# Dual coalgebra (HW3 problem 2)

I don’t understand how information from the algebra gets into the dual coalgebra.

Here’s what I think I know: You dualize the multiplication to get a comultiplication (a trick I understand diagrammatically but not constructively) , and elements of the coalgebra are linear functionals in bijection with the elements of the algebra.

Do we know based on element a of algebra A how a* maps A* to the field?

1. The notation a* is confusing. There is no way to assign to each a in A an element a* of A*. So usually, even when a already stands for an element of A, the notation a* can be used for any *arbitrary* element of A*, which may be totally unrelated to a.

There is one exception to this rule, and that appears when you have a basis $\left(e_i\right)_{i\in I}$ of A. Then, there is a “dual basis” of A*, and this basis is usually denoted by $\left(e_i^{\ast}\right)_{i \in I}$. (Caveat: it is only a basis if A is finite-dimensional.) But even in this case, it can’t be said that each $e_i^{\ast}$ depends only on the respective $e_i$; it depends on the whole basis.

How to understand the dual coalgebra? The easiest way is probably this one:

Assume that A is finite-dimensional (otherwise, the notion of the dual coalgebra is harder to define – it is not the whole A* anymore). Let $\left(e_1,e_2,...,e_n\right)$ be a basis of A. Write down the multiplication table of A; this is the $n\times n$ matrix whose (i, j)-th entry is the product $e_i e_j$, written as a vector with respect to the basis $\left(e_1,e_2,...,e_n\right)$. Now, for every k, you can compute the coproduct of $e_k^{\ast}$ in the dual coalgebra A* by

$\Delta\left(e_k^{\ast}\right) = \sum_{i, j} \left(\text{the }e_k\text{-coordinate of }e_ie_j\right)\cdot e_i^{\ast}\otimes e_j^{\ast}$.

2. Brian Cruz

This is also when I ended up getting, using $\Delta=\rho^-1m^*$ to show it. Start out with

$\Delta e_k^*=\rho^-1m^*e_k^*=\sum_{i,j}\lambda_{ijk}e_i^*\otimes e_j^*$

move the $\rho$ to the other side and the magic happens when you start evaluating on basis of the tensor space. The key is that since our dim is finite, we have a dual basis such that $e_k^*(e_j)=\delta_kj$ so that it’s one when they are the same and zero everywhere else.

I got this idea by looking at the matrices on http://haskellformaths.blogspot.com/2011/04/what-is-coalgebra.html (which does not quite present things in the exact way Darij does above, but it’s equivalent)

3. Brian Cruz

Oh yeah, and the $\Delta=\rho^{-1}m^*$ part came from Maria and Brian from a conversation we had earlier today!

4. From the responses, it seems we think that there is a strong association of the commultiplication of an algebra’s dual with its multiplication- roughly, that the coproduct of a dual-basis element $e_k^*$ is the sum of tensors $e_i^*$ and $e_j^*$ over all i’s and j’s such that $e_i \otimes e_j=\lambda_{ij} e_k$? Does anybody understand how this happens, mechanically, during the dualization? It makes mystical sense to me…