Take vector spaces $V, W$ over the field $\mathbb{K}$. I've come across two different definitions of tensor product $V \otimes W$ and was wondering whether they are the same thing.

Definition 1. $V \otimes W$ is the quotient space $F(V\times W) / \sim$ where $F(V\times W)$ is the free vector space over $V \times W$ and $\sim$ denotes the following relations:

$$(v_1+v_2, w)\sim (v_1, w)+(v_2,w), \qquad (v, w_1+w_2)\sim (v, w_1) + (v, w_2), $$ $$(\lambda v, w) \sim (v, \lambda w)\sim \lambda (v, w).$$

In other words $V \otimes W$ is the space of formal linear combinations

$$\sum_{j=1}^n \alpha_j v_j \otimes w_j, \qquad \alpha_j \in \mathbb{K}, v_j \in V, w_j \in W$$

where we explicitly require $\otimes$ to be bilinear.

Definition 2 (When $V, W$ are finite-dimensional.) $V \otimes W$ is the space of mappings from $V^\star\times W^\star$ into $\mathbb{K}$ that are linear in each variable.

Question. Under what circumstances do those two definitions coincide (up to canonical isomorphism)? I guess that this happens if and only if both $V$ and $W$ are finite-dimensional.


2 Answers 2


For each pair $(V,W)$ of vector spaces (over a fixed ground field), let $T(V,W)$ be their tensor product, and $F(V,W)$ the vector space of bilinear forms on $V^*\times W^*$. One checks that

(a) there is a unique linear map $e(V,W)$ from $T(V,W)$ to $F(V,W)$ satisfying $$ \big(e(V,W)(v\otimes w)\big)(f,g)=f(v)g(w) $$ for all $v\in V,w\in W,f\in V^*,g\in W^*$,

(b) $e(V,W)$ is injective,

(c) $T$ and $F$ are functors,

(d) $e$ is a natural transformation from $T$ to $F$,

(e) $T,F$ and $e$ are compatible (in an obvious sense) with finite direct sums.

Claim 1: $e(V,W)$ is surjective $\iff$ the cardinal number $\dim(V)\dim(W)$ is finite.

In view of (b), implication "$\Leftarrow$" follows by dimension counting. It suffices thus to prove the non-surjectivity when $V$ is infinite dimensional and $W$ nonzero. Writing $W$ as $W_1\oplus W_2$ with $\dim W_1=1$ and using (b), we are reduced to

Claim 2: if $V$ is infinite dimensional, then the canonical embedding $V\to V^{**}$ is not surjective.

To prove this, we'll use an embedding of $V$ in $V^*$, and an embedding of $V^*$ in $ V^{**}$. None of these two embeddings will be canonical, but their composition will.

Choose a basis $B$ of $V$, and identify $V$ to the space $K^{(B)}$ of finitely supported $K$-valued functions on $B$. Then $V^*$ can be identified to the space $K^B$ of all $K$-valued functions on $B$. Similarly, we can identify $V^*$ to $K^{(B\sqcup C)}$, where $C$ is a set and $\sqcup$ means "disjoint union". As $B$ is infinite, $C$ is nonempty. Using the same trick once more, we can identify $V^{**}$ to $K^{(B\sqcup C\sqcup D)}$, where $D$ is a nonempty set. Then the natural embedding of $K^{(B)}$ in $K^{(B\sqcup C\sqcup D)}$, which is clearly not surjective, corresponds to the natural embedding of $V$ in $V^{**}$. This completes the proof.


They always coincide because they both fulfill the universal property of the tensor product.

  • 7
    $\begingroup$ The asker wants to go beyound the finite-dimensional case. $\endgroup$ Oct 9, 2011 at 13:26

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