Let $H$ be a finite-dimensional Hopf algebra. Then it has a right cointegral $\lambda \in H^*$ and a left integral $c \in H$, characterized uniquely (up to scalar) by \begin{align} (\lambda \otimes id)(\Delta(h)) = \lambda(h) 1 \quad \text{and} \quad hc = \varepsilon(h) c \end{align} for any $h \in H$. Similarly one can define left cointegrals and right integrals.
There are now grouplike elements $\gamma \in G(H^*), a \in G(H)$, called the modulus and the comodulus, which capture the failure of (co)integrals to be two-sided: \begin{align} (id \otimes \lambda)(\Delta(h)) = \lambda(h) a \quad \text{and} \quad ch = \gamma(h) c \end{align}
The modulus has a nice interpretation in $H\text{-mod}$: It is an algebra morphism to the ground field, and thus specifies an invertible object, which is precisely the socle of the projective cover of the tensor unit --- also known as the distinguished invertible object of $H\text{-mod}$.
- What is the interpretation of $a$ in $H\text{-mod}$?
Or, more generally,
- Is there an analog of $a$ in any finite tensor category?
It is grouplike which, to me, suggests that it might play a part in the monoidal structure of some functor, but I'm not sure.
Edit: Now follows some progress, but that my element $a$ (which I define below) indeed satisfies the above equation is not clear to me yet. Probably overlooking something obvious.
In An analogue of Radford's $S^4$ formula for finite tensor categories, Etingof, Nikshych, and Ostrik prove that in any finite tensor category $\mathcal{C}$ there is a (monoidal) natural isomorphism \begin{align} \psi : - ^{\vee\vee} \Rightarrow D \otimes {}^{\vee\vee}- \otimes D^{-1} \ , \end{align} where $D$ is the distinguished invertible object of $\mathcal{C}$. Piecing everything together from that paper as well as the Tensor Categories-book, one find $D = \gamma^{-1}$ for $\mathcal{C} = H\text{-mod}$.
I now claim that the comodulus is \begin{align} a = \psi_H(1), \end{align} where $H$ is the unit of $H$. This I guess is in some sense implied in the paper, but it is not immediately obvious to me. Let's first note that $\psi$ is completely determined by $a$, since \begin{align} \psi_V(v) = \psi_V(1.v) = -.v(\psi_H(1)) = a.v . \end{align} The statement that $\psi$ exist generalizes Radford's famous $S^4$-formula to finite tensor categories: the fact that $\psi_V$ is an intertwiner simply means \begin{align} a S^2(h) . v = S^{-2}(\gamma \rightharpoonup h \leftharpoonup \gamma^{-1}) a . v \end{align} which in particular implies \begin{align} S^4(h) = a^{-1} (\gamma \rightharpoonup h \leftharpoonup \gamma^{-1}) a . \end{align} Here the hook notation for $f\in H^*$ and $h\in H$ is standard and means e.g. $h \leftharpoonup f = f(h') h''$, while the other version eats the other leg.
So this is indeed Radford's $S^4$-formula, and $a$ is grouplike since $\psi$ is monoidal.
But why does it satisfy the "capture difference between left and right cointegrals" equation from above?
