## Some Natural Identifications (Pt. IV)

**Point of Post: **This is a continuation of this post.

Now, I will assume it is old news to you that for finite dimensional -spaces one has that is naturally isomorphic to its double dual via the map which takes a vector to the evaluation map –a linear functional . So, now if we denote, for each fixed -space , the isomorphism gotten by taking and sending it to we see that is a natural isomorphism in both variables.

*Remark: *For those more concerned with the nitty gritty details, let me elaborate on this point. Now, in general, if are covariant functors which are isomorphic by the natural transformations and is a contravariant functor, then the contravariant functors are naturally isomorphic by –this you can check on your own (now that you are uncluttered with the specificities of our particular problem).

Thus, considering the composition of the natural isomorphisms and gives us a natural isomorphism . To be more specific we have the following chain of isomorphisms

and since the composition of natural isomorphisms are a natural isomorphism we may concatenate to get where, as we have already stated, . So:

**Theorem:** *Let be a field and finite dimensional -spaces. Then, the map given on simple tensors by is a natural isomorphism in both variables.*

*Remark: *One could have just, outright, defined instead of going through the steps (the multiple compositions) but this really breaks down everything and shows you the composite parts. If you don’t like this, you can just define and just show it is a natural isomorphism.

Ok, so this is all fine and dandy, but it remains to be revealed what this has to do with the trace map. This comes from the above theorem in a fairly neat way. To begin we note that this bilinear form that we have been using induces a linear map . Now, we know that and so we get from this a map defined by precomposing with the isomorphism .

Now, note that each time we fix an ordered basis of we get an isomorphism which is just the inverse of the isomorphism .

Let’s try to figure out what this map is. Fix a basis of . We know then that we have the dual basis of and so is a basis for . Moreover, if we define by then is a basis for . Finally, it’s easy to see that the isomorphism satisfies .

So, to see what our map does it suffices to check what it does on the basis . But, by definition is just acting on the image of in , or . That said, by definition we have that . Thus, we see that . That said, consider the normal trace map by summing the diagonal entries of a matrix. We then can see that the choice of (ordered) basis induces a map defined by (where, of course, is the matrix of with respect to ). But, note that –the same as our map . Thus, we see that . From this we see the following:

**Theorem: ***Let be a finite dimensional vector space and fix an ordered basis . Then one has that the following diagram commutes*

*commutes.*

This allows us to conclude the following:

**Corollary: ***Let and be two ordered bases of a finite dimensional vector space . Then, the maps are equal.*

This allows us to conclude that the *trace map* defined by taking an operator , taking its matrix with respect to any ordered basis, and then summing the diagonal entries, is well-defined.

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. *Module Theory.* Clarendon, 1990. Print.

[4] Lang, Serge. *Algebra*. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. *Abstract Algebra*. New York: Springer, 2007. Print.

[6] Conrad, Keith. “Exterior Powers.” *Www.math.uconn.edu/~kconrad*. Web. <http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf>.

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