Fun with k-Algebras

Continuing from last time, suppose we have a bilinear map \( f\) from \(V \times W\) to \(L\) where V, W, and L are vector spaces. Then there is a universal property function \(\Phi \) from \(V \times W\)  to \(V \otimes W \) and there is a unique linear map \( g\) from \(V \otimes W\) to \(L\) such that the diagram below commutes:

               \(\Phi\)

\( V \times W \)-------> \(V \otimes W\)

    \                        |

       \                     |

           \                 |

       \(f\)    \              | \( g \)

                  \          |

                     \       |

                      _|   \/

                           \( L \)

The proof is trivial: "f" is used to define a function from the free vector space \( F (V \times W) \) to \( L \) and then we make a descent by modding by the usual equivalence relation of the tensor product to define the map \( g \).

This all looks a bit pedantic, but the point is that any multiplication rule in an algebra \( A \) is a bilinear map from \( A \times A \) to \( A\) and we can now put it in tensor formalism.

In particular consider the algebra \( A \) of matrices over a field \( k \). Matrix multiplication is associative, and we also have a unit of the algebra: the diagonal matrix with the where the elements are the identity of \( k \).  This is a prototypical example of what is called a k-algebra.

Can we formalize the associativity and the unit using the tensor product language? Indeed we can and here is the formal definition:

A k-algebra is a k-vector space \( A \) which has a linear map \(m : A\otimes A \rightarrow A\) called the multiplication and a unit \(u: k \rightarrow A \) such that the following diagrams commute:

                     \( m \otimes 1\)

\( A \otimes A \otimes A \) ----------> \(A \otimes A\)

              |                             |

              |                             |

  \( 1 \otimes m\) |                             |  \( m \)

              |                             |

              |                             |

             \/                            \/

          \(A \otimes A\)   ---------->    \(A \)

                           \( m\)

and

                          \(A \otimes A\)

                    _                    _

                      |         |         |

\( u\otimes 1\)       /              |                \ \( 1\otimes u\)

              /                 |                    \

\( k \otimes A \)                      | \(m \)               \(A \otimes k\)

              \                 |                     /

                  \              |                 /

                       _|      \/          |_

                               \( A \)

Please excuse the sloppiness of the diagrams, it is a real pain to draw them.

So what are those commuting diagrams really saying? 

The first one states that:

\( m(m (a\otimes b) \otimes c) = m(a \otimes m(b \otimes c)) \)

In other words: associativity of the multiplication: (a b) c = a (b c)

The second one defines the algebra unit:

\( u(1_k ) a = a u(1_k )\)

which means that \( u (1_k) = 1_A \)

So why do we torture ourselves with this fancy pictorial way of representing trivial properties of algebra? Because now we can do a very powerful thing: reverse the direction of all the arrows. What do get when we do that? We get a brand new concept: the coproduct. Stay tuned next time to explore the wondeful properties of this new mathematical concept.