# What is the comultiplication of a matrix frobenius algebra?

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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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. –  Simon Wadsley Oct 27 '09 at 12:21
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>). –  Aleks Kissinger Oct 29 '09 at 12:19
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. –  Theo Johnson-Freyd Oct 31 '09 at 23:37
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. –  Aleks Kissinger Nov 1 '09 at 10:45

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 XR = 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 XX*, 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 RV) is the directed cup, and multiplication and comultiplication are both given by trivalent vertices.

In ASCII (ignore the weird coloring):

       |            |                        | |
|            |                        | |
X  =  ^ ,   X*  =  v ,  R = [empty],  V  =  ^ v
|            |                        | |
|            |                        | |

->-            |     |
/   \           ^     v
Tr  =  |     |    I  =  |     |
^     v           \   /
|     |            -<-

| |                  | |   | |
^ v                  ^ v   ^ v
| |                  | |   | |
mu  =     / _ \       delta  =   \ \_/ /
/ / \ \                  \   /
| |   | |                  | |
^ v   ^ v                  ^ v
| |   | |                  | |


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 =     / / \ \
| |   | |
^ v   ^ v
| |   | |


Which is just the canonical fact that (XX*)* = XX*. This ability to rotate XX* is why δ = μ*.

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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]. –  Aleks Kissinger Oct 29 '09 at 12:37