One of the easiest examples I can think of for frobenius algebras is a plain ol' matrix algebra with tr : V → k as the co-unit (or equivalently, tr(a⋅b) as the frobenius form). This is enough data to generate a comultiplication δ : V → V ⊗ V. This turns out to be μ†, for multiplication μ. Is there any intuition for what this map does (aside from the obvious "do multiplication on the dual space")?
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$\begingroup$ I don't quite see why you aren't happy with the intuition that you give. It seems to me that it cleanly describes what the comultiplication is and how it arises. $\endgroup$– Simon WadsleyCommented Oct 27, 2009 at 12:21
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$\begingroup$ Maybe this is all there really is to say about this co-multiply. I was just wondering if there's something else there, like this example: Define a frobenius algebra on any FD vector space by making comultiply "copy" a basis. delta :: |i> |-> |ii> and counit "delete" a basis. epsilon :: |i> |-> 1. Mult. and unit are just the daggers. For delta_X defined on the eigenvectors of Pauli X (|+>, |->), it's a (happily coincidental?) fact that the induced multiply delta^dag is actually logical XOR on the Pauli Z basis (|0>, |1>). $\endgroup$– Aleks KissingerCommented Oct 29, 2009 at 12:19
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$\begingroup$ Incidentally, your proscription for defining a frobenius algebra on a finite-dimensional vector space requires a basis. Otherwise your comultiplication and counit are not linear. $\endgroup$– Theo Johnson-FreydCommented Oct 31, 2009 at 23:37
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$\begingroup$ That's the point! In fact, this type of frobenius algebra (called special FA) uniquely picks out a basis in the underlying object. We often take this as a pure categorical way to define basis. See eg Coecke et al's "Bases" paper. $\endgroup$– Aleks KissingerCommented Nov 1, 2009 at 10:45
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1 Answer
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Here's how I live to think about matrices. Penrose (1971) figured out that you can draw linear algebra diagrammatically. A picture in the Penrose notation is a directed labeled graph with external leaves. The edges are labeled by vector spaces (changing the direction on an edge has the same effect as swapping the label X with the dual vector space X*), and vertices by multilinear maps. In this way, placing two edges next to each other is the tensor product. The ground field R should be drawn as an invisible edge, so that X ⊗ R = X.
So, pick your favorite finite-dimensional vector space X, and think about the types of diagrams you can draw using just it. Well, the space of matrices (what you call V) is X ⊗ X*, so it looks like two parallel lines pointed in opposite directions. Then you can check that the trace is the directed cap, the identity element (thought of as a map R → V) is the directed cup, and multiplication and comultiplication are both given by trivalent vertices.
In ASCII (ignore the weird coloring):
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X = ^ , X* = v , R = [empty], V = ^ v
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->- | |
/ \ ^ v
Tr = | | I = | |
^ v \ /
| | -<-
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^ v ^ v ^ v
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mu = / _ \ delta = \ \_/ /
/ / \ \ \ /
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^ v ^ v ^ v
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Not only does the notation "explain" the comultiplication, it "proves" all the associativity and unital properties you might want. Mostly, though, I think it makes it totally clear what the Frobenius pairing (a,b) → Tr(ab) is doing. It's just the map:
->-
/ _ \
pair = / / \ \
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^ v ^ v
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Which is just the canonical fact that (X ⊗ X*)* = X ⊗ X*. This ability to rotate X ⊗ X* is why δ = μ*.
answered Oct 27, 2009 at 20:59
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$\begingroup$ This is a good way of thinking about these things! Also, it justifies the existence of what I previously thought was a cute but somewhat pointless construction of turning a compact structure (cap and cup) into a frobenius algebra. This is exactly the matrix frobenius algebra, when you think of linear maps as their "names". I.e. express M as "[M] := (1 (x) M) o cup". The frobenius multiplication "mu ([M] (x) [N])" reduces by compact structure "string pulling" to [MN]. Cool! Defining trace as cap also unifies the "internal" notion of trace of a matrix with the "self-loop" one: Tr(M) = cap[M]. $\endgroup$ Commented Oct 29, 2009 at 12:37