Tensorial product and representations of finished groups

In Mathematical and more precisely within the framework of the Representations of a group finished, the tensorial Produit is a technique making it possible to build a representation from two others.

If a Groupe finished G is produced direct of two groups, then all irreducible Représentation of G is isomorphous with a tensorial product of the two groups.

Context

Tensorial product

Tensorial product of two vector spaces

Are V and W two vector spaces on a body Commutatif K of finished Dimension S respective N and m . The tensorial product of V and W , noted V

\ otimes

W corresponds to the vector space of the forms of V X W in K linear in V and W . I.e.:

\ forall \ varphi \ in V \ otimes W \; \ forall \ lambda_1, \ lambda_2 \ in \ mathbb K \; \ forall v_1, v_2 \ in V \; \ forall w_1, w_2 \ in W \ quad \ begin {boxes} \ varphi (\ lambda_1 v_1+ \ lambda_2 v_2, w_1) = \ lambda_1 \ varphi (v_1, w_1) + \ lambda_2 \ varphi (v_2, w_1) \ \
\ varphi (v_1, \ lambda_1 w_1+ \ lambda_2 w_2) = \ lambda_1 \ varphi (v_1, w_1) + \ lambda_2 \ varphi (v_1, w_2) \ end {boxes}

That is to say (a_i) (resp. (b_j)) a base of V (resp. W ), it is then possible to identify V (resp. W ) with its dual Space. Indeed, U (resp is . v ) a vector of V (resp. W ) of coordinates (α_i) (resp. (β_j)) in the preceding base, then a_i (resp. b_j) is identified with the following linear form: $= = \ alpha_i \ quad and \ quad = =\beta_j$ This identification establishes a canonical application $\ otimes$ of V X W towards V $\ otimes$ W defined, with the preceding notations by: $a_i \ otimes b_j (U, v) = \ alpha_i. \ beta_j$ The family (a_i $\ otimes$ b_j) is then the canonical base of V $\ otimes$ W . By using the three bases, one obtains the equality: $u \ otimes v = \ sum_ {ij} \ alpha_i \ beta_j \. \, a_i \ otimes b_j$
Passage aux endomorphisms
The preceding application $\ otimes$ indicates also an application of L ( V ) X L ( W ) in L ( V $\ otimes$ W ). Here L ( E ) indicates the vector space of the Endomorphisme S of E . that is to say φ (resp. ψ) a endomorphism L ( V ) (resp. L ( W )), then, with the preceding notations, one obtains: $\ varphi \ otimes \ psi (U, v) = \ varphi (U) \ otimes \ psi (v)$ If (α_ik) (resp (β_jl)) indicate the matrix φ (resp. ψ), then the matrix M of φ $\ otimes$ ψ in the base a_i $\ otimes$ b_j is the following one: $M= (\ alpha_ {ik}. \ beta_ {jl}) \;$ Let us determine the trace φ $\ otimes$ ψ. Let us calculate the image of a_i $\ otimes$ b_j by the endomorphism: $\ varphi \ otimes \ psi (a_i \ otimes b_j) = \ sum_ {kl} \ alpha_ {ki} \ beta_ {lj} .a_k \ otimes b_l$ The diagonal term corresponding to the vector a_i $\ otimes$ b_j of the base is thus α_ii.β_jj. The sum of all the diagonal terms of the matrix M is thus: $Tr (\ varphi \ otimes \ psi) = \ sum_ {ij} \ alpha_ {II}. \ beta_ {jj} =Tr (\ varphi). Tr (\ psi)$ Here, Tr indicates the application which with a endomorphism associates its trace. The trace of a tensorial product of two endormorphisms is thus equal to the product of the traces of the endomorphisms.

Case where W is equal to V
If W is equal to V , V $\ otimes$ V is equal to the space of dimension N ² of the bilinear form of V . This space has two additional S the space of the symmetrical bilinear forms of dimension N . ( N + 1) /2 and has the vector space of the alternate bilinear forms of dimension N . ( N - 1) /2.
The projector on S parallel to has associates with has _i $\ otimes$ has _j the symmetrical bilinear form: 1/2 ( has _i $\ otimes$ has _j + has _j $\ otimes$ has _i).

Direct product of two groups

See also: Produces direct (groups)
One of the objectives of the representation of the finished groups is classification. The classical theory uses the techniques of the extension to generate classes of groups. The direct product is one of the two methods of extension, it corresponds to simplest.

* the unit G , Produces Cartesian of the two groups G ₁ and G ₂ and provided with the following law * , is called produced direct of the groups G ₁ and G ₂:
$\ forall (x_1, x_2), (y_1, y_2) \ in G=G_1 \ times G_2 \ quad (x_1, x_2) * (y_1, y_2) = (x_1*_1y_1, x_2*_2y_2) \;$ The direct product allows a simple classification of the abelian groups finished:

* If the order of an abelian group finished G is different from a , there exists one and a single decomposition of G in product of cycles of order a power different of zero of a prime number.

Representations of a finished group

See also: Representations of a group finished
Let us point out the definition of a representation and fix the notations for the remainder of the article. G indicates here a Groupe finished of order G . Its neutral element is noted 1, and if S and T is two elements of G the internal law of composition of the group on S and T is noted St . V indicates a vector Space on a body noted K of characteristic first with G or null.

* a representation of the group G is the data of a vector space V of Dimension finished noted N and of a morphism of group ρ of G towards the linear group GL ( V ). A representation is noted ( V , ρ) or sometimes and wrongly V .

I.e. the ρ application is with value in the space of the bijective linear applications and preserves the law of the group, which is equivalent to:
$\ rho_1=Id \ quad and \ quad \ forall S, T \ in G \ quad \ rho (S) \ circ \ rho (T) = \ rho (St)$

Definitions

Tensorial product of two representations of the same group
The tensorial product is compatible with the representations, which means that one can define the tensorial product of two representations. Either ( V ₁, ρ¹) and ( V ₂, ρ²) two representations of a group of respective degree N ₁ and N ₂, or ( E ¹_i) a base of V ₁, ( E ²_j) a base of V ₂, one considers the base ( E ¹_i $\ otimes$ E ²_j) of V ₁ $\ otimes$ V ₂.
The tensorial product makes it possible to associate with the two representations a representation ρ on the tensorial product of two spaces and defined by:
$\ forall S \ in G \ quad \ rho_s= \ rho^1_s \ otimes \ rho^2_s \ quad or \; still \ quad \ forall v_1 \ in V_1 \ quad \ forall v_2 \ in V_2 \ quad \ rho_s (v_1 \ otimes v_2) = \ rho^1_s (v_1) \ otimes \ rho^2_s (v_2)$
Let us show that ( V ₁ $\ otimes$ V ₂, ρ) is a representation. It is enough for that to show that:
$(I) \; \ rho_ {1} =Id_ {V_1 \ otimes V_2} \ quad and \ quad (II) \; \ forall S, T \ in G \ quad \ rho_s \ circ \ rho_t= \ rho_ {St}$
It is enough to show the equalities (I) and (II) on an unspecified element of the base ( E ¹_i $\ otimes$ E ²_j)
$(I) \ quad \ rho_ {1} (e_i^1 \ otimes e_j^2) = \ rho^1_1 (e_i^1) \ otimes \ rho^2_1 (e_j^2) =e_i^1 \ otimes e_j^2$ $(II) \ quad \ rho_ {S} \ circ \ rho_ {T} (e_i^1 \ otimes e_j^2) = \ rho_ {S} \ Big (\ rho^1_t (e_i^1) \ otimes \ rho^2_t (e_j^2) \ Big) =
\ rho^1_s \ circ \ rho^1_t (e_i^1) \ otimes \ rho^2_s \ circ \ rho^2_t (e_j^2) = \ rho^1_ {St} (e_i^1) \ otimes \ rho^2_ {St} (e_j^2) = \ rho_ {St} (e_i^1 \ otimes e_j^2)$

Symmetrical and alternate square
In the general case the tensorial product of two irreducible representations is not irreducible.
There exists an important particular case, that where V ₁ is equal to V ₂, notes then V these two spaces and N their dimension, and supposes that the two associated representations is identical. The tensorial product has two additional subspaces Sym ( V ) and Alt ( V ) corresponding to the symmetrical and alternate bilinear forms . These two subspaces are stable for the representation tensorial. Let us show for the symmetrical bilinear forms: that is to say S a symmetrical bilinear form. Then, the expression of S in the base ( E _i $\ otimes$ E _j) is following form:
$\ exists s_ {ij} \ in K \; with \ quad I, J \ in \; and \; J \ Ge I \ quad S = \ sum_ {I \ the J} s_ {ij} (e_i \ otimes e_j + e_j \ otimes e_i)$ If U is element of G then: $(\ rho \ otimes \ rho) _u (S) \ sum_ {I \ the J} s_ {ij} \ left (\ rho_u (e_i) \ otimes \ rho_u (e_j) + \ rho_u (e_j) \ otimes \ rho_u (e_i) \ right)$ The last expression is well that of a symmetrical form bilinear, which shows the stability of Sym ( V ). A similar reasoning shows that of Alt ( V ).

* the subspace Sym ( V ) (resp. Alt ( V )) is a stable subspace of the tensorial representation ( V $\ otimes$ V , ρ $\ otimes$ ρ) called square symmetrical (resp. square alternate ).

Tensorial product of two groups
That is to say ( V ₁, ρ¹) a representation of a group G ₁ and ( V ₂, ρ²) a representation of a group G ₂. Let us consider the application ρ¹ $\ otimes$ ρ², it is an application of G ₁ X G ₂ in L ( V ₁ $\ otimes$ V ₂). This application is a morphism. Indeed, by using the preceding notations, one obtains: $\ forall S, T \ in G_1 \; \ forall U, v \ in G_2 \ quad \ rho_ {(S, U)}\ circ \ rho_ {(T, v)}(e_i^1 \ otimes e_j^2) =
\ rho_ {(S, U)}\ Big (\ rho_t^1 (e_i^1) \ otimes \ rho_v^2 (e_j^2) \ Big) = \ rho_ {St} ^1 (e_i^1) \ otimes \ rho_ {UV} ^2 (e_j^2) = \ rho_ {(St, UV)}(e_i^1 \ otimes e_j^2)$ The tensorial product of two representations of two groups thus corresponds to a representation of the group produced in the tensorial product of two spaces. Notice, the notation is ambiguous. Indeed, if G ₁ and G ₂ represents the same group G , ( V ₁ $\ otimes$ V ₂, ρ) indicates at the same time a representation of G and G X G . The first representing the restriction of the second on the diagonal, the context requires a precision.

Properties

tensorial product and direct product
Either G a group containing two sub-groups G ₁ and G ₂ such as G or isomorph with the direct product of G ₁ and G ₂. There exist two manners of obtaining representations of G from those of G ₁ and those G ₂, either starting from a direct sum of a representation of G ₁ and of a representation of G ₂ or using a tensorial product.
The direct sum is hardly effective, indeed, only the irreducible representations are truly interesting, they indeed form the base of all the others (cf Théorème of Maschke). However a direct sum never provides an irreducible representation, indeed, V ₁ X {0} and {0} X V ₂ are always subspaces invariants.
The tensorial product provides on the other hand the irreducible representations of G . Any irreducible representation is the tensorial product of irreducible representations of G ₁ and of G ₂ and the tensorial product of two irreducible representations of G ₁ and G ₂ is an irreducible representation of G .

Character

Case of the irreducible representations
One supposes here that the irreducible characters form an orthonormal base, for example because the body K is that of the complex numbers.

* the tensorial product ( V , ρ) of two irreducible representations ( V ₁, ρ¹) and ( V ₂, ρ²) of two groups G ₁ and G ₂ is irreducible.
Character of ρ (resp. ρ¹) (resp. ρ²) is noted χ (resp. χ₁) (resp. χ₂). To show that ρ is irreducible, it is enough to check that <χ|χ> is equal to a . The trace of a tensorial product of two endormorphisms is equal to the product of the traces of the two endomorphisms. In the case of the representations, the formula is obtained: $\ forall (s_1, s_2) \ in G_1 \ times G_2 \ quad \ chi (s_1, s_2) = \ chi_1 (s_1). \ chi_2 (s_2)$ Moreover, <χ₁|χ₁> and <χ₂|χ₂> is equal to a because the two associated representations irreducible, one are deduced from it if G _i indicates the order of G _i: $< \ chi|\chi>=<\chi_1.\ chi_2|\ chi_1. \ chi_2>= \ frac {1} {g_1g_2} \ sum_ {s_1 \ in G_1} \ sum_ {s_2 \ in G_2} \ chi_1 (s_1). \ chi_1 (s_1) ^*. \ chi_2 (s_2). \ chi_2 (s_2) ^*=< \ chi_1|\chi_1>.<\chi_2|\ chi_2>=1 \;$ What finishes the demonstration.
Case of a representation of a produced group
It is supposed here that G contains two sub-groups G ₁ and G ₂ such as G is isomorphous with the direct product G ₁ X G ₂. There exists a Bijection between the Cartesian Produit irreducible representations of G ₁ and G ₂ and those of G given by the tensorial product of the representations. The preceding paragraph shows that the tensorial product of the irreducible representations is well with value in the irreducible representations of G .
Let us show that the application is surjective. That is to say ( V , ρ) an irreducible representation of G .

* There exists an irreducible representation ( V ₁, ρ¹) of G ₁ and one ( V ₂, ρ²) of G ₂ such as ( V , ρ) is isomorphous with the tensorial product of the two preceding representations.
For that, it is enough to show that the whole of the tensorial products of irreducible representations of G ₁ and G ₂ form an orthonormal base of the central functions of G . Either F is a central function of G orthogonal to very produced tensorial of irreducible natures. To show that F is the null function shows the proposal. One has, with the preceding notations: $\ forall (S, T) \ in G_1 \ times G_2 \ quad \ sum_ {S \ in G_1} \ sum_ {T \ in G_2} F (S, T). \ chi_1 (S) ^*. \ chi_2 (T) ^*=0$ Let us note F _s the function of G ₂ in K which with T associates F ( S , T ) χ₁ ( S ). F _s is an orthogonal central function with any irreducible character of G ₂, F _s is thus the null function. Consequently: $\ forall (S, T) \ in G_1 \ times G_2 \ quad F (S, T). \ chi_1 (S) ^*=0$ Either T an unspecified element of G ₂, the function of G ₁ in K , which with S associates F ( S , T ) is orthogonal with any irreducible character of G ₁, it is thus also the null function. The proposal is then shown.
To show that the application of the irreducible couples of representations of G ₁ X G ₂ in that of G given by the tensorial product is injective, two remarks are enough. The numbers of representations irreducible of a finished group is equal to the number of classes of conjugation of the group. As the number of classes of conjugation of G is equal to the product of the numbers of classes of conjugation of G ₁ and G ₂, the number of irreducible natures of G is equal to the product of the number of irreducible natures of G ₁ and G ₂. The equality of the cardinals of the starting whole and of arrived thus that the surjectivity shows the injectivity.

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