Mathematics.

field theory

Tensor Products

Abstract Algebra II40 minDifficulty8 out of 10

You should know: vector spaces over general fields

Overview

The tensor product V ⊗_F W of two vector spaces over a field F is the universal home for bilinear maps: rather than building a new space directly, it is defined by the property that every bilinear map out of V × W factors uniquely through a single linear map out of V ⊗ W. This 'linearize the bilinear' trick converts multilinear problems into linear-algebra problems, underlies the definition of tensors in physics and differential geometry, lets scalars be extended (base change) from F to a bigger field, and generalizes cleanly from vector spaces to modules over a ring, where it becomes one of the central constructions of commutative algebra.

Intuition

A bilinear map is linear in each variable separately but not jointly, which makes it awkward to manipulate with ordinary linear-algebra tools. The tensor product manufactures a new vector space V⊗W and a fixed bilinear map into it so that every other bilinear map factors through this one — meaning any bilinear question about V × W can be answered by an ordinary linear question about V⊗W instead. Concretely, if V has basis {e_i} and W has basis {f_j}, then V⊗W has basis {e_i⊗f_j}, so tensoring literally multiplies dimensions: an m-dimensional space tensored with an n-dimensional space gives an mn-dimensional space, matching how bilinear forms are described by m×n matrices.

Formal Definition

Definition

Let V, W be vector spaces over a field F. The tensor product V ⊗_F W is a vector space equipped with a bilinear map ⊗ : V × W → V⊗W satisfying the universal property: for every vector space U and bilinear map β : V × W → U, there is a unique linear map β̃ making the diagram commute.

⊗:V×W→V⊗FW,(v,w)↦v⊗w\otimes : V \times W \to V \otimes_F W, \qquad (v,w) \mapsto v \otimes w
The canonical bilinear map
∀ bilinear β:V×W→U, ∃! linear β~:V⊗W→U with β~(v⊗w)=β(v,w)\forall \text{ bilinear } \beta: V\times W \to U,\ \exists!\ \text{linear}\ \tilde\beta: V\otimes W \to U \ \text{with}\ \tilde\beta(v\otimes w)=\beta(v,w)
Universal property
(v1+v2)⊗w=v1⊗w+v2⊗w,v⊗(w1+w2)=v⊗w1+v⊗w2,(av)⊗w=a(v⊗w)(v_1+v_2)\otimes w = v_1\otimes w + v_2\otimes w, \quad v\otimes(w_1+w_2)=v\otimes w_1+v\otimes w_2, \quad (av)\otimes w = a(v\otimes w)
Bilinearity relations defining ⊗
dim⁡F(V⊗FW)=dim⁡F(V)⋅dim⁡F(W)\dim_F(V\otimes_F W) = \dim_F(V)\cdot \dim_F(W)
Dimension count for finite-dimensional spaces

Properties

Basis of a tensor product

If {ei}i=1m is a basis of V and {fj}j=1n a basis of W, then {ei⊗fj} is a basis of V⊗W.\text{If } \{e_i\}_{i=1}^m \text{ is a basis of } V \text{ and } \{f_j\}_{j=1}^n \text{ a basis of } W, \text{ then } \{e_i \otimes f_j\} \text{ is a basis of } V \otimes W.

Symmetry

V⊗FW≅W⊗FVvia v⊗w↦w⊗v.V \otimes_F W \cong W \otimes_F V \quad \text{via } v \otimes w \mapsto w \otimes v.

Associativity

(U⊗V)⊗W≅U⊗(V⊗W).(U \otimes V) \otimes W \cong U \otimes (V \otimes W).

Identity element

F⊗FV≅Vvia a⊗v↦av.F \otimes_F V \cong V \quad \text{via } a \otimes v \mapsto av.

Base change (scalar extension)

If F⊆K is a field extension, K⊗FV is a K-vector space with dim⁡K(K⊗FV)=dim⁡F(V).\text{If } F \subseteq K \text{ is a field extension, } K \otimes_F V \text{ is a } K\text{-vector space with } \dim_K(K\otimes_F V) = \dim_F(V).

Not every element is a simple tensor

A general element of V⊗W is a finite sum ∑ivi⊗wi, not necessarily a single v⊗w.\text{A general element of } V \otimes W \text{ is a finite sum } \sum_i v_i \otimes w_i, \text{ not necessarily a single } v\otimes w.

Worked Examples

  1. ℝ² has basis {e₁,e₂} and ℝ³ has basis {f₁,f₂,f₃}.

    dim⁡(R2)=2,dim⁡(R3)=3\dim(\mathbb{R}^2)=2,\quad \dim(\mathbb{R}^3)=3

  2. The tensor product basis consists of all pairwise products e_i⊗f_j, giving 2×3 = 6 basis vectors.

    {e1⊗f1,e1⊗f2,e1⊗f3,e2⊗f1,e2⊗f2,e2⊗f3}\{e_1\otimes f_1, e_1\otimes f_2, e_1\otimes f_3, e_2\otimes f_1, e_2\otimes f_2, e_2\otimes f_3\}

  3. So the dimension multiplies as expected.

    dim⁡(R2⊗RR3)=2×3=6\dim(\mathbb{R}^2\otimes_\mathbb{R}\mathbb{R}^3) = 2\times 3 = 6

Answer: ℝ² ⊗ ℝ³ is 6-dimensional, with basis {eᵢ⊗fⱼ : i=1,2, j=1,2,3} — matching ℝ^{2×3}, the space of 2×3 matrices.

Practice Problems

Difficulty 6/10

What is dim_ℚ(ℚ³ ⊗_ℚ ℚ⁵)?

Difficulty 7/10

Using the universal property, explain why bilinear maps V × W → U correspond exactly to linear maps V ⊗ W → U.

Difficulty 8/10

Explain how base change K ⊗_F V lets you 'complexify' a real vector space, and compute dim_ℂ(ℂ ⊗_ℝ ℝ³).

Common Mistakes

Common Mistake

Thinking V ⊗ W is the same as the Cartesian product V × W or the direct sum V ⊕ W.

V×W and V⊕W both have dimension dim(V)+dim(W) and consist of literal pairs; V⊗W has dimension dim(V)·dim(W) and its general element is a sum of simple tensors, not a pair.

Common Mistake

Assuming every element of V⊗W can be written as a single simple tensor v⊗w.

Simple tensors only span the space; a generic element such as e₁⊗f₂+e₂⊗f₁ in ℝ²⊗ℝ² requires a genuine sum and cannot be reduced to one simple tensor.

Quiz

The tensor product V ⊗_F W is characterized by the property that:
For finite-dimensional V, W over a field F, dim_F(V ⊗_F W) equals:
Which statement about elements of V ⊗ W is correct?

Summary

  • V ⊗_F W is defined by a universal property: bilinear maps out of V×W correspond exactly to linear maps out of V⊗W.
  • For finite-dimensional spaces, dim(V⊗W) = dim(V)·dim(W), with basis {eᵢ⊗fⱼ} built from bases of V and W.
  • Not every element of V⊗W is a simple tensor v⊗w — general elements are finite sums ∑vᵢ⊗wᵢ.
  • Base change K⊗_F V extends scalars from F to a larger field K, e.g. complexifying a real vector space via ℂ⊗_ℝ V.
  • Tensor products generalize from vector spaces to modules over a ring, becoming a core tool in commutative algebra and algebraic geometry.

References

  1. WebsiteWikipedia — Tensor product
  2. WebsiteWikipedia — Tensor product of modules