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I started self-learning category theory where I encountered the universal property of tensor products. Then there are a few problems that I have no ideas on how to get started on:

Question: Let $U,V,W$ be vector spaces, use universal property to show that $U\otimes(V\otimes W)\cong(U\otimes V)\otimes W$ and $U\otimes V\cong V\otimes U.$

I learnt a few ways of constructing tensor products, which can help proving these statements but I want to use the universal property instead.

Universal Property: Let $U$, $V$ be two vector spaces over the same field then there exists a unique vector space $U\otimes V$ and a bilinear map $b:U\times V\to U\otimes V$ such that for any bilinear map $f:U\times V\to X$, where $X$ is a vector space, there exists a linear map $\tilde f:U\otimes V\to X$ such that $f=\tilde f b.$

My some initial thoughts: Say we focus on $U\otimes(V\otimes W)\cong(U\otimes V)\otimes W$ first. To construct an isomorphism, we can find linear maps going in both directions. To get such linear map, we can use Universal Property, this means that we need to use $U\times (V\otimes W)$ and $(U\otimes V)\times W$ instead? But I am not really sure how to construct a bilinear map and indeed I am not sure if I am on the right path also.

Thank you so much in advance!!

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2 Answers 2

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The easiest way to show associativity is using the tensor-hom adjunction. $\DeclareMathOperator{Hom}{Hom}$

$\begin{equation} \begin{split} \Hom(U \otimes (V \otimes W), C) &\cong \Hom(U, \Hom(V \otimes W, C)) \\ &\cong \Hom(U, \Hom(V, \Hom(W, C))) \\ &\cong \Hom(U \otimes V, \Hom(W, C)) \\ &\cong \Hom((U \otimes V) \otimes W, C) \end{split} \end{equation}$

This isomorphism is natural in $U, V, W,$ and $C$.

Then by the Yoneda lemma, we have an isomorphism $U \otimes (V \otimes W) \cong (U \otimes V) \otimes W$, natural in $V, W, U$.

This can be translated into a proof using only the universal property of the tensor product. The key is to note that maps $U \otimes (V \otimes W) \to C$ and maps $(U \otimes V) \otimes W \to C$ both correspond to tri-linear maps $U \times V \times W \to C$.

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You could easily show that by using the Hom-Adjunction, as shown in the other answer. However, you can use the universal property directly to show the associativity as well. For that method you are on the right track.

You need to show that the LHS fulfils the universal property of the RHS or the other way round, i.e. that $U\otimes (V\otimes W)$ fulfils the universal property of $(U\otimes V)\otimes W$.

Take the map $$ b: (U\otimes V) \times W \to U \otimes (V\otimes W) \\ (u \otimes v), w \mapsto u \otimes (v\otimes w) $$ you now need to show that it fulfils the universal property. Take any bilinear map $f:(U\otimes V) \times W \to X$, we can construct the bilinear map $$ \tilde{f}:U \otimes (V\otimes W) \to X \\ u\otimes (v\otimes w) \mapsto f(u\otimes v, w) $$

Obviously $f = \tilde{f} \circ b$, we need to prove that $\tilde{f}$ is well defined and indeed linear. Both properties come directly from the bilinearity of $f$: For any $\lambda \in R$ we have: $$ \tilde{f}((\lambda u)\otimes(v\otimes w))=f((\lambda u)\otimes v, w)=\lambda f(u\otimes v, w) = \lambda \tilde{f}(u\otimes(v\otimes w)) $$ and $$ =\lambda f(u\otimes v,w) = f(u\otimes v, \lambda w) = \tilde{f}(u \otimes (v\otimes (\lambda w))) $$ so the tensor product $(v\otimes w)$ doesn't make a problem.

As you can see it's basically just writing things down as the linearity of $\tilde{f}$ is directly coming from the linearity of $f$.

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