Some universal properties

v1⊗v2−v2⊗v1 | (v1, v2)∈V²

⊆

pq−qp | (p, q)∈(⊗V)² =

pq−qp | (p, q)∈(⊗V)² . Hence,

(⊗V)·(K₂(V))·(⊗V)⊆(⊗V)·

pq−qp | (p, q)∈(⊗V)²

·(⊗V). Since (⊗V)·(K2(V))·(⊗V) =K(V), this rewrites as

K(V)⊆(⊗V)·

pq−qp | (p, q)∈(⊗V)²

·(⊗V). This proves parta).

b) Now we will prove that (⊗V)·

pq−qp | (p, q)∈(⊗V)²

·(⊗V)⊆K(V).

Proof. Every (p, q) ∈ (⊗V)² satisfy pq − qp ∈ K(V) ³⁰. In other words, pq−qp | (p, q)∈(⊗V)² ⊆ K(V). Hence, Proposition 35 (a) (applied to ⊗V, pq−qp | (p, q)∈(⊗V)² and K(V) instead of M,S and Q) yields

pq−qp | (p, q)∈(⊗V)² ⊆K(V). Thus,

pq−qp | (p, q)∈(⊗V)²

=

pq−qp | (p, q)∈(⊗V)² ⊆K(V). Hence,

(⊗V)·

pq−qp | (p, q)∈(⊗V)²

·(⊗V)⊆(⊗V)·(K(V))·(⊗V)⊆K(V) (since K(V) is a two-sided ideal of ⊗V). This proves part b).

c) Combining K(V) ⊆ (⊗V)·

pq−qp | (p, q)∈(⊗V)²

·(⊗V) (which we know from parta)) with (⊗V)·

pq−qp | (p, q)∈(⊗V)²

·(⊗V)⊆K(V) (which we know from part b)), we obtain K(V) = (⊗V)·

pq−qp | (p, q)∈(⊗V)²

·(⊗V). This proves Corollary 85.

0.18. Some universal properties

We shall next discuss some universal properties for the pseudoexterior powers ExterⁿV, the symmetric powers SymⁿV and the exterior powers∧ⁿV.

Let us first recall the definition of a multilinear map:

30Proof. Let (p, q)∈(⊗V)². Then,

sym_V (pq) = sym_V (p)·sym_V (q) (since sym_V is ak-algebra homomorphism)

= sym_V (q)·sym_V(p) (since SymV is commutative by Theorem 83)

= sym_V (qp) (since sym_V is a k-algebra homomorphism).

In other words,pq≡qpmodK(V) (since sym_V is the projection⊗V →(⊗V)K(V)). In other words,pq−qp∈K(V), qed.

Definition 86. Let k be a commutative ring. Let n ∈ N. Let V₁, V₂, . . . , V_n be k-modules.

Let W be any k-module, and let f : V₁×V₂× · · · ×V_n →W be a map. We say that the map f is multilinear if and only if for each i∈ {1,2, . . . , n} and each

(v1, v2, . . . , vi−1, vi+1, vi+2, . . . , vn)∈V1×V2× · · · ×Vi−1 ×Vi+1×Vi+2× · · · ×Vn, the map

V_i →W,

v 7→f(v₁, v₂, . . . , vi−1, v, v_i+1, v_i+2, . . . , v_n) is k-linear.

Now, we can state the classical universal property of a tensor product:

Proposition 87. Let k be a commutative ring. Let n ∈ N. Let V₁, V₂, . . . , V_n be k-modules.

LetW be anyk-module, and letf :V1×V2× · · · ×Vn→W be a multilinear map.

Then, there exists a uniquek-linear mapf⊗:V₁⊗V₂⊗ · · · ⊗V_n→W such that every (v₁, v₂, . . . , v_n)∈V₁×V₂×· · ·×V_nsatisfiesf⊗(v₁⊗v₂⊗ · · · ⊗v_n) =f(v₁, v₂, . . . , v_n).

Proposition 87 is the classical result that allows one to construct maps from a tensor product comfortably.

The particular case of Proposition 87 when all of V₁, V₂, . . . , V_n are identical will be the most useful to us:

Corollary 88. Let k be a commutative ring. Let n ∈N. Let V be ak-module.

LetW be any k-module, and let f :Vⁿ →W be a multilinear map. Then, there exists a unique k-linear map f_⊗ : V^⊗n → W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies f⊗(v₁⊗v₂⊗ · · · ⊗v_n) =f(v₁, v₂, . . . , v_n).

Proof of Corollary 88. The map f is a multilinear mapVⁿ→W. In other words, the map f is a multilinear map V ×V × · · · ×V

| {z }

ntimes

→ W (since Vⁿ = V ×V × · · · ×V

| {z }

ntimes

).

Thus, Proposition 87 (applied to V_i = V) shows that there exists a unique k-linear map f⊗ : V ⊗V ⊗ · · · ⊗V

| {z }

ntimes

→ W such that every (v₁, v₂, . . . , v_n) ∈ V ×V × · · · ×V

| {z }

ntimes

satisfiesf⊗(v1⊗v2⊗ · · · ⊗vn) = f(v1, v2, . . . , vn). Since V ⊗V ⊗ · · · ⊗V

| {z }

ntimes

=V^⊗n and V ×V × · · · ×V

| {z }

ntimes

= Vⁿ, this rewrites as follows: There exists a unique k-linear map f⊗:V^⊗n →W such that every (v1, v2, . . . , vn)∈Vⁿ satisfiesf⊗(v1⊗v2 ⊗ · · · ⊗vn) = f(v₁, v₂, . . . , v_n). This proves Corollary 88.

We shall now use Corollary 88 to derive a universal property for the pseudoexterior powers ExterⁿV. We first state an almost obvious fact:

Lemma 89. Letkbe a commutative ring. Letn ∈N. LetV be ak-module. LetW be any k-module, and letf :Vⁿ→W be any map. Then, there existsat most one k-linear map f_Exter : ExterⁿV → W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies f_Exter(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) =f(v₁, v₂, . . . , v_n).

Proof of Lemma 89. Letαandβbe twok-linear mapsf_Exter : ExterⁿV →W such that every (v₁, v₂, . . . , v_n)∈Vⁿsatisfiesf_Exter(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) =f(v₁, v₂, . . . , v_n).

We shall show that α=β.

We know thatαis ak-linear mapf_Exter : ExterⁿV →W such that every (v₁, v₂, . . . , v_n)∈ Vⁿ satisfies f_Exter(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) = f(v₁, v₂, . . . , v_n). In other words, α is a k-linear map ExterⁿV →W and has the property that every (v1, v2, . . . , vn)∈Vⁿ satisfies

α(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) =f(v₁, v₂, . . . , v_n). (68) The same argument (applied to β instead of α) shows that β is a k-linear map ExterⁿV →W and has the property that every (v₁, v₂, . . . , v_n)∈Vⁿ satisfies

β(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) =f(v₁, v₂, . . . , v_n). (69) Now, the map α − β is k-linear (since the mps α and β are k-linear). Hence, Ker (α−β) is a k-submodule of ExterⁿV.

Define a subset S of ExterⁿV by

S={exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n) | (v₁, v₂, . . . , v_n)∈Vⁿ}. (70) Then,S ⊆Ker (α−β) ³¹. Hence, Proposition 35(a) (applied to M = ExterⁿV and Q= Ker (α−β)) shows that hSi ⊆Ker (α−β).

On the other hand, define a subset S⁰ of V^⊗n by

S⁰ ={v₁⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿ}. (71) Then,

exter_V,n(S⁰) = exter_V,n({v₁⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿ}) (by (71))

={exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n) | (v₁, v₂, . . . , v_n)∈Vⁿ}

=S (by (70)). (72)

31Proof. Let s ∈ S. Thus, s ∈ S = {exterV,n(v1⊗v2⊗ · · · ⊗vn) | (v1, v2, . . . , vn)∈Vⁿ}. In other words, s = exterV,n(v1⊗v2⊗ · · · ⊗vn) for some (v1, v2, . . . , vn) ∈ Vⁿ. Consider this (v1, v2, . . . , vn).

Applying the mapαto the equalitys= exterV,n(v1⊗v2⊗ · · · ⊗vn), we obtain α(s) =α(exterV,n(v1⊗v2⊗ · · · ⊗vn)) =f(v1, v2, . . . , vn)

(by (68)). The same argument (applied to β instead of α) shows thatβ(s) = f(v1, v2, . . . , vn).

Thus, α(s) = f(v1, v2, . . . , vn) = β(s). Now, (α−β) (s) = α(s)

| {z }

=β(s)

−β(s) = β(s)−β(s) = 0, so that s∈Ker (α−β).

Now, let us forget that we fixeds. We thus have shown thats∈Ker (α−β) for eachs∈S. In other words,S ⊆Ker (α−β).

However, the tensor product V^⊗n is generated (as a k-module) by its pure tensors.

In other words,

V^⊗n=hv₁⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿi

=

*

{v₁⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿ}

| {z }

=S⁰

+

=hS⁰i.

Applying the map exterV,n to both sides of this equality, we obtain exter_V,n V^⊗n

= exter_V,n(hS⁰i) =

*

exter_V,n(S⁰)

| {z }

=S (by (72))

+

by Proposition 35 (b) (applied to

V^⊗n, S⁰, ExterⁿV and exter_V,n instead of M,S, R and f)

=hSi.

But the map exter_V,n is surjective. Hence, ExterⁿV = exter_V,n(V^⊗n) = hSi ⊆ Ker (α−β). In other words, α−β = 0. Hence, α=β.

Now, forget that we fixed α and β. We thus have shown that if α and β are two k-linear maps f_Exter : ExterⁿV → W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ sat-isfies f_Exter(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) = f(v₁, v₂, . . . , v_n), then α = β. In other words, there exists at most one k-linear map f_Exter : ExterⁿV →W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies f_Exter(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) = f(v₁, v₂, . . . , v_n).

This proves Lemma 89.

We shall furthermore need a definition:

Definition 90. Letn ∈N. Let V be a set.

Let W be a Z-module. Let f : Vⁿ → W be a map. We say that the map f is antisymmetric if and only if each (v₁, v₂, . . . , v_n)∈Vⁿ and γ ∈S_n satisfy

f v_γ(1), v_γ(2), . . . , v_γ(n)

= (−1)^γf(v₁, v₂, . . . , v_n).

Now, we can state a universal property for the pseudoexterior powers ExterⁿV: Corollary 91. Let k be a commutative ring. Let n ∈N. Let V be ak-module.

Let W be any k-module, and let f : Vⁿ → W be an antisymmetric multilinear map. (The notion of “antisymmetric” makes sense here because the k-moduleW is clearly aZ-module.) Then, there exists a uniquek-linear mapf_Exter : ExterⁿV →W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies f_Exter(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) = f(v1, v2, . . . , vn).

Before we prove this, let us recall a classical fact from abstract algebra – viz. the universal property of quotient modules (also known as the homomorphism theorem for k-modules):

Proposition 92. Let k be a commutative ring. Let V be a k-module. Let I be a k-submodule of V. Let πI be the canonical projectionV →VI.

LetW be anyk-module, and letf :V →W be ak-linear map satisfyingf(I) = 0.

Then, there exists a unique k-linear mapf⁰ :VI →W satisfying f =f⁰◦π_I. Proof of Corollary 91. Corollary 88 shows that there exists a uniquek-linear map f_⊗ : V^⊗n → W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies f⊗(v₁⊗v₂⊗ · · · ⊗v_n) = f(v₁, v₂, . . . , v_n). Consider this f⊗. The map f⊗ is k-linear; thus, Ker (f⊗) is a k-submodule ofV^⊗n.

We know that every (v₁, v₂, . . . , v_n)∈Vⁿ satisfies

f⊗(v₁⊗v₂⊗ · · · ⊗v_n) =f(v₁, v₂, . . . , v_n). (73) The map f is antisymmetric. In other words, each (v₁, v₂, . . . , v_n)∈Vⁿ and γ ∈ S_n satisfy

f v_γ(1), v_γ(2), . . . , v_γ(n)

= (−1)^γf(v₁, v₂, . . . , v_n) (74) (by the definition of “antisymmetric”).

Define a subset T of V^⊗n by T =

v1⊗v2⊗ · · · ⊗vn−(−1)^σv_σ(1)⊗v_σ(2)⊗ · · · ⊗v_σ(n) | ((v1, v2, . . . , vn), σ)∈Vⁿ×Sn . Thus,

hTi

=

v₁⊗v₂⊗ · · · ⊗v_n−(−1)^σv_σ(1)⊗v_σ(2)⊗ · · · ⊗v_σ(n) | ((v₁, v₂, . . . , v_n), σ)∈Vⁿ×S_n

=

v₁ ⊗v₂⊗ · · · ⊗v_n−(−1)^σv_σ(1)⊗v_σ(2)⊗ · · · ⊗v_σ(n) | ((v₁, v₂, . . . , v_n), σ)∈Vⁿ×S_n

=Q_n(V)

(since Qn(V) is defined to be

v₁⊗v₂⊗ · · · ⊗v_n−(−1)^σv_σ(1)⊗v_σ(2)⊗ · · · ⊗v_σ(n) | ((v₁, v₂, . . . , v_n), σ)∈Vⁿ×S_n ).

Recall that exter_V,n is the canonical projectionV^⊗n →V^⊗nQ_n(V).

Now, T ⊆Ker (f⊗) ³². Thus, f⊗



 T

|{z}

⊆Ker(f⊗)



⊆f⊗(Ker (f⊗)) = 0, so that f⊗(T) = 0. But Proposition 35 (b) (applied to V^⊗n, T, W and f_⊗ instead of M, S, R and

32Proof. Lett∈T. Then, t∈T =

v1⊗v2⊗ · · · ⊗vn−(−1)^σvσ(1)⊗vσ(2)⊗ · · · ⊗vσ(n) | ((v1, v2, . . . , vn), σ)∈Vⁿ×Sn . In other words, t has the form t=v1⊗v2⊗ · · · ⊗vn−(−1)^σv_σ(1)⊗v_σ(2)⊗ · · · ⊗v_σ(n) for some ((v1, v2, . . . , vn), σ)∈Vⁿ×Sn. Consider this ((v1, v2, . . . , vn), σ).

It is known that (−1)^σ ∈ {1,−1}. But each g ∈ {1,−1} satisfies g² = 1. Applying this to g= (−1)^σ, we obtain ((−1)^σ)²= 1.

From ((v1, v2, . . . , vn), σ)∈Vⁿ×Sn, we obtain (v1, v2, . . . , vn)∈Vⁿ andσ∈Sn. From (73), we obtain f_⊗(v1⊗v2⊗ · · · ⊗vn) = f(v1, v2, . . . , vn). From (73) (applied to v_σ(1), v_σ(2), . . . , v_σ(n) instead of (v1, v2, . . . , vn)), we obtain

f_⊗ v_σ(1)⊗v_σ(2)⊗ · · · ⊗v_σ(n)

=f v_σ(1), v_σ(2), . . . , v_σ(n)

= (−1)^σf(v1, v2, . . . , vn) (by (74) (applied toγ=σ)).

f) yields f⊗(hTi) =

* f⊗(T)

| {z }

=0

+

= h0i = 0. Since hTi = Q_n(V), this rewrites as f⊗(Q_n(V)) = 0.

Hence, Proposition 92 (applied to V^⊗n, Q_n(V), exter_V,n, W and f⊗ instead of V, I, π_I, W and f) yields that there exists a unique k-linear map f⁰ : V^⊗nQ_n(V) → W satisfying f⊗=f⁰◦exter_V,n. Consider this f⁰.

The map f⁰ is a k-linear map V^⊗nQ_n(V)→ W. In other words, the map f⁰ is a k-linear map ExterⁿV →W (sinceV^⊗nQ_n(V) = ExterⁿV). Every (v₁, v₂, . . . , v_n)∈ Vⁿ satisfies

f⁰(exterV,n(v1⊗v2⊗ · · · ⊗vn))

= (f⁰◦exter_V,n)

| {z }

=f⊗

(v₁⊗v₂⊗ · · · ⊗v_n) =f⊗(v₁⊗v₂ ⊗ · · · ⊗v_n)

=f(v₁, v₂, . . . , v_n) (by (73)).

Thus,f⁰is ak-linear map ExterⁿV →W and has the property that every (v1, v2, . . . , vn)∈ Vⁿ satisfiesf⁰(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) =f(v₁, v₂, . . . , v_n). Hence, there existsat least one k-linear map f_Exter : ExterⁿV → W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies fExter(exterV,n(v1⊗v2⊗ · · · ⊗vn)) = f(v1, v2, . . . , vn) (namely, fExter = f⁰). Since we also know that there exists at most one such map (in fact, this follows from Lemma 89), we can therefore conclude that there exists a unique k-linear map fExter : ExterⁿV → W such that every (v1, v2, . . . , vn) ∈ Vⁿ satisfies f_Exter(exter_V,n(v₁⊗v₂⊗ · · · ⊗v_n)) =f(v₁, v₂, . . . , v_n). This proves Corollary 91.

We can similarly deal with symmetric powers. First, we state an analogue to Lemma 89:

Lemma 93. Let k be a commutative ring. Let n ∈N. Let V be a k-module. Let W be any k-module, and let f :Vⁿ →W be any map. Then, there exists at most onek-linear map fSym : SymⁿV →W such that every (v1, v2, . . . , vn)∈Vⁿ satisfies f_Sym sym_V,n(v₁ ⊗v₂⊗ · · · ⊗v_n)

=f(v₁, v₂, . . . , v_n).

Multiplying both sides of this equality by (−1)^σ, we obtain (−1)^σf_⊗ v_σ(1)⊗v_σ(2)⊗ · · · ⊗v_σ(n)

= (−1)^σ(−1)^σ

| {z }

=((−1)^σ)²=1

f(v1, v2, . . . , vn) =f(v1, v2, . . . , vn).

Now, applying the map f_⊗ to both sides of the equalityt=v₁⊗v₂⊗ · · · ⊗v_n−(−1)^σv_σ(1)⊗ v_σ(2)⊗ · · · ⊗v_σ(n), we find

f_⊗(t) =f_⊗ v1⊗v2⊗ · · · ⊗vn−(−1)^σv_σ(1)⊗v_σ(2)⊗ · · · ⊗v_σ(n)

=f_⊗(v1⊗v2⊗ · · · ⊗vn)

| {z }

=f(v1,v2,...,vn)

−(−1)^σf_⊗ v_σ(1)⊗v_σ(2)⊗ · · · ⊗v_σ(n)

| {z }

=f(v₁,v₂,...,v_n)

(since the map f_⊗ isk-linear)

=f(v₁, v₂, . . . , v_n)−f(v₁, v₂, . . . , v_n) = 0.

In other words,t∈Ker (f_⊗).

Now, forget that we fixed t. We thus have proven that t ∈ Ker (f_⊗) for each t ∈T. In other

Proof of Lemma 93. The proof of Lemma 93 is completely analogous to the proof of Lemma 89, and thus is omitted.

Next, we state a definition (which is analogous to Definition 90, but works in a greater generality, since W no longer needs to be a Z-module):

Definition 94. Letn ∈N. Let V be a set.

LetW be a set. Letf :Vⁿ →W be a map. We say that the map f is symmetric if and only if each (v1, v2, . . . , vn)∈Vⁿ and γ ∈Sn satisfy

f v_γ(1), v_γ(2), . . . , v_γ(n)

=f(v₁, v₂, . . . , v_n).

Now, we can state a universal property for the symmetric powers SymⁿV: Corollary 95. Let k be a commutative ring. Let n ∈N. Let V be ak-module.

Let W be any k-module, and let f : Vⁿ → W be a symmetric multilinear map.

Then, there exists a unique k-linear map f_Sym : SymⁿV → W such that every (v₁, v₂, . . . , v_n)∈Vⁿ satisfiesf_Sym sym_V,n(v₁ ⊗v₂⊗ · · · ⊗v_n)

=f(v₁, v₂, . . . , v_n).

Proof of Corollary 95. The proof of Corollary 95 is completely analogous to the proof of Corollary 91 (up to some replacing of + signs by−signs and some removal of powers of −1), and thus is omitted.

We shall next derive similar results for exterior powers. First of all, we can again easily obtain an analogue to Lemma 89:

Lemma 96. Let k be a commutative ring. Let n ∈N. Let V be a k-module. Let W be any k-module, and let f :Vⁿ →W be any map. Then, there exists at most one k-linear map f∧ : ∧ⁿV → W such that every (v1, v2, . . . , vn) ∈ Vⁿ satisfies f∧ wedge_V,n(v₁⊗v₂⊗ · · · ⊗v_n)

=f(v₁, v₂, . . . , v_n).

Proof of Lemma 96. The proof of Lemma 96 is completely analogous to the proof of Lemma 89, and thus is omitted.

Next, we define a notion of “weakly alternating” which is (in some weak sense) similar to Definition 90 (but at this point, there is no direct analogy any more):

Definition 97. Letn ∈N. Let V be a set.

Let W be a Z-module. Let f : Vⁿ → W be a map. We say that the map f is weakly alternating if and only if each i∈ {1,2, . . . , n−1} and (v₁, v₂, . . . , v_n)∈Vⁿ satisfying v_i =v_i+1 satisfy

f(v₁, v₂, . . . , v_n) = 0.

Now, we can state a universal property for the pseudoexterior powers ∧ⁿV: Corollary 98. Let k be a commutative ring. Let n ∈N. Let V be ak-module.

LetW be anyk-module, and letf :Vⁿ→W be a weakly alternating multilinear map. (The notion of “weakly alternating” makes sense here because the k-module W is clearly a Z-module.) Then, there exists a unique k-linear map f∧ :∧ⁿV →W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies f∧ wedge_V,n(v₁⊗v₂ ⊗ · · · ⊗v_n)

= f(v1, v2, . . . , vn).

Proof of Corollary 98. Corollary 88 shows that there exists a uniquek-linear map f_⊗ : V^⊗n → W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies f⊗(v₁⊗v₂⊗ · · · ⊗v_n) = f(v₁, v₂, . . . , v_n). Consider this f⊗. The map f⊗ is k-linear; thus, Ker (f⊗) is a k-submodule ofV^⊗n.

We know that every (v₁, v₂, . . . , v_n)∈Vⁿ satisfies

f⊗(v₁⊗v₂⊗ · · · ⊗v_n) =f(v₁, v₂, . . . , v_n). (75) The map f is weakly alternating. In other words, each i ∈ {1,2, . . . , n−1} and (v₁, v₂, . . . , v_n)∈Vⁿ satisfying v_i =v_i+1 satisfy

f(v₁, v₂, . . . , v_n) = 0. (76) (by the definition of “weakly alternating”).

Fix i∈ {1,2, . . . , n−1}. Define a subsetT of V^⊗n by

T ={v₁⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿ; v_i =v_i+1}. Thus,

hTi

=h{v₁⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿ; v_i =v_i+1}i

=hv₁⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿ; v_i =v_i+1i. (77) Recall that wedge_V,n is the canonical projectionV^⊗n→V^⊗nR_n(V).

Now, T ⊆Ker (f⊗) ³³. Thus, f⊗



 T

|{z}

⊆Ker(f⊗)



⊆f⊗(Ker (f⊗)) = 0, so that f⊗(T) = 0. But Proposition 35 (b) (applied to V^⊗n, T, W and f⊗ instead of M, S, R and f) yieldsf⊗(hTi) =

* f⊗(T)

| {z }

=0

+

=h0i= 0. In view of (77), this rewrites as

f⊗(hv₁⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿ; v_i =v_i+1i) = 0. (78) Now, forget that we fixed i. We thus have proven (78) for eachi∈ {1,2, . . . , n−1}.

But Proposition 67 yields Rn(V) =

n−1

X

i=1

hv1⊗v2⊗ · · · ⊗vn | (v1, v2, . . . , vn)∈Vⁿ; vi =vi+1i.

33Proof. Lett∈T. Then,

t∈T ={v1⊗v2⊗ · · · ⊗vn | (v1, v2, . . . , vn)∈Vⁿ; vi=vi+1}.

In other words, t has the form t = v₁⊗v₂⊗ · · · ⊗v_n for some (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfying v_i=v_i+1. Consider this (v₁, v₂, . . . , v_n).

From (75), we obtain f_⊗(v₁⊗v₂⊗ · · · ⊗v_n) =f(v₁, v₂, . . . , v_n) = 0 (by (76)).

Now, applying the mapf_⊗ to both sides of the equalityt=v₁⊗v₂⊗ · · · ⊗v_n, we findf_⊗(t) = f_⊗(v₁⊗v₂⊗ · · · ⊗v_n) = 0. In other words,t∈Ker (f_⊗).

Now, forget that we fixed t. We thus have proven that t ∈ Ker (f_⊗) for each t ∈T. In other

Applying the map f_⊗ to both sides of this equality, we obtain f⊗(R_n(V))

=f⊗ n−1

X

i=1

hv₁ ⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿ; v_i =v_i+1i

!

=

n−1

X

i=1

f⊗(hv₁⊗v₂⊗ · · · ⊗v_n | (v₁, v₂, . . . , v_n)∈Vⁿ; v_i =v_i+1i)

| {z }

=0 (by (78))

(since the map f⊗ isk-linear)

=

n−1

X

i=1

0 = 0.

Hence, Proposition 92 (applied to V^⊗n, R_n(V), wedge_V,n, W and f_⊗ instead of V, I,π_I,W andf) yields that there exists a unique k-linear map f⁰ :V^⊗nR_n(V)→W satisfying f⊗=f⁰◦wedge_V,n. Consider this f⁰.

The map f⁰ is a k-linear map V^⊗nR_n(V) → W. In other words, the map f⁰ is a k-linear map ∧ⁿV → W (since V^⊗nR_n(V) = ∧ⁿV). Every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies

f⁰ wedge_V,n(v₁⊗v₂⊗ · · · ⊗v_n)

= f⁰◦wedge_V,n

| {z }

=f⊗

(v₁⊗v₂⊗ · · · ⊗v_n) = f⊗(v₁⊗v₂⊗ · · · ⊗v_n)

=f(v₁, v₂, . . . , v_n) (by (75)).

Thus,f⁰ is ak-linear map∧ⁿV →W and has the property that every (v₁, v₂, . . . , v_n)∈ Vⁿ satisfies f⁰ wedge_V,n(v₁⊗v₂⊗ · · · ⊗v_n)

= f(v₁, v₂, . . . , v_n). Hence, there exists at least onek-linear mapf_∧ :∧ⁿV →W such that every (v₁, v₂, . . . , v_n)∈Vⁿsatisfies f∧ wedge_V,n(v₁⊗v₂⊗ · · · ⊗v_n)

= f(v₁, v₂, . . . , v_n) (namely, f∧ =f⁰). Since we also know that there exists at most one such map (in fact, this follows from Lemma 96), we can therefore conclude that there exists a unique k-linear map f_∧ : ∧ⁿV → W such that every (v₁, v₂, . . . , v_n) ∈ Vⁿ satisfies f∧ wedge_V,n(v₁⊗v₂⊗ · · · ⊗v_n)

= f(v₁, v₂, . . . , v_n). This proves Corollary 98.

References

[1] Tom (Thomas) Goodwillie, MathOverflow post #65716 (answer to “Commutator tensors and submodules”).

http://mathoverflow.net/questions/65716

[2] Darij Grinberg, The Clifford algebra and the Chevalley map - a computational ap-proach (summary version).

http://www.cip.ifi.lmu.de/~grinberg/algebra/chevalleys.pdf

Darij Grinberg, The Clifford algebra and the Chevalley map - a computational ap-proach (detailed version).

http://www.cip.ifi.lmu.de/~grinberg/algebra/chevalley.pdf

[3] Darij Grinberg, Poincar´e-Birkhoff-Witt type results for inclusions of Lie algebras, 2011.

http://www.cip.ifi.lmu.de/~grinberg/algebra/algebra.html#pbw

[4] VorlesungHopfalgebren von Hans-J¨urgen Schneider, mitgeschrieben von Darij Grin-berg.

https://sites.google.com/site/darijgrinberg/hopfalgebren

Im Dokument A few classical results on tensor, symmetric and exterior powers (Seite 80-89)