# Abstract Nonsense

## 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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commutes.

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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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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.