Weil representations associated to finite fields

1 Addenda to mathlib

This chapter introduces results that weren’t in mathlib (or that were in it but needed reformulation to stuck to our situation) and that are mandatory for our goal.

1.1 Group theory

In this section, we fix \(G\) group and \(H\) a commutative sugroup of \(G\). We denote by \(\mathcal{Z}_G\) the center of \(G\) and by \(e_G\) the neutral of \(G\).

1.1.1 Two lemmas about the center of a group

Proposition 1
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If \(h\) is an element of the center of \(G\), it commutes with every element of \(G\).

Proof

Trivial, just a reformulation in terms of type instead of memebership, useful for the \(LEAN\) part.

Proposition 2
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Let \(h\in \mathcal{Z}_G\) such that \(h=ab\) for some \((a,b)\in G^2\). Then, we also have \(h=ba\).

Proof

Suppose \(h=ab\). Then \(a=hb^{-1}=b^{-1}h\) because \(h\in \mathcal{Z}_G\). Thus \(ba=h\).

1.1.2 Quotient of a group by its center

Definition 3 Representatives system
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We define the system of representatives of \(G/\mathcal{Z}_G\) by picking up exactly one element in every classes. We denote it by \(\mathcal{S}_{G/\mathcal{Z}_G}\) from now on and denote by \(C_s\) the classe of \(s\).

Proof

To do that in \(LEAN\), we take the image of \(G\) by the map \(G \rightarrow G/\mathcal{Z}_G\to G\).

Proposition 4
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If \(G\) is finite, then the system of representatives of \(G/\mathcal{Z}_G\) is finite too.

Proof

trivial

Proposition 5
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Given \(g\) and \(g'\) in the set of representatives of \(G/\mathcal{Z}_G\), if \(g\ne g'\) then the classes of \(g\) and \(g'\) are disjoint.

Proof

Suppose that the classes aren’t disjoints. Then there exists \(y\) such that \(y\sim g\) and \(y\sim g'\). Thus \(g \sim g'\) and their classes are equal. But \(g\) and \(g'\) belongs to the set of representatives. Thus \(g=g'\).

Proposition 6

We have \(\bigcup \limits _{s\in \mathcal{S}_{G/\mathcal{Z}_G}} C_s= G\).

Proof

If \(x\) is in the union, in particular it belongs to \(G\). Let now \(g\) be an element of \(G\) and let show that it belongs to one of the classes. We apply the map defines in 3 to \(g\) and check that \(g\) belongs to the class of this element.

Proposition 7
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We have a bijection between \(\mathcal{S}_{G/\mathcal{Z}_G}\) and \(\{ \bar{g} \in G/\mathcal{Z}_G\} \) given by the map \(s\to \bar{s}\).

Proof

We check it is a bijection.

Definition 8
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We define a map \(\varphi _{G\mathcal{S}} : G \rightarrow \mathcal{S}_{G/\mathcal{Z}_G}\) that send every \(g\in G\) to its representative.

Proposition 9
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For every \(g\in G\) and \(h\in \mathbb {Z}_G\), we have \(\varphi _{G\mathcal{S}}(gh)=\varphi _{G\mathcal{S}} (g)\).

Proof

By definition \(g*h\) belongs to the orbit of \(g\), thus they have the same representative.

Definition 10
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We define a map \(\psi _{G\mathcal{Z}_G} : G \rightarrow \mathcal{Z}_G\) that send every \(g\in G\) to the corresponding \(h\in \mathcal{Z}_G\) such that \(g=sh\) where \(s\) is the representative of \(g\).

Proposition 11

For every \(g\in G\) the following identity holds : \(g=\varphi _{G\mathcal{S}}(g) \psi _{G\mathcal{Z}_G}(g)\).

Proof

By definition of \(G/\mathcal{Z}_G\).

Proposition 12

For every \(g\in G\) the following identity holds : \(\psi _{G\mathcal{Z}_G}(g)=g\varphi _{G\mathcal{S}}(g)^{-1}\).

Proof

Trivial with 11

Proposition 13

For every \(g\in \mathcal{S}_{G/\mathcal{Z}_G}\), we have \(\psi _{G\mathcal{Z}_G}(g)=e_G\).

Proof

By definition of the map \(\psi _{G\mathcal{Z}_G}\).

Proposition 14

For every \(g\in \mathcal{S}_{G/\mathcal{Z}_G}\), we have \(\varphi _{G\mathcal{S}}(g)=g\).

Proof

We have \(g=\varphi _{G\mathcal{S}}(g) \psi _{G\mathcal{Z}_G}(g)\) by 11. But \(\psi _{G\mathcal{Z}_G}(g)=e_G\) by 13. Thus we get the result.

Proposition 15

For every \(g\in G\) and \(h\in \mathcal{Z}_G\), we have \(\psi _{G\mathcal{Z}_G}(gh)=h\psi _{G\mathcal{Z}_G}(g)\).

Proof

With 12 we have \(\psi _{G\mathcal{Z}_G}(g)=g\varphi _{G\mathcal{S}}(g)^{-1}\) and \(\psi _{G\mathcal{Z}_G}(gh)=gh\varphi _{G\mathcal{S}}(gh)^{-1}\). But with 9, we have \(\varphi _{G\mathcal{S}}(gh)=\varphi _{G\mathcal{S}} (g)\). Thus, \(\psi _{G\mathcal{Z}_G}(gh)=gh\varphi _{G\mathcal{S}}(gh)^{-1}=gh\varphi _{G\mathcal{S}} (g)^{-1}= hg\varphi _{G\mathcal{S}} (g)^{-1}=h\psi _{G\mathcal{Z}_G}(g)\) with the first equality.

Definition 16

We define a bijection from \(G\) to \(\mathcal{Z}_G\times \mathcal{S}_{G/\mathcal{Z}_G}\) by \(g\mapsto (\psi _{G\mathcal{Z}_G}(g),\varphi _{G\mathcal{S}} (g))\).

Proof

We check the axioms of a bijection.

Definition 17

The bijection 16 empacked as Sigma type instead of cartesian product. Useful for \(LEAN\).

Proposition 18

We have \(G = \{ gh,\ g\in \mathcal{S}_{G/\mathcal{Z}_G},\ h\in \mathcal{Z}_G\} \).

Proof

The inclusion \(\{ gh,\ g\in \mathcal{S}_{G/\mathcal{Z}_G},\ h\in \mathcal{Z}_G\} \) is trivial. The converse is given by 12.

1.2 Direct sums and tensor products

1.2.1 Direct sums

Definition 19
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If we have two families \((\beta _i)_{i\in I}\) and \((\gamma _i)_{i\in I}\) of additive commutative monoids such that for every \(i\in I\), we have an additive bijection \(\varphi _i\) between \(\beta _i\) and \(\gamma _i\), then we have an additive bijection between \(\bigoplus \limits _{i\in I}\beta _i\) and \(\bigoplus \limits _{i\in I}\gamma _i\).

Proof

We send \(\sum \limits _{i\in I}x_i\) on \(\sum \limits _{i\in I}\varphi (x_i)\) and we check that it’s an additive bijection.

Definition 20
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Let \(A\) be a semiring. If we have two families \((\beta _i)_{i\in I}\) and \((\gamma _i)_{i\in I}\) of additive commutative monoids such that for every \(i\in I\), \(\beta _i\) and \(\gamma _i\) are \(A-\)module and we have a \(A-\)linear bijection \(\varphi _i\) between \(\beta _i\) and \(\gamma _i\), then we have a \(A\) linear bijection between \(\bigoplus \limits _{i\in I}\beta _i\) and \(\bigoplus \limits _{i\in I}\gamma _i\).

Proof

We take the map defined in 19 which became \(A-\)linear by the new properties of the \(\beta _i\) and \(\gamma _i\).

Proposition 21
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Let \(I\) be a finite set and \((\beta _i)_{i\in I}\) a family of additive commutative monoids. Let \(\Phi \) be the natural map sending \(\beta _{i_0}\) to \(\bigoplus \limits _{i\in I}\beta _i\). Then, for every \(x:=(x_i)_{i\in I}\) such that \(x_i\in \beta _i\) for all \(i\in I\), then for every \(j\in I\), the following equality holds : \(\left(\sum \limits _{i\in I} \Phi (x_i)\right)_j=x_j\).

Proof

We obvioulsy have \(\left(\sum \limits _{i\in I} \Phi (x_i)\right)=\sum \limits _{i\in I}x_i\), which immediately gives the result.

1.2.2 Tensor products

Definition 22
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Let \(A\) be ring, \(B\) an \(A-\)algebra, \(M\) an \(A-\)module and \(N\) a \(B-\)module. Then, \(\text{Hom}_B((B\otimes _AM),N)\cong \text{Hom}_A(M,N)\).

Proof

We consider the map sending \(\varphi \in \text{Hom}_B((B\otimes _AM),N)\) to the \(A-\)linear map \(\Phi _\varphi : M \rightarrow N\) defined by \(\Phi _\varphi (x)=\varphi (1\otimes _A x)\) for every \(x\in M\). It is injective : if \(\Phi _{\varphi _1}=\Phi _{\varphi _2}\), then \(\varphi _1(1\otimes _A x)=\varphi _2(1\otimes _A x)\) for every \(x\in M\). Thus \(\varphi _1=\varphi _2\) by \(B-\)linearity. It is surjective : let \(\varphi \) be a \(A\)-linear map from \(M\) to \(N\). Let consider the \(B-\)linear map \(\psi \) from \(B\otimes _AM\) to \(N\) define by \(\psi (b\otimes _A m) = b \varphi (m)\). We have then \(\Psi _\psi (x)=\psi (1\otimes _A x)= \varphi (x)\).

1.3 Group algebra

A lot of the results in this section wouldn’t really appear in classical mathematics papers, but they are needed to ensure that LEAN understand the operations we will do later.

From now on, \(\mathbb {K}\) is a field, \(G\) is a group and \(H\) is a subgroup of \(G\). We define \(mathcal{Z}_G\) as the center of \(G\).

1.3.1 Setting up operations, coercions and instances in LEAN

Definition 23
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Given \(\mathbb {K}\) a field, \(G\) a group and \(H\) a subgroup of \(G\), we have a trivial ring homomorphism \(\varphi _{kHkG}\) from \(\mathbb {K}[H]\) to \(\mathbb {K}[G]\).

Proof

Trivial.

Proposition 24
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The map defined in 23 is injective.

Proof

Trivial.

Proposition 25
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We have an equality between \(h\in H\) seen as an element of \(\mathbb {K}[H]\) and \(h\) (seen as an element of \(G\)) seens as an element of \(\mathbb {K}[G]\).

Proof

Some LEAN stuff.

Proposition 26
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The map defined in 23 is \(k\) linear.

Proof

Trivial.

Definition 27
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We define a coercion from elements of \(\mathbb {K}[H]\) to \(\mathbb {K}[G]\) by the map defined in 23.

Proof

Some LEAN stuff.

Definition 28
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We define a coercion from sets of elements of \(\mathbb {K}[H]\) to sets of elements of \(\mathbb {K}[G]\) by the map defined in 23.

Proof

Some LEAN stuff.

Definition 29
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We define a multiplication between elements of \(\mathbb {K}[H]\) and elements of \(\mathbb {K}[G]\) by \(kH*kG=\varphi _{kHkG}(kH)\times kG\).

Proof

Some LEAN stuff.

Proposition 30
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\(\mathbb {K}[G]\) is a \(\mathbb {K}[\mathcal{Z}_G]\) algebra.

Proof

We check the axiom of an algebra.

Proposition 31
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If there exists a morphism from \(H\) to \(\mathcal{Z}_G\), then \(\mathbb {K}[G]\) is a \(\mathbb {K}[H]\) algebra.

Proof

We check the axiom of an algebra with the action of \(\mathbb {K}[H]\) on \(\mathbb {K}[G]\) given by the morphism.

Proposition 32
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Let \(x\in \mathbb {K}[\mathcal{Z}_G]\) and \(g\in G\). Then \(g\times x = x \times g\).

Proof

We have \(g\times x = g\times \sum \limits _{h\in \mathcal{Z}_g}a_hh=\sum \limits _{h\in \mathcal{Z}_g}a_hgh =\sum \limits _{h\in \mathcal{Z}_g}a_hhg=\left(\sum \limits _{h\in \mathcal{Z}_g}a_hgh\right)\times g\).

Definition 33
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\(\mathbb {K}[\mathcal{Z}_G]\) defines a \(\mathbb {K}[G]\) submodule.

Proof

We check the axioms.

Definition 34
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We define the multiplciation of elements \(g\in G\) and \(kH\in \mathbb {K}[\mathcal{Z}_G]\) in \(\mathbb {K}[G]\) by \(g\times kH=g\times \varphi _{kHkG}\)

Proof

Some LEAN stuff.

Definition 35
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We define the multiplciation of elements \(g\in G\) and \(kG\in \mathbb {K}[G]\) in \(\mathbb {K}[G]\) by \(g\times kG\).

Proof

Some LEAN stuff.

Proposition 36
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Elements of \(G\) are distributive over \(\mathbb {K}[\mathcal{Z}_G]\)

Proof

Trivial, LEAN stuff.

1.3.2 Splitting of a group algebra as a direct sum

We use the notation of the section 1 concerning \(G/\mathcal{Z}_G\) The main goal of this part is to formalize the following result : \(\mathbb {K}[G] \cong \bigoplus \limits _{g\in \mathcal{S}_{G/\mathcal{Z}_G} } g \mathbb {K}[\mathcal{Z}_G]\).

Definition 37
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Let \(g\in G\) be fixed. The morphism \(\varphi _g : \mathcal{Z}_G\rightarrow G\) defined by \(\varphi _G(x)=gx\) induced a \(\mathbb {K}\)-linear map \(\Gamma _g\) from \(\mathbb {K}[\mathcal{Z}_G]\) to \(\mathbb {K}[G]\).

Proof

Trivial.

Proposition 38
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The map \(\Gamma _g\) defined in 37 is injective.

Proof

Trivial.

Proposition 39
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For all \(x\in \mathbb {K}[\mathcal{Z}[G]]\), we have \(\Gamma _g(x)=g\times x\)

Proof

LEAN stuff to setup a simp lemma.

Definition 40
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We define the set \(\Omega _g\) to be the image of \(\mathbb {K}[\mathcal{Z}_G]\) by \(\Gamma _G\).

Proof

Nothing useful mathematically, it’s just simpler for translating LEAN stuff.

Definition 41

The map \(\Gamma _g\) defines a \(k\)-linear bijection between \(\Omega _g\) and \(\mathbb {K}[\mathcal{Z}_G]\).

Proof

It’s injective like we saw before, and \(g^{-1}\times x\) is a preimage for \(x\).

We will now put some structure on \(\Omega _g\) to makes it understand by LEAN as a \(\mathbb {K}[\mathcal{Z}_G]\) module.

Definition 42

We define a multiplication between \(x\in \mathbb {K}[\mathcal{Z}_G]\) and \(y\in \Omega _g\) by \(x\times y\).

Proof

LEAN stuff.

Definition 43
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The multiplication defined in 42 induced an action of \(\mathbb {K}[\mathcal{Z}_G]\) on \(\Omega _g\).

Proof

We check the axioms.

Proposition 44
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The action defined in 43 is indeed distributive.

Proof

We check the axioms.

Proposition 45
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\(\Omega _g\) is a \(\mathbb {K}[\mathcal{Z}_G]\) module for the action defined in 43.

Proof

We check the axioms.

Definition 46

The bijection defined in 41 is indeed a \(\mathbb {K}[\mathcal{Z}_G]\)-linear map.

Proof

We check the linearity.

Elements of \(\mathcal{S}_{G/\mathcal{Z}_G}\) seen as elements of \(\mathbb {K}[G]\) defined a basis of \(\mathbb {K}[G]\) as a \(\mathbb {K}[\mathcal{Z}_G]\) algebra.

Proof

We show the independence of the family and then that it generates the whole algebra.
We suppose that there exists a family \((a_i)_{i\in \mathcal{S}_{G/\mathcal{Z}_G}}\) of elements of \(\mathbb {K}[\mathcal{Z}_G]\) such that

\begin{equation*} \sum \limits _{i\in \mathcal{S}_{G/\mathcal{Z}_G}}a_i g_i=0 \end{equation*}

Using the fact that the family \((h)_{h\in \mathcal{Z}_G}\) seen as a family of elements of \(\mathbb {K}[\mathcal{Z}_G]\) is indeed a basis of \(\mathbb {K}[\mathcal{Z}_G]\) as a \(\mathbb {K}\) vector space we get :

\begin{equation*} \sum \limits _{i\in \mathcal{S}_{G/\mathcal{Z}_G}}\left(\sum \limits _{h\in \mathcal{Z}_G}a_{ih}h\right) g_i=0 \end{equation*}

where the \(a_{ih}\) are elements of \(\mathbb {K}\). Those sums are finite, and moreover, the family \((ih)_{(i,h)\in \mathcal{S}_{G/\mathcal{Z}_G}\times \mathcal{Z}_G}\) is a partition of \(G\).
Thus, we get :

\begin{align*} \sum \limits _{i\in \mathcal{S}_{G/\mathcal{Z}_G}}\left(\sum \limits _{h\in \mathcal{Z}_G}a_{ih}h\right) g_i & = \sum \limits _{i\in \mathcal{S}_{G/\mathcal{Z}_G}}\sum \limits _{h\in \mathcal{Z}_G}a_{ih}g_ih\\ & =\sum \limits _{g\in G}a_g g \end{align*}

which is equal to 0. But \((g)_{g\in G}\) seen as a family of elements of \(\mathbb {K}[G]\) is a basis of \(\mathbb {K}[G]\) as a \(\mathbb {K}\) vector space. Finally, \(\forall g\in G,\ a_g=0\) and the family is independent.

Let shows the family generates the whole algebra. Let \(x\in \mathbb {K}[G]\). Using the natural basis \(G\) of \(\mathbb {K}[G]\) and that \((ih)_{(i,h)\in \mathcal{S}_{G/\mathcal{Z}_G}\times \mathcal{Z}_G}\) is a partition of \(G\), we get :

\begin{align*} x& =\sum \limits _{g\in G}a_gg\\ & =\sum \limits _{i\in \mathcal{S}_{G/\mathcal{Z}_G}}\sum \limits _{h\in \mathcal{Z}_G} a_{ih} ih\\ & =\sum \limits _{i\in \mathcal{S}_{G/\mathcal{Z}_G}}\left(\underbrace{\sum \limits _{h\in \mathcal{Z}_G} a_{ih} h}_{\in \mathbb {K}[\mathcal{Z}_G]}\right) i \end{align*}

So it generates the whole algebra and finally we get a basis.

Definition 48
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We define a map on \(\mathcal{S}_{G/\mathcal{Z}_G}\) that associates to every elements \(g\in \mathcal{S}_{G/\mathcal{Z}_G}\) the natural element of \(\bigoplus \limits _{s\in \mathcal{S}_{G/\mathcal{Z}_G}}\mathbb {K}[\mathcal{Z}_G]\), that is \(1\) of \(\mathbb {K}[\mathcal{Z}_G]\) on the \(g-\)th component and 0 elsewhere.

Definition 49

We construct a \(\mathbb {K}[\mathcal{Z}_G]\)-linear map on \(\bigoplus \limits _{g\in \mathcal{S}_{G/\mathcal{Z}_G}}\mathbb {K}[\mathcal{Z}_G]\) by using the images of the element of the basis 47 by the map 48.

Definition 50
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The map defined in 49 is in fact an isomorphism.

Proof

We give an explicit inverse which is \((x_1,\dots ,x_g)\mapsto \sum \limits _{g\in \mathcal{S}_{G/\mathcal{Z}_G}} x_gg\).

We have an isomorphism \(\mathbb {K}[G]\cong \bigoplus \limits _{g\in \mathcal{S}_{G/\mathcal{Z}_G}}g\mathbb {K}[\mathcal{Z}_G]\) which is \(\mathbb {K}[\mathcal{Z}_G]\)-linear.

Proof

We compose the maps defined in 50 and 46 and use the bijection 20.

1.4 Representation theory

The main goals of this part is to build the representation induced by the center \(\mathcal{Z}_G\) of a group \(G\) and its basic properties, and to formalize its character.

From now on, \(\mathbb {K}\) is a field and \(G\) is a group. We denote by \(\mathcal{Z}_G\) its center. \(W\) is a \(\mathbb {K}-\)vector space. Finally, \(\theta \) is a representation of \(\mathcal{Z}_G\) in \(W\).

1.4.1 Building the induced representation

Definition 52 Induced representation as module
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Given \(G\) a group, \(\mathbb {K}\) a field, \(W\) a \(k\) vector space and \(\theta \) a representation of \(\mathcal{Z}_G\) on \(W\), we define the tensor product \(V:=\mathbb {K}[G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta \), where \(V_\theta \) is the \(\mathbb {K}[\mathcal{Z}_G]\) module associated to \(\theta \).

Proposition 53

The \(V\) defined in 52 is an additive commutative monoid.

Proof

It comes from the fact that \(\mathbb {K}[G]\) and \(\mathbb {K}[\mathcal{Z}_G]\) are additive commutative monoids.

Proposition 54

The \(V\) defined in 52 is a \(\mathbb {K}[G]\) module.

Proof

We do mettre la preuve.

Proposition 55

The \(V\) defined in 52 is a \(\mathbb {K}[\mathcal{Z}_G]\) module.

Proof

We do mettre la preuve.

Definition 56 Induced representation by the center
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The \(V\) defined in 52 defined a representation of \(G\) called the induced representation by \(\mathcal{Z}_G\).

Definition 57 Subrepresentation of the induced
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We define the subrepresentation of 56 by \(\mathbb {K}[\mathcal{Z}_G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta \), where \(V_\theta \) is the \(\mathbb {K}[\mathcal{Z}_G]\) module associated to \(\theta \).

Proposition 58

The tensor product defined in 57 is an additive commutative monoid.

Proof

It comes from general properties of tensor products.

Proposition 59

The tensor product defined in 57 is a \(\mathbb {K}[\mathcal{Z}_G]\) module.

Proof

It comes from general properties of tensor products.

Proposition 60

The induced representation defined in 56 is a \(\mathbb {K}[\mathcal{Z}_G]\) module.

Proof

It comes from general properties of tensor products.

Proposition 61

We have an isomorphism between \(\mathbb {K}[\mathcal{Z}_G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta \) and \(V_\theta \).

Proof

It comes from a special case of a theorem à ajouter.

Proposition 62 Coercion

We have a coercion from element of type \(\mathbb {K}[\mathcal{Z}_G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta \) to element of type \(\mathbb {K}[G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta \).

Proof

It comes from 23.

Proposition 63 Coercion set

We have a coercion from element of type \(Set : \mathbb {K}[\mathcal{Z}_G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta \) to element of type \(Set : \mathbb {K}[G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta \).

Proof

It comes from 23.

Proposition 64 \(V_\theta \) as submodule
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The set of elements of \(V_\theta \) defines a \(\mathbb {K}[\mathcal{Z}_G]\)-submodule of itself.

Proof

Trivial.

Proposition 65 \(V_\theta \) as submodule isomorphic to \(V_\theta \)

The submodule defined in 64 is isomorphic to \(V_\theta \).

Proof

Trivial.

Proposition 66 Subrepresentation of the induced one as submodule

The image of the map sending 33 and 65 to their tensor product defines a \(\mathbb {K}[\mathcal{Z}_G]\)-submodule of \(\mathbb {K}[G]\otimes _{\mathbb {k}[\mathcal{Z}_G]}V_\theta \).

Proof

Trivial.

Proposition 67 Image of \(V_\theta \) as submodule
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The image of \(V_\theta \) by 61 defines a \(\mathbb {K}[\mathcal{Z}_G]\)-submodule of \(\mathbb {K}[G]\otimes _{\mathbb {k}[\mathcal{Z}_G]}V_\theta \), ie \(V_\theta \) is a subrepresentation of the induced.

Proof

Compute the axioms.

Proposition 68 Induced reprensentation property

Let \(E\) be a \(\mathbb {K}[G]\) module. We have an isomorphism \(Hom_{\mathbb {K}[G]} \left(\mathbb {K}[G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta ,\ E\right)\simeq Hom_{\mathbb {K}[\mathcal{Z}_G]} \left(\mathbb {K}[\mathcal{Z}_G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta ,\ E\right)\).

Proof

We use ?? two times.

1.4.2 Character of the induced representation

Definition 69 Central function
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Given \(G\) a group, a function \(f\) over \(G\) is called central if it is constant on the conjugacy classes of \(G\) : \(f(g^{-1}xg)=f(x)\) for all \(g\in G\) and \(x\in G\).

Definition 70 Induced central function
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If \(H\) is a subgroup of \(G\) a finite group, and if \(f\) is a function over \(H\), we define a central function \(f_G\) over \(G\) (called the induced central function on \(f\)) by the formula :

\begin{equation*} f_G(x)=\frac{1}{\text{Card}(H)}\sum \limits _{g\in G\ \wedge \ g^{-1}xg\in H}f(g^{-1}xg) \end{equation*}
Proof

We check the axiom by reordering the sum with the bijection \(x\mapsto g^{-1}x\).

Definition 71 Character as central function

A character is of course a central function over \(G\). We empacked this definition in Lean.

Proof

Trivial.

Proposition 72 Induced representation is semisimple

If \(|G|\nmid Char(k)\), then the module defined in 52 is semisimple.

Proof

Consequences of .

Definition 73

We define the \(\mathbb {K}[\mathcal{Z}_G]\) module \(W_g:= g\mathbb {K}[\mathcal{Z}_G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta \) for \(g\in \mathcal{S}_{G/\mathcal{Z}_G}\).

Definition 74

We have an isomorphism between \(\mathbb {K}[G]\otimes _{\mathbb {K}[\mathcal{Z}_G]}V_\theta \) and \(\bigoplus \limits _{g\in \mathcal{S}_{G/\mathcal{Z}_G}}W_g=\bigoplus \limits _{g\in \mathcal{S}_{G/\mathcal{Z}_G}} g \mathbb {K}[\mathcal{Z}_G] \otimes _{\mathbb {K}[\mathcal{Z}_G]} V_\theta \).

Proof

We get the resultat with the isomorphism \(\mathbb {K}[G] \cong \otimes _{\mathbb {K}[\mathcal{Z}_G]} g\mathbb {K}[\mathcal{Z}_G]\) defined in 51.