# Abstract Nonsense

## Some Natural Identifications (Pt. III)

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

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We begin by defining a map $V\times W\to\text{Hom}(V^\ast,W)$ by having $(v,w)$ map to the map $M(u,v)$ defined by the following: $M(v,w)(\varphi)=\langle\varphi,v\rangle w$. In words the map associated to $(v,w)$ takes a linear functional $\varphi$ and maps it to the element of $W$ gotten by evaluating $\varphi$ at $v$ and then multiplying this by $w$. This is map is clearly bilinear and so by the universal characterization of tensor products we can lift this to a map $\Gamma_{V,W}:V\otimes W\to\text{Hom}(V^\ast,W)$ with the property that $\Gamma_{V,W}(v\otimes w)=M(v,w)$.

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Remark: We obviously index $\Gamma$ with $V,W$ to show which spaces the map is defined on and mapped to (since we have such a map for every pair of spaces), but when the context is clear we shall omit this and just write $\Gamma$.

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Now, to prove this is an isomorphism it suffices to prove that $\Gamma$ is surjective since both spaces are of the same finite dimension. To do this it suffices to show that a basis of $\text{Hom}(V^\ast,W)$ is hit. To do this we let $e_1,\cdots,e_n$ be a basis for $V$ and $b_1,\cdots,b_m$ a basis for $W$. We know then that a basis for $\text{Hom}(V^\ast,W)$ is the set of maps $\{T_{i,j}\}_{(i,j)\in[n]\times[m]}$ defined by $T_{i,j}(e_k^\ast)=\delta_{i,k}b_j$. That said, note that

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$\Gamma(e_i\otimes b_j)(e_k^\ast)=\Gamma(e_i,b_j)(e_k^\ast)=\langle e_k^\ast,e_i\rangle b_j=\delta_{i,k}b_j$

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Thus we see that $\Gamma(e_i\otimes b_j)$ and $T_{i,j}$ agree on a basis and thus must be equal. Thus, we see that $\Gamma$ hits a basis of $\text{Hom}(V^\ast,W)$ and thus (as previously mentioned) must be an isomorphism.

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It remains to show that this isomorphism is natural in both variables. Let’s show that it is natural in $V$. This amounts to showing that if $f:V\to U$ is a linear map the following diagram commutes

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$\begin{array}{ccc} V\otimes W & \overset{\Gamma_{V,W}}{\longrightarrow} & \text{Hom}(V^\ast,W)\\ ^{f\otimes 1}\big\downarrow & & \big\downarrow^{\ell}\\ U\otimes W & \underset{\Gamma_{U,W}}{\longrightarrow} & \text{Hom}(U^\ast,W)\end{array}$

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where $f\otimes 1$ is the usual map (which acts on simple tensors as $(f\otimes 1)(v,w)=f(v)\otimes w$) and $\ell$ is the God-awful map defined by taking $h:V^\ast\to W$ and mapping it to $h\circ f^\ast$ where $(f^\ast(\varphi))(v)=\langle\varphi,f(v)\rangle$. Ok, so we take an arbitrary $v\otimes w\in V\otimes W$ (it suffices to check commutativity on simple tensors since they are a generating set)and $\varphi\in U^\ast$ and check that

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$(\Gamma_{U,W}\circ (f\otimes 1)(v\otimes w)(\varphi)=\Gamma_{U,W}(f(v)\otimes w)(\varphi)=\langle \varphi,f(v)\rangle w$

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and

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\begin{aligned}(\ell\circ\Gamma_{V,W})(v\otimes w)(\varphi) &=(\Gamma_{V,W}(v\otimes w)\circ f^\ast)(\varphi)\\ &=\Gamma_{V,W}(v\otimes w)(\varphi\circ f)\\ &=(\varphi\circ f)(v)w\\ &=\langle \varphi,f(v)\rangle w\end{aligned}

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Yay! So naturality in the more difficult variable is verified. Now, to verify that there is naturality in the $W$-variable we need to suppose that we have map $g:W\to U$ and check that the following diagram commutes

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$\begin{array}{ccc}V\otimes W & \overset{\Gamma_{V,W}}{\longrightarrow} & \text{Hom}(V^\ast,W)\\ ^{1\otimes g}\big\downarrow & & \big\downarrow^{k}\\ V\otimes U & \underset{\Gamma_{V,U}}{\longrightarrow} & \text{Hom}(V^\ast,U)\end{array}$

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where $1\otimes g$ is the usual map which acts on simple tensors as $(1\otimes g)(v\otimes w)=v\otimes g(w)$ and $k$ takes a map $h:V^\ast\to W$ and maps it to $g\circ h:V^\ast\to U$. To check this we let (for the same reasons) $v\otimes w\in V\otimes W$ and $\varphi\in V^\ast$ be arbitrary, we see then that

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$(\Gamma_{V,U}\circ(1\otimes g))(v\otimes w)(\varphi)=\Gamma_{V,U}(v\otimes g(w))(\varphi)=\langle \varphi,v\rangle g(w)$

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and

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\begin{aligned}(k\circ\Gamma_{V,W})(v\otimes w)(\varphi) &=(g\circ \Gamma_{V,W}(v\otimes w))(\varphi)\\ &=g(\Gamma_{V,W}(v\otimes w)(\varphi))\\ &=g(\langle \varphi,v\rangle w)=\langle \varphi,v\rangle g(w)\end{aligned}

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where the last step follows from the fact that $g$ is linear!

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Thus, summing this all up we may conclude the following:

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Theorem: Let $F$ be a field and $V$ and $W$ finite dimensional $F$-spaces. Then, $V\otimes_F W$ is naturally isomorphic to $\text{Hom}_F(V^\ast,W)$ (in both variables) via the map $\Gamma_{V,W}$ which acts on simple tensors as $\Gamma_{V,W}(v\otimes w)(\varphi)=\langle \varphi,v\rangle w$

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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.

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