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$\require{AMScd}$Let $\mathcal{C}$ be a tensor category and let $M$ be a Frobenius algebra object in $\mathcal{C}$. 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).

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 the Frobenius algebra object $M$.

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 answers to this post.

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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    $\begingroup$ What is definitely not true is that if you have a finite dimensional Frobenius algebra then the intersection of two Frobenius subalgebras is a Frobenius subalgebra. Even in the commutative case. $\endgroup$
    – Dave Benson
    Commented Sep 7 at 9:40
  • $\begingroup$ @DaveBenson Could the counterexamples you have in mind be realized as Frobenius algebra objects in a semisimple tensor category over $\mathbb{C}$? $\endgroup$
    – Sebastien Palcoux
    Commented Sep 7 at 10:15
  • $\begingroup$ Well, it's a finite dimensional commutative $\mathbb{C}$-algebra. Is that good enough for you? The smallest example is five dimensional. Woulld you like me to write out the details? $\endgroup$
    – Dave Benson
    Commented Sep 7 at 10:16
  • $\begingroup$ @DaveBenson Yes, please. $\endgroup$
    – Sebastien Palcoux
    Commented Sep 7 at 10:20
  • $\begingroup$ Here is the connected version: mathoverflow.net/q/478418/34538 $\endgroup$
    – Sebastien Palcoux
    Commented Sep 8 at 3:17

1 Answer 1

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Let $k$ be any field. Consider the commutative $k$-algebra $k[x,y,z]/(x^2,y^2,xz,yz,xy-z^2)$. This is a five dimensional algebra with basis $1$, $x$, $y$, $z$, $u$ where $u=xy=z^2$. It is clearly a Gorenstein ring, so it is a finite dimensional Frobenius algebra. Let $A$ be the subalgebra generated by $x$ and $y$, so that it has basis $1$, $x$, $y$ and $u$; and let $B$ be the subalgebra generated by $x$ and $y+z$, so that it has basis $1$, $x$, $y+z$, $u$. These are both Frobenius subalgebras. But their intersection has basis $1$, $x$ and $u$, with $x^2=u^2=xu=0$; this is not Frobenius because the socle is two dimensional.

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    $\begingroup$ The point here is that you can have a finite dimensional vector space with a non-degenerate quadratic form, and two non-degenerate subspaces whose intersection is degenerate. $\endgroup$
    – Dave Benson
    Commented Sep 7 at 10:33
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    $\begingroup$ Let me propose some possible additional assumptions. Please let me know if you find them relevant. Suggestion 1: Assume that $i_A^* \circ i_A = \mathrm{id}_A$ and $i_B^* \circ i_B = \mathrm{id}_B$. Suggestion 2: If the notations are coherent, Frobenius algebras are precisely the Frobenius algebra objects in the tensor category $\mathrm{Vec}$. Thus, assuming the Frobenius algebra to be "connected" (i.e., $\mathrm{Hom}_{\mathcal{C}}(1, M)$ is one-dimensional) allows us to avoid $\mathrm{Vec}$. Would these additional assumptions be relevant in addressing the problem? $\endgroup$
    – Sebastien Palcoux
    Commented Sep 7 at 11:46
  • $\begingroup$ What is $i^*_A$? $\endgroup$
    – Dave Benson
    Commented Sep 7 at 13:09
  • $\begingroup$ Let $f: X \to Y$ be a morphism in a rigid monoidal category. Then $f^*: Y^* \to X^*$ is the dual morphism. $\endgroup$
    – Sebastien Palcoux
    Commented Sep 7 at 13:55
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    $\begingroup$ You're confused about the term "zero-dimensional". It refers to the Krull dimension, not the dimension as an algebra. $\endgroup$
    – Dave Benson
    Commented Sep 22 at 6:56

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