Rotational Symmetries in Polynomial Rings
aa r X i v : . [ m a t h . R T ] J a n ROTATIONAL SYMMETRIES IN POLYNOMIAL RINGS
KEITH CONRAD AND AMBAR N. SENGUPTA
Abstract.
We obtain results describing the behavior of the actionof rotation generators on polynomials over a commutative ring. Wealso explore harmonic polynomials in a purely algebraic setting.
1. Introduction and summary of results
The purpose of this paper is to study the action of rotation generatorson a polynomial ring. We establish numerous algebraic results, manyof which have classical analytic counterparts for functions on Euclideanspace.Let R be a ring (always assumed commutative with identity) and con-sider the polynomial ring P N = R [ X , . . . , X N ] with N ≥
2. Classically R is R or C . For us, R may be a ring of polynomials over further in-determinates with coefficients in some commutative ring. Thus, for ourpurposes, it is necessary to build the framework of results with R being aring rather than a field. Other examples for R of interest include the ringof p -adic integers and the ring of formal power series over some integraldomain.By a rotation generator we mean an operator on P N of the form M jk = X j ∂ k − X k ∂ j , (1.1)where j, k ∈ { , . . . , N } with j = k and ∂ j = ∂/∂X j . The R -linear spanof these operators is a Lie algebra over R . The skew-symmetry M jk = − M kj , implies M jk is independent of the ordering of j and k .We also work with the Laplacian operator∆ N = ∂ + · · · + ∂ N (1.2)and sometimes we will drop the subscript N from ∆ N . The “quadraticCasimir” operator M · M is given by M · M = X { j,k }∈ P ( N ) M jk , (1.3) where P ( N ) is the set of all 2-element subsets of { , . . . , N } . Lengthybut straightforward computations produce the following commutator re-lations: on P N ,[ M jk , M lm ] = 0 if { j, k } , { l, m } ∈ P ( N ) are disjoint,[ M jk , M kl ] = M jl if { j, k } , { k, l } ∈ P ( N ) with j = l , [ M jk , M · M ] = 0 if { j, k } ∈ P ( N ) . (1.4) We present a sample of our results, indi-cating their overall narrative role in the development. The version ofthe results stated below may have a weaker hypothesis on R when it isproved later.The following result, which is established later (with slightly weakerhypotheses on R ) as Propositions 2.3 and 2.5, is the algebraic counterpartof the geometric result that if a function on R N is invariant under allrotations around the origin then it is a function of the Euclidean norm. Proposition 1.1.
Suppose the ring R contains Q . A polynomial p ∈ R [ X , . . . , X N ] is annihilated by all the rotation generators M jk if andonly if p is a polynomial in X · X = X + · · · + X N with R -coefficients.An R -linear derivation on R [ X , . . . , X N ] annihilates X · X if and onlyif it is an R [ X , . . . , X N ] -linear combination of the operators M jk . For a more general picture, let D A be the set of all derivations on an R -algebra A ; then D A is a Lie algebra over A . If D is any Lie subalgebraof D A then the set N ( D ) def = \ L ∈ D ker L of common zeros of all L ∈ D is a subalgebra of A . Conversely for anysubalgebra B ⊂ A the annihilator A ( B ) def = { L ∈ D A : B ⊂ ker L } is a Lie subalgebra of D A . In this framework, Proposition 1.1 is thecase where A = R [ X , . . . , X N ], and B = R [ X · X ] is the subalgebragenerated over R by the quadratic form X · X : it says The derivations on R [ X , . . . , X N ] that annihilate X · X form the Lie algebra spanned by all the operators M jk with coefficients in R [ X , . . . , X N ] , and every polynomialin R [ X , . . . , X N ] annihilated by that Lie algebra is in thesubalgebra B . OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 3
We summarize the major points of the paper, mentioning some analyticcounterparts of the algebraic results.(1) We use a purely algebraic approach, working mostly with a poly-nomial ring R [ X , . . . , X N ] and we assume that N ≥
2. A fewresults need
N >
2. Whether results hold for N = 1 can bechecked by the reader in each case.(2) Summary of notation . For simplicity we will assume in thissection that R is a ring containing Q , such as any field of char-acteristic 0 or a polynomial ring over Q . The R -algebra P N = R [ X , . . . , X N ] is the direct sum P N = M d ∈ Z P N,d , (1.5)where P N,d is the R -submodule of P N consisting of all homoge-neous degree- d polynomials together with 0. Set P N,d = { } if d <
0. The Laplacian operator on P N is∆ N = N X j =1 ∂ j : P N → P N − , and the restriction of ∆ N to P N,d is ∆
N,d : P N,d → P
N,d − . Apolynomial is harmonic if it is in ker ∆ N ; the space of harmonicpolynomials is H N = ker ∆ N = M d ∈ Z H N,d , where H N,d = ker ∆
N,d . We work with the rotation generators M jk = X j ∂ k − X k ∂ j , which map P N,d to P N,d , and the quadratic Casimir M · M = X { j,k }∈ P ( N ) M jk , where P ( N ) is the 2-element subsets { j, k } of { , , . . . , N } . Lastly, X · X = X + · · · + X N , and for each c ∈ R we set Z N ( c ) = ideal in R [ X , . . . , X N ] generated by X · X − c . KEITH CONRAD AND AMBAR N. SENGUPTA (3) (Prop. 2.1) We often use the following operator identity on P N :( X · X )∆ N = ( r∂ r ) + ( N − r∂ r + M · M , where we define the
Euler operator r∂ r = N X j =1 X j ∂ j . (4) (Remark 2.4) If S is a subset of { , . . . , N } with at least elementsand p ∈ P N is in ker M jk for all distinct j, k ∈ S , then p is apolynomial in P j ∈ S X j with coefficients in R [ X l ; l / ∈ S ]. Analyticcounterpart: if a function p on R n is invariant under all rotationsin the coordinates x , . . . , x k , then p is a function of x + · · · + x k and the remaining coordinate variables.(5) (Prop. 2.17(iii)) If a nonzero homogeneous polynomial p ∈ P N,d issuch that M jk p ∈ Z N ( c ) for some c ∈ R that is not a zero divisorand all distinct j, k ∈ { , . . . , N } then d is even and p is an R -multiple of ( X · X ) d/ . Analytic counterpart: if a homogeneousfunction p on R N restricted to some sphere centered at the originis rotation invariant then p is a function of the norm on R N .(6) (Prop. 2.8) The only harmonic polynomial in Z N ( c ) is . Ana-lytic counterpart: a harmonic function on R N that vanishes on asphere centered at the origin is 0.(7) (Prop. 2.10) Every polynomial p ∈ R [ X , . . . , X N ] can be ex-pressed uniquely in the form p = p + ( X · X ) p + · · · + ( X · X ) s p s where each p j is harmonic. If p ∈ P N,d ( homogeneous of degree d ) then each p j is also homogeneous, with p j ∈ H N,d − j . (8) (Prop. 5.3) For each c ∈ R and p ∈ P N there is a unique har-monic polynomial p ∗ ∈ H N such that p − p ∗ ∈ Z N ( c ) . Analyticcounterpart: On each sphere in R N centered at the origin, everypolynomial in R [ X , . . . , X N ] is equal to some harmonic polyno-mial.(9) (Prop. 2.12) Let p ∈ P N,d be homogeneous of degree d ≥ : p = p + p X N + · · · + p d X dN , where p , . . . , p d ∈ R [ X , . . . , X N − ] . Then p is harmonic if andonly if the coefficients p , p , . . . , p d are determined by p and p OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 5 through the relations p k = ( − k k )! ∆ k p p k +1 = ( − k k + 1)! ∆ k p (1.6) for k ≥ . This provides a dimension formula if R is a field :dim R H N,d = (cid:18) N + d − N − (cid:19) + (cid:18) N + d − N − (cid:19) . (10) (Prop. 3.2) If N = 2 n and i = √− ∈ R then common eigenvec-tors of M , M , . . . , M n − , n , and M · M in R [ X , . . . , X N ] areof the form Y a q, where Y a = ( X + iε X ) | a | · · · ( X n − + iε n X n ) | a n | , with a = ( a , . . . , a n ) ∈ Z n , ε j ∈ {± } the sign of a j , and q is apolynomial in X + X , X + X , . . . , X n − + X n , with coefficientsin R , satisfying a certain differential equation; if N = 2 then q is just an element of R . If N = 2 n + 1 then similar results areobtained except that q has coefficients in R [ X N ].(11) (Prop. 5.6) A spherical harmonic is a homogeneous polynomialthat is harmonic. A zonal harmonic is a spherical harmonic that,modulo Z N ( c ), is a polynomial in t · X = t X + · · · + t N X N ,where t = ( t , . . . , t N ) ∈ R N . For simplicity suppose c = 1 and t · t = 1 here. For each q ( Y ) ∈ R [ Y ] the polynomial q ( t · X ) in R [ X , . . . , X n ] / Z N (1) is congruent to a spherical harmonic ofdegree n if and only if q satisfies the differential equation (1 − Y ) q ′′ ( Y ) − ( N − Y q ′ ( Y ) + n ( n + N − q ( Y ) = 0 . This equation has, up to scaling by any nonzero element of R ,a unique solution; this solution is of degree n , and it is an evenpolynomial if n is even and it is an odd polynomial if n is odd.(12) (Prop. 4.1, Corollary 4.8) We define a spherical mean to be an R -linear map λ : P N → R that vanishes on all polynomials of the form M jk p with p ∈ P N . There exists a unique spherical mean λ that is equal to on KEITH CONRAD AND AMBAR N. SENGUPTA ( X · X ) n for all integers n ≥ . Moreover, λ ( p ) = p (0) for everyharmonic polynomial p in P N . Analytic counterpart: λ ( p ) = Z S N − (1) p ( x , . . . , x N ) dσ ( x ) , where σ is the normalized surface measure on the unit sphere S N − (1) in R N .(13) (Equation 4.23) With λ as in (12) and p ∈ P N, n for n ≥ λ ( p ) = 1 n !2 n (2 n + N − n + N − . . . (2 n + N − n ) ∆ n p, where on the right ∆ n p , being of degree , is just an element of R .Analytic counterpart: for a homogeneous polynomial p : R N → R of degree 2 n ≥ Z S N − (1) p ( x , . . . , x N ) dσ ( x )= 1 n !2 n (2 n + N − . . . (2 n + N − n ) ∆ n p, where the terms multiplied in the denominator go down by 2 ateach step. This is not a standard formula but may be verified byanalytic methods.(14) (Proposition 4.12) The spherical mean λ is invariant under ro-tations: if p ( X ) ∈ R [ X , . . . , X N ], viewed as a ‘column vector’,where X = ( X , . . . , X N ), and A is an N × N orthogonal ma-trix with entries in R , then λ (cid:0) p ( XA ) (cid:1) = λ (cid:0) p ( X ) (cid:1) . We alsoprove the identity (4.19), which says that for any homogeneouspolynomial p of degree d : λ ( X p ) = 1 N + d − λ ( ∂ p ) . (1.7)(15) (Proposition 4.9) If a polynomial p satisfies the mean-value prop-erty λ (cid:0) p ( X + t ) (cid:1) = p ( t ) , (1.8)where X = ( X , . . . , X N ) and t = ( t , . . . , t N ) are N -tuples ofindeterminates, then p ( X ) is harmonic. OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 7
A motivation for this work was todevelop some of basic results concerning spherical harmonics in a mannerthat demonstrates that their validity is purely algebraic and to explainwhy some of these special functions have rational coefficients.Algebraic studies of spherical harmonics seem to be largely in thephysics literature, with a focus on atomic physics; these works includeOgawa [10]. Mathematical studies include Axler and Ramey [1] andReznick [13]. Spherical harmonics, originating in work of Legendre andlater Laplace’s series expansion of the gravitational potential, are a clas-sical subject; work on them from the 19th and early 20th century in-clude Heine [6], Kellogg [7], Mehler [9], Thomson and Tait [15, 16], andWhittaker and Watson [18, section 18.31]. The relationship between spe-cial functions and representations of Lie groups is well-known (see, forexample Vilenkin and Klimyk [17] and, closer in spirit to our context,Macdonald [8]).Our study focuses on polynomials over general rings, typically contain-ing the rationals, rather than complex-valued special functions that ariseas matrix elements for representations of classical Lie groups. Instead ofbehavior under rotations and using integration over the sphere we use theLie algebra of the rotation group and we devise purely algebraic counter-parts of traditionally analytic notions such as an algebraic counterpartof integration over spheres (Section 4).
2. Polynomials and the action of rotation generators
For a ring R , each p in R [ X , . . . , X N ] is written as p = X ~ ∈ Z N p ~ X ~ = X ~ ∈ Z N p ~ X j . . . X j N N , indexed by vectors ~ = ( j , . . . , j N ) ∈ Z N , with p ~ = 0 if any componentof ~ is negative. So p is homogeneous of degree d when the monomials X j . . . X j N N appearing in p with nonzero coefficients have j + · · · + j N = d . Proposition 2.1.
With ∆ N being the Laplacian on R [ X , . . . , X N ] , ( X · X )∆ N = ( r∂ r ) + ( N − r∂ r + M · M , (2.1) where X · X = X + · · · + X N and r∂ r = N X j =1 X j ∂ j . KEITH CONRAD AND AMBAR N. SENGUPTA
The identity (2.1) is verified by a straightforward but tedious compu-tation left to the reader. It will play a central role for us.The Euler operator r∂ r on a degree- n homogeneous polynomial p in R [ X , . . . , X N ] multiplies it by n : ( r∂ r )( p ) = np . Just as the operator M jk arises in the traditional setting as a generator of rotations in the X j - X k ‘plane’, the Euler operator arises as a generator of scaling each X j by a common constant.A simple computation shows that M jk ( X · X ) = 0 , (2.2)and hence, using the derivation property of M jk , we also have M jk (( X · X ) a ) = 0 , (2.3)for every integer a ≥
0. Consequently,( M · M )(( X · X ) a ) = 0 . (2.4)We will make repeated use of the fact that X · X , being a monic poly-nomial in X , is not a zero divisor. A more general form of this observa-tion is in Lemma 2.15. Proposition 2.2.
With notation as before, for each positive integer j ∆ N (( X · X ) j ) = 2 j (2 j + N − X · X ) j − . (2.5) Proof.
Applying (2.1) to ( X · X ) j , and using the annihilation property(2.4), we have( X · X )∆ N (( X · X ) j ) = (4 j + ( N − j )( X · X ) j , and then (2.5) follows because X · X is not a zero divisor. (cid:3) Geometrically, it is clear thata function on R N that is invariant under all rotations around the originis a function of the radial distance from the origin. The following resultis an algebraic counterpart of this observation. Proposition 2.3.
Suppose that R is a ring in which positive integermultiples of the identity R are not zero divisors, and N is an integer ≥ . A polynomial p ∈ R [ X , . . . , X N ] satisfies M jk p = 0 for all distinct j, k ∈ { , . . . , N } if and only if p = q ( X + · · · + X N ) , for some q ( T ) ∈ R [ T ] . If p is homogeneous of degree d , then d is evenand q ( T ) = cT d/ for some c ∈ R . OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 9
The zero divisor condition on R implies R has characteristic 0 and thecondition is satisfied if R contains Q . A ring of characteristic 0 not fittingthe zero divisor condition is Z [ x ] / (2 x ). Proof.
As noted in (2.3), M jk (( X + · · · + X N ) a ) = 0 for all integers a ≥
0, so each polynomial in R [ X + · · · + X N ] is annihilated by all M jk .For the converse, suppose M jk p = 0 for all distinct j and k . To prove p is a polynomial in X · X = X + · · · + X N , we may assume that p is homo-geneous since each M jk maps homogeneous polynomials to homogeneouspolynomials of the same degree.Let d = deg p . The result is obvious if d = 0. If d = 1 then p is of theform a X + · · · + a N X N , so M jk p = a k X j − a j X k . This being 0 for all distinct j, k ∈ { , . . . , N } implies p is 0, contradicting p having degree 1, so we can’t have d = 1.Let d ≥
2. From all M jk p = 0 we get ( M · M ) p = 0. Applying theidentity (2.1) to p ,( X · X )∆ p = ( d + ( N − d ) p + ( M · M ) p = d ( d + N − p, (2.6)so d ( d + N − p = ( X · X ) F, (2.7)where F = ∆ p ∈ R [ X , . . . , X N ] . Equating the coefficients of X j . . . X j N N on both sides of (2.7), d ( d + N − p j ,...,j N = F j − ,j ,...,j N + F j ,j − ,...,j N + · · · + F j ,j ,...,j N − (2.8)for all j , . . . , j N ≥
0, where a coefficient on the right side of (2.8) is 0 ifone of its indices is negative. We will use induction on | ~ | ′ def = j + · · · + j N ( j is not included in this sum) to show each coefficient of F ~j = F ( j ,...,j N ) is d ( d + N −
2) times a linear combination of coefficients of p .If | ~ | ′ = 0 then ~ = ( j , , . . . , F ~ = F j , ,..., = d ( d + N − p j +2 , ,..., . If | ~ | ′ = 1 then ~ = ( j , , . . . , , . . . , F ~ = F j , ,..., ,..., = d ( d + N − p j +2 , ,..., ,..., . Suppose, inductively, that for some integer n ≥ F ~ is d ( d + N − p whenever | ~ | ′ < n ; we have just seen that this is true if n = 1 or 2. For an N -tuple ~ where | ~ | ′ = n ,replacing j with j + 2 in (2.8) gives us F ~ = d ( d + N − p j +2 ,j ,...,j N − ( F j +2 ,j − ,...,j N + · · · + F j +2 ,j ,...,j N − ) , and the coefficients F ~ı on the right have | ~ı | ′ = | ~ | ′ − < n , so by induction F ~ is d ( d + N −
2) times a linear combination of coefficients of p when | ~ | ′ = n .Thus the polynomial F itself is d ( d + N −
2) times a polynomial F ∗ ∈ R [ X , . . . , X N ] whose coefficients are linear combinations of thecoefficients of p : F = d ( d + N − F ∗ . (2.9)Substituting this into (2.7), we have d ( d + N − (cid:0) p − ( X · X ) F ∗ (cid:1) = 0 . (2.10)Finally we use the zero divisor condition on R . If any coefficient in p − ( X · X ) F ∗ were nonzero then d ( d + N − R would be a zero divisorin R , which contradicts the zero divisor condition on R since d ≥ p = ( X · X ) F ∗ . (2.11)Since p is homogeneous of degree d , the polynomial F ∗ is homogeneousof degree d −
2, because X · X is not a zero divisor. Moreover, since M jk ( X · X ) = M jk ( X j + X k ) = 0, by the product rule for the first-orderdifferential operator M jk , for each g ∈ R [ X , . . . , X N ] we have M jk (( X · X ) g ) = M jk ( X · X ) g + ( X · X ) M jk g = ( X · X ) M jk g. (2.12)Thus M jk p = M jk (( X · X ) F ∗ ) = ( X · X ) M jk F ∗ . (2.13)The left side is 0 by hypothesis, so ( X · X ) M jk F ∗ = 0 for all M jk . Since X · X is not a zero divisor in R [ X , . . . , X N ], M jk F ∗ = 0 (2.14)for all M jk .To summarize, if p is homogeneous of degree d ≥ M jk p being 0, then p = ( X · X ) F ∗ where F ∗ is homogeneous of degree d − M jk F ∗ are 0. By induction on the degree, d − F ∗ = c ( X · X ) ( d − / for some c ∈ R . Thus d is even and p = c ( X · X )( X · X ) ( d − / = c ( X + · · · + X N ) d/ , which completes the proof. (cid:3) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 11
Remark 2.4.
If we replace R by R [ Y , . . . , Y m ] using new indetermi-nates Y , . . . , Y m , we get a result for polynomials with ‘partial’ rotationalsymmetry: if S is a subset of { , . . . , N } containing at least 2 elementsand M jk p = 0 for all distinct j, k ∈ S then p = q (cid:16)P j ∈ S X j , { X l } l / ∈ S (cid:17) ,where q is a polynomial with coefficients in R . Proposition 2.3 is the case S = { , . . . , N } . Proposition 2.5.
Suppose is not a zero divisor in R , and N anypositive integer. An R -linear derivation on R [ X , . . . , X N ] annihilates X · X if and only if it is an R [ X , . . . , X N ] -linear combination of theoperators M jk .Proof. We saw in (2.2) that X · X is annihilated by every M jk , so X · X isalso annihilated by each R [ X , . . . , X N ]-linear combination of the M jk ’s.We turn now to proving the converse.Each R -linear derivation on R [ X , . . . , X N ] is determined by its effecton X , . . . , X N , so if L is an R -linear derivation on R [ X , . . . , X N ] then L = N X j =1 a j ∂ j (2.15)where a j = L ( X j ) ∈ R [ X , . . . , X N ]: both sides of (2.15) are R -linearderivations that agree on each X k .The condition L ( X · X ) = 0 is equivalent to2 N X j =1 a j X j = 0 . (2.16)Since 2 is assumed not to be a zero divisor in R , (2.16) is equivalent to a X + · · · + a N X N = 0 . (2.17)To deduce from (2.17) that L in (2.15) is an R [ X , . . . , X N ]-linear com-bination of all M jk , we induct on N . We will verify the cases N = 1 and N = 2 separately.For N = 1, the condition (2.17) implies that a = 0, and so L = 0,which is indeed a linear combination of M = 0, trivially.If N = 2 then (2.17) says a X + a X = 0in R [ X , X ]. Thus a is a multiple of X : a = bX where b ∈ R [ X , X ].Since X is not a zero divisor in R [ X , X ], we get a = − bX . Therefore L = a ∂ + a ∂ = bX ∂ − bX ∂ = bM , which proves the case N = 2.Now assume N > R in fewer than N indeterminates. Write a j as a oj + b j X N wherethe polynomial a oj does not involve X N . Extracting the terms on the leftside of (2.17) that do not involve X N , we have a o X + · · · + a oN − X N − = 0 . (2.18)Then, inductively, a o ∂ + · · · + a oN − ∂ N − is an R [ X , . . . , X N − ]-linearcombination of the operators M jk for distinct j, k ∈ { , . . . , N − } .Since M kj = − M jk , we may take j < k : a o ∂ + · · · + a oN − ∂ N − = X ≤ j A polynomial p is harmonic if ∆ p = 0;thus the harmonic polynomials in R [ X , . . . , X N ] form the kernel of theLaplacian ∆ N . Using the algebraic relationship between ∆ N and theoperator of multiplication by X · X , principally coming from (2.1), wewill see how R [ X , . . . , X N ] decomposes in terms of these two operatorswhen R contains Q .In this section, as before, we use the notation P N,d to denote the R -module of all polynomials in R [ X , . . . , X N ] that are homogeneous ofdegree d . The restriction of ∆ N to P N,d is denoted ∆ N,d , and the R -module of all harmonic polynomials that are homogeneous of degree d will be denoted H N,d = ker(∆ N,d ) = ker(∆ N |P N,d ) . (2.20)Let us note that ∆ N ( P N,d ) ⊂ P N,d − . (2.21)Thus ∆ N carries homogeneous polynomials into homogeneous polynomi-als, and so if p is harmonic then each homogeneous component of p isharmonic.The following simple property of the second order differential operator∆ N will be used repeatedly:∆ N ( pq ) = (∆ N p ) q + 2 N X j =1 ( ∂ j p )( ∂ j q ) + p ∆ N q, (2.22)for all p, q ∈ P N . For Proposition 2.7 we will use the following observa-tion. Lemma 2.6. Let m be a non-negative integer and q ∈ P N,d . Then ∆ N p = (2 m (2 m + N − 2) + 4 md )( X · X ) m − q + ( X · X ) m ∆ N q, (2.23) where p = ( X · X ) m q , and the first term on the right side of (2.23) istaken to be when m = 0 .Proof. Using the product formula (2.22), we have∆ N p = ∆ N (cid:0) ( X · X ) m q (cid:1) = ∆ N (( X · X ) m ) q + 2 N X j =1 m ( X · X ) m − X j ∂ j q + ( X · X ) m ∆ N q. (2.24) The first term is 2 m (2 m + N − X · X ) m − q by Proposition 2.2. Wehave P Nj =1 X j ∂ j q = dq since q is homogeneous, so the second term is4 md ( X · X ) m − q . Hence we have (2.23). (cid:3) Now we can prove that nonzero multiples of X · X cannot be harmonic. Proposition 2.7. Suppose that in the ring R positive integer multiplesof R are not zero divisors. The only harmonic multiple of X · X in R [ X , . . . , X N ] is .Proof. Even though R might not be an integral domain, for nonzero h in R [ X , . . . , X N ] we havedeg(( X · X ) h ) = 2 + deg h. (2.25)Indeed, this is reduced to the case of homogeneous h by writing h as asum of homogeneous parts, and when h is homogeneous every product ofmonomials in ( X · X ) h has the same degree. So the only way (2.25) failsfor homogeneous h is if ( X · X ) h = 0, which is impossible by thinking of X · X and h as polynomials in X N since X · X is monic in X N .From (2.25) we have deg(( X · X ) m h ) = 2 m + deg h for all m ≥ 0, soa nonzero polynomial in R [ X , . . . , X N ] isn’t arbitrarily highly divisibleby X · X .Let p be a harmonic multiple of X · X in R [ X , . . . , X N ]. To prove p =0 we will show p is arbitrarily highly divisible by X · X : if ( X · X ) m | p for some m ≥ X · X ) m +1 | p .Let us recall that ∆ N maps homogeneous polynomials to homogeneouspolynomials, and we note also that any homogeneous component of apolynomial multiple of X · X is also a multiple of X · X since multipli-cation by X · X maps any homogeneous polynomial to a homogeneouspolynomial with degree raised by 2. We may then assume p is homoge-neous. Set p = ( X · X ) m q with m ≥ q ∈ R [ X , . . . , X N ]. We mayassume q = 0. The polynomial q has to be homogeneous, because if ithas nonzero homogeneous terms of different degrees then so does p by(2.25), which contradicts the homogeneity of p .As seen in (2.23),∆ N p = (2 m (2 m + N − 2) + 4 md )( X · X ) m − q + ( X · X ) m ∆ N q, (2.26)where q ∈ P N,d .Since ∆ N p = 0 we have2 m (2 m + N − d ) q = − ( X · X )∆ N q. (2.27) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 15 We want to deduce from this that ( X · X ) | q in R [ X , . . . , X N ], so( X · X ) m +1 | p in R [ X , . . . , X N ] and we’d be done.Since X · X is monic in X N , we can use the division algorithm bymonics in R [ X , . . . , X N − ][ X N ] to write q = ( X · X ) g + r where r = 0or deg X N ( r ) ≤ 1. Multiplying both sides by 2 m (2 m + N − d ) andusing (2.27) we get X · X | m (2 m + N − d ) r , which for degreereasons implies 2 m (2 m + N − d ) r = 0, so r = 0 from the zero divisorhypothesis about R . (cid:3) Proposition 2.8. Suppose that in the ring R positive integer multiplesof R are not zero divisors. For c ∈ R , the only harmonic multiple of X · X − c in R [ X , . . . , X N ] is . This result with R = R is standard in analysis but the statement hereis purely algebraic, and the proof will be as well. Proof. Suppose p = 0 is harmonic and a multiple of ( X · X − c ). Thenwe can write p as p = ( X · X − c ) q, with q = 0. Write q as a sum of homogeneous terms: q = q + q + · · · + q m , with each q k ∈ R [ X , . . . , X N ] homogeneous of degree k , and q m = 0.Therefore the highest-degree homogeneous term in p is ( X · X ) q m , and( X · X ) q m is nonzero since q m = 0 and X · X is not a zero divisor (forexample, it is monic in X ). From looking at the effect of ∆ N on ho-mogeneous parts of ( X · X ) q , we get ∆ N (( X · X ) q m ) = 0. By Propo-sition 2.7 it follows that ( X · X ) q m = 0, but we already explained why( X · X ) q m = 0, so we have a contradiction. (cid:3) Lemma 2.9. Let A : V → W and B : W → V be linear mappings be-tween finite-dimensional vector spaces over a field. Suppose also that ker( A ) ∩ Im( B ) = { } and ker B = 0 . Then A is surjective, AB is bijective, and V = ker( A ) ⊕ Im( B ) . (2.28) Moreover, v = ( I − B ( AB ) − A ) v + B ( AB ) − Av, (2.29) for all v ∈ V , where the first term on the right is in ker A . Proof. The kernel of AB : W → W is the set of all w ∈ W for which Bw ∈ ker A , which means Bw ∈ ker( A ) ∩ Im( B ) = { } ; since ker B = 0it follows then that ker( AB ) is also 0. Since W is finite-dimensionalover a field, we conclude that AB is bijective, so A is surjective. In theidentity (2.29), the first term on the right side clearly lies in ker( A ) andthe second term is in Im( B ). Hence ker( A ) + Im( B ) = V . This sum isdirect because ker( A ) ∩ Im( B ) = { } . (cid:3) Proposition 2.10. Suppose that the ring R contains Q . Then everypolynomial p ∈ R [ X , . . . , X N ] can be expressed uniquely in the form p = p + ( X · X ) p + · · · + ( X · X ) s p s , (2.30) where p , . . . , p s are harmonic polynomials. Moreover, if p is homoge-neous of degree d then p j is or homogeneous of degree d − j for eachnon-negative integer j ≤ d/ . This result is widely known in the analytic context; for example, Axlerand Ramey [1] (Theorem 1.7 and Corollary 1.8) give an explicit formulafor p m . Reznick [13](Theorem 4.7) proves a more general expansion andalso traces some earlier works on the expansion (2.30). See also Petersonand Sengupta [11](equation (3.12)). The earliest reference we could findto this result is to Gauss [5] (page 630). Further work was done by Prasad[12]. Proof. First we treat the case R = Q . We use Lemma 2.9, with V = P N,j , W = P N,j − , the operator A being ∆ N,j = ∆ N |P N,j and B beingmultiplication by X · X . We apply repeatedly, to obtain P N,d = ker ∆ N,d ⊕ ( X · X ) P N,d − (2.31)= ker ∆ N,d ⊕ ( X · X ) ker ∆ N,d − ⊕ ( X · X ) P N,d − . Continuing in this way we get the direct sum decomposition P N,d = H N,d ⊕ ( X · X ) H N,d − ⊕ . . . , (2.32)and hence (2.30) for polynomials with rational coefficients.In particular, each monomial in Q [ X , . . . , X N ] is the sum of a har-monic polynomial and a multiple of X · X with Q -coefficients (preservinghomogeneity as the proposition describes). Taking R -linear combinationsof monomials, when R contains Q , shows every element of R [ X , . . . , X N ]is the sum of a harmonic polynomial and a multiple of X · X (preservinghomogeneity). This decomposition is unique by Proposition 2.7, and thedegree part follows from (2.25). (cid:3) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 17 The decomposition (2.31), or more simply the surjectivity of ∆ N,d : P N,d → P N,d − given by Lemma 2.9, provides the following dimensionformula for harmonic polynomials in case R is a field:dim R H N,d = dim R P N,d − dim R P N,d − = (cid:18) N + d − N − (cid:19) − (cid:18) N + d − N − (cid:19) , (2.33)where the second formula is based ondim R P N,d = (cid:18) N + d − N − (cid:19) , (2.34)which is the number of monomials of degree d in N indeterminates. For-mula (2.33) has appeared in the literature before; for example, [2] (The-orem 5.8). We will obtain another expression for the dimension of H N,d below in (2.41).We will see after Proposition 2.12 that H N,d is a free R -module when R is a ring containing Q , not necessarily a field, so (2.33) is true as arank formula for H N,d as an R -module. However, we can’t take R = Z :some denominators need to be allowed. For example, taking N = d = 2in (2.31) we have x = ( x − y ) + ( x, y ) · ( x, y ) , where ( x − y ) isharmonic. Even though x on the left is in Z [ x, y ], on the right ( x − y )is not in Z [ x, y ]. ∆ N . The Laplacian ∆ N,d sends P N,d to P N,d − ,so the product ( X · X )∆ N,d sends P N,d to itself. From (2.32) we get aneigenspace decomposition for this operator on P N,d , as follows. Proposition 2.11. Suppose that the ring R contains Q . The eigenvaluesof ( X · X )∆ N,d acting on P N,d are λ m = 2 m ( N − d − m ) forintegers m where ≤ m ≤ d/ , and the λ m -eigenspace of ( X · X )∆ N,d is ( X · X ) m H N,d − m . In particular, when R is a field of characteristic ,the characteristic polynomial of ( X · X )∆ N,d acting on P N,d is det (cid:0) tI − ( X · X )∆ N,d (cid:1) = [ d/ Y m =0 ( t − λ m ) dim R H N,d − m . (2.35) Proof. The case d = 0 is clear, and so we assume d ≥ 1. For h ∈ ker ∆ N,d − m , where 0 ≤ m ≤ d/ 2, we have deg h = d − m . Therefore by(2.23) with d there replaced by d − m ,∆ N,d (cid:0) ( X · X ) m h (cid:1) = 2 m (2 m + N − d − m )) ( X · X ) m − h. (2.36) Therefore ( X · X )∆ N,d acts on ( X · X ) m H N,d − m as multiplication by λ m .Each ( X · X ) m H N,d − m is an eigenspace for ( X · X )∆ N,d and these are allof its eigenspaces in P N,d by the direct sum decomposition in Proposition2.10. (cid:3) We show that a har-monic polynomial p ( X , . . . , X N ), when viewed as a polynomial in X N ,is completely determined by the ‘constant term’ that depends only onthe indeterminates X , . . . , X N − . Proposition 2.12. Suppose R is a ring containing Q , and N and d are non-negative integers with N ≥ . Pick p ∈ R [ X , . . . , X N ] that ishomogeneous of degree d ≥ and write it as a polynomial in X N : p = p + p X N + · · · + p d X dN , (2.37) where p , . . . , p d ∈ R [ X , . . . , X N − ] . Then p is harmonic if and only ifthe X N -coefficients p , p , . . . , p d are determined by p and p through therelations p k = ( − k k )! ∆ k p for 1 ≤ k ≤ d/ ,p k +1 = ( − k k + 1)! ∆ k p for 1 ≤ k ≤ ( d − / . (2.38) Proof. Let us note first that for any polynomial q ∈ R [ X , . . . , X N ] thatis independent of X N , we have ∆ N q = ∆ N − q .Since p is homogeneous of degree d , in (2.37) each nonzero p k is homo-geneous of degree d − k .Since p j and X jN are polynomials in disjoint sets of indeterminates,using the product formula (2.22), we have:∆( p j X jN ) = p j ∆( X jN ) + (∆ p j ) X jN = j ( j − p j X j − N + (∆ p j ) X jN OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 19 (the middle term P di =0 ∂ i p j )( ∂ i X jN ) is 0 since each summand is 0).Therefore ∆ p = d X j =0 ∆( p j X jN )= d X j =0 (cid:0) j ( j − p j X j − N + (∆ p j ) X jN (cid:1) = d X j =0 (( j + 2)( j + 1) p j +2 + (∆ p j )) X jN . (2.39)We have ∆ p = 0 if and only if every coefficient of X N in (2.39) is 0, whichis equivalent to p j = − (∆ p j − ) / ( j ( j − j ≥ 2. This is the same as(2.38). (cid:3) Since p and p determine p j for j ≥ p is harmonic,and there is no constraint on p and p other than their degrees (andbeing homogeneous if they are not 0) the mapping H N,d → P N − ,d ⊕ P N − ,d − given by p ( p , p ) (2.40)is an R -module isomorphism. Both P N − ,d and P N − ,d − are free R -modules for all R , so if R is a ring containing Q the R -module H N,d isfree with a basis of cardinalityrank R H N,d = (cid:18) N + d − N − (cid:19) + (cid:18) N + d − N − (cid:19) , (2.41)where for the right side we use the rank formula (2.34) (that formuladoes not need R to be a field). The formula in (2.41) agrees with the onein (2.33) by using a standard binomial coefficient identity twice. Corollary 2.13. When R contains Q , there is a basis of H N,d with co-efficients in Q .Proof. Both P N − ,d and P N − ,d − have monomial bases with coefficient1. By (2.38), if p and p have Q -coefficients then the associated homo-geneous harmonic polynomial (2.37) has Q -coefficients. (cid:3) Example 2.14. Let us look at the case N = 2: from (2.41), when thecoefficient ring R contains Q the R -module H ,d of homogeneous har-monic polynomials in x and y of degree d has a basis of size 2. General polynomials in P N − ,d and P N − ,d − are p = ax d and p = bx d − , respec-tively, where a and b are in R . Using (2.38), the homogeneous harmonicpolynomial in x and y of degree d with such p and p is a X k ≤ d ( − k (cid:18) d k (cid:19) x d − k y k + b X k +1 ≤ d ( − k d (cid:18) d k + 1 (cid:19) x d − k − y k +1 . The first sum is (( x + iy ) d + ( x − iy ) d ) / x + iy ) d ) and the secondsum is (( x + iy ) d − ( x − iy ) d ) / (2 id ) = Im(( x + iy ) d ) /d , which recoversthe classical fact that Re(( x + iy ) d ) and Im(( x + iy ) d ) are a basis of H ,d when R = R . If a polynomial p is in theideal of R [ X , . . . , X N ] generated by X · X − c , for some c ∈ R , then wethink of p as being ‘equal to zero’ on the ‘sphere’ given by the equation X · X = c . Similarly, we say that p is ‘constant’, equal to c ′ ∈ R , on thesphere given by X · X − c if p − c ′ is zero on this sphere.We will apply the harmonic polynomial decomposition in Proposition2.10 to determine polynomials that are ‘constant on a sphere’ using thefollowing two lemmas. Lemma 2.15. For a commutative ring R , p ∈ R [ X , . . . , X N ] and a ∈ R that is not a zero divisor, suppose ap is divisible by X · X − c , where c ∈ R . Then p is divisible by X · X − c .Proof. Since X · X − c is monic as a polynomial in X N we can write p = ( X · X − c ) q + r where q, r ∈ R [ X , . . . , X N − ][ X N ] with r = 0 or deg X N ( r ) ≤ X N ( X · X ) = 2. Multiplying both sides by a , ap = ( X · X − c ) aq + ar, so from ap being divisible by X · X − c we get ( X · X − c ) | ( ar ) in R [ X , . . . , X N ]. Write ar = ( X · X − c ) g . Since X · X − c is monic in X N , if r = 0 then deg X N ( ar ) = 2 + deg X N ( g ) ≥ 2, but deg X N ( ar ) =deg X N ( r ) ≤ a is not a zero divisor. This is a contradiction, so r = 0 and thus ( X · X − c ) | p . (cid:3) Lemma 2.16. Let R be a commutative ring, and c , . . . , c m for m ≥ bedistinct elements of R such that no difference c i − c j for i = j is a zero-divisor. If p ∈ R [ X , . . . , X N ] is divisible by X · X − c j for j = 1 , . . . , m then p is divisible by Q mj =1 ( X · X − c j ) . OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 21 Proof. First we will show that if X · X − a divides ( X · X − b ) g for a, b ∈ R and a − b is not a zero divisor in R then X · X − a divides g .For some polynomial h we have ( X · X − a ) h = ( X · X − b ) g . Then bg = ( X · X )( g − h ) + ah, (2.42)so ( b − a ) g = ( X · X − a )( g − h ) . (2.43)By Lemma 2.15, X · X − a divides g .Now assume p is divisible by X · X − c j for j = 1 , . . . , m with c i − c j notbeing a zero divisor for all i = j . Inductively we can suppose p is divisibleby ( X · X − c ) · · · ( X · X − c m − ), say p = ( X · X − c ) · · · ( X · X − c m − ) q in R [ X , . . . , X N ]. We will show X · X − c m divides q and then we’ll bedone. Since X · X − c m | ( X · X − c )( X · X − c ) · · · ( X · X − c m − ) q ,we get X · X − c m | ( X · X − c ) · · · ( X · X − c m − ) q since c m − c is nota zero divisor. Then X · X − c m | ( X · X − c ) · · · ( X · X − c m − ) q since c m − c is not a zero divisor. Eventually we are left with ( X · X − c m ) | q and we are done. (cid:3) Proposition 2.17. Suppose that R is a ring containing Q , and p ∈ R [ X , . . . , X N ] , where N ≥ . (i) If, for some c ∈ R , all M jk p are divisible by X · X − c then p isequal to a constant modulo X · X − c . (ii) If p is nonzero of degree d and for an m ≥ d/ there are c , . . . , c m in R whose pairwise differences are not zero divisors with each M jk p divisible by all X · X − c i then p is a polynomial in X · X . (iii) If p is nonzero and homogeneous of degree d and there is c ∈ R that is not a zero divisor such that all M jk p are divisible by X · X − c then d is even and p = a ( X · X ) d/ where a ∈ R . If p is equal to a ‘constant’ mod X · X − c (that is, for some c ′ ∈ R , p − c ′ belongs to the ideal generated by X · X − c ) then every M jk p is0 mod X · X − c . The proposition provides statements in the conversedirection to this observation. Note that in (iii) the hypothesis is that,for p homogeneous, M jk p are divisible by X · X − c and the conclusionimplies that M jk p are in fact all 0. Proof. By Proposition 2.10 we can write p = p + ( X · X ) p + · · · + ( X · X ) s p s , (2.44) with p , . . . , p s all harmonic polynomials. By (2.12), M jk (( X · X ) m p m ) =( X · X ) m M jk ( p m ) for all m ≥ 0, so M jk p = M jk p + ( X · X ) M jk p + · · · + ( X · X ) s M jk p s . Now let p ∗ = p + cp + · · · + c s p s . (2.45)This is harmonic (since each p k is) and equal to p mod ( X · X − c ).(i) Since M jk ( X · X − c ) = 0, from p ≡ p ∗ mod X · X − c we get M jk p ∗ ≡ M jk p mod ( X · X − c ), and so, by the hypothesis in (i), M jk p ∗ ≡ X · X − c ). Since p ∗ is harmonic and M jk commutes with ∆, M jk p ∗ is harmonic. Thus by Proposition 2.8, M jk p ∗ = 0. This being truefor all distinct j, k ∈ { , . . . , N } , it follows by Proposition 2.3 that p ∗ isa polynomial in X · X with coefficients in R . Therefore p ∗ − p ∗ (0) is amultiple of X · X . Since p ∗ is harmonic so is p ∗ − p ∗ (0), so p ∗ − p ∗ (0) = 0by Proposition 2.7: p ∗ is a constant. Thus p is equal to a constant modulo X · X − c .(ii) First we give a proof when m > d/ 2. By Lemma 2.16, each M jk p is divisible by Q mi =1 ( X · X − c i ), so if M jk ( p ) = 0 then deg( M jk p ) ≥ m by (2.25). Since p has degree d , M jk ( p ) has degree at most d , so 2 m ≤ d .Thus if m > d/ M jk ( p ) = 0, so p is a polynomial in X · X by Proposition 2.3.The argument above does not apply if m = d/ 2, but the followingargument does (and also applies to the case m > d/ p = c + ( X · X − c ) q, (2.46)where c ∈ R and q ∈ R [ X , . . . , X N ]. Then deg q = d − q = 0. If d = 2 then q is constant and (2.46) shows p is a polynomialin X · X . Suppose that d > m = d/ 2. Since M jk ( X · X ) = 0,applying M jk to both sides of (2.46) gives us M jk p = ( X · X − c ) M jk q. As seen in the proof of Lemma 2.16 (the first paragraph in the proof), M jk q is a multiple of X · X − c i for i = 2 , . . . , m ; note that m − d − / 2. Inductively, we conclude that q is a polynomial in X · X , so p is as well.(iii) When p is homogeneous of degree d , each p j in (2.44) is 0 or ishomogeneous of degree d − j by Proposition 2.10. It follows then thatfor each j , the term c j p j is the homogeneous part of p ∗ of degree d − j .By the proof of (i), p ∗ in (2.45) is constant; hence each c j p j is 0 unless d − j = 0. Since c is not a zero divisor, if c j p j = 0 then p j = 0. Since p OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 23 is not 0, we get p = p j ( X · X ) j , with d − j = 0. Therefore d = 2 j iseven and p = a ( X · X ) d/ where a = p j is in R . (cid:3) Next we note a consequence of Proposition 2.17. Proposition 2.18. Let R be a ring containing Q and N ≥ . Supposethat p ∈ R [ X , . . . , X N ] is such that M p is in the ideal generated by X + X − c for some c ∈ R . Then p = a + ( X + X − c ) q where a ∈ R and q ∈ R [ X , . . . , X N ] .Proof. Since M p and M p = − M p are divisible by X + X − c ,by Proposition 2.17(i) with R replaced by R [ X , . . . , X N ] we have p ≡ a mod ( X + X − c ) for some a ∈ R . Therefore p = a + ( X + X − c ) q where q is in R [ X , . . . , X N ][ X , X ] = R [ X , . . . , X N ]. (cid:3) We turn now to showing(Proposition 2.20) that, under some conditions on the ring R , the idealof polynomials in R [ X , . . . , X N ] that vanish on the ‘sphere of radius’ a in R N is the ideal generated by X · X − a .For a ring R , any positive integer k and any a ∈ R we define the k -dimensional open ball B k ( a ) to be the set of all ( c , . . . , c k ) ∈ R k suchthat c + · · · + c k + y = a , for some nonzero y ∈ R . (This is motivated by the geometry of the case R = R .)For the following results we will impose a certain property on R thatensures that open balls B N ( a ) contain infinitely many points on each‘slice’ specified by fixing the first few coordinates. This includes assumingthat B ( a ) itself contains infinitely many points. The condition on R ,with any nonzero a ∈ R , holds if R is the field of algebraic numbers, thereals R , or the complexes C . Lemma 2.19. Let R be an integral domain and suppose that a ∈ R issuch that the open ball B ( a ) is infinite and, for every integer k ≥ andevery ( c , . . . , c k ) ∈ B k ( a ) , there are infinitely many c k +1 ∈ R for which ( c , . . . , c k +1 ) ∈ B k +1 ( a ) . If p ∈ R [ X , . . . , X N ] is zero on B N ( a ) then p = 0 .Proof. The result holds for N = 1 because a nonzero polynomial in R [ X ] cannot have infinitely many zeros in R : if p is 0 at distinct points α , . . . , α m then, using the fact that R is an integral domain and repeat-edly applying the division algorithm, we have p = ( X − α ) . . . ( X − α m ) q for some polynomial q ∈ R [ X ] and so m can be at most the degree of p .Now suppose N > 1. Write p as a polynomial in X N with coefficientsin R [ X , . . . , X N − ]: p = m X k =0 p k X kN . For each ( c , . . . , c N − ) ∈ B N − ( a ) the polynomial p ( c , . . . , c N − , X N )has infinitely many zeros, one for every point c N for which ( c , . . . , c N ) ∈ B N ( a ). Hence each coefficient p k is zero at each point of B N − ( a ). Thenby induction each polynomial p k is 0, so p is 0. (cid:3) Proposition 2.20. Let R be an integral domain and suppose that a ∈ R is nonzero and such that the open ball B ( a ) is infinite and, for everyinteger k ≥ and every ( c , . . . , c k ) ∈ B k ( a ) , there are infinitely many c k +1 ∈ R for which ( c , . . . , c k +1 ) ∈ B k +1 ( a ) . Suppose also that R doesnot have characteristic . Let p ( X , . . . , X N ) ∈ R [ X , . . . , X N ] be suchthat p ( c , . . . , c N ) = 0 whenever c + · · · + c N = a . Then p ( X , . . . , X N ) = ( X · X − a ) q, (2.47) for some q ∈ R [ X , . . . , X N ] .Proof. For N = 1, the hypotheses imply that p ( X ) is divisible by X − a and X + a , which are distinct because 2 a = 0. Hence (2.47) holds.Suppose N > 1. By the division algorithm by monic polynomials inthe ring of polynomials in X N with coefficients in R [ X , . . . , X N − ] wehave p ( X , . . . , X N ) = (cid:0) X N + X + · · · + X N − − a (cid:1) q + r ( X , . . . , X N − ) X N + r ( X , . . . , X N − ) (2.48)for some polynomials q ∈ R [ X , . . . , X N ], and r , r ∈ R [ X , . . . , X N − ].We will show that r and r are zero by using Lemma 2.19.Let ( c , . . . , c N − ) ∈ B N − ( a ). Then there is a non-zero t ∈ R suchthat c + · · · + c N − − a = − t . Evaluating (2.48) at ( c , . . . , c N − , t ), and noting that the left side is 0by the assumption on p , we have0 = r ( c , . . . , c N − ) t + r ( c , . . . , c N − ) . OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 25 This holds for a nonzero value of t and also for − t = t , since R does nothave characteristic 2. It follows that both r and r are 0 when evaluatedat ( c , . . . , c N − ). Then by Lemma 2.19, the polynomials r and r areboth 0. (cid:3) 3. Simultaneous eigenvectors for commuting generators In this section we study the common eigenvectors in R [ X , . . . , X N ] ofthe commuting operators M , M , . . . , M n − , n , where n is the largestinteger for which 2 n ≤ N .Most, but not all, results in this section will use the hypothesis thatin R , no positive integer multiples of 1 R are zero divisors. Some resultswill also use the assumption that i = √− ∈ R . We will repeat theseassumptions on R in the statements of the results.The following result gives the eigenvectors of each operator M jk . Proposition 3.1. Let R be a ring containing i = √− . Then anypolynomial p ∈ R [ X , X ] of the form p = ( X + iε X ) a q ( X + X ) , (3.1) where a is a non-negative integer, ε ∈ {± } , and q ( T ) ∈ R [ T ] , satisfies M p = iε a p. (3.2) Conversely, suppose that R is an integral domain with i = √− ∈ R and is invertible in R . Then any eigenvector p of M is of the form (3 . .Proof. Straightforward computation shows that any polynomial p of theform (3.1) satisfies the eigenvector equation (3.2) with λ = iε a .Now suppose R is an integral domain containing i in which 2 is in-vertible. Then every p ∈ R [ X , X ] can be expressed uniquely in theform p = X b,c k bc ( X + iX ) b ( X − iX ) c , (3.3)where b, c run over all non-negative integers and k bc ∈ R . By straightfor-ward computation, M p = X b,c i ( b − c ) k bc ( X + iX ) b ( X − iX ) c . Now suppose p is an eigenvector: M p = λp. Since R is an integral domain, we have λ = i ( b − c )whenever k bc = 0. We focus now only on those b, c for which k bc = 0.The value a = b − c is the same for all such pairs ( b, c ). If a ≥ X + iX ) b ( X − iX ) c = ( X + iX ) a ( X + X ) c , whereas if a ≤ X + iX ) b ( X − iX ) c = ( X − iX ) − a ( X + X ) b . Thus (3.3) is a sum of terms of the form k ( X + iε X ) a ( X + X ) d ,where a = | a | ≥ ε is the sign of a (if a = 0 we can just set ε = +1for definiteness), and d ≥ 0. The eigenvalue for p is λ = ia = iε a . (cid:3) Now we can readily obtain eigenvectors that are common to the oper-ators M , M , . . . , M n − , n , which are also eigenvectors of M · M . Proposition 3.2. Let R be a ring containing i = √− and let N be aninteger ≥ . Let n be the largest integer ≤ N/ . For ε , . . . , ε n ∈ {± } and a , . . . , a n non-negative integers, set Y ε,a = ( X + iε X ) a . . . ( X n − + iε n X n ) a n , (3.4) where we use ε to denote ( ε , . . . , ε n ) , and a = ( a , . . . , a n ) . Then M j − , j ( Y ε,a ) = iε j a j Y ε,a , (3.5) for j ∈ { , . . . , n } .Conversely, suppose R is an integral domain containing i and isinvertible in R . Then every Y ∈ P N that is an eigenvector of M , M ,..., M n − n is of the form Y = Y ε,a q ( X + X , . . . , X n − + X n ) (3.6) for some ε and a as above, and some q ∈ R ∗ [ T , . . . , T n ] , where R ∗ = R if n = N and R ∗ = R [ X n +1 ] if n + 1 = N . In particular, Y is an R -multiple of Y ε,a if Y is homogeneous of degree | a | = a + · · · + a n .Moreover, Y ε,a ∈ ker ∆ N, | a | , and Y ε,a is an eigenvector of M · M . OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 27 Proof. The identity (3.5) is readily verified by computation. For theconverse we apply Proposition 3.1 with R [ X , . . . , X N ] in place of R anduse induction. This leads to Y = ( X + iε X ) a . . . ( X n − + iε n X n ) a n q, (3.7)where q is a polynomial in X + X , . . . , X n − + X n , with coefficientsin R ∗ . If | a | = d then q must be a constant, an element of R .It is readily verified that Y ε,a is harmonic. Since it is also homogeneous,it follows by Proposition 2.1 that it is an eigenvector of M · M as well. (cid:3) In Proposition 3.2 we did not determine any specific form of the polyno-mial q . Next we obtain a precise description of all harmonic polynomialsthat are common eigenvectors of the operators M , . . . , M n − , n , where n = [ N/ Proposition 3.3. Suppose R is an integral domain containing i and is invertible in R . Let N = 2 n be a positive even integer.Let p ∈ R [ X , . . . , X N ] be a harmonic polynomial, homogeneous ofdegree d , and suppose p is an eigenvector for the operators M , M ,. . . , M n − n . Then there exist non-negative integers a , . . . , a n , and ε , . . . , ε n ∈ {± } such that M p = iε a p, M p = iε a p, . . . , M n − n p = iε n a n p, (3.8) and p = Y ε,a q ( X + X , . . . , X n − + X n ) , (3.9) where q ( T , . . . , T n ) ∈ R [ T , . . . , T n ] satisfies n X j =1 ( T j ∂ j + ( a j + 1) ∂ j ) q ( T , . . . , T n ) = 0 , (3.10) where ∂ j = ∂ T j .Conversely, if p ∈ R [ X , . . . , X N ] is homogeneous of degree d and sat-isfies (3 . for some non-negative integers a , . . . , a n , some ε , . . . , ε n ∈{± } , and some q ∈ R [ T , . . . , T n ] satisfying (3 . then p ∈ ker ∆ N,d and p satisfies the eigenvalue relations (3 . .Proof. By lengthy but straightforward computation, using the productformula (2.22), we have∆ p = 4 Y ε,a Lq ( X + X , . . . , X n − + X n ) , (3.11) if p is of the form (3.9), and L = n X j =1 ( T j ∂ j + ( a j + 1) ∂ j ) . (3.12)Thus p is harmonic if and only if q ∈ ker L . (cid:3) In the special case N = 2, we can work out the condition for q to bein ker L , and it shows that q is of degree 0. Thus the form of p given in(3.9) reduces to just p = q ( X ± iX ) a for N = 2, q ∈ R , and a ∈ { , , , . . . } .Let us assume for simplicity that R is a field and contains Q and i .From (2.41) the dimension of H ,d over R is 2, and so the two elements( X ± iX ) d form a basis of H ,d , as we already saw in Example 2.14.The following result is along the lines of Proposition 2.12. We use thenotation R [ T , . . . , T n ] d for the R -module of all homogeneous polynomialsin T , . . . , T n of degree d , with coefficients in R . For d = − { } . Proposition 3.4. Suppose R is a ring containing Q , and n , d , and a , . . . , a n are non-negative integers. Let L d be the operator L d = n X j =1 ( T j ∂ j + ( a j + 1) ∂ j ) : R [ T , . . . , T n ] d → R [ T , . . . , T n ] d − , (3.13) where ∂ j = ∂ T j . Then a polynomial q in R [ T , . . . , T n ] d , written as q = q T dn + · · · + q d − T n + q d , is in ker L d if and only if the coefficients q , . . . , q d − ∈ R [ T , . . . , T n − ] are related to q d ∈ R [ T , . . . , T n − ] by q d − k = ( − k k !(1 + a n ) . . . ( k + a n ) L kd q d (3.14) for all integers k ∈ { , , . . . , d } . In particular, the mapping ker L d → R [ T , . . . , T n − ] d : q q d is an R -linear isomorphism. In the case n = 1 , ker L d = 0 if d ≥ and ker L = R . OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 29 Proof. Although L is a second-order differential operator it satisfies theproduct rule of first-order differential operators when applied to polyno-mials that depend on different sets of indeterminates: L ( f g ) = ( Lf ) g + f ( Lg ) if f ∈ R [ T , . . . , T m ] and g ∈ R [ T m +1 , . . . , T n ].Applying L to q , and using the above property, we obtain Lq = [ d ( d + a n ) q + Lq ] T d − n + [( d − d + a n − q + Lq ] T d − n + [( d − d + a n − q + Lq ] T d − n + · · · + [2(2 + a n ) q d − + Lq d − ] T n + (1 + a n ) q d − + Lq d So q ∈ ker L if and only if q d − k = ( − k k !(1 + a n ) . . . ( k + a n ) L k q d for k running up from 1 to d . (cid:3) As we noted in (3.11), if a polynomial p ∈ R [ X , . . . , X N ] is of the form p = q ( X + X , . . . , X n − + X n ) n Y j =1 ( X j − ± iX j ) a j , (3.15)where q is a polynomial in R [ T , . . . , T n ], then∆ p = 4( Lq )( X + X , . . . , X n − + X n ) n Y j =1 ( X j − ± iX j ) a j , (3.16)where on the right we have the evaluation of the polynomial Lq with theindeterminates set to ( X + X , . . . , X n − + X n ).Thus the R -module of all harmonic polynomials p ∈ ker ∆ N,d that areof the form (3.15) is isomorphic to P n, ( d −| a | ) / .Next we have a counterpart of Proposition 3.3 for odd values of N . Proposition 3.5. Suppose R is an integral domain containing i = √− and is invertible in R . Let N = 2 n + 1 be a positive odd integer, with n ≥ .Let p ∈ R [ X , . . . , X N ] be a harmonic polynomial, homogeneous ofdegree d , and suppose p is an eigenvector for the operators M , M ,. . . , M n − n . Then there exist non-negative integers a , . . . , a n , and ε , . . . , ε n ∈ {± } such that M p = iε a p, M p = iε a p, . . . , M n − n p = iε n a n p, (3.17) and p = Y ε,a q ( X + X , . . . , X n − + X n , X N ) , (3.18) where q ( T , . . . , T n , Y ) ∈ R [ T , . . . , T n , Y ] is a homogeneous polynomialof the form q ( T , . . . , T n , Y ) = d −| a | X k =0 q k ( T , . . . , T n ) Y k , (3.19) where q k ( T , . . . , T n ) ∈ R [ T , . . . , T n ] is homogeneous for each k , and ( k + 2)( k + 1) q k +2 + 4 Lq k = 0 , (3.20) for all k ∈ { , , . . . , d − | a | − k } . If d − | a | is even then q is of degree ( d − | a | ) / and q is . If d − | a | is odd then q = 0 and q has degree ( d − | a | − / .Conversely, if p ∈ R [ X , . . . , X N ] is homogeneous of degree d and sat-isfies (3 . and (3 . for some non-negative integers a , . . . , a n , some ε , . . . , ε n ∈ {± } , and some q ∈ R [ T , . . . , T n , Y ] as in (3 . , then p ∈ ker ∆ N,d and p satisfies the eigenvalue relations (3 . .Proof. The argument is similar to the proof of Proposition 3.3 exceptthat in place of (3.11) we use∆ (cid:16) Y ε,a q ( X + X , . . . , X n − + X n , X N ) (cid:17) = Y ε,a d −| a | X k =0 s k X kN , (3.21)where s k = 4 Lq k ( X + X , . . . , X n − + X n ) + ( k + 2)( k + 1) q k +2 . (3.22) (cid:3) 4. Spherical means In this section we prove an algebraic counterpart (Corollary 4.8 below)of the classical analytic result that the mean of a harmonic function overa sphere equals the value of the function at the center of the sphere.We would like to identify the integral of a polynomial over a spherecentered at 0 by using purely algebraic notions. Let λ be such an integral,viewed as a linear functional on the space of polynomials. Applying M jk to a polynomial q is the effect of taking the derivative of a one-parametergroup of rotations of q in the X j - X k -plane. Thus λ ( M jk q ) should be 0,since the spherical integral would remain invariant under rotations. Thus,for any integer N ≥ 2, we define a spherical mean to be an R -linear map λ : R [ X , . . . , X N ] → R OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 31 for which λ ◦ M jk = 0 for all j, k ∈ { , . . . , N } with j = k .For any polynomials p and q , M jk ( pq ) = pM jk ( q ) + qM jk ( p ), so applying λ to M jk ( pq ) shows that λ ( pM jk q ) = − λ ( qM jk p ) . For N = 1, when there are no nonzero operators M jk , we requireinstead that λ vanish on X n for all odd integers n ≥ { x, − x } . λ . Our next goal is to show that there existsa unique spherical mean λ on R [ X , . . . , X N ] for which λ (cid:0) ( X · X ) n (cid:1) =1 for all integers n ≥ 0, and, moreover, that all spherical means on R [ X , . . . , X N ] are R -multiples of λ when restricted to homogeneouspolynomials of any fixed degree.We will use the double factorial on nonnegative integers: b !! = b ( b − · · · · b is odd and ≥ b ( b − · · · · b is even and ≥ b = 0. (4.1)It will also be convenient to set ( − n !! = ( n + 2)!! / ( n + 2). Proposition 4.1. Let M be an R -module, where the ring R contains Q ,and let N be an integer ≥ . For an R -linear map λ : R [ X , . . . , X N ] → M (4.2) the following two properties are equivalent: • for all j, k ∈ { , . . . , N } with j = k , λ ◦ M jk = 0 , (4.3) • for each even d ≥ there is v d ∈ M such that for all monomials X a · · · X a N N , λ ( X a · · · X a N N ) = ( ( a − · · · ( a N − v a + ··· + a N , if all a j are even , , if some a j is odd . Moreover, there exists a unique spherical mean λ on R [ X , . . . , X N ] whose value on ( X · X ) n is for all integers n ≥ . Let us note that the last paragraph above holds even for N = 1: weset λ ( X k ) equal to 0 for odd k and equal to 1 for even k . We alsonote that the value λ ( X a · · · X a N N ) is invariant under permutations ofthe indeterminates X , . . . , X N , and so, using λ ( X · X ) = 1, it followsthat λ ( X j ) = 1 N , (4.4)for all j ∈ { , . . . , N } . Proof of Proposition 4.1. First we assume (4.3) holds and derive the for-mula for λ on monomials. Let X and Y be two distinct indeterminatesamong X , . . . , X N , and let q be a polynomial in the other indetermi-nates. Then for any integers a ≥ b ≥ 0, we have( X∂ Y − Y ∂ X ) (cid:0) X a − Y b +1 q (cid:1) = ( b + 1) X a Y b q − ( a − X a − Y b +2 q, (4.5)where the second term on the right is defined to be 0 if a = 1. Applying λ to both sides, we get by (4.3)0 = ( b + 1) λ (cid:0) X a Y b q (cid:1) − ( a − λ (cid:0) X a − Y b +2 q (cid:1) . (4.6)Taking a = 1 we see that λ ( Xr ) = 0 for all polynomials r in the in-determinates other than X . Moreover, if a is an odd positive integer,repeated application of (4.6) reduces λ ( X a Y b q ) to a rational multiple of λ ( XY b + a − q ), which is 0. Conversely, if λ ( X a Y b q ) = 0 whenever a isodd and q does not involve X or Y then (4.6) holds for odd a ≥ λ ( X a · · · X a N N ) = 0 if some a j is odd.Now suppose a and b are both even positive integers. Dividing bothsides of (4.6) by the product of double factorials ( a − b + 1)!!, weobtain 1( a − b − λ (cid:0) X a Y b q (cid:1) = 1( a − b + 1)!! λ (cid:0) X a − Y b +2 q (cid:1) . (4.7)It follows that λ has a common value on all monomials of the form1( a − b − X a Y b q, where a and b are even non-negative integers with a fixed value for thesum a + b . Of course, this applies to X and Y being any two of theindeterminates X , . . . , X N . It follows that λ has a common value on allmonomials of the form 1( a − · · · ( a N − X a · · · X a N N (4.8) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 33 for all even integers a , . . . , a N ≥ | a | .Conversely, if λ has a common value on all monomials of the form (4.8),where the a j are all even, then (4.7) and hence (4.6) hold for X and Y running over distinct indeterminates among X , . . . , X N . Then by (4.5), λ ( M XY p ) is 0 for all polynomials p and hence λ ◦ M jk = 0 for all j, k .Thus an M -valued spherical mean λ is obtained by setting the value of λ to be 0 on monomials having an odd exponent, and to have a commonvalue v d ∈ M on the monomials in (4.8) having all non-negative evenexponents and being of total degree d .Focusing now on the case M = R , let λ be an R -valued sphericalmean on R [ X , . . . , X N ] and set s n,N ∈ R to be its common value on themonomials in (4.8) of total degree 2 n , where n ≥ 0. Then λ ( X a · · · X a N N ) = ( a − · · · ( a N − s n,N , (4.9)where the exponents a , . . . , a N are all even with a + · · · + a N = 2 n .Then expanding ( X · X ) n with the multinomial theorem and applying λ , λ (cid:0) ( X · X ) n (cid:1) = X b ,...,b N ≥ b + ··· + b N = n n ! b ! . . . b N ! (2 b − . . . (2 b N − s n,N = s n,N X b ,...,b N ≥ b + ··· + b N = n n ! b ! . . . b N ! (2 b − . . . (2 b N − . The summations are really over ordered N -tuples ( b , . . . , b N ). Since thesummation is positive, for each n ≥ s n,N that makes λ (( X · X ) n ) = 1. This determines a unique R -valued spherical mean λ on R [ X , . . . , X N ], by (4.9), and that λ is whatwe take as λ . (cid:3) Let us note, using (4.9), that for even 2 b , . . . , b N ≥ λ ( X b · · · X b N N ) = (2 b − . . . (2 b N − s n,N , (4.10)where b + · · · + b N = n ; the multiplier s n,N , which is chosen to make λ (cid:0) ( X · X ) n (cid:1) = 1, is given below in (4.12). Remark 4.2. We observe, as a consequence of Proposition 4.1, that thevalue of a spherical mean on X a · · · X a N N , with each a i even, is a multipleof ( a − · · · ( a N − a + · · · + a N . Thus, a mapping λ : R [ X , . . . , X N ] → R is a spherical meanif and only if it is an R -multiple of λ when restricted to homogeneous polynomials of fixed degree. For this reason we can, without loss of muchgenerality, restrict our attention to λ rather than on all spherical means.For a polynomial p with real coefficients, viewed as a function on R N ,the value of λ ( p ) is the normalized integral of p over the unit spherein R N with center at the origin. The following result reflects the factthat multiplying by a factor of X · X , which would correspond to justmultiplying by 1 over the geometric sphere in R N , does not affect thevalue of λ on any polynomial. Proposition 4.3. For a commutative ring R containing Q , λ (cid:0) ( X · X ) p (cid:1) = λ ( p ) (4.11) for all p ∈ R [ X , . . . , X N ] .Proof. The case N = 1 follows by direct verification. We assume thenthat N ≥ 2. We will show the R -linear map µ : R [ X , . . . , X N ] → R givenby µ ( p ) = λ (( X · X ) p ) has the properties that uniquely characterize p λ ( p ).For each M jk , µ ( M jk p ) = λ (( X · X ) M jk ( p )) = λ ( M jk (( X · X ) p )) by(2.12). Thus µ ( M jk p ) = ( λ ◦ M jk )(( X · X ) p ) = 0, so µ is a sphericalmean.We have µ (( X · X ) n ) = 1 for all n ≥ µ (( X · X ) n ) = λ (( X · X ) n +1 ) = 1 . Hence µ = λ . (cid:3) As an application of Proposition 4.3 we next give a formula for s n,N when n ≥ N -tuples at the end of the proof of Proposition 4.1. Proposition 4.4. For each n ≥ , s n,N = ( N − N + 2 n − N + 2 n − N + 2 n − · · · N , (4.12) where successive terms in the denominator drop by .Proof. Use Proposition 4.3 on p = X a · · · X a N N where a , . . . , a N are non-negative even integers with sum 2 n . Since ( X · X ) p is homogeneous of OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 35 degree 2 n + 2, writing X · X as P ni =1 X i and using linearity of λ implies λ (( X · X ) p ) = (( a + 1)!!( a − · · · ( a N − · · · + ( a − · · · ( a N − − a N + 1)!!) s n +2 ,N = ( a + 1 + · · · + a N + 1) λ ( p ) s n,N s n +2 ,N = ( N + 2 n ) λ ( p ) s n +2 ,N s n,N . (4.13)Division by s n,N is okay since it is nonzero by the formula for it at the endof the proof of Proposition 4.1. Multiplying both sides by n !( a / ··· ( a N / and summing over all ordered N -tuples of nonnegative even integers a , . . . , a N with a + · · · + a N = 2 n , we obtain λ (cid:0) ( X · X ) n +1 (cid:1) = ( N + 2 n ) λ (cid:0) ( X · X ) n (cid:1) s n +2 ,N s n,N . (4.14)Thus s n +2 ,N = 1 N + 2 n s n,N , (4.15)which together with the value s ,N = 1 shows for n ≥ s n,N = 1( N + 2 n − N + 2 n − · · · N , (4.16)where successive terms in the denominator drop by 2 each time (andthe empty product equals 1). To get a formula for s n,N that is moretransparently valid at n = 0, we multiply the right side of (4.16) by( N − / ( N − (cid:3) Recall from the end of the proof of Proposition 4.1 that the condition λ (( X · X ) n ) = 1 says X b ,...,b N ≥ b + ··· + b N = n n ! b ! · · · b N ! (2 b − · · · (2 b N − s n,N = 1 . (4.17) Combining this with (4.16) and the formula (cid:0) bb (cid:1) = b (2 b − b ! for b ≥ 0, weget X b ,...,b N ≥ b + ··· + b N = n (cid:18) b b (cid:19) · · · (cid:18) b N b N (cid:19) = 2 n n ! ( N + 2 n − N + 2 n − · · · N. (4.18)For example, when n = 2 the left side of (4.18) is N (cid:0) (cid:1) + (cid:0) N (cid:1)(cid:0) (cid:1)(cid:0) (cid:1) =2 N + 4 N and the right side (2 / N + 2) N = 2( N + 2) N = 2 N + 4 N .The identity (4.18) can be derived in a second way using generatingfunctions: the generating function of the left side is ( P b ≥ (cid:0) bb (cid:1) X b ) N ,and the generating function of the central binomial coefficients (cid:0) bb (cid:1) is(1 − X ) − / , whose N th power has coefficients given by the right side of(4.18) using the binomial theorem with exponent − N/ 2. A probabilisticinterpretation of (4.18) is in [4]. When N = 2, (4.18) is the identity P nb =0 (cid:0) bb (cid:1)(cid:0) n − b ) n − b (cid:1) = 4 n , which has a combinatorial interpretation [14].The identity (4.18) is valid at N = 1 if the right side is written as n ( N +2 n − n !( N − .We conclude this discussion with a formula relating λ ( X q ) to λ ( ∂ q ). Proposition 4.5. Suppose R is a commutative ring containing Q , and N ≥ an integer. Then for any p ∈ R [ X , . . . , X N ] , which is homoge-neous of degree d , we have λ ( X p ) = 1 N + d − λ ( ∂ p ) . (4.19) Proof. We may assume first that p is a monomial of degree d . If any X j ,other than X , appears with odd degree in p then both sides of (4.19)are 0, whereas if X appears with even degree in p then again both sidesof (4.19) are 0. Hence we assume that p = X b − X b · · · X b N N , where b , . . . , b N are integers ≥ , and b ≥ 1. Then λ ( X p ) = (2 b − · · · (2 b N − s n,N , (4.20)where 2 n = 2 b + · · · + 2 b N = d + 1 , and λ ( ∂ p ) = (2 b − · (2 b − b − · · · (2 b N − s n − ,N . (4.21) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 37 Thus λ ( X p ) = s n,N s n − ,N λ ( ∂ p ) = 1 N + 2 n − λ ( ∂ p ) (4.22)where we used (4.12). The multiplier on the right here is 1 / ( N + d − (cid:3) Using Proposition 4.3 wecan now describe the normalized spherical mean on homogeneous poly-nomials in terms of iterations of the Laplacian. Proposition 4.6. Let R be a commutative ring containing Q . The nor-malized spherical mean λ on R [ X , . . . , X N ] can be described on homo-geneous polynomials p as follows. If p has odd degree then λ ( p ) = 0 .For a constant r , λ ( r ) = r . If p has even degree n ≥ then λ ( p ) = 1 n !2 n (2 n + N − n + N − · · · (2 n + N − n ) ∆ n p. (4.23)On the right side of (4.23), ∆ n p is in R since it has degree 0.By Remark 4.2, formula (4.23) holds for a general spherical mean λ if a scaling term, depending only on the degree of p , is inserted on theright hand side. Proof. If p has odd degree then every monomial in p has an odd exponent,so λ ( p ) = 0. If p = r is constant in R , then λ ( r ) = rλ (1) = r .If p has even degree d ≥ 2, then apply λ to the identity (2.1) on p .We get λ (cid:0) ( X · X )∆ p (cid:1) = d ( d + N − λ ( p ) (4.24)since λ ◦ M jk = 0 for all M jk . Therefore by Proposition 4.3, λ ( p ) = 1 d ( d + N − λ (∆ p ) . (4.25)Writing d = 2 n ≥ 2, apply (4.25) n times to obtain (4.23). (cid:3) Example 4.7. For N = 4 and p = X X = X X X X , so n = 10 / λ ( X X ) = 15!2 (12 · · · · 4) ∆ ( X X )The value of ∆ ( X X ) is (cid:0) (cid:1) · λ ( X X ) = 10 · (12 · · · · 4) = 12 . This agrees with the value computed using (4.9) and the formula for s n,N in Proposition 4.4: λ ( X X ) = 3!!5!! s , = 3!!5!! 112 · · · · . (4.26) Inclassical analysis it is known that the average value of a harmonic func-tion over a sphere is the value of the function at the center of the sphere.In this subsection we establish purely algebraic results concerning thismean value property and harmonic polynomials. The first result is adirect corollary of Proposition 4.6. Corollary 4.8. If p is a harmonic polynomial in R [ X , . . . , X N ] then thenormalized spherical mean of p is the value of p at : λ ( p ) = p (0) . (4.27) Proof. Both sides of (4.27) are linear in p , so we can assume p is homo-geneous. If p is constant then (4.27) is true by linearity of λ and thecondition λ (1) = 1. If p is not constant, then by (4.25) we get λ ( p ) = 0(the derivation of (4.25) works for all nonconstant homogeneous p ). (cid:3) For t = ( t , . . . , t N ), any N -tuple of indeterminates such that( t, X ) = ( t , . . . , t N , X , . . . , X N )is a (2 N )-tuple of algebraically independent indeterminates over R , wehave the normalized spherical mean λ : R [ t, X , . . . , X N ] → R [ t ] , (4.28)simply by using the ring R [ t ] = R [ t , . . . , t N ] in place of R . Now considera harmonic polynomial p ( Y , . . . , Y N ) ∈ R [ Y , . . . , Y N ]; then the polyno-mial p ( X + t , . . . , X N + t N ) ∈ R [ t ][ X , . . . , X N ]is harmonic as a polynomial in X , . . . , X N with coefficients in R [ t ].Hence λ (cid:0) p ( X + t ) (cid:1) = p ( t ) , (4.29)by Corollary 4.8.The following result is the converse to the observation in (4.29). Proposition 4.9. Let R be a commutative ring containing Q , and sup-pose p ( Y ) ∈ R [ Y , . . . , Y N ] , where Y = ( Y , . . . , Y N ) , satisfies the mean-value property: λ (cid:0) p ( X + t ) (cid:1) = p ( t ) (4.30) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 39 where X = ( X , . . . , X N ) and t = ( t , . . . , t N ) are N -tuples of indetermi-nates. Then p ( X ) is harmonic.Proof. Let us first note that the result holds for p of degree ≤ p is of degree 2. Then,writing p ( Y ) = N X m,n =1 a mn Y m Y n + N X m =1 a m Y m + a , we have λ (cid:0) p ( X + t ) (cid:1) = λ N X m,n =1 a mn X m X n + N X m =1 a m X m ! + N X m,n =1 a mn (cid:0) t n λ ( X m ) + t m λ ( X n ) (cid:1) + p ( t )= 1 N N X m =1 a mm + 0 + 0 + p ( t ) , (4.31)and so N X n =1 a nn = 0 , which implies that p ( Y ) is harmonic, because ∆ p ( Y ) = 2 P Nn =1 a nn .We assume then that p has degree > 2. Let s = ( s , . . . , s N ) be another N -tuple of indeterminates. Then from (4.30) we have λ (cid:0) p ( X + s + t ) (cid:1) = p ( s + t ) . (4.32)Expanding both sides in powers of the indeterminates t j , and comparingcoefficients of t j = t j · · · t j N N , where j = ( j , . . . , j N ) ∈ Z N ≥ , we have λ (cid:0) ∂ j p ( X + s ) (cid:1) = ∂ j p ( s ) . (4.33)We can focus just on those j that have | j | = P Nk =1 j k equal to 1 (that is,we focus on the first partial derivatives of p ). Then ∂ j p ( Y ) is a polynomialof degree 1 less than that of p , and so the condition (4.33) implies, bythe induction hypothesis, that ∂ j p ( Y ) is harmonic. Hence ∂ j (cid:0) ∆ p ( Y ) (cid:1) = ∆ ∂ j p ( Y ) = 0 . (4.34)Thus ∆ p ( Y ) is constant, an element of R . Writing p ( Y ) as p ( Y ) = p + p ( Y ) + · · · + p d ( Y ) , where each p k ( Y ) is homogeneous of degree k , we observe that∆ p ( Y ) = ∆ p ( Y ) + d X k =3 ∆ p k ( Y ) , where ∆ p k ( Y ) is homogeneous of degree k − 2. Since ∆ p ( Y ) is constant,an element of R , it then follows that p k ( Y ) is harmonic for k ≥ q ( Y ) = p + p ( Y ) + p ( Y ) = p ( Y ) − Σ dk =3 p k ( Y ) , (4.35)satisfies the mean-value property (4.30) because it is the difference of twopolynomials that satisfy (4.30). Since it is of degree ≤ q ( Y ) must thenbe harmonic. Hence p ( Y ), being the sum of q ( Y ) and Σ dk =3 p k ( Y ) is alsoharmonic. (cid:3) Let Matr N ( R ) be the R -algebraof all N × N matrices with entries in the commutative ring R . For an N -tuple of indeterminates X = ( X , . . . , X N ) and any A ∈ Matr N ( R ) wedenote by XA the N -tuple whose k -th component is( XA ) k = N X k =1 X m A mk . (4.36)This determines a natural action of Matr N ( R ) on R [ X , . . . , X N ]: A [ p ( X )] = p ( XA ) , (4.37)for all A ∈ Matr N ( R ) and p ( X ) ∈ R [ X , . . . , X N ]. We note that this isindeed a left action:( AB )[ p ( X )] = p ( XAB ) = A [ Bp ( X )] , (4.38)for any A, B ∈ Matr N ( R ). Of interest to us are the orthogonal matrices:O N ( R ) = { A ∈ Matr N ( R ) : A ⊤ A = I } . (4.39)It is a fact that in any commutative ring R , the condition A ⊤ A = I implies that A is invertible and hence A ⊤ = A − is both a left and aright inverse for A . OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 41 We work, as before,with a commutative ring R that contains Q . A spherical mean λ on R [ X , . . . , X N ] is an algebraic counterpart to integration over the unit( N − R N with respect to the uniformmeasure. Since that measure on the ( N − R N around the origin, integration over that sphere is invari-ant under an orthogonal change of variables. We want to demonstratean algebraic analogue: λ ( p ( XA )) = λ ( p ( X )) for all p in R [ X , . . . , X N ]and A in O N ( R ) = { A ∈ Matr N ( R ) : A ⊤ A = I N } . This is Proposition4.12 below and will be proved in two ways. The first proof is based onthe commutativity of orthogonal matrices and the Laplacian, and thesecond proof is based on a formula for the spherical mean of a product ofhomogeneous linear polynomials in terms of spherical means of productof two homogeneous linear polynomials at a time. Lemma 4.10. For A ∈ O N ( R ) , ∆ m ( p ( XA )) = (∆ m p )( XA ) for all p in R [ X , . . . , X N ] and m ≥ .Proof. The case m = 0 is obvious, and after we establish m = 1 the cases m ≥ m = 1 when R = R is a special case of the classical fact[2, p. 3] that the Laplacian on R N commutes with an orthogonal changeof variables. The proof carries over to R [ X , . . . , X N ]. By the chain rule, ∂ k ( p ( XA )) = N X i =1 ( ∂ i p )( XA ) a ki (4.40) for all p , where A = ( a ij ). Therefore∆( p ( XA )) = N X k =1 ∂ k ( ∂ k ( p ( XA )))= N X k =1 ∂ k N X i =1 a ki ( ∂ i p )( XA ) ! by (4 . X i,k a ki N X j =1 a kj ( ∂ j ( ∂ i p ))( XA ) by (4 . X i,j X k a ki a kj ! ( ∂ j ( ∂ i p ))( XA )= X i ( ∂ i p )( XA ) since A ∈ O N ( R )= (∆ p )( XA ) . (cid:3) Lemma 4.11. Let p , . . . , p n be polynomials in X , . . . , X N that are ho-mogeneous and linear. Then λ ( p · · · p n ) = N n s n,N X π ∈ P n Y { a,b }∈ π λ ( p a p b ) , (4.41) where P n is the set of partitions of { , . . . , n } into two-element subsets. This result is the spherical counterpart of a feature of the standardGaussian measure on R N (for the relationship with Gaussian measuresee [11, Theorem 2.1]).By Remark 4.2, formula (4.41) holds for a general spherical mean λ if the right hand side is multiplied by a scaling term in R that dependsonly on n . Proof. Since both sides of (4.41) are multilinear in ( p , . . . , p n ), checkingthe identity is reduced to checking it when each p j is one of X , . . . , X N .If some X i appears an odd number of times among the p j ’s then the leftside is 0 by Proposition 4.1 and the right side is 0 since in each productover a partition π , X i appears in some factor λ ( p a p b ) as p a , with p b notbeing X i (then use Proposition 4.1 again).Now suppose 2 b of the p j ’s are X and so on up to 2 b N of the p j ’s being X N . If j = k then λ ( X j X k ) = 0 by Proposition 4.1 and λ ( X j ) = 1 /N OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 43 by (4.4). Thus a product over a partition π on the right side in (4.41)is 0 unless p a = p b for each pair { a, b } ∈ π , and in this case the productassociated to π is 1 /N n . This value is independent of π , so the right sideof (4.41) is s n,N times the number of partitions of { , . . . , n } into pairs { a, b } where p a = p b , with { X j , X j } appearing b j times for j = 1 , . . . , N .A set of size 2 k can be partitioned into pairs in (2 k − s n,N (2 b − · · · (2 b N − , which is λ ( X b · · · X b N N ) = λ ( p · · · p n ) by (4.10). (cid:3) Using each of the previous two lemmas we will prove the rotation-invariance of λ . Proposition 4.12. For a commutative ring R containing Q and every N × N matrix A with entries in the ring R that is orthogonal, in the sensethat A ⊤ A = I N , λ (cid:0) p ( XA ) (cid:1) = λ (cid:0) p ( X ) (cid:1) for all p ∈ R [ X , . . . , X N ] . In view of Remark 4.2, this result holds for any spherical mean λ . Proof. Both sides of the desired formula λ (cid:0) p ( XA ) (cid:1) = λ (cid:0) p ( X ) (cid:1) arelinear in p , so it suffices to verify this formula when p is a monomial.If p is a monomial of odd degree then λ ( p ( X )) = 0 by Proposition4.1. Since p ( XA ) is a linear combination of odd-degree monomials, also λ ( p ( XA )) = 0, so the proposition is established in this case.Now suppose p is a monomial of even degree.Method 1. By Proposition 4.6, when p is homogeneous of even degree2 n , λ ( p ( X )) is a multiple of ∆ n p ∈ R , where the the multiplier is deter-mined by N and n . Since p ( XA ) is homogeneous of the same degree as p ( X ), λ ( p ( XA )) is the same scalar multiple of ∆ n ( p ( XA )), so it sufficesto show ∆ n ( p ( X )) = ∆ n ( p ( XA )). By Lemma 4.10, ∆ n ( p ( XA )) equals(∆ n p )( XA ), which is (∆ n p )( X ) since ∆ n p ∈ R .Method 2. In view of the pair-product formula (4.41) in Lemma 4.11,it suffices to assume that the polynomial p ( X ) is a product of two homo-geneous linear factors. Thus, without loss of generality, we may assumethat p ( X ) = X j X k , for some j, k ∈ { , . . . , N } that are not necessarily distinct. In this case, we have λ (cid:0) ( XA ) j ( XA ) k (cid:1) = N X ℓ,m =1 A ℓj A mk λ ( X ℓ X m )= N X ℓ,m =1 A ℓj A mk δ ℓm N = N X m =1 A mj A mk N = 1 N δ jk = λ ( X j X k ) . (4.42)Thus λ (cid:0) p ( XA ) (cid:1) is indeed equal to λ (cid:0) p ( X ) (cid:1) for all p . (cid:3) 5. Spherical harmonics In this section R is a ring such that positive integer multiples of 1 R are not zero divisors ( e.g. , R ⊃ Q ). We follow some of the ideas in [11].By a spherical harmonic we mean a homogeneous harmonic polynomial.Denote by Z N ( c ) the ideal generated by X · X − c : Z N ( c ) = ideal generated by X · X − c in R [ X , . . . , X N ]. (5.1)Recall P N = R [ X , . . . , X N ] Proposition 5.1. Let R be a commutative ring, c ∈ R not a zero-divisor,and Z N ( c ) the ideal in P N = R [ X , . . . , X N ] generated by X · X − c .Then the quotient map P N → P N / Z N ( c ) is injective when restricted tohomogeneous polynomials of the same degree. The analytic counterpart of this result is that a homogeneous functionon R n of a specified degree is completely determined by its restriction toa sphere of nonzero radius centered at the origin. Proof. The difference of homogeneous polynomials of the same degree isstill homogeneous, so it suffices to prove that 0 is the only homogeneouspolynomial in Z N ( c ).Suppose p ∈ Z N ( c ) is homogeneous of degree m . Then p = ( X · X − c ) q = ( X · X ) q − cq, (5.2) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 45 for some q ∈ R [ X , . . . , X N ]. If q = 0 then, since c is not a zero-divisor,the two terms on the right side are of different degrees, contradicting thehomogeneity of p . Thus q = 0 and hence p = 0. (cid:3) L c operators. The following re-sult and its proof are from [11]. Proposition 5.2. The following commutation relations hold among op-erators on R [ X , . . . , X N ] : M X · X M jk = M jk M X · X , (5.3) where M X · X is the operator that multiplies polynomials by X · X , and ∆ N M jk = M jk ∆ N , (5.4) so each M jk maps H N,d into itself, and M jk ( M · M ) = ( M · M ) M jk . (5.5) Proof. By (2.12), M jk (( X · X ) p ) = ( X · X ) M jk p . Thus (5.3) holds.The identity (5.5) is readily verified by computation, as already notedin (1.4).To prove (5.4), it suffices to check both sides are equal on homogeneouspolynomials of a fixed degree. Consider the expression for M X · X ∆ N in (2.1): r∂ r acts as a scalar on homogeneous polynomials of a fixeddegree and therefore it commutes with the action of each M jk on thosepolynomials. Similarly, M · M commutes with each M jk . Thus (2.1)tells us that M X · X ∆ N commutes with each M jk . Since M jk commuteswith M X · X by (5.3), it follows that for each homogeneous polynomial p , M jk ( X · X )(∆ N ( p ))) = ( X · X )(∆ N ( M jk ( p ))), or( X · X )( M jk (∆ N ( p ))) = ( X · X )(∆ N ( M jk ( p ))) . Dividing by X · X , we get M jk (∆ N ( p )) = ∆ N ( M jk ( p )).From (5.4), M jk p is harmonic whenever p is harmonic. (cid:3) Proposition 5.3. Suppose R is a ring containing Q , and c ∈ R . Let H N = ker ∆ N denote the submodule of harmonic polynomials in P N .Then the inclusion map H N → P N induces an isomorphism of R -modules: H c : H N → P N / Z N ( c ) . (5.6) Moreover, there is a unique R -linear mapping L c : P N → H N (5.7) which restricts to the identity map on H N and is on Z N ( c ) . Further-more, (i) ker L c = Z N ( c ) ; (ii) H c ◦ L c is the projection map P N → P N / Z N ( c ) ; (iii) L c commutes with each operator M jk . Item (ii), combined with H c ( p ) = p + Z N ( c ), means that L c p ≡ p mod Z N ( c ). (5.8) Proof. By Proposition 2.8, if p ∈ H N is a polynomial multiple of X · X − c then p = 0; thus ker H c = 0. Next, if p ∈ P N then we can write p as p = p d + ( X · X ) p d − + · · · + ( X · X ) n p d − n , (5.9)where each p j is harmonic, and so p is equal, modulo X · X − c , to theharmonic polynomial p d + cp d − + · · · + c n p d − n . Thus H c is surjective.With this notation, we can write p in the form p = ( p d + cp d − + · · · + c n p d − n ) + ( X · X − c ) q (5.10)and so L c is given uniquely by L c ( p ) = p d + cp d − + · · · + c n p d − n . (5.11)Then H c ( L c p ) = L c ( p ) + Z N ( c ) = p + Z N ( c ) . (5.12)Applying M jk to the expansion (5.9) and using the product rule M jk ( f g ) =( M jk f ) g + f ( M jk g ) and the fact that M jk ( X · X ) = 0, we have M jk p = M jk p d + ( X · X ) M jk p d − + · · · + ( X · X ) n M jk p d − n . (5.13)Here each of the terms M jk p s is harmonic and of the same degree as p s .Hence, (5.13) is the expansion of M jk p in ‘base’ X · X , the counterpartof (5.9) for M jk p ; as seen in Proposition 2.10, this expansion is unique.Then, by the definition of L c in (5.7), we have: L c M jk p = M jk p d + cM jk p d − + · · · + c n M jk p d − n . (5.14)Rewriting the right side of (5.14) with M jk on the left in each term, weget L c M jk p = M jk L c p. (5.15)Since each M jk maps H N,d into itself (Proposition 5.2), for every d ≥ L c commutes with each M jk . (cid:3) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 47 By a zonal harmonic we mean a spherical har-monic that is congruent modulo Z N ( c ) to a polynomial of the form q ( t X + · · · + t N X N ), with q ∈ R [ T ] and t = ( t , . . . , t N ) ∈ R N . When N = 3 this agrees with the terminology in classical analysis. In higherdimensions there are other ways of defining this notion, but for our pur-poses we choose this more restrictive meaning.In the following we use the notation t · X = t X + · · · + t N X N and t = t + · · · + t N , (5.16)so t = t · t .A case of interest is a spherical harmonic that coincides, modulo Z N ( c ),with a polynomial in the variable X . These are spherical harmonics thathave rotational symmetry around the axis (1 , , . . . , N ≥ 3. By Remark 2.4, if R is a ring in whichpositive integer multiples of 1 R are not zero divisors, N ≥ 3, and apolynomial p ∈ R [ X , . . . , X N ] is annihilated by the operators M jk fordistinct j, k ∈ { , . . . , N } then p = q ∗ ( X + · · · + X N ) , (5.17)where q ∗ is a polynomial in one variable with coefficients in R [ X ]. Thus, p ≡ q ( X , c − X ) mod Z N ( c ) , (5.18)for some polynomial q in two indeterminates, with coefficients in R . Thus,for N > 2, if p is a spherical harmonic that is annihilated by the operators M jk for all j, k ∈ { , . . . , N } with j = k then p is a zonal harmonic,corresponding to t being (1 , , . . . , Proposition 5.4. Suppose R is a ring containing Q . Let L denote anarbitrary R -linear derivation on P N = R [ X , . . . , X N ] of degree n ≥ − ,in the sense that L maps P N,m to P N,m + n for all m . Suppose p = t · X asin (5 . with rational coefficients that are not all . If q ∈ P N has theproperty that Lp = 0 ⇒ Lq = 0 as L runs over all R -linear derivationsof P N as described above, then there is a polynomial f ∈ R [ T ] such that q = f ( p ) , and if q is homogeneous then q is an R -multiple of a power of p .Proof. By focusing on the homogeneous components of q separately, wemay assume that q is homogeneous. Moreover, inductively, we assumethat the result holds for all homogeneous polynomials q of lower degree(the degree-0 case being automatically true). We write L as L = N X j =1 L j ∂ j , (5.19)where L j = L ( X j ) is a homogeneous polynomial of degree n + 1 withcoefficients in R . Thus the given condition is:if P Nj =1 L j ∂ j p = 0 then P Nj =1 L j ∂ j q = 0 (5.20)for all choices of homogeneous polynomials L , . . . , L N of degree n + 1.Pick k such that t k = 0. For any j = k we take L i to be 0 for i = j or k , L j = Y ∂ k p = Y t k and L k = − Y ∂ j p = − Y t j , where Y is a fixedmonomial of degree n + 1; then the hypothesis of (5.20) is satisfied, so ∂ k p∂ j q = ∂ j p∂ k q. (5.21)This holds for all j, k ∈ { , . . . , N } . Multiplying both sides by X j andsumming over j we obtain, by homogeneity of p and q , Aq = Bp, (5.22)where A = (deg q ) ∂ k p = (deg q ) t k ∈ Q × and B = (deg p ) ∂ k q = ∂ k q . So q = q p, where q is also homogeneous, of degree one less than q . Now if Lp = 0for L as in the proposition then we have Lq = ( Lq ) p + q ( Lp ) = 0so Lq = 0. Since the degree of q is one less than the degree of q , itfollows by the induction hypothesis that q is an R -multiple of a powerof p . Hence so is q . (cid:3) Remark 5.5. If p = q then L ( p ) = 2 qL ( q ) for any first-order differentialoperator L , and thus L ( p ) = 0 ⇒ L ( q ) = 0 even though q is not apolynomial in p . This does not contradict the proposition since this p isnot of degree 1.For a single-variable polynomial q ( Y ) ∈ R [ Y ], let q t = q ( t · X ) ∈ R [ X , . . . , X N ] (5.23)These polynomials are annihilated by the differential operators that an-nihilate t · X . The converse also holds. OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 49 Proposition 5.6. Let R be an integral domain of characteristic . Let q ( Y ) ∈ R [ Y ] be non-zero and t = ( t , . . . , t N ) ∈ R N with coordinates notall , and N ≥ . Let c ∈ R and L c : P N → H N be the linear map as in (5 . . Then the harmonic polynomial L c q t is homogeneous of degree n ifand only if q satisfies the differential equation [( t · t ) c − Y ] q ′′ ( Y ) − ( N − Y q ′ ( Y ) + n ( n + N − q ( Y ) = 0 . (5.24)Let us note that the equation (5.24) involves t only through t · t ; thusthe same polynomial q works for all t ∈ R N having a fixed value of t · t .For example, for t = (1 , , . . . , 0) and c = 1 the zonal harmonics are ofthe form q ( X ), with q satisfying (5.24) with t · t = c = 1.Suppose k ≥ q ( Y ). Then, by focusing on the highestdegree term on the left side of (5.24), we see that k = n . Proof of Theorem 5.6 . We have M · M q ( t · X )= X { j,k }∈ P ( N ) M jk (( t k X j − t j X k ) q ′ ( t · X ))= X { j,k }∈ P ( N ) [ − ( t j X j + t k X k ) q ′ ( t X + · · · + t N X N )+ (cid:0) t k X j − t j X k ) q ′′ ( t X + · · · + t N X N ) (cid:1)(cid:3) = − ( N − t · X ) q ′ ( t · X ) + (cid:2) t X · X − ( t · X ) (cid:3) q ′′ ( t · X ) , (5.25)where t = t · t = P Nj =1 t j .Since L c q t is harmonic we have by (2.1)0 = X · X ∆ L c q t = [( r∂ r ) + ( N − r∂ r + M · M ] L c q t = [( r∂ r ) + ( N − r∂ r ] L c q t + L c M · M q t , (5.26)where we used Proposition 5.6(iii) in commuting L c and M · M .Now suppose that the harmonic polynomial L c q t is homogeneous ofdegree n . Then the first term in the last line in (5.26) is n ( n + N − L c q t .Hence, in this case, (5.26) is equivalent to n ( n + N − q t + M · M q t ∈ ker L c . (5.27)By Proposition 5.3(i) then n ( n + N − q t + M · M q t is a polynomialmultiple of X · X − c . The expression in the last line of (5.25) then shows that n ( n + N − q t − ( N − t · X ) q ′ ( t · X ) + [ t c − ( t · X ) ] q ′′ ( t · X )is a polynomial multiple of X · X − c . By Lemma 5.7 (proved below) thisis only possible if n ( n + N − q t − ( N − t · X ) q ′ ( t · X ) + [ t c − ( t · X ) ] q ′′ ( t · X ) = 0 , so when L c q t is homogeneous of degree n this second-order differentialequation is satisfied.Conversely, if the above second-order differential equation is satisfiedthen using (5.26) and then (5.25) we see that[( r∂ r ) + ( N − r∂ r ] L c q t = − L c M · M q t = L c (cid:2) ( N − t · X ) q ′ ( t · X ) − (cid:2) t X · X − ( t · X ) (cid:3) q ′′ ( t · X ) (cid:3) = L c (cid:2) ( N − t · X ) q ′ ( t · X ) − (cid:2) t c − ( t · X ) (cid:3) q ′′ ( t · X ) (cid:3) = n ( n + N − L c q t . (5.28)Now if we let p k be the homogeneous degree- k component of L c q t , andcompare the degree k terms on both sides of (5.28), we obtain k ( k + N − p k = n ( n + N − p k , and so k ( k + N − 2) = n ( n + N − R is an integral domain ofcharacteristic 0. Thus k = n or k = − ( N − − n and the second optionis impossible since k ≥ (cid:3) We turn to two results used in the preceding proof. Lemma 5.7. Let R be an integral domain of characteristic . Let q ( Y ) ∈ R [ Y ] be a polynomial in one variable and suppose for some c ∈ R and ( t , . . . , t N ) ∈ R N , where N ≥ , with at least one t j = 0 , that q ( t · X ) ∈ R [ X , . . . , X N ] is divisible by X · X − c in the ring R [ X , . . . , X N ] . Then q = 0 .Proof. Suppose q ( t · X ) = ( X · X − c ) r (5.29)for some non-zero r ∈ R [ X , . . . , X N ]. Set n = deg q ( Y ) and let q Y n be the degree- n term in q ( Y ), with q = 0 in R . The degree of r in R [ X , . . . , X N ] is necessarily n − X · X − c is monic in each X i .Focusing on the degree- n terms on both sides, we have q ( t · X ) n = X · Xr n − , (5.30) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 51 where r n − is the (necessarily nonzero) term of degree n − r andthe right side of (5.30) is nonzero. Then we have a contradiction fromLemma 5.8 (proved below). (cid:3) Lemma 5.8. Let R be an integral domain of characteristic , and t =( t , . . . , t N ) ∈ R N , with at least one nonzero component. Let p ( Y ) ∈ R [ Y ] be such that p ( t X + · · · + t N X N ) is divisible by X + · · · + X N in thepolynomial ring R [ X , . . . , X N ] and N ≥ . Then p = 0 .Proof. Any nonzero polynomial multiple of Q = X + · · · + X N is ofdegree at least 2 in each X j . Suppose that p ( t X + · · · + t N X N ) = Qf, (5.31)where f ∈ R [ X , . . . , X N ]. Evaluating at X j = 0 for all j > 2, we canfocus on the case of just two indeterminates p ( t X + t X ) = ( X + X ) f ( X , X ) . (5.32)Let n = deg p ( Y ) and p Y n be the degree- n term in p ( Y ); so p ∈ R isnon-zero and n ≥ n terms on both sides of (5.32). On the left sideof (5.32) we get p ( t X + t X ) n .( t X + t X ) n = ( t X ) n + n ( t X ) n − t X + [ n/ X j =1 (cid:18) n j (cid:19) t j ( X ) j ( t X ) n − j + X n − / X j =1 (cid:18) n j + 1 (cid:19) t j +11 ( X ) j ( t X ) n − (2 j +1) . (5.33)Upon writing X as Q − X , we have( t X + t X ) n = ( t X ) n + n ( t X ) n − t X + [ n/ X j =1 (cid:18) n j (cid:19) t j ( Q − X ) j ( t X ) n − j + X n − / X j =1 (cid:18) n j + 1 (cid:19) t j +11 ( Q − X ) j ( t X ) n − (2 j +1) = aX + b + cQ, (5.34) where a, b ∈ R [ X ] and c ∈ R [ X , X ]; specifically, a = [( n − / X j =0 (cid:18) n j + 1 (cid:19) ( − j t j +11 t n − (2 j +1)2 X n − b = [ n/ X j =0 (cid:18) n j (cid:19) ( − j t j t n − j X n . (5.35)Hence, ( t X + t X ) n p = p aX + p b + cp Q. (5.36)Since the left side is a multiple of Q as in (5.32), we have( f − p c ) Q = p aX + p b. (5.37)The left side is either 0 or of degree ≥ X , whereas the right side iseither 0 or of degree 1 in X . Thus both sides must be 0. But neither a nor b is 0, and so p = 0, a contradiction. (cid:3) The solutions to equation (5.24) are traditionally understood in termsof Gegenbauer polynomials. We conclude with the following observation. Proposition 5.9. Suppose R is an integral domain containing Q , α ∈ R , and q ( Y ) ∈ R [ Y ] a nonzero polynomial that satisfies the differentialequation [ α − Y ] q ′′ ( Y ) − ( N − Y q ′ ( Y ) + n ( n + N − q ( Y ) = 0 , (5.38) where n ≥ and N ≥ are integers. Then the degree of q is n . If α = 0 then q ( Y ) = Y n is, up to multiplication by any element of R , the uniquesolution to (5.38). If α = 0 then a solution q ( Y ) , unique up to scalingby any element of R , to (5 . exists, with coefficients in R [ α − ] , and isan even polynomial if n is even and an odd polynomial of n is odd.Proof. Substituting q ( Y ) = m X k =0 q k Y k into (5.38), and examining the coefficient of Y k , (5.38) is equivalent to αq k +2 = k ( k − 1) + k ( N − − n ( n + N − k + 2)( k + 1) q k = ( k − n )( k + n + N − k + 2)( k + 1) q k , (5.39) OTATIONAL SYMMETRIES IN POLYNOMIAL RINGS 53 holding for all k ∈ { , , , . . . } . The condition N ≥ k + n + N − = 0 since k and n are non-negative integers.If α = 0 then q k = 0 unless k = n . Now suppose α = 0. If n is evenand q were nonzero then (5.39) would imply that q k is nonzero for allodd k ; this would contradict the fact that q ( Y ) is a polynomial. Thus q ( Y ) is an even polynomial. Moreover, (5.39) also implies that q m = 0for all even m > n . A similar argument works for n odd. (cid:3) Acknowledgments . 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