# Two definitions of tensor product: when are they equivalent?

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.

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.

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