Timeline for Intersection of Frobenius subalgebra objects
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 30 at 16:05 | comment | added | Sebastien Palcoux | Here is a dedicated mathoverflow post: mathoverflow.net/q/481529/34538 | |
| Oct 30 at 8:05 | comment | added | Sebastien Palcoux | Does the poset of Frobenius subalgebras form a lattice? Based on your counterexample, it appears that the meet, if it exists, is not always determined by the intersection. | |
| Oct 30 at 8:01 | vote | accept | Sebastien Palcoux | ||
| Sep 22 at 6:56 | comment | added | Dave Benson | You're confused about the term "zero-dimensional". It refers to the Krull dimension, not the dimension as an algebra. | |
| Sep 22 at 6:37 | comment | added | Sebastien Palcoux | Could you provide your reference for the claim that a Gorenstein ring is a Frobenius algebra? The last link you cited indicates that a zero-dimensional Gorenstein ring (under additional assumptions) is a Frobenius algebra. However, the example you mentioned is five-dimensional. Could you clarify this discrepancy? | |
| Sep 11 at 13:23 | comment | added | Dave Benson | Then you should be able to express your question in terms of the usual axioms. en.wikipedia.org/wiki/Frobenius_algebra | |
| Sep 11 at 13:18 | comment | added | Sebastien Palcoux | A Frobenius algebra object in a monoidal category is both an algebra and a coalgebra object that satisfies the Frobenius condition. If the notation is consistent, a Frobenius algebra should be a Frobenius algebra object in the category ${\rm Vec}$. Therefore, a Frobenius algebra should inherently possess a coalgebra structure. | |
| Sep 11 at 12:51 | comment | added | Dave Benson | I have no idea what you mean by the counit and the comultiplication. That isn't part of the structure of a Frobenius algebra. | |
| Sep 11 at 12:45 | comment | added | Sebastien Palcoux | I have realized that my proof for the semisimple case necessitates an assumption stronger than symmetric. Specifically, I need the counit composed with the multiplication to be equal to the evaluation map ($\epsilon \circ m = ev_M$), and the comultiplication composed with the unit to be equal to the coevaluation map ($\delta \circ e = coev_M$). What does this stronger assumption translate to in the context of a (usual) Frobenius algebra (in ${\rm Vec}$)? Does your example meet this stronger assumption? | |
| Sep 8 at 1:46 | comment | added | Sebastien Palcoux | Yes. I just posted the connected version: mathoverflow.net/q/478418/34538 | |
| Sep 8 at 1:20 | vote | accept | Sebastien Palcoux | ||
| Sep 22 at 6:37 | |||||
| Sep 7 at 14:56 | comment | added | Dave Benson | There seem to be several ways to interpret this for a Frobenius algebra. You're given a map $M \otimes M \to k$, which by adjunction gives a left and a right map $M \to M^*$. So $i^*_A\circ i_A$ really means $A \to M \to M^* \to A^* \to A$ using some choices for these maps? Even so, since my example is not just Frobenius but also symmetric, the map $M \otimes M \to k$ is symmetric so these choices agree. Then I think it just amounts to saying that the symmetric bilinear form on $A$ is the restriction of that on $M$. In my example this seems to already hold. | |
| Sep 7 at 13:55 | comment | added | Sebastien Palcoux | Let $f: X \to Y$ be a morphism in a rigid monoidal category. Then $f^*: Y^* \to X^*$ is the dual morphism. | |
| Sep 7 at 13:09 | comment | added | Dave Benson | What is $i^*_A$? | |
| Sep 7 at 11:46 | comment | added | Sebastien Palcoux | 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? | |
| Sep 7 at 10:33 | comment | added | Dave Benson | 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. | |
| Sep 7 at 10:20 | history | answered | Dave Benson | CC BY-SA 4.0 |