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.


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__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][1], 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?


  [1]: https://arxiv.org/abs/math/0404504