Abstract Nonsense

Crushing one theorem at a time

Some Natural Identifications (Pt. IV)

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

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Now, I will assume it is old news to you that for finite dimensional F-spaces V one has that V is naturally isomorphic to its double dual V^{\ast\ast} via the map \Theta_V:V\to V^{\ast\ast} which takes a vector v\in V to the evaluation map \langle\bullet,v\rangle–a linear functional V^\ast\to F. So, now if we denote, for each fixed F-space W, the isomorphism \Delta_{V,W}:\text{Hom}(V^{\ast\ast},W)\to \text{Hom}(V,W) gotten by taking h\in \text{Hom}(V^{\ast\ast},W) and sending it to h\circ\Theta_V we see that \Delta_{V,W} is a natural isomorphism in both variables.

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Remark: For those more concerned with the nitty gritty details, let me elaborate on this point. Now, in general, if F,G:\mathcal{D}\to\mathcal{D} are covariant functors which are isomorphic by the natural transformations \eta_X:F(X)\xrightarrow{\approx}G(X) and H:\mathcal{D}\to\mathcal{D} is a contravariant functor, then the contravariant functors H\circ F,H\circ G:\mathcal{D}\to\mathcal{D} are naturally isomorphic by H(\eta_X):H(G(X))\to H(F(X))–this you can check on your own (now that you are uncluttered with the specificities of our particular problem).

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Thus, considering the composition of the natural isomorphisms \Gamma_{V^\ast,W} and \Delta_{V,W} gives us a natural isomorphism \Pi_{V,W} V^\ast\otimes W\to\text{Hom}(V,W). To be more specific we have the following chain of isomorphisms

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V^\ast\otimes W\overset{\Gamma_{V,W}}{\cong}\text{Hom}(V^{\ast\ast},W)\overset{\Delta_{V,W}}{\cong}\text{Hom}(V,W)

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and since the composition of natural isomorphisms are a natural isomorphism we may concatenate to get V^\ast\otimes W\overset{\Pi_{V,W}}{\cong}\text{Hom}(V,W) where, as we have already stated, \Pi_{V,W}=\Delta_{V,W}\circ\Gamma_{V^\ast,W}. So:

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Theorem: Let F be a field and V,W finite dimensional F-spaces. Then, the map \Pi_{V,W}:V^\ast\otimes W\to\text{Hom}(V,W)) given on simple tensors by \Pi_{V,W}(\varphi\otimes w)(v)=\langle\varphi,v\rangle w is a natural isomorphism in both variables.

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Remark: One could have just, outright, defined \Pi 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 \Pi and just show it is a natural isomorphism.

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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 \langle -,-\rangle that we have been using induces a linear map T:V\otimes V^\ast\to \text{Hom}(V,W). Now, we know that V^\ast\otimes V\cong\text{Hom}(V,V)=\text{End}(V) and so we get from this a map \widetilde{T}:\text{End}(V)\to F defined by precomposing T with the isomorphism \text{End}(V)\to V^\ast\otimes V.

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Now, note that each time we fix an ordered basis \mathcal{B} of V we get an isomorphism \phi_\mathcal{B}:\text{Mat}_n(F)\to\text{End}(V) which is just the inverse of the isomorphism \text{End}(V)\to\text{Mat}_n(F):A\mapsto [A]_\mathcal{B}.

Let’s try to figure out what this map is. Fix a basis e_1,\cdots,e_n of V. We know then that we have the dual basis e_1^\ast,\cdots,e_n^\ast of V^\ast and so \{e_i^\ast\otimes e_j\}_{(i,j)\in[n]\times[n]} is a basis for V^\ast\otimes V. Moreover, if we define A_{i,j}:V\to V by A_{i,j}(e_k)=\delta_{i,k}e_j then \{A_{i,j}\}_{(i,j)\in[n]\times[n]} is a basis for \text{End}(V). Finally, it’s easy to see that the isomorphism \text{End}(V)\to V^\ast\otimes V satisfies A_{i,j}\mapsto e_i^\ast\otimes e_j.

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So, to see what our map \text{End}(V)\to F does it suffices to check what it does on the basis \{A_{i,j}\}. But, by definition \widetilde{T}(A_{i,j}) is just T acting on the image of A_{i,j} in V^\ast\otimes V, or T(e_i^\ast\otimes e_j). That said, by definition we have that T(e_i^\ast\otimes e_j)=\langle e_i^\ast,e_j\rangle=\delta_{i,j}. Thus, we see that \widetilde{T}(A_{i,j})=\delta_{i,j}. That said, consider the normal trace map \text{tr}:\text{Mat}_n(F)\to F by summing the diagonal entries of a matrix. We then can see that the choice of (ordered) basis \mathcal{B}=(e_1,\cdots,e_n) induces a map \text{tr}_\mathcal{B}:\text{End}(V)\to F defined by \text{tr}_\mathcal{B}(A)=\text{tr}([A]_\mathcal{B}) (where, of course, [A]_\mathcal{B} is the matrix of A with respect to \mathcal{B}). But, note that \text{tr}_\mathcal{B}(A_{i,j})=\delta_{i,j}–the same as our map \widetilde{T}. Thus, we see that \widetilde{T}=\text{tr}_\mathcal{B}. From this we see the following:

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Theorem: Let V be a finite dimensional vector space and fix an ordered basis \mathcal{B}. Then one has that the following diagram commutes

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\begin{array}{ccc}V^\ast\otimes V & \overset{\Pi_{V,V}}{\longrightarrow} & \text{End}(V)\\ & \searrow^T & \bigg\downarrow{\text{tr}_\mathcal{B}}\\ & & F\end{array}

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This allows us to conclude the following:

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Corollary: Let \mathcal{B} and \mathcal{B}' be two ordered bases of a finite dimensional vector space V. Then, the maps \text{tr}_\mathcal{B},\text{tr}_{\mathcal{B}'}:\text{End}(V)\to F are equal.

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This allows us to conclude that the trace map \text{End}(V)\to F defined by taking an operator A, taking its matrix with respect to any ordered basis, and then summing the diagonal entries, is well-defined.

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[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&gt;.


August 14, 2012 - Posted by | Algebra, Linear Algebra | , , , , ,

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