Frank Warner's definition of tensor product

Question

Frank Warner defines the tensor product between two finite dimensional vector spaces $V$ and $W$ as the quotient $F(V,W)/R(V,W)$ where $F(V,W)$ is the "free vector space over $\mathbb{R}$ whose generators are points of $V\times W$ and $R(V,W)$ is the subset whose elements ensure the bilinearity, that is, they are of one of the following forms:

$(v_1+v_2,w)-(v_1,w)-(v_2,w)$
$(v,w_1+w_2)-(v,w_1)-(v,w_2)$
$(av,w) -a(v,w)$
$(v,aw)-a(v,w)$

for $v,v_1,v_2\in V$, $w,w_1,w_2\in W$ and $a\in \mathbb{R}$.

Question: What is exactly the definition of "free vector space"? How is it constructed?

I've never heard of a "free vector space"; the author was probably trying to think of free modules.
The definition of tensor product as an algebraic quotient can be quite hard to work with. Here is a more direct definition. — avs, May 26 '19 at 19:02
Well, you can call a free module over a field a "free vector space". — Ivo Terek, May 26 '19 at 19:24
Well, what do you call the vector space that you need to construct with some specified basis $S$? — user10354138, May 26 '19 at 19:30

Ivo Terek · Accepted Answer · 2019-05-26T20:45:08.707

Given any set $S$, there is a unique (up to isomorphism) vector space which has the set $S$ as a basis. Namely, you look at the functions $\delta_s\colon S \to \{0,1\}$ defined by $$\delta_s(x) = \begin{cases} 1, \mbox{ if }x=s \\ 0, \mbox{ else} \end{cases}. $$Then consider $F(S) = \{f\colon S \to \Bbb R \mid f(s) = 0 \mbox{ for all but finitely many }s \in S \}$. This is a vector space with operations inherited from the codomain $\Bbb R$. The catch is that $\{\delta_s \mid s \in S\}$ is a basis for $F(S)$. And we identify this basis with $S$ itself. Proof:

$S$ is linearly independent: if $$\lambda_1\delta_{s_1}+\cdots +\lambda_n \delta_{s_n} = 0,$$evaluate both sides at $s_i \in S$ to get $\lambda_i = 0$.
$S$ generates $F(S)$: let $f \in F(S)$, and let $s_1,\ldots, s_n \in S$ be all the elements of $S$ for which $f$ does not vanish. Then $$f = f(s_1)\delta_{s_1}+\cdots + f(s_n)\delta_{s_n}.$$

score 1 · Answer 2 · answered May 26 '19 at 19:23

The free vector space $F(S)$ on some set $S$ is just a vector space (in this case, over $\mathbb{R}$) with a (Hamel) basis $S$. Although $S$ might had prior additional structures (as in this case, when $S=V\times W$ is a vector space in its own right), we do not impose them on $F(S)$.

So one more concrete way of thinking about $F(S)$ is the set of all functions $f\colon S\to \mathbb{R}$ which is $0$ except for finitely many points, equipped with the pointwise addition and multiplication by $\mathbb{R}$. Your $F(V,W)$ is a huge vector space, $R(V,W)$ is also very large, but the quotient is manageable.

This construction of the tensor product is really only of interest for showing its existence.

Frank Warner's definition of tensor product

2 Answers2