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MIT 18 769 - Frobenius- Perron dimensions

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86 1.45. Tensor categories with finitely many simple objects. Frobenius-Perron dimensions. Let A be a Z+-ring with Z+-basis I. Definition 1.45.1. We will say that A is transitive if for any X, Z ∈ I there exist Y1, Y2 ∈ I such that XY1 and Y2X involve Z with a nonzero coefficient. Proposition 1.45.2. If C is a ring category with right duals then Gr(C) is a transitive unital Z+-ring. Proof. Recall from Theorem 1.15.8 that the unit object 1 in C is simple. So Gr(C) is unital. This implies that for any simple objects X, Z of C, the object X ⊗ X∗ ⊗ Z contains Z as a composition factor (as X ⊗ X∗ contains 1 as a composition factor), so one can find a simple object Y1 occurring in X∗ ⊗ Z such that Z occurs in X ⊗ Y1. Similarly, the object Z ⊗ X∗ ⊗ X contains Z as a composition factor, so one can find a simple object Y2 occurring in Z ⊗ X∗ such that Z occurs in Y2 ⊗ X. Thus Gr(C) is transitive. � Let A be a transitive unital Z+-ring of finite rank. Define the group homomorphism FPdim : A → C as follows. For X ∈ I, let FPdim(X) be the maximal nonnegative eigenvalue of the matrix of left multiplication by X. It exists by the Frobenius-Perron theorem, since this matrix has nonnegative entries. Let us extend FPdim from the basis I to A by additivity. Definition 1.45.3. The function FPdim is called the Frobenius-Perron dimension. In particular, if C is a ring category with right duals and finitely many simple objects, then we can talk about Frobenius-Perron dimensions of objects of C. Proposition 1.45.4. Let X ∈ I. (1) The number α = FPdim(X) is an algebraic integer, and for any algebraic conjugate α� of α we have α ≥ |α�|. (2) FPdim(X) ≥ 1. Proof. (1) Note that α is an eigenvalue of the integer matrix NX of left multiplication by X, hence α is an algebraic integer. The number α� is a root of the characteristic polynomial of NX , so it is also an eigenvalue of NX . Thus by the Frobenius-Perron theorem α ≥ |α�|. (2) Let r be the number of algebraic conjugates of α. Then αr ≥N(α) where N(α) is the norm of α. This implies the statement since N(α) ≥ 1. �� � � 87 Proposition 1.45.5. (1) The function FPdim : A C is a ring →homomorphism. (2) There exists a unique, up to scaling, element R ∈ AC := A⊗ZC such that XR = FPdim(X)R, for all X ∈ A. After an appro-priate normalization this element has positive coefficients, and satisfies FPdim(R) > 0 and RY = FPdim(Y )R, Y ∈ A. (3) FPdim is a unique nonzero character of A which takes non-negative values on I. (4) If X ∈ A has nonnegative coefficients with respect to the basis of A, then FPdim(X) is the largest nonnegative eigenvalue λ(NX ) of the matrix NX of multiplication by X. Proof. Consider the matrix M of right multiplication by X∈I X in A in the basis I. By transitivity, this matrix has strictly positive entries, so by the Frobenius-Perron theorem, part (2), it has a unique, up to scaling, eigenvector R ∈ AC with eigenvalue λ(M) (the maximal posi-tive eigenvalue of M). Furthermore, this eigenvector can be normalized to have strictly positive entries. Since R is unique, it satisfies the equation XR = d(X)R for some function d : A C. Indeed, XR is also an eigenvector of M with→eigenvalue λ(M), so it must be proportional to R. Furthermore, it is clear that d is a character of A. Since R has positive entries, d(X) = FPdim(X) for X ∈ I. This implies (1). We also see that FPdim(X) > 0 for X ∈ I (as R has strictly positive coefficients), and hence FPdim(R) > 0. Now, by transitivity, R is the unique, up to scaling, solution of the system of linear equations XR = FPdim(X)R (as the matrix N of left multiplication by X∈I X also has positive entries). Hence, RY = d�(Y )R for some character d�. Applying FPdim to both sides and using that FPdim(R) > 0, we find d� = FPdim, proving (2). If χ is another character of A taking positive values on I, then the vector with entries χ(Y ), Y ∈ I is an eigenvector of the matrix N of the left multiplication by the element X∈I X. Because of transitivity of A the matrix N has positive entries. By the Frobenius-Perron theorem there exists a positive number λ such that χ(Y ) = λ FPdim(Y ). Since χ is a character, λ = 1, which completes the proof. Finally, part (4) follows from part (2) and the Frobenius-Perron the-orem (part (3)). � Example 1.45.6. Let C be the category of finite dimensional repre-sentations of a quasi-Hopf algebra H, and A be its Grothendieck ring. Then by Proposition 1.10.9, for any X, Y ∈ C dim Hom(X ⊗ H, Y ) = dim Hom(H, ∗X ⊗ Y ) = dim(X) dim(Y ),� � � 88 where H is the regular representation of H. Thus X ⊗ H = dim(X)H, so FPdim(X) = dim(X) for all X, and R = H up to scaling. This example motivates the following definition. Definition 1.45.7. The element R will be called a regular element of A. Proposition 1.45.8. Let A be as above and ∗ : I I be a bijection →which extends to an anti-automorphism of A. Then FPdim is invariant under ∗. Proof. Let X ∈ I. Then the matrix of right multiplication by X∗ is the transpose of the matrix of left multiplication by X modified by the permutation ∗. Thus the required statement follows from Proposi-tion 1.45.5(2). � Corollary 1.45.9. Let C be a ring category with right duals and finitely many simple objects, and let X be an object in C. If FPdim(X) = 1 then X is invertible. Proof. By Exercise 1.15.10(d) it is sufficient to show that X ⊗ X∗ = 1. This follows from the facts that 1 is contained in X ⊗ X∗ and FPdim(X ⊗ X∗) = FPdim(X) FPdim(X∗) = 1. � Proposition 1.45.10. Let f : A1 → A2 be a unital homomorphism of transitive unital Z+-rings of finite rank, whose matrix in their Z+-bases has non-negative entries. Then (1) f preserves Frobenius-Perron dimensions. (2) Let I1, I2 be the Z+-bases of A1, A2, and suppose that for any Y in I2 there exists X ∈ I1 such that the coefficient of Y in f(X) is non-zero. If R is a regular element of A1 then f(R) is a regular element of A2. Proof. (1) The function X �→ FPdim(f(X)) is a nonzero character of A1 with nonnegative values on the basis. By Proposition 1.45.5(3), FPdim(f(X)) = FPdim(X) for all X in I. (2) By part (1) we have (1.45.1) f( X)f(R1) = FPdim(f(


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