MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 11 2.1.1. direct sum. The trace of a direct sum of matrices is the sum of traces: A 0 Tr(A B) = Tr = Tr(A) + Tr(B) ⊕ 0 B ! " Theorem 2.2. If V, W are two G-modules then χV W = χV + χW ⊕ Proof. If ρV ,ρW ,ρV W are the corresponding representations then ⊕ ρV W (σ)=ρV (σ) ρW (σ) ⊕ ⊕ The theorem follows. ! 2.1.2. character formula using dual basis. Instead of using traces of matrices, I prefer the following equivalent formula for characters using bases and dual bases. If V is a G-module, we choose a basis v , ,v for V as a vector { 1 ··· d} space over C. Then recall that the dual basis for V ∗ = HomC(V,C) consists of the dual vectors v1∗, ,v∗ : V C given by ··· d → d vj∗ aivi = aj # i=1 % $ I.e., vj∗ picks out the coefficient of vj. Proposition 2.3. d χV (σ)= vi∗(σvi) i=1 $ Proof. The matrix of the linear transformation ρ(σ) has (i, j) entry vi∗(σvj) Therefore, its trace is vi∗(σvi). ! For example, the trace of the& identity map is d Tr(idV )= vi∗(vi)=d i=1 $ Theorem 2.4. The value of the character at 1 is the dimension of the representation: χV (1) = d = dimC(V ) 12 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 2.1.3. tensor product. If V, W are two G-modules then the tensor prod- uct V W is defined to be the tensor product over C with the following action⊗ of G: σ(v w)=σv σw ⊗ ⊗ Theorem 2.5. The character of V W is the product of the characters of V and W . I.e., ⊗ χV W (σ)=χV (σ)χW (σ) ⊗ for all σ G. ∈ Proof. Choose bases vi , wj for V, W with dual bases vi∗ , wj∗ . Then the tensor product{ }V{ }W has basis elements v w{ with} { dual} ⊗ i ⊗ j basis elements v∗ w∗. So, the character is: i ⊗ j χV W (σ)= (vi∗ wj∗)σ(vi wj)= (vi∗ wj∗)(σvi σwj) ⊗ ⊗ ⊗ ⊗ ⊗ i,j i,j $ $ = vi∗(σvi)wj∗(σwj)= vi∗(σvi) wj∗(σwj)=χV (σ)χW (σ) i,j i j $ $ $ ! 2.1.4. dual reprentation. The dual space V ∗ is a right G-module. In order to make it a left G-module we have to invert the elements of the group. I.e., for all f V ∗ we define ∈ 1 (σf)(v) := f(σ− v) Lemma 2.6. 1 χV ∗ (σ)=χV (σ− ) Lemma 2.7. 1 χV (σ− )=χV (σ) Proof. The trace of a matrix A is equal to the sum of its eigenvalues λi. m If A has finite order: A = Id then its eigenvalues are roots of unity. Therefore, their inverses are equal to their complex conjugates. So, 1 1 Tr(A− )= λi− = λi = Tr(A) Since G is finite, the lemma$ follows. $ ! Theorem 2.8. The character of the dual representation V ∗ is the com- plex conjugate of the character of V : χV ∗ (σ)=χV (σ) MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 13 2.2. Irreducible characters. Definition 2.9. A representation ρ : G AutC(V ) is called irre- ducible if V is a simple G-module. The character→ χρ = χV : G C → of an irreducible representation is called an irreducible character. Theorem 2.10. Every character is a nonnegative integer linear com- bination of irreducible characters. Proof. Since C[G] is semisimple, any G-module V is a direct sum of simple modules V ∼= Sα. So, the character of V is a sum of the corresponding irreducible characters: χV = χSα . ! ' If we collect together multiple copies of the& same simple module we get V = niSi and r ' χV = niχi i=1 $ where χi is the character of Si. This makes sense only if we know that there are only finitely many nonisomorphic simple modules Si and that the corresponding characters χi are distinct functions G C. In fact we will show the following. → Theorem 2.11. (1) There are exactly b (the number of blocks) ir- reducible representations Si up to isomorphism. (2) The corresponding characters χi are linearly independent. This will immediately imply the following. Corollary 2.12. The irreducible characters χ , ,χ form a basis for 1 ··· b the b-dimensional vector space of all class functions G C. → 2.2.1. regular representation. This is a particularly elementary repre- sentation and character which in contains all the simple modules. Definition 2.13. The free module C[G]C[G] is called the regular rep- resentation of G. The corresponding character is called the regular character: χreg = χ [G] : G C. C → n = G if σ =1 Theorem 2.14. χ (σ)= | | reg 0 if σ =1 ( & Proof. I used the basis-dual basis formula for characters. The regular representation V = C[G] has basis elements σ G and dual basis ∈ elements σ∗ given by σ∗ aτ τ = aσ )$ *.