I don't know why should we need coalgebra?What is the motivation?By changing all the arrows of algebra structure, it seems strange.What is the application of coalgebra? What is the relation with Lie algebra, representation theory, derived category? I am very confused about this definition.

Thank you.

  • $\begingroup$ There are useful examples. For instance, group algebras are coalgebras as well as algebras. $\endgroup$ – Lord Shark the Unknown Apr 11 at 15:47
  • $\begingroup$ Roya and his students have written much on the combinatorial motivations. $\endgroup$ – Bill Dubuque Apr 11 at 16:17
  • 1
    $\begingroup$ The existence of a bialgebra structure (or Hopf algebra structure) on certain algebras (such as symmetric algebras over a vector space or group algebras) gives more tools and gives us information about the algebra structure iself. For representation theory, the map $A\to A\otimes A$ allows the construction of tensor products in the category of $A$-representations $\endgroup$ – Max Apr 11 at 16:22
  • $\begingroup$ Roughly speaking, the motivation for comultiplication is rooted in trying to understand how things break up opposing to multiplication which is an operation that merges and combine different things. $\endgroup$ – Dac0 Apr 13 at 4:47

Though there are many reasons why one could be interested in coalgebras, many researchers are interested in bialgebras (or Hopf algebras). Before you can understand these, a basic understanding of coalgebras is very advisable.

Let's consider some natural examples first.

Let $G$ be a finite group and consider the complex group algebra $\mathbb{C}[G]$. Consider the category $\mathcal{C}=\text{Rep}(G)$ of finite-dimensional $G$-representations ($\mathcal{C}=\mathbb{C}[G]-\text{mod}$). As you probably learned in a group representation theory course, one can take the tensor product of two $\mathbb{C}[G]$-modules $M$ and $N$. The tensor product $M\otimes_{\mathbb{C}}N$ is again a $\mathbb{C}[G]$-module and the action of $g\in G$ is given by $g\cdot (m\otimes n):=g\cdot m\otimes g\cdot n$.

Now consider a linear map $\Delta\colon\mathbb{C}[G]\to \mathbb{C}[G]\otimes_\mathbb{C} \mathbb{C}[G]$ determined by $\Delta(g)=g\otimes g$ for all $g\in G$. Note that this map encodes the action of $\mathbb{C}[G]$ on $M\otimes N$ in a natural way. Associativity of the tensor product naturally yields coassociativity for $\Delta$. Moreover, associativity of the $G$ action on $M\otimes_\mathbb{C} N$ forces the map $\Delta$ to be a algebra morphism, i.e. $\Delta(gh)=\Delta(g)\Delta(h)$. Hence the fact that $\mathcal{C}$ is a tensor category forces the algebra $\mathbb{C}[G]$ to be a bialgebra.

We can repeat te same story for Lie algebras.

Let $L$ be a nice complex finite-dimensional semisimple Lie algebra. Let $\mathcal{C}$ be the category of finite-dimensional Lie algebra representations of $L$. Equivalently, $\mathcal{C}=U(L)-\text{mod}$ where $U(L)$ is the universal enveloping algebra of $L$. Given two $L$-representations $M$ and $N$, the tensor product $M\otimes_\mathbb{C} N$ becomes a $L$-representation and the action is determined by $l\cdot (m\otimes n)=l\cdot m\otimes n+m\otimes l\cdot n$. Equivalenty, the action is encoded in the comultiplication $\Delta\colon U(L)\to U(L)\otimes_\mathbb{C} U(L)$ determined by $\Delta(l)=l\otimes 1+1\otimes l$ for all $l\in L\subset U(L)$. Once again, the tensor structure on $\mathcal{C}$ forces $U(L)$ to be a bialgebra.

After understanding these examples, you will quickly realize that a bialgebra $B$ yields a monoidal structure on its representation category. The converse is true as well to some extent (see the Tannaka reconstruction for more details).

As representation categories and monoidal structures have proven to be fundamental in many branches of physics and mathematics, I'd argue that these are strong reasons to be interested in coalgebras and bialgebras.


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.