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Dec 15, 2021 at 20:33 comment added Ben A-S Okay, thank you.
Dec 15, 2021 at 8:36 comment added Matthew Daws I also find it hard to work "just with diagrams". Vicary showed in arxiv.org/abs/0805.0432 that SSFA's are just finite-dimensional $C^*$-algebras equipped with a trace satisfying a certain normalisation condition (which I always forget...) The trace turns the $C^*$-algebra into a Hilbert space, and $\delta$ is the the Hilbert space adjoint of the multiplication map. This perspective makes it clear that $\delta(1)=1\otimes 1$ only for trivial $A$.
Dec 15, 2021 at 6:37 history edited YCor CC BY-SA 4.0
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Dec 15, 2021 at 6:30 history edited David Roberts CC BY-SA 4.0
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Dec 15, 2021 at 2:58 comment added Ben A-S Is it true that a Frobenius algebra is trivial if and only if $\delta(1) = 1\otimes 1$?
Dec 15, 2021 at 2:34 comment added Ben A-S Oh, wow, yes, you're absolutely right. I'm mixing up the co-algebra and algebra structures. Thanks for your help!
Dec 14, 2021 at 23:28 comment added Jules Lamers .. And I suppose that in that way Thm 2.6 gives an inner product, allowing us to switch between elements and their duals. I think that on the right in (16) the author omitted the little 'popsicle diagram' for the unit dangling at the bottom in (15). So for Q2 I think that the relation says: $f^\dag(x) = (f(x^\dag))^\dag$.
Dec 14, 2021 at 23:13 comment added Jules Lamers For Q1: a Frobenius algebra is not a bialgebra, so I'm not sure if $\delta(1)$ equals $1\otimes1$ (which the mixes algebra and coalgebra structures). In Sweedler notation $\delta(1)=\sum 1_{(1)} \otimes 1_{(2)}$ so we seem to get $a\mapsto \sum a^\dag(1_{(2)}) \, 1_{(1)}$ which still $\in A$. [Also, I would think the last diagrams in (12) and (17) ought to be reflected]
Dec 14, 2021 at 22:52 history asked Ben A-S CC BY-SA 4.0