CHAPTER 8 Toolbox

9

4. Tensor Products

Definition and Remark 8.4.1. Let M_{R} and _{R}N be R-modules, and let A be an abelian group. A map f : M × N ⟶ A is called R-bilinear if

1. f(m + m′, n) = f(m, n) + f(m′, n),

2. f(m, n + n′) = f(m, n) + f(m, n′),

3. f(mr, n) = f(m, rn) for all r ∈ R, m, m′ ∈ M, n, n′ ∈ N.

Let Bil_{R}(M, N; A) denote the set of all R-bilinear maps f : M × N ⟶ A.

Bil_{R}(M, N; A) is an abelian group with (f + g)(m, n) := f(m, n) + g(m, n).

Definition 8.4.2. Let M_{R} and _{R}N be R-modules. An abelian group M ⊗_{R} N together with an R-bilinear map

⊗ : M × N ∋ (m, n) ↦ m ⊗ n ∈ M ⊗_{R} N

is called a tensor product of M and N over R if for each abelian group A and for each R-bilinear map f : M × N ⟶ A there exists a unique group homomorphism g : M ⊗_{R} N ⟶ A such that the diagram

[figure: M × N; M ⊗_{R} N; ⊗; f; A; g]

commutes. The elements of M ⊗_{R} N are called tensors, the elements of the form m ⊗ n are called decomposable tensors.

Warning: If you want to define a homomorphism f : M ⊗_{R} N ⟶ A with a tensor product as domain you must define it by giving an R-bilinear map defined on M ×N.

Lemma 8.4.3. A tensor product (M ⊗_{R} N, ⊗) defined by M_{R} and _{R}N is unique up to a unique isomorphism.

Proof. Let (M ⊗_{R} N, ⊗) and (M ⊠_{R} N, ⊠) be tensor products. Then

[figure: M × N; ⊗; ⊠; ⊗; ⊠; M ⊗_{R} N; M ⊠_{R} N; h; k; M ⊗_{R} N; M ⊠_{R} N; h]

implies k = h^{−1}.

Because of this fact we will henceforth talk about the tensor product of M and N over R.

Proposition 8.4.4. (Rules of computation in a tensor product) Let (M ⊗_{R} N, ⊗) be the tensor product. Then we have for all r ∈ R, m, m′ ∈ M, n, n′ ∈ N 1. M ⊗_{R} N = {∑_{i} m_{i} ⊗ n_{i} | m_{i} ∈ M, n_{i} ∈ N},

2. (m + m′) ⊗ n = m ⊗ n + m′ ⊗ n,

3. m ⊗ (n + n′) = m ⊗ n + m ⊗ n′,

4. mr ⊗ n = m ⊗ rn (observe in particular, that ⊗ : M × N ⟶ M ⊗ N is not injective in general),

5. if f : M × N ⟶ A is an R-bilinear map and g : M ⊗_{R} N ⟶ A is the induced

homomorphism, then

g(m ⊗ n) = f(m, n).

Proof. 1. Let B := ⟨m ⊗ n⟩ ⊆ M ⊗_{R} N denote the subgroup of M ⊗_{R} N generated by the decomposable tensors m⊗n. Let j : B ⟶ M ⊗_{R}N be the embedding homomorphism. We get an induced map ⊗′ : M × N ⟶ B. In the following diagram

[figure: M × N; B; ⊗′; M ⊗_{R} N; j; B; M ⊗_{R} N; j; ⊗′; id_{B}; jp; p]

we have id_{B} ∘⊗′ = ⊗′, p with p ∘ j ∘ ⊗′ = p ∘ ⊗ = ⊗′ exists since ⊗′ is R-bilinear. Because of jp ∘ ⊗ = j ∘ ⊗′ = ⊗ = id_{M⊗_{R}N} ∘⊗ we get jp = id_{M⊗_{R}N}, hence the embedding j is surjective and thus the identity.

2. (m + m′) ⊗ n = ⊗(m + m′, n) = ⊗(m, n) + ⊗(m′, n) = m ⊗ n + m′ ⊗ n.

3. and 4. analogously.

5. is precisely the definition of the induced homomorphism.

Remark 8.4.5. To construct tensor products, we use the notion of a free module. Let X be a set and R be a ring. An R-module RX together with a map ι : X ⟶

RX is called a free R-module generated by X, if for every R-module M and for every map f : X ⟶ M there exists a unique homomorphism of R-modules g : RX ⟶ M such that the diagram

[figure: X; RX; ι; f; M; g]

commutes.

Free R-modules exist and can be constructed as RX := {α : X ⟶ R| for almost all x ∈ X : α(x) = 0}.

Proposition 8.4.6. Given R-modules M_{R} and _{R}N. Then there exists a tensor product (M ⊗_{R} N, ⊗).

Proof. Define M ⊗_{R} N := ℤ{M × N}/U where ℤ{M × N} is a free ℤ-module over M × N (the free abelian group) and U is generated by

ι(m + m′, n) − ι(m, n) − ι(m′, n)
ι(m, m + n′) − ι(m, n) − ι(m, n′)
ι(mr, n) − ι(m, rn)

for all r ∈ R, m, m′ ∈ M, n, n′ ∈ N. Consider

[figure: M × N; ℤ{M × N}; ι; M ⊗_{R} N; ν; = ℤ{M × N}/U; A; ψ; ρ; g]

Let ψ be given. Then there is a unique ρ ∈ Hom(ℤ{M × N}, A) such that ρι = ψ. Since ψ is R-bilinear we get ρ(ι(m + m′, n) − ι(m, n) − ι(m′n)) = ψ(m + m′, n) − ψ(m, n) − ψ(m′, n) = 0 and similarly ρ(ι(m, n + n′) − ι(m, n) − ι(m, n′)) = 0 and ρ(ι(mr, n) − ι(m, rn)) = 0. So we get ρ(U) = 0. This implies that there is a unique g ∈ Hom(M ⊗_{R} N, A) such that gν = ρ (homomorphism theorem). Let ⊗ := ν ∘ ι. Then ⊗ is bilinear since (m + m′) ⊗ n = ν ∘ ι(m + m′, n) = ν(ι(m + m′, n)) = ν(ι(m + m′, n) − ι(m, n) − ι(m′, n) + ι(m, n) + ι(m′, n)) = ν(ι(m, n) + ι(m′, n)) = ν ∘ ι(m, n) + ν ∘ ι(m′, n) = m ⊗ n + m′ ⊗ n. The other two properties are obtained in an analogous way.

We have to show that (M ⊗_{R} N, ⊗) is a tensor product. The above diagram shows that for each abelian group A and for each R-bilinear map ψ : M × N ⟶ A there is a g ∈ Hom(M ⊗_{R} N, A) such that g ∘ ⊗ = ψ. Given h ∈ Hom(M ⊗_{R} N, A) with h ∘ ⊗ = ψ. Then h ∘ ν ∘ ι = ψ. This implies h ∘ ν = ρ = g ∘ ν hence g = h.

Proposition and Definition 8.4.7. Given two homomorphisms

f ∈ Hom_{R}(M., M′.) and g ∈ Hom_{R}(.N, .N′).

Then there is a unique homomorphism

f ⊗_{R} g ∈ Hom(M ⊗_{R} N, M′ ⊗_{R} N′)

such that f ⊗_{R} g(m ⊗ n) = f(m) ⊗ g(n), i.e. the following diagram commutes

[figure: M′ × N′; M′ ⊗_{R} N′; ⊗; M × N; M ⊗_{R} N; ⊗; f × g; f ⊗_{R} g]

Proof. ⊗ ∘ (f × g) is bilinear.

Notation 8.4.8. We often write f ⊗_{R} N := f ⊗_{R} 1_{N} and M ⊗_{R} g := 1_{M} ⊗_{R} g. We have the following rule of computation:

f ⊗_{R} g = (f ⊗_{R} N′) ∘ (M ⊗_{R} g) = (M′ ⊗_{R} g) ∘ (f ⊗_{R} N)

since f × g = (f × N′) ∘ (M × g) = (M′ × g) ∘ (f × N).

Proposition 8.4.9. The following define covariant functors

1. - ⊗ N : Mod-R ⟶ Ab;

2. M ⊗ - : R-Mod ⟶ Ab;

3. - ⊗ - : Mod-R × R-Mod ⟶ Ab.

Proof. (f × g) ∘ (f′ × g′) = ff′ × gg′ implies (f ⊗_{R} g) ∘ (f′ ⊗_{R} g′) = ff′ × gg′. Furthermore 1_{M} × 1_{N} = 1_{M×N} implies 1_{M} ⊗_{R} 1_{N} = 1_{M⊗_{R}N}.

Definition 8.4.10. Let R, S be rings and let M be a left R-module and a right Smodule. M is called an R-S-bimodule if (rm)s = r(ms). We define Hom_{R-S}(.M., .N.) := Hom_{R}(.M, .N) ∩ Hom_{S}(M., N.).

Remark 8.4.11. Let M_{S} be a right S-module and let R × M ⟶ M a map. M is an R-S-bimodule if and only if

1. ∀r ∈ R : (M ∋ m ↦ rm ∈ M) ∈ Hom_{S}(M., M.),

2. ∀r, r′ ∈ R, m ∈ M : (r + r′)m = rm + r′m,

3. ∀r, r′ ∈ R, m ∈ M : (rr′)m = r(r′m),

4. ∀m ∈ M : 1m = m.

Lemma 8.4.12. Let_{R}M_{S} and_{S}N_{T} be bimodules. Then_{R}(M⊗_{S}N)_{T} is a bimodule by r(m ⊗ n) := rm ⊗ n and (m ⊗ n)t := m ⊗ nt.

Proof. Obviously we have 2.-4. Furthermore (r ⊗_{S} id)(m ⊗ n) = rm ⊗ n = r(m ⊗ n) is a homomorphism.

Corollary 8.4.13. Given bimodules _{R}M_{S}, _{S}N_{T}, _{R}M′_{S}, _{S}N′_{T} and homomorphisms f ∈ Hom_{R-S}(.M., .M′.) and g ∈ Hom_{S-T} (.N., .N′.). Then we have f ⊗_{S} g ∈ Hom_{R-T} (.M ⊗_{S} N., .M′ ⊗_{S} N′.).

Proof. f ⊗_{S} g(rm ⊗ nt) = f(rm) ⊗ g(nt) = r(f ⊗_{S} g)(m ⊗ n)t.

Remark 8.4.14. Every module M over a commutative ring 𝕂 and in particular every vector space over a field 𝕂 is a 𝕂-𝕂-bimodule by λm = mλ. So there is an embedding functor ι : 𝕂-Mod ⟶ 𝕂-Mod-𝕂. Observe that there are 𝕂-𝕂-bimodules that do not satisfy λm = mλ. Take for example an automorphism α : 𝕂 ⟶ 𝕂 and a left 𝕂-module M and define mλ := α(λ)m. Then M is such a 𝕂-𝕂-bimodule.

The tensor product M ⊗_{𝕂} N of two 𝕂-𝕂-bimodules M and N is again a 𝕂-𝕂-bimodule. If we have, however, 𝕂-𝕂-bimodules M and N arising from 𝕂-modules as above, i.e. satisfying λm = mλ, then their tensor product M ⊗_{𝕂} N also satisfies this equation, so M ⊗_{𝕂} N comes from a module in 𝕂-Mod. Indeed we have λm ⊗ n = mλ ⊗ n = m ⊗ λn = m ⊗ nλ. Thus the following diagram of functors commutes:

[figure: 𝕂-Mod; 𝕂-Mod-𝕂.; ι; 𝕂-Mod × 𝕂-Mod; 𝕂-Mod-𝕂 × 𝕂-Mod-𝕂; ι × ι; ⊗_{𝕂}; ⊗_{𝕂}]

So we can consider 𝕂-Mod as a (proper) subcategory of 𝕂-Mod-𝕂. The tensor product over 𝕂 can be restricted to 𝕂-Mod.

We write the tensor product of two vector spaces M and N as M ⊗ N.

Theorem 8.4.15. In the category 𝕂-Mod there are natural isomorphisms

1. Associativity Law: α : (M ⊗ N) ⊗ P ≅ M ⊗ (N ⊗ P).

2. Law of the Left Unit: λ : 𝕂 ⊗ M ≅ M.

3. Law of the Right Unit: ρ : M ⊗ 𝕂 ≅ M.

4. Symmetry Law: τ : M ⊗ N ≅ N ⊗ M.

5. Existence of Inner Hom-Functors: Hom(P ⊗ M, N) ≅ Hom(P, Hom(M, N)).

Proof. We only describe the corresponding homomorphisms.

1. Use (8.4.45.) to define α((m ⊗ n) ⊗ p) := m ⊗ (n ⊗ p).

2. Define λ : 𝕂 ⊗ M ⟶ M by λ(r ⊗ m) := rm.

3. Define ρ : M ⊗ 𝕂 ⟶ M by ρ(m ⊗ r) := mr.

4. Define τ(m ⊗ n) := n ⊗ m.

5. For f : P ⊗ M ⟶ N define ϕ(f) : P ⟶ Hom(M, N) by ϕ(f)(p)(m) := f(p ⊗ m).

Usually one identifies threefold tensor products along the map α so that we use M ⊗ N ⊗ P = (M ⊗ N) ⊗ P = M ⊗ (N ⊗ P). For the notion of a monoidal or tensor category, however, this natural transformation is of central importance.

Problem 8.4.1. 1. Give an explicit proof of M ⊗ (X ⊕ Y) ≅ M ⊗ X ⊕ M ⊗ Y.

2. Show that for every finite dimensional vector space V there is a unique element ∑_{i=1}^{n} v_{i} ⊗ v_{i}^{∗} ∈ V ⊗ V^{∗} such that the following holds

∀v ∈ V : ∑_{i} v_{i}^{∗}(v)v_{i} = v.

(Hint: Use an isomorphism End(V) ≅ V ⊗ V^{∗} and dual bases {v_{i}} of V and {v_{i}^{∗}} of V^{∗}.)

3. Show that the following diagrams (coherence diagrams or constraints) commute in 𝕂-Mod:

[figure: ((A ⊗ B) ⊗ C) ⊗ D; (A ⊗ (B ⊗ C)) ⊗ D; α(A,B,C)⊗1; A ⊗ ((B ⊗ C) ⊗ D); α(A,B⊗C,D); α(A⊗B,C,D); 1⊗α(B,C,D); (A ⊗ B) ⊗ (C ⊗ D); A ⊗ (B ⊗ (C ⊗ D)); α(A,B,C⊗D)]

[figure: (A ⊗ 𝕂) ⊗ B; A ⊗ (𝕂 ⊗ B); α(A,𝕂,B); A ⊗ B; ρ(A)⊗1; 1⊗λ(B)]

4. Write τ(A, B) : A ⊗ B ⟶ B ⊗ A for τ(A, B) : a ⊗ b ↦ b ⊗ a. Show that τ is a natural transformation (between which functors?). Show that

[figure: (A ⊗ B) ⊗ C; (B ⊗ A) ⊗ C; τ(X,B)⊗1; B ⊗ (A ⊗ C); α; α; 1⊗τ(A,C); A ⊗ (B ⊗ C); (B ⊗ C) ⊗ A; τ(A,B⊗C); B ⊗ (C ⊗ A); α]

commutes for all A, B, C ∈ 𝕂-Mod and that

τ(B, A)τ(A, B) = id_{A⊗B}

for all A, B in 𝕂-Mod.

5. Find an example of M, N ∈ 𝕂-Mod-𝕂 such that M ⊗_{𝕂} N ≅ N̸ ⊗_{𝕂} M.