REPRESENTATION THEORY WEEK 7

REPRESENTATION THEORY WEEK 7 1. Characters of L k and S n A character of an irreducible representation of L k is a polynomial function constant on every conjugacy class. Since the set of diagonalizable matrices is dense in L k, a character is defined by its values on the subgroup of diagonal matrices in L k. Thus, one can consider a character as a polynomial function of x 1,..., x k. Moreover, a character is a symmetric polynomial of x 1,...,x k as the matrices diag (x 1,...,x k ) and diag ( ) x s(1),...,x s(k) are conjugate for any s Sk. For example, the character of the standard representation in E is equal to x 1 + + x k and the character of E n is equal to (x 1 + + x k ) n. Let λ = (λ 1,...,λ k ) be such that λ 1 λ λ k. Let D λ denote the determinant of the k k-matrix whose i, j entry equals x λ j i. It is clear that D λ is a skew-symmetric polynomial of x 1,...,x k. If ρ = (k 1,...,1, ) then D ρ = i j (x i x j ) is the well known Vandermonde determinant. Let S λ = D λ+ρ D ρ. It is easy to see that S λ is a symmetric polynomial of x 1,...,x k. It is called a Schur polynomial. The leading monomial of S λ is the x λ 1...x λ k k (if one orders monomials lexicographically) and therefore it is not hard to show that S λ form a basis in the ring of symmetric polynomials of x 1,..., x k. Theorem 1.1. The character of W λ equals to S λ. I do not include a proof of this Theorem since it uses beautiful but hard combinatoric. The proof is much easier in general framework of Lie groups and is included in 61A course. Exercise. Check that dim W λ = ( λi λ ) j (ρ i<j i ρ j ) = i<j ( λi λ ) j (k 1)! (k )!...1!, if λ = λ + ρ. Now we use Schur-Weyl duality to establish the relation between characters of S n and L k. Recall that the conjugacy classes in S n are given by partitions of n. Let C (µ) be the class associated with the partition µ in the natural way. Let ρ denote Date: March 15, 11. 1

REPRESENTATION THEORY WEEK 7 the representation of S n L k in E n. Let r be the number of rows in µ. Then one can see that (1.1) tr(ρ s g ) = (x µ 1 1 + + x µ 1 k )...(xµr 1 + + x µr k ), for any s C (µ) and a diagonal g L k. Denote by P µ the polynomial in the right hand side of the identity. Let χ λ be the character of V λ. Since tr (ρ s g ) = λ Γ n,k χ λ (s) S λ (g), one obtains the following remarkable relation (1.) P µ = λ Γ n,k χ λ (s) S λ.. Representations of compact groups Let be a group and a topological space. We say that is a topological group if the multiplication map and the inverse are continuous maps. Naturally, is compact if it is compact topological space. Examples. The circle S 1 = {z C z = 1}. A torus T n = S 1 S 1. Unitary group U n = { X L n X t X = 1 n }. Special unitary group Orthogonal group Special orthogonal group SU n = {X U n det X = 1}. O n = { x L n (R) X t X = 1 n }. SO n = {X O n det X = 1}. Theorem.1. Let be a compact group. There exists a unique measure on such that f (ts) dt = f (t) dt, for any integrable function f on and any s, and dt = 1. In the same way there exists a measure d t such that f (st) dt = f (t) d t, d t = 1. Moreover, for a connected compact group dt = d t.

REPRESENTATION THEORY WEEK 7 3 The measure dt (d t) is called right-invariant (left-invariant) measure, or Haar measure. We do not give the proof of this theorem in general. However, all examples we consider are smooth submanifolds in L k. Thus, to define the invariant measure we just need to define a volume in the tangent space at identity T 1 and then use right (left) multiplication to define it on the whole group. More precisely, let γ Λ top T 1. Then γ s = m s (γ), where m s : is the right (left) multiplication on s and m s is the induced map Λ top T 1 Λtop T s. After this normalize γ to satisfy γ = 1. Consider a vector space over C equipped with topology such that addition and multiplication by a scalar are continuous. We always assume that a topological vector space satisfies the following conditions (1) for any v V there exist a neighborhood of which does not contain v; () there is a base of convex neighborhoods of zero. Topological vector spaces satisfying above conditions are called locally convex. We do not go into the theory of such spaces. All we need is the fact that there is a non-zero continuous linear functional on a locally convex space. A representation ρ : L(V ) is continuous if the map V V given by (s, v) ρ s v is continuous. Regular representation. Let be a compact group and L () be the space of all complex valued functions on such that f (t) dt exists. Then L () is a Hilbert space with respect to Hermitian form f, g = f (t) g (t) dt. Moreover, a representation R of in L () given by R s f (t) = f (ts) is continuous and the Hermitian form is -invariant. A representation ρ : L(V ) is called topologically irreducible if any invariant closed subspace of V is either V or. Lemma.. Every irreducible representation of is isomorphic to a subrepresentation in L (). Proof. Let ρ : L(V ) be irreducible. Pick a non-zero linear functional ϕ on V and define the map Φ : V L () which sends v to the matrix coefficient f v,ϕ (s) = ϕ, ρ s v. It is clear that a matrix coefficient is a continuous function on, therefore f v,ϕ L (). Furthermore Φ is a continuous intertwiner and KerΦ =.

4 REPRESENTATION THEORY WEEK 7 Recall that a Hilbert space is a space over C equipped with positive definite Hermitian form, complete in topology defined by the norm v = v, v 1/. We need the fact that a Hilbert space has an orthonormal topological basis. A continuous representation ρ : L (V ) is called unitary if V is a Hilbert space and v, v = ρ g v, ρ g v for any v V and g. The regular representation of in L () is unitary. In fact, Lemma. implies Corollary.3. Every topologically irreducible representation of a compact group is a subrepresentation in L (). Lemma.4. Every irreducible unitary representation of a compact group is finitedimensional. Proof. Let ρ : L(V ) be an irreducible unitary representation. Choose v V, v = 1. Define an operator T : V V by the formula Let Tx = v, x v. One can check easily that T is self-adjoint, i.e. x, Ty = Tx, y. Tx = ρ g T ( ρ 1 g x) dg. Then T : V V is an intertwiner and a self-adjoint operator. Furthermore, T is compact, i.e. if S = {x V x = 1}, then T (S) is a compact set in V. Every self-adjoint compact operator has an eigenvector. To construct an eigen vector find x S such that ( Tx, x ) is maximal. Then Tx = λx. Since Ker ( T λ Id ) is an invariant subspace in V, Ker ( T λ Id ) =. Hence T = λid. Note that for any orthonormal system of vectors e 1,..., e n V ei, Te i = ei, Te i 1, that implies λn 1. Hence dim V 1 λ. Corollary.5. Every irreducible continuous representation of a compact group is finite-dimensional.

REPRESENTATION THEORY WEEK 7 5 3. Orthogonality relations and Peter-Weyl Theorem If ρ : L (V ) is a unitary representation. Define a matrix coefficient by the formula f v,w (g) = w, ρ g v. It is easy to check that (3.1) f v,w ( g 1 ) = f w,v (g) Theorem 3.1. For an irreducible unitary representation ρ : L(V ) f v,w, f v,w = f v,w (g)f v,w (g)dg = 1 dim ρ v, v w, w. The matrix coefficient of two non-isomorphic representation are orthogonal in L (). Proof. Define T End C (V ) and Tx = v, x v T = ρ g Tρ 1 g dg. As follows from Shur s lemma, T = λid. Since we obtain Hence On the other hand, w, Tw = tr T = tr T = v, v, T = v, v dim ρ. w, Tw = 1 dim ρ v, v w, w. w, v, ρ 1 g w ρ g v ( dg = f ) w,v g 1 f v,w (g)dg = = f v,w (g)f v,w (g)dg = 1 dim ρ f v,w, f v,w. In f v,w and f v,w are matrix coefficients of two non-isomorphic representation, the T =, and the calculation is even simpler. Corollary 3.. Let ρ and σ be two irreducible representations, then χ ρ, χ σ = 1 if ρ is isomorphic to σ and χ ρ, χ σ = otherwise. Theorem 3.3. (Peter-Weyl) Matrix coefficient form a dense set in L () for a compact group.

6 REPRESENTATION THEORY WEEK 7 Proof. We will prove the Theorem under assumption that L(E), in other words we assume that has a faithful finite-dimensional representation. Let M = End C (E). The polynomial functions C [M] on M form a dense set in the space of continuous functions on (Weierstrass theorem), and continuous functions are dense in L (). On the other hand, C [M] is spanned by matrix coefficients of all representations in T (E) = n= E n. Hence matrix coefficients are dense in L (). Corollary 3.4. The characters of irreducible representations form an orthonormal basis in the subspace of class function in L (). Corollary 3.5. Let be a compact group and R denote the representation of in L () given by the formula Then R s,t f (x) = f ( s 1 xt ). L () = ρ b V ρ V ρ, where Ĝ denotes the set of isomorphism classes of irreducible unitary representations of and the direct sum is in the sense of Hilbert spaces. Remark 3.6. Note that it follows from the proof of Theorem 3.3, that if E is a faithful representation of a compact group, then all other irreducible representations appear in T (E) as subrepresentations. 4. Examples Example 1. Let S 1 = {z C z = 1}, z = e iθ. The invariant measure on S 1 is dθ The irreducible representations are one dimensional. They are given by the π characters χ n : S 1 C, where χ n (θ) = e inθ. Hence Ŝ1 = Z and L ( S 1) = n Z Ce inθ, this is well-known fact that every periodic function can be extended in Fourier series. Example. Let = SU. Then consists of all matrices a b b ā, satisfying the relations a + b = 1. One also can realize SU as the subgroup of quaternions with norm 1. Thus, topologically SU is isomorphic to the threedimensional sphere S 3. To find all irreducible representation of SU consider the polynomial ring C [x, y] with the action of SU given by the formula ( ) a b ρ g (x) = ax + by, ρ g (y) = bx + āy, if g =. b a

REPRESENTATION THEORY WEEK 7 7 Let ρ n be the representation of in the space C n [x, y] of homogeneous polynomials of degree n. The monomials x n, x n 1 y,..., y n form a basis of C n [x, y]. Therefore dim ρ n = n + 1. We claim that all ρ n are irreducible and that every irreducible representation of SU is isomorphic to ρ n. Hence Ĝ = Z +. We will show this by checking that the characters χ n of ρ n form an orthonormal basis in the Hilbert space of class functions on. Note that every unitary matrix is diagonal in some orthonormal basis, therefore every conjugacy class of SU intersects the diagonal subgroup. Moreover, ( z z) and ( z z ) are conjugate. Hence the set of conjugacy classes can be identified with S1 quotient by the equivalence relation z z. Let z = e iθ, then (4.1) χ n (z) = z n + z n + + z n = zn+1 z n 1 sin (n + 1)θ =. z z 1 sin θ Now let us calculate the scalar product in the space of class function. It is clear that the invariant measure dg on is proportional to the standard volume form on the three-dimensional sphere induced by the volume form on R 4. Let C (θ) denote the conjugacy class of all matrices with eigenvalues e iθ, e iθ. The characteristic polynomial of a matrix from C (θ) equals t cosθt + 1. Thus, we obtain a + ā = cosθ, or a = cos θ + yi for real y. Hence C (θ) satisfy the equation a + b = cos θ + y + b = 1, or y + b = sin θ. In other words, C (θ) is a two-dimensional sphere of radius sin θ. Hence for a class function φ on φ (g)dg = 1 π φ (θ)sin θdθ. π All class function are even functions on S 1, i.e. they satisfy the condition φ ( θ) = φ (θ). One can see easily from (4.1) that χ n (θ) form an orthonormal basis in the space of even function on the circle with respect to the Hermitian product ϕ, η = 1 π π ϕ(θ) η (θ) sin θdθ. Example 3. Let = SO 3. Recall that SU can be realized as the set of quaternions with norm 1. Consider the representation γ of SU in H defined by the formula γ g (α) = gαg 1. One can see that the 3-dimensional space H im of pure imaginary quaternions is invariant and (α, β) = Re ( α β ) is invariant positive definite scalar product on H im. Therefore ρ defines a homomorphism γ : SU SO 3. Check that Kerγ = {1, 1} and that γ is surjetive. Hence SO 3 = SU / {1, 1}. Thus, every representation of SO 3 can be lifted to the representations of SU, and a representation of SU factors to the representation of SO 3 iff it is trivial on 1. One can check easily that ρ n ( 1) = 1 iff n is even. Thus, an irreducible representations of SO 3 is

8 REPRESENTATION THEORY WEEK 7 isomorphic to ρ m and dim ρ m = m + 1. Below we give an independent realization of irreducible representation of SO 3. Harmonic analysis on a sphere. Consider the sphere S in R 3 defined by the equation x + y + z = 1. It is clear that SO 3 acts in the space of complex-valued functions on S. Introduce differential operators in R 3 : e = 1 ( x + y + z ), h = x x + y y + z z + 3, f = 1 ( ) x + y + z, note that e, f, and h commute with the action of SO 3 and satisfy the relations [e, f] = h, [h, e] = e, [h, f] = f. Let P n be the space of homogeneous polynomial of degree n and H n = Ker f P n. The polynomials of H n are harmonic polynomials since they are annihilated by Laplace operator. For any ϕ P n ( h (ϕ) = n + 3 ) ϕ. If ϕ H n, then ( fe (ϕ) = ef (ϕ) h (ϕ) = n + 3 ) ϕ, and by induction fe k (ϕ) = efe k 1 (ϕ) he k 1 (ϕ) = In particular, this implies that (4.) fe k (H n ) = e k 1 (H n ). We prove that (4.3) P n = H n e (H n ) e (H n 4 ) +... by induction on n. Indeed, by induction assumption P n = H n e (H n 4 ) +..., ( nk + k (k 1) + 3k ) e k 1 ϕ. then (4.) implies fe (P n ) = P n. Hence H n ep n =. On the other hand, f : P n P n is surjective, and therefore dim H n + dim P n = dim P n. Therefore (4.4) P n = H n P n, which implies (4.3). Note that after restriction on S, the operator e acts as the multiplication on 1. Hence (4.3) implies that C [ S ] = n H n. To calculate the dimension of H n use (4.4) dim H n = dim P n dim P n = (n + 1)(n + ) n (n 1) = n + 1.

REPRESENTATION THEORY WEEK 7 9 Finally, we claim that the representation of SO 3 in H n is irreducible and isomorphic to ρ n. Check that ϕ = (x + iy) n H n and the rotation on the angle θ about z axis maps ϕ to e inθ ϕ. Since this rotation is the image of e iθ/ e iθ/, under the homomorphism γ : SU SO 3, the statement follows from (4.1). Recall now the following theorem (Lecture Notes 1). A convex centrally symmetric solid in R 3 is uniquely determined by the areas of the plane cross-sections through the origin. A convex solid B can be defined by an even continuous function on S. Indeed, for each unit vector v let ϕ (v) = sup { t / tv B }. Define a linear operator T in the space of all even continuous functions on S by the formula Tϕ (v) = π ϕ (w)dθ, where w runs the set of unit vectors orthogonal to v, and θ is the angular parameter on the circle S v. Check that Tϕ (v) is the area of the cross section by the plane v. We have to prove that T is invertible. Obviously T commutes with the SO 3 -action. Therefore T can be diagonalized. Moreover, T acts on H n as the scalar operator λ n Id. We have to check that λ n for all n. Let ϕ = (x + iy) n H n. Then ϕ (1,, ) = 1 and Tϕ (1,, ) = π π (iy) n dθ = ( 1) n sin n θdθ, here we take the integral over the circle y + z = 1, and assume y = sin θ, z = cosθ. Since Tϕ = λ n ϕ, we obtain π λ n = ( 1) n sin n θdθ.