The Unapologetic Mathematician

Tensor Products and Bases

Since we’re looking at vector spaces, which are special kinds of modules, we know that $\mathbf{Vec}(\mathbb{F})$ has a tensor product structure. Let’s see what this means when we pick bases.

First off, let’s remember what the tensor product of two vector spaces $U$ and $V$ is. It’s a new vector space $U\otimes V$ and a bilinear (linear in each of two variables separately) function $B:U\times V\rightarrow U\otimes V$ satisfying a certain universal property. Specifically, if $F:U\times V\rightarrow W$ is any bilinear function it must factor uniquely through $B$ as $F=\bar{F}\circ B$. The catch here is that when we say “linear” and “bilinear” we mean that the functions preserve both addition and scalar multiplication. As with any other universal property, such a tensor product will be uniquely defined up to isomorphism.

So let’s take finite-dimensional vector spaces $U$ and $V$, and bases $\left\{e_i\right\}$ of $U$ and $\left\{f_j\right\}$ of $V$. I say that the vector space with basis $\left\{e_i\otimes f_j\right\}$, and with the bilinear function $B(e_i,f_j)=e_i\otimes f_j$ is a tensor product. Here the expression $e_i\otimes f_j$ is just a name for a basis element of the new vector space. Such elements are indexed by the set of pairs $(i,j)$, where $i$ indexes a basis for $U$ and $j$ indexes a basis for $V$.

First off, what do I mean by the bilinear function $B(e_i,f_j)=e_i\otimes f_j$? Just as for linear functions, we can define bilinear functions by defining them on bases. That is, if we have $u=u^ie_i\in U$ and $v=v^jf_j\in V$, we get the vector

$B(u,v)=B(u^ie_i,v^je_j)=u^iv^jB(e_i,f_j)=u^iv^je_i\otimes f_j$

in our new vector space, with coefficients $u^iv^j$.

So let’s take a bilinear function $F$ and define a linear function $\bar{F}$ by setting

$\bar{F}(e_i\otimes f_j)=F(e_i,f_j)$

We can easily check that $F$ does indeed factor as desired, since

$\bar{F}(B(e_i,f_j))=\bar{F}(e_i\otimes f_j)=F(e_i,f_j)$

so $F=\bar{F}\circ B$ on basis elements. By linearity, they must agree for all pairs $(u,v)$. It should also be clear that we can’t define $\bar{F}$ any other way and hope to satisfy this equation, so the factorization is unique.

Thus if we have bases $\left\{e_i\right\}$ of $U$ and $\left\{f_j\right\}$ of $V$, we immediately get a basis $\left\{e_i\otimes f_j\right\}$ of $U\otimes V$. As a side note, we immediately see that the dimension of the tensor product of two vector spaces is the product of their dimensions.

May 23, 2008 Posted by | Algebra, Linear Algebra | 5 Comments