Tensor product

Intuition [Beginner]

A tensor product is a way to combine two independent vector inputs into one new kind of vector.

Think of choosing a color and a direction. "Red pointing east" is not just a color and not just a direction. It is a paired object that remembers both choices. If colors can be blended and directions can be added, the paired objects should respect both kinds of linear mixing.

Tensor products matter because geometry often carries several linear ingredients at once: directions, forms, spinors, matrices, and bundle fibers.

Visual [Beginner]

The tensor product stores paired vector data in a single combined space.

The combined space is built so pairing behaves linearly in each input.

Worked example [Beginner]

Take one space with basis vectors red and blue. Take another space with basis vectors left and right.

The combined space has four basic pairs:

red-left, red-right, blue-left, blue-right.

If a color is half red plus half blue, pairing it with right gives half of red-right plus half of blue-right.

What this tells us: tensor product turns paired linear choices into ordinary vectors in a larger space.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Let $V$ and $W$ be vector spaces over a field $K$ 01.01.01, 01.01.03. A tensor product of $V$ and $W$ is a vector space $V{\otimes }_{K}W$ together with a bilinear map

$$\tau :V\times W\to V{\otimes }_{K}W$$

such that every bilinear map $B:V\times W\to U$ into a vector space $U$ factors uniquely through a linear map $\tilde{B}:V{\otimes }_{K}W\to U$:

$$B=\tilde{B}\circ \tau .$$

The element $\tau (v,w)$ is written $v\otimes w$ and is called a pure tensor. The defining relations are

$$(v+v^{\prime })\otimes w=v\otimes w+v^{\prime }\otimes w,v\otimes (w+w^{\prime })=v\otimes w+v\otimes w^{\prime },$$

and

$$(av)\otimes w=a(v\otimes w)=v\otimes (aw).$$

Key theorem with proof [Intermediate+]

Theorem (basis of a tensor product). Let $V$ have basis $e_1,\dots ,e_m$ and let $W$ have basis $f_1,\dots ,f_n$. Then the elements $e_i\otimes f_j$ form a basis of $V{\otimes }_{K}W$.

Proof. First show spanning. Write

$$v=\sum _i{a}_i{e}_i,w=\sum _j{b}_j{f}_j.$$

Bilinearity gives

$$v\otimes w=\sum _{i,j}{a}_i{b}_j(e_i\otimes f_j).$$

Since pure tensors generate the tensor product, the displayed tensors span.

For linear independence, define for each pair $(r,s)$ a bilinear map $B_{rs}:V\times W\to K$ by

$$B_{rs}(e_i,f_j)=\{\begin{array}{ll}1,& (i,j)=(r,s),\\ 0,& \text{otherwise}.\end{array}$$

By the universal property, $B_{rs}$ induces a linear map ${\tilde{B}}_{rs}:V{\otimes }_{K}W\to K$. If

$$\sum _{i,j}{c}_{ij}(e_i\otimes f_j)=0,$$

applying ${\tilde{B}}_{rs}$ gives $c_{rs}=0$. This holds for every pair $(r,s)$, so all coefficients vanish. $\square$

Bridge. The construction here builds toward 03.01.04 (tensor algebra), where the same data is upgraded, and the symmetry side is taken up in 03.05.02 (vector bundle). The defining pattern appears again in those units in a sharpened form, where the local data is glued or quotiented. Putting these together, the foundational insight is that the data of this unit gives the structural signature that the rest of the strand reads off.

Exercises [Intermediate+]

Lean formalization [Intermediate+]

[object Promise]

Advanced results [Master]

The tensor product is characterized up to unique isomorphism by its universal property. If $(T,\tau )$ and $(T^{\prime },{\tau }^{\prime })$ both represent bilinear maps out of $V\times W$, the universal property gives unique linear maps $T\to T^{\prime }$ and $T^{\prime }\to T$ carrying one universal bilinear map to the other. The two composites preserve the universal maps and therefore equal the identity maps.

The construction is functorial: linear maps $f:V\to V^{\prime }$ and $g:W\to W^{\prime }$ induce a linear map

$$f\otimes g:V\otimes W\to V^{\prime }\otimes W^{\prime }$$

with $(f\otimes g)(v\otimes w)=f(v)\otimes g(w)$. Associativity and symmetry constraints make tensor product the monoidal structure on vector spaces ^{[Lang §XVI §1-§2]}.

For modules over a commutative ring, the same universal property defines $M{\otimes }_{R}N$. Exactness properties become subtler: tensoring is right exact in general and exact when the module is flat. The present unit uses vector spaces, where all modules are free and tensoring is exact.

Synthesis. This construction generalises the pattern fixed in 01.01.01 (field), with the symmetric data replaced by its skew or twisted analogue. Read in the opposite direction, the construction is dual to the metric story: complements and orthogonality are taken with respect to the bilinear datum of this unit, not a metric, and the resulting category of subobjects is the one the rest of the strand classifies. The central insight is that this datum identifies algebra with geometry: functions become vector fields, subspaces become quotients, and invariants become cohomology classes — and that identification is the engine driving every theorem downstream.

Full proof set [Master]

Proposition (uniqueness). Tensor products satisfying the universal property are uniquely isomorphic.

Let $(T,\tau )$ and $(T^{\prime },{\tau }^{\prime })$ be two tensor products of $V$ and $W$. Since ${\tau }^{\prime }$ is bilinear, the universal property of $T$ gives a unique linear map $\alpha :T\to T^{\prime }$ with $\alpha \circ \tau ={\tau }^{\prime }$. Since $\tau$ is bilinear, the universal property of $T^{\prime }$ gives a unique linear map $\beta :T^{\prime }\to T$ with $\beta \circ {\tau }^{\prime }=\tau$.

Then $\beta \alpha \circ \tau =\beta \circ {\tau }^{\prime }=\tau$. The identity map on $T$ also has this property, so uniqueness gives $\beta \alpha ={\mathrm{id}}_T$. Similarly, $\alpha \beta ={\mathrm{id}}_{T^{\prime }}$. Thus $\alpha$ is an isomorphism.

Connections [Master]

• Tensor algebra 03.01.04 is the direct sum of all tensor powers of a vector space.

• Quotient algebra 03.01.05 uses tensor algebras and ideals to impose relations.

• Vector bundles 03.05.02 have tensor products fiberwise, producing new bundles from old ones.

• Clifford algebra 03.09.02 starts from tensor algebra and then quotients by quadratic relations.

Historical & philosophical context [Master]

Tensor products grew out of multilinear algebra and invariant theory, where one needed a linear object representing bilinear and multilinear operations. Bourbaki and Cartan-Eilenberg helped fix the universal-property formulation in modern algebraic language ^{[Bourbaki Algebra I Ch. III]}.

Lang's treatment emphasizes tensor products as representing objects for bilinear maps, the formulation used in algebra, geometry, and topology ^{[Lang §XVI §1-§2]}.

Bibliography [Master]

[object Promise]