# 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.

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Posted by John Armstrong | Algebra, Linear Algebra

## 5 Comments	»

1. [...] Given two finite-dimensional vector spaces and , with bases and respectively, we know how to build a tensor product: use the basis [...]

Pingback by | May 26, 2008 | Reply

2. [...] Like we saw with the tensor product of vector spaces, the dual space construction turns out to be a functor. In fact, it’s a [...]

Pingback by | May 28, 2008 | Reply

3. [...] a bilinear multiplication, which is just a linear map from the tensor square to . And we know that the basis for a tensor product consists of pairs of basis elements. So we can specify this linear map on a basis and extend by linearity — bilinearity — [...]

Pingback by | July 28, 2008 | Reply

4. [...] on what such a thing looks like by assuming has finite dimension and picking a basis . Then we have bases for tensor powers: a basis element of the th tensor power is given by an -tuple of basis elements for . We’ll [...]

Pingback by | December 22, 2008 | Reply

5. [...] go. You can also review tensor products in the context of vector spaces and linear transformations here. What we want to think of here is that the matrix shuffles around the two copies of the irrep , [...]

Pingback by | October 5, 2010 | Reply

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