Intuition for Coproduct of Grouplike vs Primitive elements in a Coalgebra?

Question

I'm trying to understand Hopf Algebras as a physicist with a limited background in abstract algebra, and I might be in a little over my head.

In particular I'm trying to wrap my head around the fact that the coproduct in a group algebra $k[G]$ is given by

$$\Delta(g) = g \otimes g$$

But the coproduct in e.g. a tensor algebra $\otimes V$ is given by

$$\Delta(x) = x \boxtimes 1 + 1 \boxtimes x$$

My understanding is that in general elements of a coalgebra are either "grouplike" or "primitive" depending on whether they act like the first or second example. And I am (in theory) able to show that these two definitions are correct based on the axioms for the coproduct.

But I'm having trouble wrapping my head around conceptually why these things are defined like this. Why shouldn't the group coproduct be $e_g \otimes g + g \otimes e_g$? Or is that that would define a coalgebra, but the structure just isn't compatible with the group algebra such that you can construct a bialgebra / Hopf algebra with it?

Answer 1

This isn't written with the precision or the comprehensiveness of a QY, but hopefully helps with intuition (which is my goal here). It is written from the perspective of quantum groups (and when I talk about Hopf algebras that are not quantum groups I am liable to stray in accuracy).

In quantum groups, a Hopf algebra $A$ can be thought of as an algebra of functions on an object that has a multiplication, identity, and an inverse map.

For example, let $G$ be a finite group, and $C(G)$ the algebra of complex-valued continuous functions (in the finite case all functions $G\to\mathbb{C}$ are continuous). The group $G$ has a multiplication $m:G\times G\to G$, $(g,h)\to g\cdot h$, an identity $e\in G$, and an inverse $g\mapsto g^{-1}$.

We want the comultiplication to be the transpose of the multiplication. That is:

$$\Delta(f)(g,h)=f(m(g,h))=f(g\cdot h)\qquad (g,h\in G,\,f\in C(G)).$$

This means that $f\mapsto \Delta f$ is going from a function on $G$ to a function on $G\times G$.

For multiplicative and linear:

For compact Hausdorff spaces $X$, $Y$, for every continuous function $\phi:Y\rightarrow X$, the map $\pi:C(X)\rightarrow C(Y)$: $$\pi(f)=f\circ \phi\qquad (f\in C(X)).$$ is a unital $*$-homomorphism, in particular linear and multiplicative.

So the transpose of the continuous $m:G\times G\to G$ is a $*$-homomorphism $C(G)\mapsto C(G\times G)$. Now we envoke the isomorphism:

$$C(G\times G)\cong C(G)\otimes C(G) \qquad(\text{ algebraic tensor product}).$$

This gives a map $\Delta:C(G)\to C(G)\otimes C(G)$.

Now that we have linearity we can consider $\Delta$ on the basis of delta functions, $\delta_g\in C(G)$. It shouldn't take you long to show that:

$$\Delta(\delta_g)=\sum_{t\in G}\delta_{gt^{-1}}\otimes \delta_t,$$

and e.g. coassociativity of $\Delta$ 'up' in $C(G)$ is inherited from associativity 'down' in $G$.

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This is the one I don't know so much:

Consider the additive group $(\mathbb{C}^N,\cdot)$: $$(a_1,a_2,\dots,a_N)\cdot (b_1,b_2,\dots,b_N)=(a_1+b_1,a_2+b_2,\dots,a_N+b_N)\qquad (a_i,b_i\in\mathbb{C}).$$ Consider the algebra $\mathcal{O}(\mathbb{C}^N)$ of polynomial functions $\mathbb{C}^N\to \mathbb{C}$. This is generated by the coordinate functions $x_i\in\mathcal{O}(\mathbb{C}^N)$:

$$x_i(a_1,a_2,\dots,a_N)=a_i.$$

We want the comultiplication $\Delta$ to be the transpose of the multiplication $\cdot$ so that:

$$\Delta(x_i)(a,b)=x_i(a\cdot b)=a_i+b_i.$$

Being very fast and loose here (!!) we can see where the formula $\Delta(x_i)=x_i\otimes 1+1\otimes x_i$ comes from:

$$\Delta(x_i)(a\otimes b)=(x_i\otimes 1+1\otimes x_i)(a\otimes b)=x_i(a)\otimes 1(b)+1(a)\otimes x_i(b)\cong a_i+b_i.$$

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With the group algebra... let's play a little fast and loose here and consider the group algebra $\mathbb{C}G$ as an algebra of functions on the algebra of functions $C(G)$. I guess that isn't so bad... but the multiplication we are transposing is not the multiplication in the group... but in the algebra of functions... and this is just the pointwise multiplication (this can make more sense because the pointwise multiplication is a multiplication $C(G)\otimes C(G)\to C(G)$). $$m(u,v)(g)=u(g)\,v(g)\qquad (u,v\in C(G),\,g\in G).$$

We are viewing $\mathbb{C}G$ as functions on $C(G)$ by duality, by the evaluation:

$$g(f):=f(g)\qquad (f\in C(G),\,g\in \mathbb{C}G),$$

by seeing $g\in \mathbb{C}G$ also as an element of $G$.

So the comultliplication here should be:

$$\Delta(g)(u\otimes v)=g(m(u\otimes v))=g(uv) =uv(g)=u(g)v(g)\qquad (g\in \mathbb{C}G,\,u,v\in C(G)).$$ And we can check this against:

$$\Delta(g)(u\otimes v)=(g\otimes g)(u\otimes v)=(g(u)\otimes g(v))=u(g)\otimes v(g)\cong u(g)v(g).$$