2
$\begingroup$

I was hoping someone might be able to shed some light on the choice of indices for expressing the coaction using Sweedler notation.

For example, in the paper of Andruskiewitsch About finite-dimensional Hopf algebras, (see page 3) http://cel.archives-ouvertes.fr/docs/00/37/43/83/PDF/2000Andrusk.pdf, $C$ is a coalgebra, $V$ is a left comodule, and the structure map is expressed as

$$\delta: V \longrightarrow C \otimes V$$ $$\delta(x) = x_{(-1)} \otimes x_{(0)}$$

Can someone explain this choice of indices and how it is meaningful for understanding the coaction? I am working on building an intuition for comodules apart from the fact that they are dual to modules, and understanding this notation would be helpful.

The use of Sweedler notation for the comultiplication is familiar and clear to me:

$$\Delta(x)= x_{(1)} \otimes x_{(2)}$$

but I am unsure of the significance of the $(-1)$ and $(0)$.

$\endgroup$
  • 1
    $\begingroup$ Maybe it is helpful: Sweddler's book on Hopf algebras has several exercises about this notation. $\endgroup$ – Leandro Vendramin Nov 15 '13 at 22:56
4
$\begingroup$

I would start with considering the case of a right $C$-comodule $U$, for which the structure map $\delta\colon U\to U\otimes C$ would be expressed as $\delta(u) = u_{(0)}\otimes u_{(1)}$. The double (iterated) coaction map $U\to U\otimes C\otimes C$ would then be denoted by $u\mapsto u_{(0)}\otimes u_{(1)}\otimes u_{(2)}$, with the coassociativity equiation (for the coaction) being $u_{(0)}\otimes u_{(1)(1)}\otimes u_{(1)(2)} = u_{(0)}\otimes u_{(1)}\otimes u_{(2)} = u_{(0)(0)}\otimes u_{(0)(1)}\otimes u_{(1)}$.

Now this looks quite natural, with the index ${}_{(0)}$ distinguishing the factors belonging to $U$, while the factors with positive indices (or in more complicated formulas, with at least one positive index) belonging to $C$.

If this is clear, notice that the indices always go in the order of succession of the integers in our tensor notation, that is one has $j=i+1$ whenever there are two successive factors like $x_{(i)}\otimes x_{(j)}$ in our expressions. Given that, and the desire to reserve the index ${}_{(0)}$ for the factors belonging to comodules (as opposed to coalgebras), the notation $\delta(v)= v_{(-1)}\otimes v_{(0)}$ for a left coaction map $\delta\colon V\to C\otimes V$ comes out pretty naturally, too.

The factors with zero indices (or in more complicated formulas, with all zero indices) still belong to the comodule, while the ones with at least one nonzero index belong to the coalgebra.

$\endgroup$

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.