Abstract Nonsense

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.