Intuition for left Hopf-modules

Question

I'm reading Radford's text on Hopf Algebras right now and I'm a bit confused about the definition of a left $A$-Hopf algebra. The definition given in the book is:

Let $A$ be a $\Bbbk$-bialgebra. A left $A$-Hopf module is a triple $(M, \mu, \rho)$ where $(M, \mu)$ is a left $A$-module, $(M, \rho)$ is a left $A$-comodule, and $$ \rho(a m) = a_{(1)}m_{(-1)} \otimes a_{(2)}m_{(0)}.$$

The book uses the convention of sumless Sweedler notation that

1. $\Delta (a) = a_{(1)} \otimes a_{(2)}$
2. $\rho (m) = m_{(-1)} \otimes m_{(0)}$.

I'm just looking to gain some intuition about what the axiom for the $\rho$ map in the definition is trying to convey. It's not clear to me what the meaning of this axiom is.

Answer 1

Another way to view Hopf modules is through the fundamental theorem of Hopf modules: Take a $\mathbb{K}$-vector space $V$, and form the tensor proudct $$ A \otimes_{\mathbb{K}} V, $$ endowed with the left $A$-action $$ A \times A \otimes_{\mathbb{K}} V \to A \otimes_{\mathbb{K}} V, ~~~~ (a_1,a_2 \otimes v) \mapsto a_1a_2 \otimes v, $$ and the left $A$-coaction $$ A \otimes_{\mathbb{K}} V \to A \otimes A \otimes_{\mathbb{K}} V, ~~~~ a \otimes v \mapsto a_{(1)} \otimes a_{(2)} \otimes v. $$ It is easy to check that $A \otimes V$ endowed with this $A$-action and $A$-coaction is in fact a Hopf module. Moreover, the fundamental theorem of Hopf modules says that every $A$-Hopf module arises in this way. In fact we have an equivalence of categories between the category of Hopf modules ${}^A_A\mathrm{Mod}$ (the morphisms are left $A$-module left $A$-comodule maps) and the category of vector spaces: The equivalence is given by the functors $$ A \otimes_{\mathbb{K}} - : \mathrm{Vect}_{\mathbb{K}} \to {}^A_A\mathrm{Mod}, ~~~~ V \mapsto A \otimes_{\mathbb{K}} A $$ and $$ {}^{\mathrm{co(A)}}(-): {}^A_A\mathrm{Mod} \to \mathrm{Vect}_{\mathbb{K}}, ~~~~ M \mapsto {}^{\mathrm{co(A)}}M := \{m \in M ~|~ \rho(m) = 1 \otimes m\}, $$ where $\rho$ is the coaction of $M$. (I'll let you figure what the unit and counit are and how the functors act on morphisms.)

So in summary, a Hopf module is just an abstraction of the properties of $A \otimes V$, and in fact every Hopf module is of this form. It proves useful to do this because often ``in nature'' Hopf modules $M$ arise that are not obviously of the form $A \otimes V$. However, once one has verified the Hopf module axioms, one can use the fundamental theorem of Hopf modules to find such a presentation $M \simeq A \otimes_{\mathbb{K}} {}^{\mathrm{co}(A)}M$.

Finally, the notion of a Hopf module generalises to the notion of a relative Hopf module, which can be considered a noncommutative generalisation of the notion of a homogeneous vector bundle. The fundamental theorem of Hopf modules generalises to Takeuchi's equivalence (under the a faithful flatness assumption).

Answer 2

I think the axiom might be thought of as a compatibility condition between the $A$-action $\rho$ and the $A$-coaction $\mu$. Since they take the form $$ \begin{align*} \rho &\colon M \longrightarrow M\otimes_{R}A,\\ \mu &\colon M\otimes_{R}A \longrightarrow M, \end{align*} $$ one might wonder what happens if you combine them, say by considering the composition $$ M\otimes_{R}A\xrightarrow{\mu}M\xrightarrow{\rho}M\otimes_{R}A. $$ If I understood it correctly, perhaps another way of stating it is by saying that the diagram commutes (the notation here is different, as it is formulated for $B$ a bialgebra in a general monoidal category $\mathcal{C}$. Taking $\mathcal{C}=\mathsf{Mod}_{R}$, $R=\mathbb{k}$, and $B$ to be a Hopf algebra over $\mathbb{k}$ should recover the case you're considering. I hope nevertheless that its meaning is clear!).

That is, the following operations give you the same result:

• Following the bottom path:
  • Apply the action, mapping $(a,m)$ to $am$;
  • Apply the coaction, mapping $am$ to $\rho(am)$;
• Following the top path:
  • Apply $\mathrm{id}\times$ the coaction, mapping $(a,m)$ to $(a,\rho(m))=(a,m_{(-1)}\otimes m_{(0)})$;
  • Apply the comultiplication, mapping $(a,\rho(m))$ to $$ (\Delta(a),\rho(m)) = (a_{(1)}\otimes a_{(2)},m_{(-1)}\otimes m_{(0)}) $$
  • Reorder the $H$ factors in $H\otimes H\otimes H\otimes M$, so that we can apply the action and multiplication;
  • Apply the multiplication and action, which maps $(a_{(1)}\otimes a_{(2)},m_{(-1)}\otimes m_{(0)})$ to $a_{(1)}m_{(-1)} \otimes a_{(2)}m_{(0)}$.

So "action then coaction" is the same as "coaction, then comultiplication, (then reorder), then multiplication+action".

Answer 3

Let me put my two cents in here too, as on MSE, together with Ré and Emily answers.

Directly concerning your question about the significance of the axiom involving $\rho$, consider the following: if $A$ is a bialgebra over the field $\Bbbk$, then the category ${{}_A\mathfrak{M}}$ of left $A$-modules is a monoidal category in such a way that the forgetful functor $\omega\colon {{}_A\mathfrak{M}} \to \mathfrak{M}$ to the category of vector spaces is a monoidal functor. That is to say, $\Bbbk$ is a left $A$-module with action provided by $$A \otimes \Bbbk \to \Bbbk, \qquad a \otimes \lambda \mapsto \varepsilon(a)\lambda,$$ and if $M,N$ are left $A$-modules, then $M \otimes N$ is a left $A$-module with action $$A \otimes (M \otimes N) \to M \otimes N, \qquad a \otimes (m \otimes n) \mapsto a_{(1)}\cdot m \otimes a_{(2)} \cdot n.$$ By the compatibility between the coalgebra and the algebra structure on $A$, $(A,\Delta,\varepsilon)$ is a comonoid in the monoidal category $\left({{}_A\mathfrak{M}},\otimes,\Bbbk\right)$ of left $A$-modules.

In this framework, a left Hopf module over $A$ can be defined as a comodule over the comonoid $(A,\Delta,\varepsilon)$ in the monoidal category $\left({{}_A\mathfrak{M}},\otimes,\Bbbk\right)$ and the axiom satisfied by $\rho$ states exactly that $\rho\colon M \to A \otimes M$ is a morphism of left $A$-modules.

Equivalently, you may check that also the category ${{}^A\mathfrak{M}}$ of left $A$-comodules is a monoidal category with $\otimes$ and $\Bbbk$ and that a left Hopf module over $A$ can be also defined as a left module over the monoid $(A,m,u)$ in the monoidal category $\left({{}^A\mathfrak{M}},\otimes,\Bbbk\right)$. In this case, the condition you are looking at is a condition on the action $\mu$ rather than on $\rho$ and it is telling you that $\mu\colon A \otimes M \to M$ is a morphism of left $A$-comodules.