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…

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