Relating Two Notions of Tensor Product

[Newtime]

When I was first introduced to the tensor product, I was actually introduced to a special case: the tensor product of vector spaces over $$\mathbb{C}$$, which was explained to be as the space of multilinear maps on the cross product of the dual spaces, for example. At the time I wasn't aware this was a special case, but now that I'm working through chapter 4 of Hungerford's Algebra, it's clear.

I feel like I understand the general construction in terms of the universality of the tensor product of two modules in the category of middle linear maps (I think this terminology may be exclusive to Hungerford). My problem is relating this more general construction to the one above. That is, if I replace the ring $$R$$ with $$\mathbb{C}$$, and the modules with vector spaces, I don't get what I want. This probably also implies that I don't really understand the tensor product in the more general setting as much as I think I do.

So my question is this: How does one get from the more general construction to the special case mentioned above?

 

[mathwonk]

finite dimensional vector spaces are isomorphic to their double duals. presumably the special case you learned was the double dual of the general definition. A tensor product of spaces A,B is a space X plus a bilinear map AxB-->X such that every bilinear map out of AxB factors through this one, as composing with a linear map out of X.

In particular the dual of X is isomorphic by this construction to Bil{(AxB);k}, and hence X is itself isomorphic to the dual of Bil{(AxB);k}, which presumably is isomorphic to
Bil{(AxB)*;k}, and also to your construction Bil{(A*xB*),k}.


So your first job is to find a bilinear map from AxB to Bil{(A*xB*),k}. I suppose it takes (a,b) to the product of the evaluation maps on A* and B*. I.e. if f is in A* and g is in B*, you get f(a)g(b).

anyway this sort of nonsense becomes routine after a while. I.e. the basic thing to remember is to show things are isomorphic find a map, and in cases like this there is usually only one possible map, so whatever you can think of usually works.