Abstract Nonsense

Crushing one theorem at a time

Some Natural Identifications (Pt. III)


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

\text{ }

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).

\text{ }

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

\text{ }

\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

\text{ }

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.

\text{ }

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}

\text{ }

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

\text{ }

(\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

\text{ }

and

\text{ }

\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}

\text{ }

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

\text{ }

\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}

\text{ }

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

\text{ }

(\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)

\text{ }

and

\text{ }

\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}

\text{ }

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

\text{ }

\text{ }

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&gt;.

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

1 Comment »

  1. […] Some Natural Identifications (Pt. IV) Point of Post: This is a continuation of this post. […]

    Pingback by Some Natural Identifications (Pt. IV) « Abstract Nonsense | August 14, 2012 | Reply


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