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Recall from linear algebra that a matrix M ∈ GLn(k) is diagonalisable if and only if there exist basis {vi} such that M(vi) = λivi for some λi ∈ k. So with Vi = Span{vi} one-dimensional subspaces of V, we know that we can decompose V = L

Vi such that

M(Vi)⊆Vi for allVi. Taking this idea a bit further, we can have a subgroupM ⊆GLn(k) with all M ∈ M diagonalisable under the same basis. So if M is the image of some representationG → GL(V) ∼= GLn, this says exactly that our representation V is a direct sum of one dimensional subrepresentations Vi. Motivated by this, we make the following definition: A representation Φ :G → GL(V) is diagonalisable if it is a direct sum of one- dimensional subrepresentations.

We shall for a moment switch our focus to one-dimensional representations. For an affine groupG, we call homomorphismsξ :G →_Gm (the one-dimensional representations) thecharacters of G. Sincek[Gm]∼=k[x, x−1], by transporting the Hopf structure onk[Gm], every character corresponds to invertible elementsa∈k[G] such that

∆(a) =a⊗a, S(a) =a−1, (a) = 1.

We call thesegroup-like elements of k[G].

Proposition 5.3.1. The set of group-like elements are linearly independent.

Proof. Assume to the contrary that they are not, then for some group like element a we can write it as a linear combination of other group like elements:

a=Xciai, a6=ai and ci ∈k for all i.

Note that we can always reduce the set{ai}to a linearly independent set and rewrite aas a linear combination of linearly independent group-like elements, so WLOG assume that {ai} are linearly independent. Being group-like, we have that

X ciai⊗ai= ∆ X ciai = ∆(a) =a⊗a=Xcicjai⊗aj.

Since{ai} are linearly independent, so is {ai⊗aj}. Thus by comparing coefficients we get

ci=c2i and cicj = 0 for i6=j. Furthermore,

X ci = X ciai =(a) = 1.

5.3. DIAGONALISABLE GROUPS 59

Thus{ai} is also a set of complete idempotents in k, where kbeing a field shows that one (and only one) suchai is equal to 1 and the rest are all 0. Thena=ai for some i, which is a contradiction.

Given any abelian group G, we can form the group algebrak[G], i.e. the free module

L

g∈Gkeg (we are assumingkis a field, so a vector space) with multiplication defined on the basis by egeh =egh expanded linearly. By making the basis elements group-like, it would have a Hopf algebra structure since each basis element would satisfy the three required commutative diagrams (as we know it does fork[Gm]). Thus, Spec(k[G]) is an affine group

and we call such affine groupsdiagonalisable groups, denoted asD(G).

Before we explain the name chosen, we shall prove a theorem that characterises these affine groups:

Theorem 5.3.2. An affine groupG is diagonalisable if and only ifk[G]is spanned by group- like elements. Furthermore there is an anti-equivalence between the category of abelian groups and the category of diagonalisable affine groups given by G7→ D(G).

Proof. The direction (⇒) is true just by definition of diagonalisable. Conversely, assume thatk[G] is spanned by group-likes. Then the group-likes form a basis as shown in Propo- sition 5.3.1. Since the group-like elements are invertible, they form a group XG under multiplication (the identity 1 is obviously a group-like element too). With k[G]→ k[XG] sending each group-like a7→ ea, it clearly preserves the Hopf structures and therefore we have a Hopf algebra isomorphism, showingG is diagonalisable.

The functor sending G 7→ D is essentially surjective by definition. Every group homo- morphism G → H uniquely determines a unique Hopf algebra map k[G]→ k[H] induced on the basis elements, which thus corresponds to an affine group morphism Spec(k[H])→ Spec(k[G]). Thus the (contravariant) functor is also fully faithful.

Example 5.3.3. The affine group Gm is diagonalisable. Just note that k[Gm]∼=k[x, x−1] is

spanned by X ={xn}_n∈Z as a k-vector space and we know that these elements are group

like since x is group-like. SoX forms a group under multiplication and we can see that it is isomorphic to the group Zsincexaxb =xa+b.

The following theorem establishes the connection between diagonalisable groups and diagonalisable representations:

Theorem 5.3.4. Let G ∼=Spec(A) be an affine group. The following conditions are equiv- alent:

(i) G is diagonalisable;

(ii) Every finite dimensional representation of G is diagonalisable.

Proof. Assume that G is diagonalisable. Let Φ : G → GL(V) be a representation with

ρ : V → V ⊗A the corresponding comodule. Since the group-like elements {ai} form a basis of A, we can write ρ(v) =P

vi⊗ai. Being a comodule gives us the equality

X

ρ(vi)⊗ai= (ρ⊗id)(ρ(v)) = (id⊗∆)(ρ(v)) =

X

With{ai}being a basis, by comparing coefficients we obtain

ρ(vi) =vi⊗ai.

Thus each Span(vi) corresponds to a one-dimensional subrepresentation ofV. By reducing the setW ={vi}if necessary, we can make it into a linearly independent set. So WLOG we may assume thatW is linearly independent. Pickv0 ∈V\Span(W), and repeat the process above. This gives us another set {v0_i} ⊆ V such that each Span(v0_i) is a one-dimensional subrepresentation ofV, which we can again assume that the set{v0_i}is linearly independent. As in Remark 5.1.3, we have identity (id⊗)◦ρ= id, which tells us that

v0 = (id⊗)(ρ(v0)) =Xv0_i.

By virtue of v0 ∈/ Span(W), we can remove enough but not all v0_i such that Span{v0_i} ∩ Span(W) = 0. Now Span(W ∪ {v_i0}) has dimension strictly more than Span(W), so by induction we will achieve a set of basis of V such that each of them corresponds to a one-dimensional subrepresentation ofV, which shows thatV is diagonalisable.

Now assume (ii) instead. The regular representation of G is therefore diagonalisable, which says that there is a basis {mi} of A such that each Mi := Span(mi) is a one- dimensional subrepresentation. So we have one-dimensional subcomodules ρ = ∆ : Mi →

Mi⊗Asuch that ∆(mi) =mi⊗ai for someai ∈A. The equation (10) at the end of Section 5.1 tells us that ∆(ai) = ai⊗ai, so these ai are group-like elements. Finally the equation id = (◦id)◦∆ gives us that

mi = (◦id)(∆(mi)) =(mi)ai,

which shows thatmiis a multiple ofai. Thus the set of group-like elements{ai}also spans

A. By Proposition 5.3.2 we get thatG is diagonalisable.