Higiene y seguridad en el trabajo

2. MARCO TEÓRICO

2.2. Marco Conceptual

2.2.11. Higiene y seguridad en el trabajo

5.6 Jordan Decomposition for Abelian Affine Groups

For a Hopf algebraAoverk, we can consider the vector space spanned by all of its group-like elements, which we shall denote byχ(A). Note this is actually a Hopf subalgebra ofA. The product of two group like elementsf andgis group-like since ∆(f g) = ∆(f)∆(g) =f g⊗f g, therefore a product of two linear sums of group-like elements is still a linear sum of group- like elements. Obviously the multiplicative identity, 1, is group-like. This shows that χ(A) is a subalgebra. Furthermore, ∆(P

cifi) =Pcifi⊗fi∈χ(A)⊗χ(A). SimilarlyS(f) =f−1 is group-like since ∆(f−1) = ∆(f)−1 = (f ⊗f)−1 = f−1⊗f−1, and with S a linear map we have that S(χ(A))⊆χ(A). Thusχ(A) is a Hopf subalgebra of A. We shall denote the quotient group corresponding to χ(A) by χ(G). With i:χ(A)→ A the inclusion, we have an extension of G:

1→keri#→ G

→χ(G)→1.

In general this might not split, but for abelian affine groups over algebraically closed field, this turns out to be a splitting exact sequence. To do so we start off with a proposition about irreducible representations of abelian affine groups over algebraically closed fields.

Proposition 5.6.1. Let G ∼=Spec(A) be an abelian affine group overk=kalg. Then every irreducible representation of G is one-dimensional.

Proof. LetV be an irreducible representation ofG andρ:V →V ⊗Abe the corresponding comodule. Since every representation is a sum of finite-dimensional subrepresentations, V

must be finite-dimensional. So let{vi}be a basis forV withρ(vj) =Pvi⊗aij. This gives us ∆(aij) = Paik⊗akj, which shows that the finite-dimensional vector space C spanned by{aij} is a subcoalgebra. Taking the linear dual we obtain a finite-dimensionalk-algebra

CD, which is commutative since ∆ is cocommutative. ThusCD ∼=Q

Ri with eachRi local. Note that we now have ρ(V)⊆V ⊗C and ∆(C)⊆C⊗C. By restricting the respective domains and codomains, the following diagrams obtained from V being a comodule still commute: V V ⊗C V V ⊗C V ⊗C V ⊗C⊗C V V ⊗k. ρ ρ _ρ⊗id ρ = id⊗ id⊗∆ ∼= Dualising, we obtain V V ⊗CD V V ⊗CD V ⊗CD V ⊗CD⊗CD V V ⊗k. ρD ρD _ρD_⊗_id ρD = _id⊗D id⊗∆D ∼=

Thus ρD defines an action of CD on V that turns V into a CD-module. Since CD is finite-dimensional, it is a product of local rings Q

Bi. Thus we have a corresponding set of complete idempotents {ei} ⊆ CD. This allows us to write each v ∈ V uniquely as

v = P

an A-subcomodule of V since CD ·V0 ⊆ V0 if and only if ρ(V0) ⊆ V0 ⊗A. With each

ei·V being aCD-submodule, all but one of theei must act as 0 by irreducibility ofV as a comodule. LetBi be the factor with the non-zero action and let M be its unique maximal ideal. The action of CD on V factors through Bi, makingV a Bi-module. Since M·V is a Bi-submodule, it must either be 0 or V. But Mn = 0 for some n ∈ _N, so M ·V must be 0. The action therefore factors further throughBi/M, making V a Bi/M-vector space. The composition k → Bi → Bi/M is a field extension, with k algebraically closed we get that Bi/M ∼=k. So V is a k-vector space with no non-trivial subspace, which shows that dimk(V) = 1.

Corollary 5.6.2. Let G be an abelian affine group over k =kalg. If it has no non-trivial characters, then it is unipotent.

Proof. Since every representation of G ∼= Spec(A) contains an irreducible representation, WLOG we can assume ρ:V →V ⊗A is irreducible. By the previous proposition it is one dimensional and therefore sendsv 7→v⊗a for somea∈A. But if a6= 1, then ρ defines a non-trivial character Φ : G → GL(V) ∼=Gm. So by assumption we must have a = 1 and

thereforeG is unipotent.

We can now prove the corresponding multiplicative Jordan decomposition for abelian affine groups over algebraically closed fields:

Theorem 5.6.3. Let G ∼=Spec(A) be an abelian affine group over an algebraically closed field k. Then the exact sequence

1→keri#→ G

→χ(G)→1

is split, giving us G ∼= keri#×χ(G). Furthermore, keri# is unipotent and χ(G) is diago-

nalisable.

Proof. With χ : χ(A) → k the counit of A restricted to χ(A), we know that k[keri#] ∼=

A/Iχ. Since every group-like element inAis equal to 1 moduloIχ, this shows thatk[keri#]

has no non-trivial group-like elements. By Corollary 5.6.2 we get that keri# is unipotent.

It is by definition ofχ(G) that it is diagonalisable. We are left to show that there exists a section mappingA back toχ(A).

By Theorem 5.2.3, we get that A is a directed union of finite-dimensional coalgebras

C. Taking the dual, we obtain (commutative) finite-dimensional k-algebras CD sinceA is cocommutative. Sincek is algebraic and is therefore perfect, Corollary 1.9.9 tells us that

π0(CD)

CD CD/Nil(CD).

∼

So we have an algebra section mapping CD/Nil(CD) back into π0(CD) ⊆ CD. Dualising,

with Cs denoting π0(CD)

5.6. JORDAN DECOMPOSITION FOR ABELIAN AFFINE GROUPS 69

the inclusion π0(CD) ,→ CD, where we identified C ∼= (CD)D. Now suppose E is some

larger subcoalgebra containing C. WithED CD induced by the injection C ,→ E, the compositionED CD CD/Nil(CD) factors uniquely throughED/Nil(ED):

CD CD/Nil(CD)

ED _ED_/_Nil(_ED₎_.

∃!

WithEs and pE :E→Es the coalgebra projection as constructed for C, we get that

C Cs

E Es

commutes, which shows that the projection maps are compatible. So with As := ∪Cs the union over all such Cs, we have a unique coalgebra projection p : A → As induced by

pC :C →Cs for each C.

We first show that the projection p : A → As defined preserves multiplication, i.e.

p(ab) =p(a)p(b) with multiplication onAs induced by multiplication onA. For any finite- dimensional subcoalgebra C of A, we can find another finite-dimensional subcoalgebra E

large enough so that the multiplication mapµ on Amaps C⊗C toE. Since the comultiplication onA is an algebra morphism, i.e.

A⊗A A

(A⊗A)⊗(A⊗A) A,

∆⊗∆ ∆

µ⊗µ

this also shows that µ is a coalgebra morphism. So taking the dual of the µ restricted to

C⊗C we obtain an algebra map δ : ED → CD ⊗CD. Since an etale algebra has etale image, we obtain

π0(ED) π0(CD⊗CD)

ED CD ⊗CD.

Let qE : ED → π0(ED) denote the map ED ED/Nil(ED) preceded by the canonical

section ED/Nil(ED)−→=∼ π0(ED); similarly for qC :CD →π0(CD) and qC⊗C :CD⊗CD →

qC⊗C =qC ⊗qC. Our commutative diagram extends to π0(ED) π0(CD)⊗π0(CD) ED CD⊗CD. δ qC⊗qC δ qE

Since eachpC was the dual map of the inclusionπ0(CD) ,→CD, which is equal toqC, we have thatqC(f) =f◦p for any f ∈CD. This gives us

f(p(ab)) =qE(f)(ab) = (qE(f)◦µ)(a⊗b) =δ(qE(f))(a⊗b) = ((qC ⊗qC)◦δ)(f)(a⊗b) = (δ(f)◦(p⊗p))(a⊗b) = (f◦µ)(p(a)⊗p(b)) =f(p(a)p(b))

for alla, b∈C, all f ∈ED.

This shows that p(ab) = p(b)p(c) for all a, b ∈A, which also implies thatAs is closed under multiplication. Consideringp as a map fromA→Ainstead, withp preserving both multiplication and comultiplication structures, we have that p :A → A is a Hopf algebra map. This shows thatp preserves the antipode, i.e. S◦p=p◦S. So we get that

S(As) =S(p(As)) =p(S(As))⊆As.

Since the f.d. subcoalgebraC =k contains 1, we also have that 1∈As. We can therefore conclude thatAs is a Hopf subalgebra ofA, andp:A→As is a Hopf algebra projection.

Lastly, we claim that the As constructed is actually χ(A). With kalgebraically closed we have that π0(CD) is a product of copies of k with a set of complete idempotents {ei}. Thus HomAlgk(π0(C

D₎_{, k}_{) =}_{_e∗ i}with eache ∗ i defined bye ∗ i(ej) =δ j

i. It is clear that these are exactly the basis elements spanning (π0(CD))D =Cs. With (Cs)D ∼=π0(CD), Lemma

5.5.2 tells us that these elements in HomAlg_k(π0(CD), k) corresponds to elements c ∈ Cs

such that ∆(c) =c⊗c. But withCs identified as a subcoalgebra of A, these are group-like elements inA. So allCs are spanned by group-like elements ofA, which implies thatAs is also spanned by group-like elements ofA. We are left to show that all group-like elements of

Aare inAs. Letabe a group-like element ofAand take the one-dimensionalk-linear space

V spanned bya. It is clearly a subcoalgebra ofAand soVD is a one-dimensionalk-algebra spanned by a∗ defined by a∗(a) = 1. But ∆D(λa∗⊗λ0a∗)(a) = (λa∗⊗λ0a∗)(∆(a)) =λλ0, which is equal to zero if and only if λa∗ ⊗λ0a∗ = 0. So VD is reduced, implying that

π0(V)∼=VD/Nil(VD)=∼VD. This shows that Vs∼= (VD)D ∼=V. In particular, the group- like elementais in Vs ⊆As. Asawas arbitrary,As contains all the group-like elements of

Chapter 6

Algebraic Envelope of the Group

of Integers

We know that for abelian affine groups over an algebraically closed field, the exact sequence

1→keri#→ G

→χ(G)→1

splits. As one would expect, the algebraic envelope of an abelian group is always abelian, as we will now show:

Proposition 6.0.1. For Gabelian, G is also abelian. Proof. Just note that for ∆(f) =P

fi⊗gi, we get

fi(y)gi(x) = (f)(y+x) = (f)(x+y) =Xfi(x)gi(y).

Therefore (π◦twist◦∆)(f)(x, y) =P

gi(x)fi(y) =Pfi(x)gi(y) = (π◦∆)(f)(x, y). Since

π is injective this proves our statement.

So Zis also abelian, which over an algebraically closed field the exact sequence

1→keri#→Z

→χ(Z)→1

is split. We will determine concretely both the unipotent and diagonalisable part ofZ. For

the rest of this section we shall takek to be an algebraically closed field of characteristic 0.

Proposition 6.0.2. The group-like elements of Rk(Z) are elements g with g(n) = (g(1))n. Therefore the mapχ(Rk(Z))→k[k×]given byg7→eg(1)is an isomorphism of Hopf algebras.

Proof. We know that group-like elementsf ∈R_k(Z) have the property ∆(f) =f⊗f. This

says that

f(n) =f(n−1 + 1) =f(n−1)f(1) =f(n−2)f(1)f(1) =· · ·= (f(1))n

for alln∈_Z. Thusf being a group like element is uniquely determined by f(1). The map sending the group-like elements f 7→e_f₍₁₎ is clearly an isomorphism of Hopf algebras.

So the diagonalisable part of Zis given by D(k×). To understand the unipotent part,

we shall go through some theory about (linear homogeneous) recurrence relations. With

d∈_Nfixed, we say

an+d=c0an+c1an+1+...+cd−1an+d−1, for all n∈Z

is a recurrence relation and we shall denote it as (ci)di=0−1. Any f ∈ kZ with f(i) = ai satisfying the recurrence relationC:= (ci)d_i₌₀−1 is called asolution to the recurrence relation

C. So f being a solution to C is therefore uniquely determined by the recurrence relation and a set ofinitial values f(i) for 0≤ i≤d−1. Note that we will allow all but c0 to be

zero: when it is, the recurrence relation reduces to (c0_i)_id₌₀−2 with c0_i = ci+1. The following

proposition relates solutions to recurrence relations with elements ofRk(Z).

Proposition 6.0.3. Let f ∈kZ_{. Then the following conditions are equivalent:} (i) f ∈R_k(Z) with d= dim(Span{t·f}t∈Z);

(ii) d·f =Pd−1

i=0 ci(i·f) for some ci ∈k with{i·f}

d−1

i=0 linearly independent;

(iii) f is a solution to some recurrence relation (ci)d_i₌₀−1.

Proof. (i) ⇐⇒ (ii) Suppose f ∈ R_k(Z) with d= dim(Span{t·f}t∈_Z). Take the minimal

M for 0≤M ≤dsuch that {i·f}M

i=0 is linearly dependent. Then

M ·f = M−1

i=0

ci(i·f) for some ci∈k (12)

We will show that Span{i·f}M_i₌₀−1 = Span{t·f}_t∈_Z by induction. If (M+j)·f ∈Span{i·

f}M_i₌₀−1 for all 0 ≤ j ≤ m, then equation (12) above tells us that (M +m + 1)·f ∈ Span{i·f}M_i₌+_mm₊₁. By the induction hypothesis we get (M+m+ 1)·f ∈Span{i·f}M_i₌₀−1. With equation (12) as the base case, we have shown that

(M+j)·f ∈Span{i·f}M_i₌₀−1.

for allj non-negative. Now assume that (−j)·f ∈Span{i·f}M_i₌₀−1 for all 0≤j≤m, then we have (−m−1)·f ∈Span{i·f}_iM₌−₋m_m−1 by rearranging (12) sincec0 is always non-zero.

The induction hypothesis then tells us that (M+m−1)·f ∈Span{i·f}M_i₌₀−1. With equation (12) as a base case, we have that

(M +j)·f ∈Span{i·f}M_i₌₀−1

for all j non-positive. Thus, this shows that Span{i·f}_iM₌₀−1 = Span{t·f}t∈Z. Since

dim(Span{t·f}t∈Z) =d,M must be at leastd, soM =d. In particular,d·f = Pd−1

i=0 ci(i·f)

for someci∈k.

Now suppose insteadf ∈kZ_satisfies_d_·_f ₌Pd−1

i=0 ci(i·f) for someci ∈kwith{i·f}di=0−1

linearly independent. Note thatd·f =Pd−1

i=0 ci(i·f) gives us ((d+m)·f)(x) = (d·f)(x+m) = d−1 X i=0 ci(i·f)(x+m) = d−1 X i=0 ci((i+m)·f)

for allm ∈Z. By induction on m we get that (d+m)·f ∈Span{i·f}d_i₌₀−1 for allm ∈Z,

which shows that t·f ∈Span{i·f}d_i₌₀−1 for allt∈Z. By linear independence of {i·f}d_i₌₀−1 we get that dim({t·f}_t∈_Z) =d.

(ii) ⇐⇒ (iii) The equationd·f =Pd−1

i=0 ci(i·f) holds if and only if

f(n+d) = (d·f)(n) = d−1 X i=0 ci(i·f)(n) = d−1 X i=0 cif(n+i)

holds for all n∈ _Z, but this is equivalent to f being a solution to the recurrence relation (ci)d_i₌₀−1.

Given a recurrence relation C = (ci)di=0−1, it is obvious that the solutions to C form a

vector subspace ofRk(Z) and we shall denote it asVC. For 0≤i≤d−1, defineσi∈VC with initial values given byσi(j) =δ_ji for 0≤j ≤d−1. It is obvious that any f ∈VC is given byf =Pd−1

i=0 f(i)σi since every solution to a recurrence relation is uniquely determined by

its initial values. So VC = Span{σi}di=0−1. By looking at the initial values of all theσi, it is obvious that theσi are linearly independent and therefore forms a basis forVC. Motivated by this, we call these σi basis solutions to the recurrence relation. Since each non-zero representative function can only be a solution to one recurrence relation, we can decompose

Rk(Z)∼= M

as a vector space.

With :χ(Rk(Z))→ k the counit ofχ(Rk(Z)), all group like elementsfa are mapped to 1 under. Therefore the augmented idealI is the ideal generated byhfa−fbia,b∈k×. So Gu := keri# being the kernel of the canonical map i#:Z→ χ(Z) has corresponding Hopf

algebra given by k[G_u]∼=Rk(Z)/hfa−fbia,b∈k×.

Letδ ∈kZ_{be defined by}_δ₍_n_{) =}_n_{. It is representative since}_δ₍_a₊_b_{) =}_a₊_b₌_δ₍_a₎₊_δ₍_b_),

giving us ∆(δ) = δ ⊗1 + 1⊗δ. Furthermore, this shows that the algebra morphism

ϕ :k[Ga] ∼= k[x] → k[Gu] sending x 7→ δ is therefore a Hopf morphism. To see that it is actually an isomorphism, we will need the following theorem on recurrence relations:

Theorem 6.0.4. Let C = (ci)d_i₌₀−1 be a recurrence relation and VC be the corresponding

vector space containing all the solutions toC. Let {ri}ki=1 be the roots of the characteristic

polynomial xd−c0xd−1−...−cd−1 with corresponding multiplicity {mi}ki=1. Then VC has

a basis consisting of ∪k i=1{fri, δfri, δ 2_f ri, ..., δ mi−1_f ri}. Proof. [Boz, pg. 6]

Since each f ∈ R_k(Z) lies in one and only one VC for some recurrence relation C, we can write f as a unique linear combination

f =X i

αi,1fri+αi,2δfri +...+αi,mi−1δ

mi−1_f

of some basis{fri, δfri, ..., δ

mi−1_f

ri}

i=1 of VC. Taking moduloI, we get

f ≡X i

αi,1+αi,2δ+...+αi,mi−1δ

mi−1 _mod _I

Thus every f mod I has a unique polynomial expression in terms ofδ, which shows that

ϕ:k[Ga]→k[Gu] is an isomorphism. We can therefore conclude that

Z∼=Ga× D(k×) . This also says thatZis not algebraic, since

Rk(Z)∼=k[k×]⊗k[x]∼=k[x, en:n∈k×],

which can not be finitely generated as that would implyk×is finitely generated (as a group). Moreover, Zis a group with sufficient representation over any field, i.e. Z∼=Z/NZ. To

see this, just consider the basis elements of the recurrence relations (ci)d_i₌₀−1 withc0 = 1 and

c1 =c2 =· · · =cd−1 = 0 for eachd∈N. These elements are sequences given by repeated

chunks of (1,0, ...,0), and by picking the appropriate ones we easily see that evn6= evm if

n6=m. Therefore the mapη_Z :Z→Z(k) sendingn7→evn is injective (cf. theorem 4.2.4). Furthermore, every subgroup ofZ is given bynZ, which has finite index, so by Propo-

sition 4.1.10 we have that

Appendix A

The Zariski Topology

We have made no mention about the Zariski Topology throughout this thesis, as it did not turn out to be crucial in most of our proofs. But it is a fundamental concept of algebraic geometry and is one of the basic building block to the other approach of affine schemes

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