Averages and Ratios of Characteristic Polynomials
References 1. Keating and Snaith, RMT and 1 2 + i t, Comm. Math. Phys. 214 (2000), no. 1, 5789. Value distribution of mirrors value distribution of characteristic polynomials. Moments computed using Selberg integral. 2. Conrey and Snaith, Applications of the L- functions ratios conjectures. Instead of moments, consider ratios. This will have many applications. http://arxiv.org/abs/math.nt/0509480 3. Conrey, Farmer and Zirnbauer, Howe pairs, supersymmetry, and ratios of random characteristic polynomials for the unitary groups U(N), Computes mean values of ratios for classical ensembles using supersymmetry. http://arxiv.org/abs/math-ph/0511024 4. Conrey, Forrester and Snaith, Averages of ratios of characteristic polynomials for the compact classical groups, Int. Math. Res. Not. 2005, no. 7, 397431. 5. Bump and Gamburd, Averages and ratios of characteristic polynomials, Comm. Math. Phys. (to appear) is the basis of this talk. Papers [4] and [5] reprove results of [3] by dierent methods. Only [3] has results for N small. 2
Representation Theory of S k Let = ( 1 ; ; N ) be a partition of k. Thus i 2 Z, 1 > > N > 0 and P i = k. Let S = S 1 S N S k : If is a partition represent it by a oung diagram. Transposing the diagram of gives the diagram of the conjugate partition 0. Example. N = 4, = (2; 2; 1; 0) = (2; 2; 1) = (2 2 1). = 0 = Thus is a partition of 5 of length 3 or into 3 parts. Theorem. S Ind k S S (1) and Ind k S 0(sgn) irreducible constitutent in common. Proof. Mackey theory + combinatorics. have a unique Call this s. Thus the irreducible representations of S k are parametrized by partitions of k. 3
The ring We dene a graded Z-algebra. The homogeneous part k of degree k is the additive group of generalized characters of S k. The multiplication is induction. Thus S k S l S k+l so if ; are characters of S k and S l then is a character of S k S l which we induce to S k+l. A Z-basis of k is fs j is a partition of kg. Let h k and e k be the trivial and sign characters of S k. Then = Z[h 1 ; h 2 ; ] = Z[e 1 ; e 2 ; ]. We have h k = s (k) (k) is the partition (k; 0; ; 0) e k = s (1 k ) (1 k ) is the partition (1; 1; ; 1) has an automorphism of degree 2 such that (h i ) = e i and (e i ) = h i. This is the involution. On k the involution amounts to tensoring with the sign character. The involution corresponds to conjugation of partitions: (s ) = (s 0) 4
The characteristic map Let s ( 1 ; ; N ) = Alt( 1 +N?1 1 2 +N?2 N 2 N ) N Alt(?1 N?2 1 2 N 0 ) be the Schur polynomial, where Alt = 2S N (? 1) sgn() : There is a ring homomorphism (due to Frobenius) ch: Z[ 1 ; ; N ] S N, ch(h i ) = h i ( 1 ; ; N ) = ch(e i ) = e i ( 1 ; ; N ) = ch(s ) = s ( 1 ; ; N ): i 1 6 6i i 1 < <i N i1 in ; N i1 in ; (Complete, elementary, Schur symmetric polynomials). The homomorphism is surjective but not injective. Thus does not induce an involution of symmetric polynomials. However it is bijective on k provided N > k. 5
Frobenius-Schur duality Let V = C N be the standard module of U (N ). We have commuting actions of U (N ) and S k on k V = V V (k times): : v 1 v k v?1 (1) v?1 (k) ; 2 S k : Thus we can decompose k V simultaneously: k V = s as U (N ) S k -modules: The irreducible representations of S k that appear are the s where = ( 1 ; ; N ) is a partition of k of length 6 N. The irreducible representations of U (N ) that appear are those whose matrix coecients are homogeneous polynomials of degree k. If g 2 U (N ) has eigenvalues 1 ; ; N and f 2 maps to f = ch(f ) 2 Z[ 1 ; ; N ] S N then (g) = f ( 1 ; ; N ); = character of : Multiplication in corresponds to tensor production of representations of U (N ). Thus The c s s = s ; =M c c : are Littlewood-Richardson coecients. 6
Weights We consider integer sequences ( 1 ; ; N ). = ( 1 ; ; N ) 2 Z N is called a weight of N. Weights correspond to rational characters of the diagonal torus of U (N ): : 0 @ t 1 t N 1 A t 1 1 tn N : If 1 > > N the weight is called dominant. The height of is the inner product h; i, = (N? 1; N? 2; ; 0): If is a dominant weight, there is a unique irreducible of U (N ) containing, no higher weight. Highest weight module (Weyl). If N > 0 the dominant weight = ( 1 ; ; N ) is a partition of k = P i into 6 N parts. So dominant weights parametrize irreducible representations of U (N )...... and partitions of k parametrize irreducible representations of S k. Weights and partitions overlap, hence give a bijection between some representations of U (N ) and some representations of S k. This is the bijection we constructed using k C N. 7
Generating functions Remember that h i ( 1 ; ; N ) = e i ( 1 ; ; N ) = i 1 6 6i i 1 < <i N i1 in ; N i1 in ; are the complete and elementary symmetric poly's. 1 t=0 1 t=0 t k h k ( 1 ; ; N ) = t k e k ( 1 ; ; N ) = N j =1 N j =1 (1? t j )?1 : (1 + t j ): These may be regarded as generating functions for the symmetric and exterior algebras on C N. The involution of interchanges 1 t=0 t k h k 1 t=0 t k e k There is no corresponding correspondence of symmetric functions (since e k = 0 for k > N) but roughly we can think of the involution as transforming the two types of generating functions into one another. 8
Correspondences If G and H are groups, a correspondence of representations is a bijection between some of the irreducible representations of G and some of the irreducible representations of H. We consider the module! = M of G H: By assumption there are no repetitions among the or. In important cases,! has a natural construction. If G = U (N ) and H = S k this module is k V. Howe discovered that the Weil representation gives correspondences for reductive dual pairs of subgroups of Sp(2n) or its double cover (the metaplectic group. GL(N;C) GL(m;C) duality Let GL(N ; C) GL(m; C) act on Mat nm (C) by left and right multiplication. There is induced an action on the polynomial ring S(Mat nm (C)). This is a correspondence. It induces a correspondence of the maximal compact subgroups U (N ) U (M ) since irreducible rep's of U (N ) correspond bijectively to analytic reps of GL(N ; C). This is a Howe correspondence. 9
Inner product formulas Theorem. (Peter-Weyl + Schur orthogonality) Assume G compact. The characters of the irreducible representations span the subspace of L 2 (G) consisting of class functions. They are an orthonormal basis. Thus given a correspondence: G H 2 L 2 (G) 2 L 2 (H) ( 2 indexing set) Let L! 2 (G) and L! 2 (H) be the span of the characters and. The correspondence determines an isometry L! 2 (G) L! 2 (H): They may allow us to transfer an inner product computation from G to H. Thus if 2 L! 2 (G) is a class function, and 2 L! 2 (H) is the corresponding class function on H we have Z G j(g)j 2 dg = Z H j (h)j 2 dh: We will give some examples where the right-hand side is easier to evaluate than the left-hand side, leading to nontrivial results in RMT. 10
Diaconis and Shashahani Let G = U (N ), H = S k,! = Frobenius-Schur duality Theorem 1. If N > k 1 + 2k 2 + + rk r then Z U (N) jtr(g)j 2k 1 jtr(g 2 )j 2k2 jtr(g r )j 2k r d g = r j =1 j k j k j! : Proof. Let k = k 1 + 2k 2 + + rk r, and let be the partition of k containing k 1 entries equal to 1, k 2 entries equal to 2, and so forth. Let C be the conjugacy class of permutations 2 S k of type (so has k j cycles of length j in its decomposition to disjoint cycles). Let p be the conjugacy class indicator on S k, p (g) = z 0 otherwise ; r where z = j =1 j k j k j!. The class function g tr(g) k 1 tr(g 2 ) k2 tr(g r ) k r in L! 2 (G) corresponds to the function p 2 L! 2 (H) and so its norm is the same as the norm of p, i.e. z. Thus Z U(N ) jtr(g)j 2k 1 jtr(g 2 )j 2k2 jtr(g r )j 2k r d g stabilizes when N is large. Asymptotically the value distribution of tr(g); tr(g 2 ); ; tr(g r ) is of independent normal (i.e. Gaussian) random variables. 11
Cauchy Identity s ( 1 ; ; n )s ( 1 ; ; m ) = i;j (1? i j )?1 Sum is over partitions of length 6 N. Fundamental in what we do next. Underlying correspondence: G; H = U (n); U (m) Take n = m. Consider the action of U (n) U (n) on L 2 (G): (g; h)f (x) = f (g?1 xh) Lemma. If G is any compact group, G G acts on L 2 (G) and L 2 (G) = M 2Irr(G) ^ ^ = contragredient rep'n: (This is again Peter-Weyl theorem). For G = U (n), there is an involution : G G namely g = t g?1 such that g ( g) is equivalent to ^. This is because g?1 is conjugate to g. So we modify the action: and in this action: (g; h)f (x) = f ( t g xh) L 2 (G) = M 2Irr(G) 12
Polynomial version In this decomposition L 2 (G) = M 2Irr(G) (Hilbert space.) we may restrict ourselves to the subspace of U (n)-nite vectors, which form the ane ring A = C[ ij ; det?1 ; ij = coordinate functions. (G is compact so continuous functions are L 2.) Then A = M 2Irr(G) (Algebraic.) runs through where = ( 1 ; ; n ) is a weight, i.e. 1 > > n, i 2 Z The weights with N > 0 are partitions. Ane ring C[ ij ] of Mat n is C[ ij ] = M partitions (Algebraic.) This is because only the matrix coecients of where is a partition are regular on the determinant locus. 13
Proof of the Cauchy Identity We have C[ ij ] = M partitions (Algebraic ) (1) as U (n) U (n) modules, where (g; h) act by (g; h)f (x) = f ( t gxh); f : Mat n (C) C: Let g; h have eigenvalues 1 ; ; n and 1 ; ; n. Taking the trace in this identity gives s ( 1 ; ; n )s ( 1 ; ; n ) = i;j (1? i j )?1 : The series is only convergent if j i j; j j j < 1. But (1) extends to GL(n; C) GL(n; C) of which U (n) U (n) is a maximal compact, and extend to analytic reps of GL(n; C). The case n > m can be deduced by specializing b m+1 ; ; n 0: At the heart of the proof is a correspondence with G; H = U (n) and! = M partitions = polynomials on Mat n (C): This a case of the Howe correspondence. 14
Dual Cauchy Identity If is a partition represent it by a oung diagram. Transposing the diagram of gives the diagram of the conjugate partition 0. Example. N = 4, = (2; 2; 1; 0) = (2; 2; 1) = (2 2 1). = 0 = Thus is a partition of 5 of length 3 or into 3 parts. 0 is a partition of 5 of length 2 or into 2 parts. The ring has a basis s which specialize to characters of irreducible rep's of S k (k = P i ) or U (N ) N > length(). The map s s 0 is an automorphism of which we will call the involution. Applying the involution to one set of variables in the Cauchy identity s ( 1 ; ; n )s ( 1 ; ; n ) = produces the dual Cauchy identity s ( 1 ; ; n )s 0( 1 ; ; n ) = i;j i;j (1? i j )?1 : (1 + i j ): 15
Keating and Snaith The following theorem was very inuential in the application of RMT to. Theorem. We have Z U(N) jdet(g? I)j 2k dg = N?1 j!(j + 2k)! j=0 (j + k)! 2 : The same constants appear in the (conjectural) 2k-th moment of. The original proof of Keating and Snaith used the Selberg integral. We will give another proof (due to Gamburd) that uses GL(N ) GL(2k) duality. The two proofs have dierent generalizations. The proof of Keating and Snaith allows interpolation of k to real numbers, while Gamburd's proof allows more general evaluations such as Z U(N) jdet(g? I)j 2k (g)dg where is the character of. 16
Proof of Keating-Snaith formula If 1 ; ; N and 1 ; ; N are complex numbers, we will show Z k U (N) i=1 n o det(i + i g)det(i +?1 i g?1 ) dg = s (N k ) ( 1 ; ; k ; 1 ; ; k ): (2) The left-hand side equals i?1z U(N ) k i=1 fdet(i + i g)det(g i + I)gdet(g) k dg: By dual Cauchy id, if t 1 ; ; t N are eigenvalues of g, k i=1 fdet( )det( )g = s ( 1 ; ; k ; 1 ; ; k )s 0(t 1 ; ; t N ); Since det(g) k = s 0(t 1 ; ; t N ), integrating over g picks o just one term, with 0 = (k N ) and so = (N k ) so. This proves (2). Taking i = i = 1 Z U (N) jdet(g? I)j 2k dg = s (N k ) 2k terms (1; ; 1); the dimension of the rep'n (N k ) of U (2k). This dimension is computed using Weyl's dimension formula, proving the theorem of Keating and Snaith. 17
Analysis of the proof Underlying this computation is the (dual) Cauchy identity. The Cauchy identity for GL(N ) GL(2k) amounts to the use of the Howe correspondence for GL(N ) GL(2k). In this correspondence, if is a dominant weight, GL(N) corresponds to GL(2k). But in the Cauchy identity GL(N) corresponds to GL(2k) 0. When = (N k ), 0 = (k N ) and the answer turns out to be the dimension of this GL(2k) 0. We used the correspondence to transfer the computation from GL(N ) to GL(2k). Similarly, moments for classical groups can be expressed in terms of characters of other groups parametrized by rectangular partitions. Thus a result of Keating and Snaith can be written: "2f1g Z k Sp(2N) j =1 (x 1 x k ) N N k k j=1 x j N(1?" j ) i6j det(i? x j g) dg = Sp(2k) (x1 1 ; ; x k 1 ) = (1? x i " i x j " j)?1 : 18
Ratios Let L;K consist of permutations 2 S K +L such that (1) < < (L); (L + 1) < < (L + K): Theorem. (Conrey, Farmer and Zirnbauer) If N > ; R and j q j; j r j < 1 we have Z U(N) 2 L;K q=1 K k=1 L l=1 q=1 K det (I?1 + l g?1 ) K k=1 det (I + L+k g) det (I? q g) R r=1 det (I? r g?1 )?1 ( (L+k) L+k ) N k=1 L?1 l=1 (1 + q (l) ) R r=1 K k=1 (1 + r (L+k) ) L l=1 (1??1 (l) (K+k) ) R r=1 : q=1 (1? q r ) dg = According to CFZ the assumption that N is large can be eliminated. Proof will depend on the generalized Cauchy identity involving Littlewood-Schur functions. These were studied by Littlewood and also by Berele and Regev, who apparently rediscovered them. 19
Generalized Cauchy identity Dene the Littlewood-Schur polynomial The c ; LS (x 1 ; ; x k ; y 1 ; ; y l ) = c s (x 1 ; ; x k )s 0( y 1 ; ; y l ): are the Littlewood-Richardson coecients. Theorem. (Berele and Regev) LS ( 1 ; ; m ; 1 ; ; n )LS ( 1 ; ; s ; 1 ; ; t ) = i;k j;k (1??1 i k ) i;l (1 + j k ) j;l (1 + i l ) (1? j l )?1 : We will assume it now and discuss the proof later. Laplace expansion Let (a ij ) be (L + K) (L + K). Then det(a ij ) = 2 L;K sgn() a 1;(1) a 1;(L) a L;(1) a L;(L) a L+1;(L+1) a L+K;(L+K) a L+K;(L+1) a L+K;(L+K) : Proof easy: 20
Laplace expansion for LS Proposition. Suppose of length 6 K such that L > L+1 +, let = [ with Then = ( 1 ; ; L ); = ( L+1 ; ; L+K ): 2 L;K LS ( 1 ; ; L+K ; 1 ; ; ) = 16l6L 16k6K ( (l)? (L+k) )?1 LS + K L ((1) ; ; (L) ; 1 ; ; ) LS ( (L+1) ; ; (L+K) ; 1 ; ; ) Proof. Induction on. If = 0, this says 2 L;K 16l6L 16k6K s ( 1 ; ; L+K ) = ( (l)? (L+k) )?1 s + K L ((1) ; ; (L) )s ( (L+1) ; ; (L+K) ): This is proved by applying the Laplace expansion to the determinant denition of the Schur function. For > 0 one adds the i one at a time using Pieri's formula (i.e. the degenerate Littlewood-Richardson rule). 21
Proof (sketch) of unitary CFZ By the dual Cauchy identity, L det (I +?1 l g?1 ) det (I + L+k g) k=1 l=1 L = det (g) L l=1 K +L l?n k=1 K det (I + k g) L = det (g) L l=1 l?n s ( 1 ; ; K +L ) 0(g) On the other hand by the Cauchy identity and q=1 R r=1 det (I? q g)?1 = det (I? r g?1 )?1 = By Schur orthogonality Z U(N) L l=1 q=1 s ( 1 ; ; ) (g) s ( 1 ; ; R ) (g): det (I?1 + l g?1 ) K k=1 det (I + L+k g) det (I? q g) R r=1 det (I? r g?1 ) dg = E D 0 ; det L L l=1 ;; l?n s ( 1 ; ; L+K )s ( 1 ; ; )s ( 1 ; ; R ): 22
We rewrite this as L L l=1 l=1 L l=1 l?n ;; l?n l?n ~ c 0 s ( 1 ; where ~ = + Laplace expansion for LS ^: ; L+K )s ( 1 ; ; )s ( 1 ; ; R ) = LS ~ ( 1 ; ; ; 1 ; ; L+K )s ( 1 ; ; R ) = 2 L;K LS ^( 1 ; ; L+K ; 1 ; ; )s ( 1 ; ; R ); L N and ^ = ~ 0 = N L [ 0. Using the LS ^( 1 ; ; L+K ; 1 ; ; ) = 16l6L 16k6K ( (l)? (L+k) )?1 LS (N +K) L ((1) ; ; (L) ; 1 ; ; ) LS 0( (L+1) ; ; (L+K) ; 1 ; ; ): Substituting this, using generalized Cauchy identity to evaluate the sum over, and Littlewood's formula LS h(l+m) k i (x 1 ; ; x k ; y 1 ; ; y l ) = gives q=1 K k=1 K k i=1 2L;K k=1 L?1 l=1 (1 + q (l) ) R r=1 L l=1 x i! m 16i6k 16j6l?1 ( (L+k) L+k ) N (1??1 (l) (K +k) ) r=1 (x i + y j ) K k=1 (1 + r (L+k) ) R q=1 (1? q r ): 23
Remarks on the proof There may be more than one way to proceed once we have an adequate set of tools. The tools, mainly the generalized Cauchy identity, Laplace expansion and Littlewood's identity are themselves of considerable interest. We will concentrate on ideas around the generalized Cauchy identity and Laplace expansion. U (p + q) U (p) U (q) branching As before, = character of, a dominant weight. We assume is a partition, so is a polynomial rep'n. Theorem. (i) We have (p+q) j U(p)U (q) g1 (p+q) g 1 g 2 g 2 = = ; M ; c (g 1 ) (g 2 ); c (g 1 ) (g 2 ): (ii) Let x 1 ; ; x p and y 1 ; ; y q be two sets of variables. ; s (x 1 ; ; x p ; y 1 ; ; y q ) = c s (x 1 ; ; x p )s (y 1 ; ; y q ): The two statements are equivalent. (Take x i ; y i to be eigenvalues of g 1 and g 2.) 24
Proof of unitary branching In the context of the unitary groups the Littlewood- Richardson rules occur in 2 distinct ways: Clebsch-Gordan coef: c is the multiplicity of in reps of U (N ) or GL(N ; C). Unitary branching rule: c is the multiplicity of GL(p) GL(q) in the restriction of GL(p+q). The See-Saw: U(p + q) U(n) U(n) U(p) U(q) U(n) Vertical lines are inclusions Diagonal lines are correspondences Let! = action of U (p + q), U (n) on symmetric algebra of Mat (p+q)n (C) (left, right translation) L! = U(p+q) U(N). Alternatively we have action! of U (p) U (q) and U (n) U (n) on same symmetric algebra. L! = ; ( U(p) U(q) ) U(n) U(n) ( ) 25
Unitary branching, continued The representation! is the action of U ((p + q)n) on the symmetric algebra on Mat (p+q)n (C). Both dual pairs can be embedded U (p + q) U (n) (U (p) U (q)) (U (n) U (n)) & % U ((p + q)n) The actions are as follows. Let = 1 2 2 2 Mat qn (C): 2 Mat (p+q)n (C); 1 2 Mat pn (C); Action of U (p + q) is by left multiplication. U (n) U (n) is by right multiplication on 1 and 2 individually. U(p + q) U(n) U(n) U(p) U(q) U(n) The unitary branching rule now follows... 26
See-Saw Let! be a representation of. Let G 1, and H 2 be its centralizer. Assume!j G1 H 2 = M i2i i (1) i (2) where (1) i and (2) i are irreducible rep's of G 1 and H 2, (1) and i (2) j is the graph of a correspondence. Let H 1 G 1. The centralizer G 2 of H 1 contains H 2. G 1 G 2 H 1 H 2 Assume!j H1 G 2 is also a correspondence.!j H1 G 2 = M j2j j (1) j (2) : Lemma 2. Assume the branching rules i (1) = c ij j (1) ; j (2) = j2j i2i d ji i (2) (3) Then the c ij = d ij. Proof. Both c ij and d ij = multiplicity of (1) j (2) i in! as H 1 H 2 modules. 27
Proof of generalized Cauchy Recall that the involution roughly interchanges the two generating functions: 1 t=0 1 t=0 t k h k ( 1 ; ; N ) = t k e k ( 1 ; ; N ) = N j =1 N j =1 (1? t j )?1 : (1 + t j ): Start with Cauchy identity, apply unitary branching: i;k (1? i k )?1 i;l (1? i l )?1 j;k (1? j k )?1 j;l (1? j l )?1 = s ( 1 ; ; m ; 1 ; ; n )s ( 1 ; ; s ; 1 ; ; t ) = ; ; Now apply in variables and : i;k (1? i k )?1 i;l c s ( 1 ; ; m )s ( 1 ; ; n ) c s ( 1 ; ; s )s ( 1 ; ; t ): (1 + i l ) j;k (1 + j k ) j;l LS ( 1 ; ; m ; 1 ; ; n )LS ( 1 ; ; s ; 1 ; ; (1? j l )?1 = t ): Thus we obtain the Generalized Cauchy identity. 28
Hopf algebra structure for The Generalized Cauchy identity is equivalent to an important fact. The ring is a graded algebra. = M k k ; k = gen. characters of S k : The multiplication (induction) is a bilinear map that induces a homomorphism. In degree k this is a map M p+q=k p q k : On p q this is induction of chars S p S q S p+q. There is a dual operation, namely restriction of chars S p+q S p S q. This gives a homomorphism of graded rings called comultiplication. Theorem. (Geissinger) The two operations of multiplication and comultiplication make a Hopf algebra. This means that comultiplication is a homomorphism of graded algebras, or (equivalently) that multiplication is a homomorphism of graded coalgebras. The Hopf algebra structure was popularized by Zelevinsky. We will show that the theorem is equivalent to the Generalized Cauchy identity! 29
The Hopf axiom Geissinger's theorem boils down to the commutativity of the following diagram: m Λ Λ m 1 τ 1 Λ Λ Λ Λ Λ Λ Λ Λ Λ m m Λ Λ m m m = multiplication; m = comult:; (x y) = y x: Start with a character of in p q and push it forward to r s, where p + q = r + s = k. Thus we are inducing a character from S p S q to S k, then restricting to S r S s. Mackey theory If G H 1 ; H 2 (nite groups) there are two ways we can get from characters of H 1 to characters of H 2. We can Induce then restrict or restrict then induce And these are the same. More exactly G Res H2 Ind H1() = M H Ind 2 2H 2 H Res H ( ) G/H1 where H = H 2 \ H 1?1 and (h) = (?1 h). For symmetric groups this gives the Hopf axiom. 30
Hopf Axiom = Generalized Cauchy The Hopf axiom reduces to the formula c c = '; c ' c c 'c ; (4) since if we apply m m to s s, the coecient of s s is the left side, (m m) (1 1) (m m ) gives the right side. To deduce (4) from the generalized Cauchy identity we note that (in obvious notation) the right side of is i;k (1? i k )?1 i;l (1 + i l ) j;k (1 + j k ) j;l LS ( 1 ; ; m ; 1 ; ; n )LS ( 1 ; ; s ; 1 ; ; while the left side is = = c s ()s 0()c s ()s 0() (1? j l )?1 = t ): s ' ()s ' ()s 0 ()s 0 ()s ()s 0()s 0()s () c ' c c ' c c 'c s ' ()s ()s 0 ()s 0()s ()s 0() Comparing, we obtain the result. s ()s 0()c s ()s 0(): 31