Hallazgos

later on, we will not be using any of these results, but they are included as extensions of results from affine schemes.

2.6. GEOMETRY OF CONNECTED COMPONENTS OF AFFINE GROUPS 31

We say that an affine group is connected (resp. etale) if its underlying algebra is con- nected (resp. etale). As mentioned at the end of Chapter 1, we will prove the following characterisation for an algebraic affine group to be connected:

Proposition 2.6.1. Let G ∼=Spec(A) be an algebraic affine group. Then the following are equivalent

(i) π0(G) is the trivial affine group,

(ii) G is connected.

Proof. (i)⇒(ii) is true for affine scheme. For the converse, first note thatAbeing connected implies thatπ0(A) has no non-trivial idempotents. Thus π0(A) can only be some separable

field extensionL ⊇k. But :A →k being ak-algebra morphism gives us k→L→ k, so

π0(A) =k.

Indeed, when G ∼= Spec(A) is connected, π0(G) does not tell us much; But when it is

not, note that the images of etale algebras are etale, so we have ∆ :π0(A)→π0(A⊗A) =

π0(A)⊗π0(A) and S :π0(A) → π0(A) (nothing to show for counit). This makes π0(A) a

Hopf subalgebra ofA, andπ0(G)∼= Spec(π0(A)) is a finite etale affine group. We callπ0(G)

thegroup of connected components ofG. Once again, since the images of etale algebras are etale, we have that:

Proposition 2.6.2. Let G → H be an algebraic affine group homomorphism with H etale. Then the map factors through π0(G).

Chapter 3

Constant Group Schemes

In this chapter we will use some results from scheme theory, which we will not cover but provide references for interested readers. For this chapter and the next, we will be working over fields.

3.1

Construction; Affine or non-Affine?

Given a finite groupG, we can easily construct an affine group scheme using the structure of

G. Take the disjoint union of|G|number of pointsG=`

g∈GSpeck(k)∼= Speck(

Q

g∈Gk)∼= Spec_k(kG), with kG denoting the algebra of functions from G to k. To understand its structure, note that with eg the function sending g to 1 and the rest to 0, Pg∈Geg = 1,

eg ·eh = 0 for allg 6=h and they are all idempotents. Since{eg}g∈G forms a basis for kG, morphisms out of kG are uniquely determined by where each eg is sent to. Thus, for the case where R ∈ Alg_k with no non-trivial idempotents, we have that each homomorphism inG(R) = HomAlgK(k

G_{, R) has to send one and only onee}

g to 1, and the rest to 0, which then uniquely determines the function. ThusG(R) andGare in bijection, given byeg 7→g. With that bijection in mind we define its group structure through its corresponding Hopf algebra: define comultiplication ∆G : kG → kG⊗kG to be the map sending eg 7→

P

h·h0_=geh⊗eh0; the counit_G:kG→kis defined byf 7→f(1G), and coinverseS :kG→kG is defined by f 7→finv withfinv defined by finv(s) = f(s−1). It is clear that they satisfy the required commutative diagrams for Hopf algebras.

But when G is infinite, there is an obvious problem: the multiplication map we define above will be an infinite sum, which does not make sense. As we will see later, it turns out that without making it a group scheme rather than an affine group scheme, the same construction would not work. We shall look at the definition of the multiplication map from another, possibly not as intuitive, but equivalent perspective, which will provide us a way of solving this.

We shall first prove a lemma that will come in handy.

Lemma 3.1.1. Let V be a finite dimensional k-subspace of kG. There exists a basis {vi}

of V with corresponding set of elements {si} ⊂ G such that vi(sj) = δj_i, where δji is the

kronecker delta.

Proof. We shall prove this by induction on the dimension of V. For the base case where dim(V) = 1, it is obvious by linearity that we can finds1 ∈Gsuch thatv1(s1) = 1 for some

v1∈V. Now assume that the statement is true for allm-dimensional subspace ofkG. LetV

be a subspace ofkGwith dimensionm+ 1 and letV0 be a subspace ofV with dimension m. By the induction hypothesis the subspaceV0has a basis{v0_i}m

i=1with corresponding elements

{si}mi=1satisfyingvi0(sj) =δji. Choosing an elementvm0 +1∈V\V0, we obtain a basis{vi0} m+1

i=1

ofV. Ifv0_m+1(s) = 0 for all elements s∈G\ {si}mi=1, then vm0 +1 =

Pm

i=1civ0i−v0i ∈V0 for someci ∈ k. So there must be an element sm+1 ∈G\ {si}m_i=1 such that v0m+1(sm+1) 6= 0

We can then define

vm+1= (v0m+1(sk+1))−1vm0 +1

and

vi=v0i−vi0(sk+1)vm+1

for all 1≤i≤m. It is obvious now that the set of vectors{vi}m_i=1+1 and the subset{si}m_i=1+1 ofS satisfy the condition vi(sj) =δ_ji. We just need to show that {vi}m_i=1 still forms a basis forV. IfPm+1 i=1 civi = 0, then 0 = m+1 X i=1 civi=cm+1 vm0 +1(sk+1) −1 v0_m+1+ m X i=1 ci v0i−v 0 i(sk+1 vm+1) = cm+1 vm0 +1(sk+1) −1 − m X i=1 civi0(sk+1) ! v_m0 ₊₁+ m X i=1 civi0,

which shows thatci = 0 for all 1≤i≤m. Hence cm+1 v0m+1(sk+1)

−1

= 0, which shows thatcm+1 is also 0. We can therefore conclude that {vi}m_i=1+1 are linearly independent and thus form a basis forV. This completes the induction proof.

Proposition 3.1.2. Let G and H be groups. The canonical map π : kG⊗kH → kG×H

defined by

πXfi⊗gi

(a, b) =Xfi(a)gi(b)

is injective, and the images are all f ∈ kG×H such that the k-vector space spanned by

(t·f)(s) =f(s, t) over all t∈H is finite dimensional. Proof. To show injectivity, letπ(P

fi⊗gi) = 0, i.e. Pfi(a)gi(b) = 0 for all a∈G, b∈H. Using lemma 3.1.1, we have a basis {vi} for the finite dimensional space Span{gi} with corresponding{si} ⊂H such thatvi(sj) =δij. Rewriting

P

fi⊗gi =Phi⊗vi in terms of the new basis, an evaluation at (a, sj) gets us

π(Xhi⊗vi)(a, sj) =

X

hi(a)vi(sj) =hj(a) = 0.

Since this works for all j and a, we know allhj = 0 and therefore Pfi⊗gi = 0, showing injectivity.

For the second statement, if f(s, t) = P

fi(s)gi(t), then obviously Span{(t·f)}t∈H ⊆ Span{fi}, which is therefore finite dimensional. Conversely suppose f ∈ kG×H such that Span{(t·f)}t∈H is finite dimensional with basis {fi}, then

3.1. CONSTRUCTION; AFFINE OR NON-AFFINE? 35

with each ci,t ∈ k. Running this over all t∈ G, we define functions gi : G→ k such that

gi(t) =ci,t. Thus we get that

f(s, t) =Xgi(t)fi(s) = X fi(s)gi(t) =π X fi⊗gi (s, t),

concluding the proof of the second statement.

Now take H = G. When G is finite, all f ∈ kG×G clearly have Span{(t·f)} finite dimensional and therefore Proposition 3.1.2 saysπis an isomorphism. So given a finite group

Gwith multiplication map m, we can define π−1◦m∗ :kG →kG⊗kG with m∗(f)(a, b) =

f(m(a, b)) =f(a·b). The following proposition shows that this comultiplication agrees with the one we discussed previously, and therefore defines the same constant group scheme of

G.

Proposition 3.1.3. The maps π−1◦m∗ and ∆G agree.

Proof. Equivalently, we can show thatm∗=π◦∆G. As noted above, we just need to show that they to agree on the image of eg for all g∈G. We compute

π◦∆G(eg)(s, t) =π   X h·h0_=g eh⊗eh0  (s, t) = X h·h0_=g eh(s)eh0(t) =δ_sg_·t =eg(s·t) =m∗(eg)(s, t),

which shows that the two maps agree.

WhenGis infinite, we see that not all the images off ∈kGunderm∗ lie in the image of

π, so again the same construction wouldn’t work. This is an indication that what we want to construct might not be affine. So we shall construct this in the larger category: category of schemes. The coproduct (disjoint union) G =

sch

`

g∈GSpeck(k) will be the coproduct in the category of schemes, which when finite, is isomorphic to the disjoint union in the subcategory affine scheme [GD71, pg. 104]. We will now explicitly construct the constant group scheme, which will unavoidably involve some knowledge in scheme theory.

Construction: Intuitively, the disjoint union

sch

`

g∈GSpeck(k) is really just a disjoint union of affine points A0, each representing an element in G. Let U ⊆G be open. The topology

of Gis given by the disjoint union topology, soU =

sch

`

s∈SSpeck(k) for some S⊆G, which shows that every open set corresponds to some subset of G. The sheaf O of G is also straight-forward: O(U) is just kS. For V ⊆U and S0 ⊆ S its corresponding subset of G, the restriction maps resU_V :kS → kS0 are given by the canonical projection maps. As one

would expect, the multiplicationµis just the map taking the component representing (g, h) inG×Gto the component ghinG. It is known that

G×G= sch a g∈GSpec(k)× sch a g∈GSpec(k) ∼ = sch a (g,h)∈G×GSpec(k).

So the sheaf onG×G is also the obvious one. SinceG×Gis a disjoint union of Spec(k), morphism out of it is uniquely determined by morphism out of each of its components. We get thatµ is uniquely determined by maps

` G×GSpec(k) ` GSpec(k) Spec(k) Spec(k) µ i(g,h) igh =

on each component (g, h), withi denoting the inclusion map to the respective component subscripted accordingly. On global section of the sheaves, we have

O ` G×GSpec(k) ∼ =Q<sub>G</sub>×Gk O( ` GSpec(k))∼= Q Gk O(Spec(k))∼=k O(Spec(k))∼=k µ# P(g,h) Pgh =

withP the canonical projections onto the components subscripted accordingly. Thus, iden- tifying kG ∼= Q_Gk and kG×G ∼= Q_G×Gk, we get that µ# on global section maps each

f ∈ kG to f0 ∈ kG×G such that f0(a, b) = f(ab). So µ# is just the map m∗ as before. This construction, whenGis finite, leads to the same construction above. As we mentioned in Section 1.4, the infinite disjoint union of affine schemes (in the category of schemes, of course) is not even affine, so constant group schemes for G infinite are non-affine group scheme. In the next section, we shall determine the affinisation of such group schemes, which will be the central mathematical object of this thesis.