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fixed a typo
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Theo Johnson-Freyd
Theo Johnson-Freyd
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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 XX with the dual vector space XX*), 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 XX ⊗ R = XX.

So, pick your favorite finite-dimensional vector space XX, and think about the types of diagrams you can draw using just it. Well, the space of matrices (what you call V) is XX ⊗ 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 R → V) is the directed cup, and multiplication and comultiplication are both given by trivalent vertices.

In ASKIIASCII (ignore the weird coloring):

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

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 ASKII (ignore the weird coloring):

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

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):

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

Source Link
Theo Johnson-Freyd
Theo Johnson-Freyd
  • 54.6k
  • 10
  • 142
  • 335

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 ASKII (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 (X ⊗ X*)* = X ⊗ X*. This ability to rotate X ⊗ X* is why δ = μ*.