$?”

• eloquently said; thank you, brian.

8. Karen Walters

I want to clarify what we talked about in office hours about HW2 #1. What I heard was that it was necessary to show the all 4 diagrams in Sweedlers book commuted with all basis in H4. That means that for two of the diagrams we need to check each possible pair of basis elements commute around the diagrams. That is for 2 of the diagrams we have 16 pairs of elements to work around for each. Did I hear that correctly?