Abstract Nonsense

Crushing one theorem at a time

Tensor Product


Point of post: In this post I will discuss the very basic, and simple minded, definition of the tensor product \mathscr{V}\otimes\mathscr{W} of finite dimensional vector spaces \mathscr{V} and \mathscr{W} and it’s consequences, as is outlined in Halmos (viz. reference 1).

Nota Bene: The following may seem to be a far-cry from the typical definition of the tensor product as \mathcal{F}\left(\mathscr{U}\times\mathscr{V}\right)/\sim where \mathcal{F}\left(\mathscr{U}\times\mathscr{V}\right) is the free vector space and \sim is the usual equivalence relation. That said, the following is a fairly large amount of theoretical buck for a fairly small complexity buck.

Motivation

In the  last post we discussed how given vector spaces \mathscr{U},\mathscr{V} over a field F there is a canonical way to form the vector space of all bilinear forms on \mathscr{U}\boxplus\mathscr{V}, denoted \text{Bil}\left(\mathscr{U},\mathscr{V}\right). But, as is fast becoming a motif in our studies we begin with a vector space \mathscr{W} and the study it’s dual space, as always denoted either \text{Hom}\left(\mathscr{W},F\right) or \mathscr{W}^*. For the case of \text{Bil}\left(\mathscr{U},\mathscr{V}\right) we make a small notational change. Instead of denoting the dual space of \text{Bil}\left(\mathscr{U},\mathscr{V}\right) by \text{Hom}\left(\text{Bil}\left(\mathscr{U},\mathscr{V}\right),F\right) we denote it by \mathscr{U}\otimes\mathscr{V} and call it the tensor product of \mathscr{U} and \mathscr{V}.

What’s the motivation for this space? Why is it important, and more importantly how can we get a feel for it. Consider \mathbb{C}^n. This is as innocuous of a vector space as one is liable to come across. But, for a second let’s think of it instead of being a set of n-tuples, but instead of a set of functions. Namely, let’s define

\mathbf{C}^n=\left\{f \mid f:\{1,\cdots,n\}\to\mathbb{C}\right\}

We can define \mathbf{C}^m similarly. Evidently \mathbf{C}^n\cong\mathbb{C}^n but, really indulging in a logical inaccuracy we can really think of them as being the same in a very natural way. That said, we can clearly consider the vector space

\mathbf{V}=\left\{f\mid f:\{1,\cdots,n\}\times\{1,\cdots,m\}\to\mathbb{C}\right\}

Once again, one can easily see that \mathbf{V}\cong\mathbb{C}^{nm}. Clearly then \mathbf{V} is intimately related to \mathbf{C}^n and \mathbf{C}^m, but the million dollar question is…how? A quick dimension argument shows that  \mathbf{V} is not isomorphic to \mathbf{C}^n\boxplus\mathbf{C}^m. Using another dimension argument we can see that \mathbf{V}\cong\text{Bil}\left(\mathbf{C}^n,\mathbf{C}^m\right) but this doesn’t seem to catch the aforementioned feel we are seeking, this isomorphism doesn’t have a natural quality about it. Similarly, what if we considered

\mathbb{C}_{n+1}[x]=\left\{a_1+\cdots+a_nx^n:a_1,\cdots,a_n\in\mathbb{C}\right\}

and

\mathbb{C}_{m+1}[y]=\left\{a_0+\cdots+a_my^m:a_1,\cdots,a_m\in\mathbb{C}\right\}

Then how do these two related to

\mathbb{C}_{n+1,m+1}[x,y]=\left\{a_{0,0}+a_{1,0}x+a_{0,1}y+a_{1,1}xy+\cdots+a_{n,m}x^ny^m:a_{0,0},a_{1,0},\cdots,a_{m,n}\in\mathbb{C}\right\}

Both of these give a feel about what the tensor product is. It takes a vector space “acting” on n object and one “acting” on m objects and produces a vector space “acting” on nm objects.

 

Tensor Product

Formally, for vector spaces \mathscr{U} and \mathscr{V} over the field F we define \mathscr{U}\otimes\mathscr{V}, called the tensor product of \mathscr{U} and \mathscr{V}, to be

\mathscr{U}\otimes\mathscr{V}=\left\{\varphi:\text{Bil}\left(\mathscr{U},\mathscr{V}\right)\to F:\varphi\left(\alpha B+\beta B'\right)=\alpha\varphi\left(B\right)+\beta \varphi\left(B'\right)\right\}=\text{Hom}\left(\text{Bil}\left(\mathscr{U},\mathscr{V}\right),F\right)

And for u\in\mathscr{U} and v\in\mathscr{V} we define the tensor product of u and v by

u\otimes v:\text{Bil}\left(\mathscr{U},\mathscr{V}\right)\to F:B\mapsto B(u,v)

from where it’s clear that u\otimes v\in\mathscr{U}\otimes\mathscr{V}.

Remark: From here on it it’s assumed that \mathscr{U} is a vector space of dimension n, \mathscr{V} a vector space of dimension n, both over a field F.

Theorem: Let u_1,u_2\in\mathscr{U}, v_1,v_2\in\mathscr{V}, and \alpha_1,\alpha_2,\beta_1,\beta_2\in F. Then, for any v\in\mathscr{V}

(\alpha_1 u_1+\alpha_2 u_2)\otimes v=\alpha_1(u_1\otimes v)+\alpha_2(u_2\otimes v)

and for any u\in\mathscr{U}

u\otimes(\beta_1v_1+\beta_2v_2)=\beta_1(u\otimes v_1)+\beta_2(u\otimes v_2)

where addition and multiplication are the usual operations with linear functionals.

Proof: Let B\in\text{Bil}\left(\mathscr{U},\mathscr{V}\right) then

\begin{aligned}\left((\alpha_1u_1+\alpha_2u_2)\otimes v\right)\left(B\right) &= B\left(\alpha_1u_1+\alpha_2u_2,v\right)\\ &= \alpha_1 B(u_1,v)+\alpha_2B(u_2,v)\\ &= \left(\alpha_1(u_1\otimes v)\right)\left(B\right)+\left(\alpha_2(u_2\otimes v)\right)\left(B\right)\end{aligned}

But, since B was arbitrary it follows that (\alpha_1u_1+\alpha_2u_2)\otimes v=\alpha_1(u_1\otimes v)+\alpha_2(u_2\otimes v) as required. The case for the second slot is done exactly the same. \blacksquare

Basis and Dimensionality

We are now prepared to show that the tensor product’s namesake is well-deserved, namely that it’s a product of some kind. Namely, we will show that \mathscr{U}\otimes\mathscr{V} admits a very natural basis.

Theorem: Let \{x_1,\cdots,x_n\} be a basis for \mathscr{U} and \{y_1,\cdots,y_m\} a basis for \mathscr{V}. Then, \left\{x_i\otimes y_j:i\in[n]\text{ and }j\in[m]\right\} is a basis for \mathscr{U}\otimes\mathscr{V}.

Proof: Recall that a basis for \text{Bil}\left(\mathscr{U},\mathscr{V}\right) is \left\{B_{i,j}:i\in[n]\text{ and }j\in[m]\right\} where B_{i,j}(x_p,y_q)=\delta_{i,p}\delta_{q,j}. Thus, we know that this basis admits a dual basis \left\{\varphi_{i,j}:i\in[n]\text{ and }j\in[m]\right\}  for \mathscr{U}\otimes\mathscr{V} by \varphi_{i,j}\left(B_{p,q}\right)=\delta_{i,p}\delta_{q,j}. Thus, it suffices to check that x_i\otimes y_j=\varphi_{i,j}. To do this we merely note that

\begin{aligned}\left(x_i\otimes y_j\right)\left(B_{p,q}\right) &= B_{p,q}\left(x_i,y_j\right)\\ &= \delta_{p,i}\delta_{j,q}\\ &= \delta_{i,p}\delta_{q,j}\\ &= \varphi_{i,j}\left(B_{p,q}\right)\end{aligned}

and since a linear functional is determined entirely on the a basis it follows that x_i\otimes y_j=\varphi_{i,j}. The conclusion follows. \blacksquare

Corollary: \dim_F\left(\mathscr{U}\otimes\mathscr{V}\right)=\dim_F\left(\mathscr{U}\right)\cdot\dim_F\left(\mathscr{V}\right)

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print.

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October 29, 2010 - Posted by | Algebra, Linear Algebra | , , ,

2 Comments »

  1. […] of post: This will be the solutions to the problems in Halmos’s book corresponding to this […]

    Pingback by Halmos Section 24 and 25: Tensor Product and Product Bases « Abstract Nonsense | November 1, 2010 | Reply

  2. […] have previously discussed the notion of the tensor product, but will be considering a slightly different (although […]

    Pingback by Representation Theory: The Tensor Product and the Tensor Product of Representations « Abstract Nonsense | February 14, 2011 | Reply


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