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This post is the connected version of this one.

$\require{AMScd}$Let $\mathcal{C}$ be a tensor category over $\mathbb{C}$ and let $M$ be a Frobenius algebra object in $\mathcal{C}$. Recall that a Frobenius algebra object is self-dual. A Frobenius subalgebra object of $M$ is a Frobenius algebra object $X$ equipped with a monomorphism $i_X: X \to M$, which is compatible with the algebra and unit structures (although not necessarily with the coalgebra and counit structures, otherwise $i_X$ would be an isomorphism); moreover, let us also assume here that $i_X^* \circ i_X = {\rm id}_X$ and $i_X^{**} = i_X$.

Following [Fr64, $\S$1.5], $(X, i_X)$ is a subobject of $M$ (more precisely, a representative of an equivalence class), and these subobjects form a poset. The intersection $A \cap B$ of two subobjects $A$ and $B$ is defined in [Fr64, $\S$2.1] as the greatest lower bound, and it is proven to be the pullback as displayed in the following diagram:

$$ \begin{CD} A \cap B @>j_A>> A \\ @VVj_BV @VVi_AV \\ B @>i_B>> M \end{CD} $$

Assume that $A$ and $B$ are Frobenius subalgebra objects of a connected Frobenius algebra object $M$ (i.e. ${\rm Hom}_{\mathcal{C}}(1, M)$ is one-dimensional).

Question: Is it true that $A \cap B$ is also a Frobenius subalgebra object of $M$?

A Frobenius algebra object is self-dual, and to prove that $A \cap B$ is a Frobenius subalgebra object of $M$, I first need to prove that it is self-dual. In the more general case where $A$, $B$, and $M$ are just assumed to be self-dual, then $A \cap B$ is self-dual in the semisimple case but does not necessarily remain so otherwise. For counterexamples, see the answer to this post.

If the notations are consistent, the Frobenius algebras are precisely the Frobenius algebra objects in the tensor category ${\rm Vec}$, where counterexamples exist, see the answer to this post. Assuming that the Frobenius algebra object is connected allows us to avoid ${\rm Vec}$.

If the above question has a negative answer, I wonder if there is a more appropriate way to define such an intersection to yield a positive answer.

Reference:

[Fr64] Freyd, Peter. Abelian Categories: An Introduction to the Theory of Functors. Harper's Series in Modern Mathematics. Harper & Row, Publishers, New York, 1964. xi+164 pp.

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  • $\begingroup$ Can you please give an example in the connected case, that isn't semisimple? $\endgroup$
    – Dave Benson
    Commented Sep 8 at 6:28
  • $\begingroup$ @DaveBenson To be honest, I didn't have a specific example in mind when I first read your question. However, after some research, I found the following reference: Section 6 in this paper. The examples of connected Frobenius algebra objects I am most familiar with are the irreducible finite index subfactors, which are semisimple and unitary. My recent questions aim to generalize some well-known results on subfactors to Frobenius algebra objects by relaxing the unitary condition, e.g. this post. $\endgroup$
    – Sebastien Palcoux
    Commented Sep 8 at 8:19
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    $\begingroup$ @DaveBenson: Here is maybe a more usual family of examples from [EGNO, Example 7.20.6]: let $H$ be a unimodular non-semisimple finite dimensional Hopf algebra, then $H^*$ is a connected Frobenius algebra object in the category ${\rm Rep}(H \otimes H^{\rm cop})$. $\endgroup$
    – Sebastien Palcoux
    Commented Sep 8 at 8:51

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