Symmetric polynomials and partitions Eugene Mukhin


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1 Symmetic polynomials and patitions Eugene Mukhin. Symmetic polynomials.. Definition. We will conside polynomials in n vaiables x,..., x n and use the shotcut p(x) instead of p(x,..., x n ). A pemutation w is a one to one map of the set {,..., n} to itself. Thee ae n! pemutations. The poduct of pemutations w w 2 is just the composition of maps. We will wite w x fo x w(),..., x w(n). An invesion in pemutation w is a pai of numbes i < j n, such that w(i) > w(j). A pemutation w is called even o odd if the numbe of invesions is even o odd. The sign of a pemutation w, sgn(w) is if w is odd and sgn(w) = if w is even. Execise: Pove that sgn(w w 2 ) = sgn(w 2 w ) = sgn(w )sgn(w 2 ). Symmetic polynomials ae polynomials which do not change values if some aguments ae switched. Definition: A polynomial p(x) is called symmetic if p(x) = p(w x) fo any pemutation w. Fo example, let n = 3, then a polynomial p(x) = x + x 2 + x 3 is symmetic, say p(3, 5, 2) = p( 5, 2, 3). The polynomial (x) = x + x 2 + x 3 x is not symmetic, (, 2, 3) (2,, 3). Note that p(x) is the sum of all vaiables, no matte how you shuffle the vaiables, but if you pemute the vaiables in, you can also obtain expessions x 2 + x + x 3 x 2, x 3 + x 2 + x x 2 and x 3 + x + x x 2. Execise: Pove that a polynomial p(x) is symmetic if and only if p(x) does not change unde the pemutations of vaiables as an expession..2. Monomial polynomials. Let λ = (λ,..., λ n ). Definition: The monomial symmetic polynomial m λ is the sum of monomial x λ... x λn n and all distinct monomials obtained fom it by a pemutation of vaiables. Fo example, if λ = (2,, ) then m λ = x 2 x 2 x 3 + x x 2 2x 3 + x x 2 x 2 3. The total degee of m λ is i λ i, the degee of m λ in each vaiable x i is λ. In ode to avoid epetitions among m λ we will always assume that λ λ n. A basis is the smallest set of polynomials though which you can expess all the othes. Definition: A set of symmetic polynomials S is called a basis, if ) any symmetic polynomial can be expessed as a sum of polynomials fom S with some coefficients. 2) No polynomial fom S can be expessed as a sum of othe polynomials fom S. Execise: The monomial polynomials {m λ, λ = (λ, λ n 0)} fom a basis.
2 2.3. Patitions. Definition: The vecto λ = (λ,..., λ n ) is called a patition of k if λ λ n 0 and λ = λ +... λ n = k. The numbe k is called length, numbes λ i ae called pats of λ. Patitions can be epesented by pictues called Young diagams (o Fees diagams). The Young diagam of λ consists of n ows of boxes aligned on the left, such that ith ow is ight on i + st ow. The length of ith ow is λ i. The conjugate patition λ is the patition with the Young diagams consisting of columns of lengths λ i. Fo example λ is the numbe of nonzeo pats of λ. If λ = (3, 3, ) then λ = (3, 2, 2). Also λ = λ. Execise: Show that the numbe of patitions of n with odd distinct pats euals to numbe of self conjugated patitions of n (that is patitions λ with the popety λ = λ ). Definition: A patition λ is said to be lage than a patition µ if λ = µ and we have λ µ λ + λ 2 µ + µ 2 λ + λ 2 + λ 3 µ + µ 2 + µ 3... The lagest patition of length k is (k, 0, 0,..., 0). If k n then the smallest patition of length k is (,,...,, 0,..., 0). Execise: Show that λ µ if and only if the Young diagams of λ can be obtained fom Young diagam of µ by aising some boxes fom lowe ows to highe ones. Execise: Find an example of two patitions of 6, none of which is geate then anothe..4. Multiplying monomial polynomials. Let µ + ν be a patition (λ + µ, λ 2 + µ 2,..., λ n + µ n ). Lemma. m λ m µ = m λ+µ + a ν λ,µm ν, a ν λ,µ Z 0. Execise: Poof the lemma. ν<λ+µ 2. Bases 2.. Elementay polynomials. Definition: The elementay polynomials e λ ae defined by the fomulas e λ = e λ e λ2..., e = m λ whee λ = (,...,, 0,..., 0) ( ones).
3 3 Lemma 2. We have e λ = m λ + µ<λ a λµ m µ. Theefoe {e λ, λ = (n λ λ m 0), m Z 0 } fom a basis of symmetic polynomials in n vaiables. Execise: Poof the lemma. Note that one can expess any symmetic polynomial as a sum of poducts of e i, i = 0,,..., n, whee e 0 =. In the mathematical language e,..., e n ae a set of geneatos of ou ing Powe sum polynomials. Definition: The powe sum polynomials p λ ae defined by the fomulas p λ = p λ p λ2..., p = m λ, whee λ = (, 0,..., 0) Lemma 3. We have p λ = a λ m λ + µ>λ b λµ m m u, b λµ Z 0, whee a λ is a natual numbe. Theefoe {p λ, λ = (λ λ n 0)} fom a basis of symmetic polynomials. Execise: Poof the lemma Complete polynomials. Definition: The complete polynomials h λ ae defined by the fomulas h λ = h λ h λ2..., h = m λ. λ = Lemma 4. Polynomials {h λ, λ = (λ λ n 0)} fom a basis of symmetic polynomials. Execise: Poof the lemma using the elation () below.
4 Schu polynomials. Definition: A Schu function s λ is the sum of function x λ x... x λn i n x i x j with all functions obtained fom it by a pemutation of vaiables. Euivalently, s λ is the antisymmetization of monomial x λ x λ x λn+n n divided by the Vandemond function i<j (x i x j ), ( ) s λ = ( ) sgn(w) x λ w() xλ x λn+n w(n) / i x j ), i<j(x w whee the sum is ove all pemutations of n elements. Execise: Show that s λ is a symmetic polynomial. Lemma 5. i<j s λ = m λ + µ<λ K λµ m µ. Theefoe {s λ, λ = (λ λ n 0)} fom a basis of symmetic polynomials. Execise: Poof the lemma. In fact K λµ ae vey impotant nonnegative integes called Kostka numbes Geneating functions and elations between diffeent bases. We have the geneating functions n n E(t) := e i t i = ( + x i t), H(t) := P (t) := h i t i = p i t i = n n x i t, x i x i t. Note that the fist euality is a vesion of Vieta theoem. We have the elations theefoe H(t)E( t) =, H (t) = P (t)h(t ), ( ) s e i h i = 0, () h = p i h i.
5 5 Execise: Use the elation P (t) = (log H(t)) to show that h = p λ n, z λ = (i m i m i!), z λ λ = whee m i is the numbe of pats of λ eual i. 3. Counting symmetic polynomials 3.. Gaussian binomial coefficients. Definition: The Gaussian binomial coefficient is given by ( ) m = ( m )( m )... ( m + ). ( )( 2 )... ( ) Execise: Pove the following identities ( ) ( m m = ( ) ( ) ( ) m m m = + n n ( ) n ( + i t) = i(i )/2 i n i t = ( ) n + i i ), (2) ( ) m = ( ) m + m, (3) t i, (4) t i. (5) The identity 2 shows that Gaussian binomial coefficients ae genealizations of usual binomial coefficients. The identities 3 ae called Pascal idenitites, the identity 4 is called Newton binomial fomula. Use one of the identities to show that Gaussian binomial coefficient is a polynomial in Main theoem via ecusion elations. Define the counting function of of symmetic polynomials by χ n,k () = a i,k i, a i,k = {λ, λ k, λ = i}. The numbe a i,k counts polynomials of total degee i, such that degee in any vaiable is at most k. Theoem 6. ) ( n + k χ k,n () = k Execise: Pove the theoem using Pascal identity (3)..
6 Main theoem via h k. Note that LHS of (5) is eual to H(t) whee x i ae substituted with i. Execise: Pove Theoem 6 by compaing monomials in h k with n + vaiables and patitions λ contibuting to a i. 4. Appendix 4.. Eule Identity. The Eule function is defined by the fomula ϕ(t) = ( t i ). The coefficient of t k in function /ϕ(t) euals the numbe of all patitions of k. Execise: Use geneating functions to pove that the numbe of patitions of n with odd pats is eual to the numbe of patitions of n with uneual pats. Lemma 7. (Eule identity) ϕ(t) = n= ( ) i t (3n2 n)/2. The numbes (3n 2 n)/2 ae called pentagon numbes. Compae to numbes n, tiangula numbes n(n + )/2, suae numbes n 2. Execise: Pove Eule identity by constucting a map fom patitions consisting of odd numbe of uneual pats to patitions consisting of even numbe of uneual pats Roges Ramanudjan Identities. Lemma = + n(n+) 8 ( )( 2 )... ( n ), 4 6 = + n2 9 ( )( 2 )... ( n ). Hee in LHS of the fist identity we have powes of which have emaindes 2 o 3 mod 5 and in LHS of the second identity the powes have emaindes o 4 mod 5. Execise: Refomulate RogesRamanudjan identities in the language of patitions A challenge. Let N k be a numbe of diffeent figues obtained by a putting two Young diagams of patitions λ, µ, such that λ + µ = k on top of each othe. Fo example, N 0 = N =, N 2 = 3, N 4 = 5, N 5 = 0, N 6 = 6. n= n= CHALLENGE. Compute the function N(t) = N it i. At the moment I know the answe but I do not know an elementay poof of it.
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