Symmetric Ideals and Invariant Hilbert Schemes
Sebastian Debus Andreas Kretschmer

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Symmetric Ideals and Invariant Hilbert Schemes








A symmetric ideal is an ideal in a polynomial ring which is stable under all permutations of the variables. In this paper we initiate a global study of zero-dimensional symmetric ideals. By this we mean a geometric study of the invariant Hilbert schemes _ρ^S_n(^n) parametrizing symmetric subschemes of ^n whose coordinate rings, as S_n-modules, are isomorphic to a given representation ρ. In the case that ρ = M^λ is a permutation module corresponding to certain special types of partitions λ of n, we prove that _ρ^S_n(^n) is irreducible or even smooth. We also prove irreducibility whenever ρ≤ 2n and the invariant Hilbert scheme is non-empty. In this same range, we classify all homogeneous symmetric ideals and decide which of these define singular points of _ρ^S_n(^n). A central tool is the combinatorial theory of higher Specht polynomials. 


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    Andreas Kretschmer
    April 29, 2024
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§ INTRODUCTION


A symmetric ideal is an ideal in a polynomial ring which is stable under all permutations of the variables.
Special classes of symmetric ideals, for instance Specht ideals and Tanisaki ideals <cit.>, have been studied intensively in the algebraic combinatorics literature and are related to Kostka–Macdonald polynomials and the famous work of Haiman on n!, see <cit.> and the references therein.
In commutative algebra, symmetric ideals are mainly studied for their asymptotic properties. We refer to <cit.> for plenty of examples as well as to <cit.> for foundational results on representation stability and FI-modules. One of the most well-known results on symmetric ideals is that an infinite chain of symmetric ideals I_n ⊆[x_1,…,x_n] in increasingly many variables with I_n-1⊆ I_n for all n eventually stabilizes in the sense that I_n = (S_n · I_n-1) for all n large enough <cit.>, where S_n is the symmetric group on n elements.

The main goal of this article is to initiate a global study of symmetric ideals, starting with the zero-dimensional case. By this we mean a geometric study of the invariant Hilbert schemes _ρ^S_n(^n) parametrizing symmetric subschemes of ^n whose coordinate rings are isomorphic to ρ as S_n-representations. This is a very special case of the invariant Hilbert schemes of Alexeev and Brion <cit.>.

Of central interest to us is the case where ρ = M^λ is the permutation module corresponding to a partition λ of n. It is a consequence of <cit.> that there is only a single homogeneous ideal in _M^λ^S_n(^n), called the Tanisaki ideal I_λ. It can be obtained as the associated graded ideal of the vanishing ideal of the S_n-orbit of a point in ^n whose entries occur with multiplicities λ_1, λ_2, and so on <cit.>. The closure of the set of these vanishing ideals is an irreducible component of _M^λ^S_n(^n), the smoothable component. A natural question is therefore whether this is all there is, i.e., whether _M^λ^S_n(^n) is irreducible. For certain types of partitions, for instance if λ is a hook partition or has only two non-zero parts, we show that _M^λ^S_n(^n) is indeed irreducible, see Corollary <ref>. In the latter case it is even smooth by Proposition <ref>. For arbitrary λ, we prove that the Hilbert–Chow morphism

    γ_M^λ^S_n(^n) ⟶^n/S_n,

mapping an ideal to its vanishing orbit, is finite and has singleton fibers over the image of the smoothable component (Theorem <ref>). From this one may deduce that the dimension of _M^λ^S_n(^n) agrees with the number m of non-zero parts of λ, and that the normalization of its smoothable component is isomorphic to ^m with the normalization map being bijective.

In addition, for all ρ satisfying ρ≤ 2n we prove that _ρ^S_n(^n) is irreducible whenever it is non-empty, and we decide when this happens (Theorem <ref>). Moreover, in the same range we explicitly classify all homogeneous ideals in _ρ^S_n(^n) and decide which of these define singular points, see Table <ref>.

Beyond these cases, it can be observed that if ρ = [S_n]^⊕ l is a direct sum of regular representations, the Hilbert–Chow morphism is actually an isomorphism _[S_n]^⊕ l^S_n(^n) ≅_l(^n/S_n) (Proposition <ref>). In particular, this shows that ^S_n_ρ(^n) is in general at least as complicated as the usual Hilbert scheme of points of affine space. An analogous statement should also hold for other types of finite reflection groups.

A peculiar observation in our study is that the geometry of _ρ^S_n(^n) does not seem to depend strongly on the ambient dimension n when n grows while ρ “stays the same,” in an appropriate sense. For instance, this is what made it possible to complete our classification in the range ρ≤ 2n for arbitrary n. We make this apparent stabilization phenomenon more precise in Subsection <ref>, see Question <ref>.

The paper is structured as follows. In Section <ref> we quickly introduce higher Specht polynomials and invariant Hilbert schemes. Section <ref> deals with the case of permutation modules, ρ = M^λ. The classification in the range ρ≤ 2n is contained in Section <ref>. Several open questions are raised in the final Section <ref>.













§ PRELIMINARIES




 §.§ Higher Specht polynomials

We encourage the reader familiar with basic notions of the characteristic zero representation theory of S_n to skip this subsection on a first reading and refer back to it when needed.

We introduce our notation and define the higher Specht polynomials of <cit.>. Foundational references on S_n-representations and related combinatorics are, e.g., <cit.>.

A partition λ=(λ_1,…,λ_m) of n is a sequence of non-increasing positive integers such that ∑_i≥1λ_i = n and λ_m ≠ 0.
We write λ⊢ n.
The size of λ is |λ|  n and its length is denoted by (λ)  m. 

If λ, μ⊢ n, then μ dominates λ if ∑_j=1^k μ_j ≥∑_j=1^k λ_j for all k, where μ_j is interpreted as zero if j > (μ). We denote domination by μλ.
A partition λ can be represented by its (Young) diagram, i.e., the ordered sequence of left-justified boxes from the left to the right and the top to the bottom, where the i-th line contains λ_i boxes. We say that the associated diagram has shape λ. The transpose partition λ' of λ is the partition whose diagram is obtained from transposing the diagram of λ. A tableau of shape λ is a bijective filling of the diagram of shape λ with the integers in [n] {1,…,n}. We write (T) = λ if T is a tableau of shape λ.
For instance,

    smalltableaux
    9     3     6     4 
    
    2     1     8
     
    5     7

is a tableau of shape (4,3,2).  
A standard tableau is a tableau in which the integers in every row and column are increasing from the left to the right and the top to the bottom.

For a sequence (i_1,…,i_l) of distinct positive integers, we define the associated Vandermonde polynomial in the variables x_i_1,…,x_i_l as 
    Δ_(i_1,…,i_l)(x) = ∏_j < k ∈  [l](x_i_j-x_i_k).
 By convention, Δ_(i) = 1.
For a tableau T of shape λ with l columns, we denote by T_i the sequence of numbers contained in the i-th column of T from above to below. Then the associated Specht polynomial _T(x) is the product of the Vandermonde polynomials of all columns T_i, i.e.,

    _T(x) ∏_i=1^l Δ_T_i.


For example, for the above tableau T of shape (4,3,2), we have 

    _T(x)   =Δ_(9,2,5)(x)Δ_(3,1,7)(x)Δ_(6,8)(x)Δ_(4)(x) 
    
       =(x_9-x_2)(x_9-x_5)(x_2-x_5)(x_3-x_1)(x_3-x_7)(x_1-x_7)(x_6-x_8).


For a group G acting linearly on a complex vector space V we denote the action of a group element σ∈ G on a vector v ∈ V by σ· v.




If G is finite, then the irreducible representations are in correspondence with the conjugacy classes. In the case of the symmetric group S_n these in turn correspond bijectively to integer partitions of n. For λ⊢ n the irreducible representation of S_n corresponding to λ is the Specht module S^λ. It can be constructed as the vector space spanned by all Specht polynomials of shape λ. It was proven by Specht <cit.> that these representations are indeed irreducible and pairwise non-isomorphic.

We will consider the action of S_n on the polynomial ring P [x_1,…,x_n] via permutations of the variables σ· x_i  x_σ(i).
A polynomial is symmetric if it is invariant under the S_n-action. The invariant ring P^S_n is generated by n algebraically independent polynomials which can be chosen as the sequence of the first n elementary symmetric polynomials e_j= ∑_I ⊂ [n], |I| = n∏_i ∈ I x_i or the power sum polynomials p_j = ∑_i=1^n x_i^j. The coinvariant algebra of P is the quotient of P by the ideal (p_1,…,p_n) generated by all homogeneous symmetric polynomials of positive degree. As an S_n-representation, the coinvariant algebra has the structure of the regular representation of S_n, i.e., each irreducible representation S^λ occurs (S^λ) times in P / (p_1,…,p_n). Moreover, we have an isomorphism of graded S_n-modules P ≅ P/(p_1,…,p_n) ⊗_[p_1,…,p_n]. An isotypic decomposition of the coinvariant algebra lifts to an isotypic decomposition of P by multiplying the coinvariants by symmetric polynomials. Terasoma and Yamada <cit.> provided an explicit decomposition of the coinvariant algebra into its irreducible subspaces by introducing higher Specht polynomials. This was extended in <cit.>, for example to certain direct products of reflection groups.

For a Young tableau T of shape λ⊢ n we denote by C(T) the column stabilizer and by R(T) the row stabilizer of T. These are the maximal subgroups of S_n which permute the entries in any column, respectively any row of T. 
The word of T is the sequence w(T) ∈^n where we read each column of the tableau T from the bottom to the top, starting from the left.
The index i(T) ∈^n of T is defined as follows. The number 1 in the word w(T) has index 0. Recursively, if the number k in w(T) has index p, then k+1 has index either p or p+1 according to whether k+1 is right or left from k in w(T). For instance, for the above tableau T we have w(T)=(5,2,9,7,1,3,8,6,4) and i(w(T))=(2,1,4,3,0,1,3,2,1)).
Next, we can associate to a tuple of tableaux (T,S) of the same shape λ a monomial x_T^i(S) in n variables by

    x_T^i(S) := x_w(T)_1^i(w(S))_1⋯ x_w(T)_n^i(w(S))_n.

The higher Specht polynomial associated with (T,S) is then defined as

    F_T^S := ∑_σ∈ C(T)
    τ∈ R(T) (σ) σ∘τ· x_T^i(S)∈ P.




The λ-part of the isotypic decomposition of the coinvariant algebra has a basis given by the higher Specht polynomials F_T^S where (S) = (T) = λ and S,T are both standard tableaux. Moreover, for a fixed standard tableau S, the representation ⟨ F_T^S : (T) = (S) = λ⟩ is isomorphic to the Specht module S^λ.


For instance, when n=3 we have 

    [x_1,x_2,x_3]/(p_1,p_2,p_3)    ≅ S^(3)⊕ 2 S^(2,1)⊕ S^(1,1,1)
        = ⟨ 1 ⟩⊕⟨ x_i - x_j ⟩⊕⟨ (x_i-x_j)x_k ⟩⊕⟨ (x_1-x_2)(x_1-x_3)(x_2-x_3) ⟩.

For any λ there exists a unique standard tableau S_0 of shape λ such that the sum of the entries of the indices of standard tableaux of shape λ is minimal. The higher Specht polynomial F_T^S_0 then coincides with the Specht polynomial _T up to a multiplicative constant.

For a partition λ⊢ n we denote by O(λ) the subset of ^n of points of orbit-type λ, i.e., the points whose stabilizer subgroup in S_n is a conjugate of the Young subgroup S_λ = S_λ_1×⋯× S_λ_m. The ideal of P generated by all Specht polynomials of shape λ is called the Specht ideal of λ and we write V_λ for its vanishing set. Then V_λ = ⋃_μλ O(μ) by <cit.>. A simple consequence is ⋂_λ⋭̸θ V_λ = ⋃_τ⊵θ O(τ). By abuse of notation, we denote by O(λ)/S_n the image of O(λ) in ^n/S_n = ([p_1,…,p_n]) and by O(λ)/S_n its closure. We emphasize that O(λ)/S_n is usually a non-normal subvariety of ^n/S_n.
Let us also recall the following well-known lemma.


Set-theoretically, the fiber over every closed point of the map (p_1,…,p_n) : ^n →^n is the S_n-orbit of a single point in ^n. In particular, the ideal (p_1,…,p_n) ⊆ P is a complete intersection.



We first consider the fiber of (a_1,…,a_n) ∈^n for the map (e_1,…,e_n): ^n →^n. By Vieta's formula, we have t^n - a_1 t^n-1 + … + (-1)^n a_n = ∏_i=1^n (t - α_i) for some α_i ∈ which are unique up to permutation. This shows that the preimage of (a_1,…,a_n) consists precisely of the permutations of (α_1,…,α_n).
The same is then true for the map (p_1,…,p_n) by the Newton identities.





 §.§ Invariant Hilbert schemes

For a reductive group G, an affine G-scheme W = (A) of finite type over , and a multiplicity-finite (but not necessarily finite-dimensional) G-representation ρ, the invariant Hilbert scheme _ρ^G(W) parametrizes all closed G-subschemes Z ⊆ W with Γ(𝒪_Z) ≅ρ as G-representations. We refer to <cit.>, noting that the second reference allows G to be disconnected.

The construction in the special case where G is a finite group is quite simple: In this case, there are only finitely many irreducible G-representations up to isomorphism, so ρ is automatically finite-dimensional. Hence, the parametrized subschemes Z ⊆ W are zero-dimensional of length (ρ). It turns out that _ρ^G(W) can be constructed as the union of some connected components of the G-fixed point subscheme of _(ρ)(W), the latter denoting the usual Hilbert scheme of (ρ) points on W. Since the latter is quasi-projective, so is _ρ^G(W).

We collect several foundational results of <cit.> valid for any reductive group G.


    Let [Z] ∈_ρ^G(W) be a closed point corresponding to a G-stable closed subscheme Z ⊆ W defined by an ideal I ⊆ A. Then
    
    T_[Z]_ρ^G(W) ≅_A^G(I,A/I),

    where the former denotes the Zariski tangent space at [Z] and the latter denotes the vector space of A-linear G-equivariant homomorphisms I → A/I.


Let Univ_ρ^G(W) ⊆_ρ^G(W) ×_ W be the universal family. If H is an algebraic group acting on W by G-automorphisms, i.e., such that the actions of G and H on W commute, then H acts on both Univ_ρ^G(W) and on _ρ^G(W) such that the projection Univ_ρ^G(W) →_ρ^G(W) is H-equivariant <cit.>.

Denote by ρ^G the maximal trivial subrepresentation of ρ and by W//G = (A^G) the categorical quotient, where A^G is the subring of G-invariants of A.


    There is a projective morphism
    
    γ_ρ^G(W) ⟶_(ρ^G)(W//G),

    called the Hilbert–Chow morphism, sending a G-stable ideal I ⊆ A to I ∩ A^G.


In the special case (ρ^G) = 1, naturally _1(W//G) = W//G.




 §.§ The torus action, reduction to fixed points

From now on, we let G = S_n and W = ^n with the action given by σ· (a_1,…,a_n) = (a_σ^-1(1),…,a_σ^-1(n)). Let 𝔾_a = (,+) denote the additive group. Then 𝔾_a^n acts on _r(^n) by translating the support of an ideal. Clearly, the actions of both GL_n = GL_n() and 𝔾_a^n on _(ρ)(^n) do not preserve S_n-stable ideals. What is left of these actions on ^S_n_ρ(^n) are the centralizers of S_n in GL_n and 𝔾_a^n.
In the case of 𝔾_a^n the centralizer is clearly 𝔾_a ↪𝔾_a^n, embedded via the diagonal; in other words, translating a symmetric ideal along a diagonal vector (a,a,…,a) ∈^n does not break the symmetry.
For GL_n, the answer is the following.

    Let n ≥ 2. The centralizer of S_n in GL_n is a 2-dimensional torus
    
    GL_n^S_n = { a1_n + b1_n ∈GL_n : a,b ∈^* },

    where 1_n is the identity matrix and 1_n is the matrix filled with ones only. An explicit isomorphism is given by
    
    (^∗)^2 ≅⟶GL_n^S_n,  (s,t) ↦ t 1_n + (s-1)t/n1_n.


For the remainder of the paper, we will write T  (^∗)^2 for the 2-dimensional torus and refer to the induced action on ^S_n_ρ(^n) simply as the T-action. In most situations that we will encounter it suffices to consider only the action of one of the two copies of ^∗ embedded into T via s=1 or t=1.

One of the most useful facts about the T-action is that every orbit closure contains a T-fixed point <cit.>. In particular, smoothness of ^S_n_ρ(^n) is equivalent to smoothness at every T-fixed point because the singular locus is closed. The same is true for every closed geometric property, for example non-reducedness.



 §.§ The associated graded

From now on, we abbreviate _ρ^S_n_ρ(^n).
Let ^∗ act on ^n by the usual scalar multiplication, inducing the standard grading on [x_1,…,x_n]. For a Hilbert function h, let H_ρ,h denote the locally closed subset of _ρ consisting of all ideals whose associated graded has Hilbert function h. We endow H_ρ,h with the reduced scheme structure. Then there is a natural map

    H_ρ,h⟶^S_n ×^∗_ρ,h(^n),  I ↦ I

where I denotes the associated graded ideal and the target Hilbert scheme denotes the disjoint union of the (finitely many) (S_n ×^∗)-invariant Hilbert schemes running over all (S_n ×^∗)-representations restricting to ρ as an ungraded S_n-representation and to h as a ^∗-representation (i.e., having Hilbert function h).

Formally, this map can be constructed as follows (this is inspired by <cit.>). Let 𝒵_ρ⊆_ρ×^n →_ρ be the universal family. Base-changing by the embedding H_ρ,h↪_ρ, we obtain a closed subscheme 𝒵_ρ,h⊆ H_ρ,h×^n, defined by a quasi-coherent ideal sheaf ℐ⊆𝒪_H_ρ,h[x_1,…,x_n], and the composition with the projection gives a finite flat family 𝒵_ρ,h→ H_ρ,h. Then the associated graded ℐ⊆𝒪_H_ρ,h[x_1,…,x_n] defines a closed subscheme 𝒵_ρ,h⊆ H_ρ,h×^n, and the finite flat family 𝒵_ρ,h→ H_ρ,h provides the desired map.




§ S_N-ORBITS AND TANISAKI IDEALS




 §.§ Radical ideals
 A first natural question is which representations ρ admit a smooth subscheme, i.e., when does ℋ_ρ contain a radical ideal? This is essentially answered by the Chinese remainder theorem.


    Let n ≥ 2 and let ρ be any S_n-module. The invariant Hilbert scheme ℋ_ρ contains a radical ideal if and only if ρ is a direct sum of permutation modules, i.e.,
    
    ρ≅ M^λ^1⊕⋯⊕ M^λ^N,

    for some partitions λ^1, …, λ^N of n.



    Let I ⊆ P = [x_1,…,x_n] be a radical symmetric ideal such that P/I ≅ρ as S_n-modules. The vanishing set of I is a symmetric point configuration, consisting of some number N of disjoint S_n-orbits with vanishing ideals I_1, …, I_N. The Chinese remainder theorem provides a canonical algebra isomorphism P/I ≅⊕_i=1^N P/I_i which is S_n-equivariant, so it suffices to prove the case of only one orbit. Let q_1, …, q_k ∈^n be the distinct points of this orbit and let λ = (λ_1 ≥λ_2 ≥⋯≥λ_m ≥ 1) ⊢ n be the corresponding partition, i.e., each q_i has m distinct coordinates which occur with multiplicities λ_1, …, λ_m. Again by the Chinese remainder theorem we obtain a canonical isomorphism P/I ≅⊕_i=1^k P/_q_i of algebras, in particular of vector spaces. We define an S_n-module structure on the direct sum by
    
    f∈ P/_q_i  ⇒  σ(f) σ(f)∈ P/_σ(q_i).

    In this way, the Chinese remainder isomorphism is also S_n-equivariant. The canonical inclusion ⊆ P/_q_i is an isomorphism, so ⊕_i=1^k P/_q_i≅ M^λ as S_n-modules.



    Let ρ be a direct sum of permutation modules. The closure of the locus of radical ideals in _ρ is called the smoothable component.


The smoothable component is indeed an irreducible component. The reason is that every radical ideal in the usual Hilbert scheme of points _ρ(^n) has an open neighborhood containing only radical ideals, so the same is true for the closed subscheme _ρ^S_n(^n). Moreover, every radical ideal is a smooth point of _ρ because the same is true for the usual Hilbert scheme of points, and taking invariants preserves smoothness by <cit.>.


    Every symmetric monomial ideal lies in the smoothable component of some _ρ where ρ is a direct sum of permutation modules. The reason is that the well-known procedure of distraction, which is used to prove that any monomial ideal in the usual Hilbert scheme of points lies in the smoothable component, leaves the symmetry intact. Distraction proves in particular that every irreducible component of the usual Hilbert scheme of points intersects the smoothable component. We do not know whether this is also true for _ρ whenever ρ is a direct sum of permutation modules. Let us note that a symmetric monomial ideal is rarely a torus-fixed point of _ρ. In fact, the only symmetric monomial ideals which are also torus-fixed points are the powers of the homogeneous maximal ideal.



    Let λ⊢ n be a partition with m non-zero parts. The invariant Hilbert scheme _M^λ is connected and has dimension m. Moreover, the Hilbert–Chow morphism γ_M^λ⟶^n/S_n is finite and all fibers over points in O(λ)/S_n are singletons. Finally,
    
    O(λ)/S_n ⊆im(γ) ⊆⋃_μ⊵λ O(μ)/S_n.



The proof of Theorem <ref> will be given in the next subsection. As a consequence, _M^λ is irreducible if and only if so is its image under the Hilbert–Chow morphism if and only if the first inclusion is an equality.


    Let λ⊢ n be a hook partition, i.e., λ = (k,1,1,…,1). Then _M^λ is irreducible.



    Since λ is a hook partition, O(λ)/S_n = ⋃_μ⊵λ O(μ)/S_n.


Consider the linear map ^m ↪^n which sends (a_1,…,a_m) to the n-tuple of which the first λ_1 entries equal a_1, the next λ_2 entries equal a_2, and so on. The image of the composition φ_λ^m ↪^n ↠^n/S_n is O(λ)/S_n, and φ_λ^m →O(λ)/S_n is a finite morphism. Moreover, φ_λ factors uniquely as

    ^m [r,"φ_λ"] [d,two heads]
                   O(λ)/S_n 
    ^m/S_m_1×⋯× S_m_k, [ur,dotted]

where k = |{λ_1, …, λ_m}| is the number of distinct parts of λ and m_i is the multiplicity of λ_i in λ, so ∑_i=1^k m_i = m. The quotient ^m/S_m_1×⋯× S_m_k is abstractly isomorphic to ^m. Moreover, the dotted arrow is a bijection on closed points, and it is a finite morphism since so is φ_λ. Therefore, the dotted arrow is a finite, bijective, birational morphism from a smooth variety and therefore the normalization of O(λ)/S_n.

Let Z_λ be the smoothable component of _M^λ, endowed with the reduced scheme structure. By Theorem <ref>, restricting the Hilbert–Chow morphism to Z_λ, we obtain a finite, bijective, birational map Z_λ⟶O(λ)/S_n. Composing it with the normalization (Z_λ)^ν of Z_λ, we obtain the normalization of O(λ)/S_n above. In particular,

    (Z_λ)^ν≅^m,

and the normalization map (Z_λ)^ν→ Z_λ is a bijection.

However, it is not unreasonable that something much stronger holds.


    Let λ⊢ n be a partition with m non-zero parts. Is _M^λ≅^m?


Note that Question <ref> has an affirmative answer if and only if _M^λ is smooth. This, in turn, can be checked at a single point of _M^λ, the Tanisaki ideal, as we shall soon see.

Evidence for a positive answer to Question <ref> is provided by Theorem <ref>, Corollary <ref> and Proposition <ref> below, the latter proving the case m=2. The answer is positive also in the cases m=1 and m=n where the proof is much simpler. In fact, for m=1 and m=n the Tanisaki ideals are, respectively, the homogeneous maximal ideal  and the complete intersection (p_1,p_2,…,p_n).



 §.§ Tanisaki ideals
 We start with the following lemma which can be found in <cit.>. We include a proof since it is of central importance for Theorem <ref>. 


    Let F_T^S be a higher Specht polynomial with T any Young tableau of shape λ⊢ n and S a standard Young tableau of the same shape. Then F_T^S is divisible by the usual Specht polynomial _T. In particular, the Specht ideal of type λ contains the entire λ-isotypic component of the polynomial ring P.



    It suffices to prove that F_T^S is divisble by (x_k_1 - x_k_2) for any two distinct indices k_1, k_2 appearing in the same column of T. For this, in turn, it is enough to see that (F_T^S) = 0, where  means setting x_k_1 and x_k_2 equal. Let α∈ S_n be the transposition that exchanges k_1 and k_2. Then clearly (F_T^S) = (α(F_T^S)). However, by definition,
    
    α(F_T^S)    = ( ∑_σ∈ R(T)∑_τ∈ C(T)(τ) (α∘τ) ∘σ) (x_T^i(S)) 
       = - ( ∑_σ∈ R(T)∑_τ∈ C(T)(α∘τ) (α∘τ) ∘σ) (x_T^i(S)) 
       = - ( ∑_σ∈ R(T)∑_τ∈ C(T)(τ) τ∘σ) (x_T^i(S)) = - F_T^S,

    so (F_T^S) = (α(F_T^S)) = - (F_T^S), hence (F_T^S) = 0.


In fact, the proof shows that there is nothing special here about x_T^i(S). Instead, applying the Young symmetrizer ∑_σ∈ R(T)∑_τ∈ C(T)(τ) τ∘σ to any polynomial results in a multiple of _T (possibly zero, of course). 
By the theory of higher Specht polynomials <cit.>, the lowest degree in which the Specht module S^λ occurs in the polynomial ring P is d(λ) ∑_i=1^n (i-1)λ_i. The multiplicity of S^λ in P_d(λ) is 1, and it is generated by the Specht polynomials of shape λ.

Recall next the apolarity pairing


    P × P ⟶ P,   ⟨ f , g ⟩ f(∂_1, …, ∂_n)(g(x_1, …, x_n)).


This pairing is S_n-equivariant in the sense that σ(⟨ f, g ⟩) = ⟨σ(f), σ(g) ⟩. For every d, it restricts to a non-degenerate symmetric bilinear form P_d × P_d ⟶ for which the monomials of degree d form an orthogonal basis. In fact, it is different from the standard scalar product only by a diagonal matrix with positive integers as diagonal entries. In particular, the restriction to -coefficients yields an inner product. Let now μ_1 and μ_2 be two distinct partitions of n and denote by V^μ_i the μ_i-isotypic component in P_d (i = 1,2). Then ⟨ V^μ_1, V^μ_2⟩ = 0 because the orthogonal complement (V^μ_1)^⊥ is an S_n-submodule of P_d which intersects V^μ_1 trivially. Considering again the entire apolarity pairing, for any subset W ⊆ P, the set

    {f ∈ P : ⟨ f, g ⟩ = 0  for all  g ∈ W}

is an ideal of P, called the ideal apolar to W, or simply the apolar ideal.

Let now I ⊆ P = [x_1,…,x_n] be a homogeneous symmetric ideal such that P/I ≅ M^λ as S_n-modules.
By Lemma <ref>, I must intersect S^λ⊆ P_d(λ) trivially since P/I ≅ M^λ contains S^λ with multiplicity 1. It follows that I_d(λ), the degree d(λ) part of I, is contained in the apolar ideal to S^λ⊆ P_d(λ). We denote this apolar ideal by I_λ and call it the Tanisaki ideal. We claim I ⊆ I_λ. This is because otherwise, if there was some homogeneous f ∈ I ∖ I_λ, then by definiton of I_λ there would be some homogeneous g ∈ S^λ⊆ P_d(λ) such that ⟨ f, g ⟩≠ 0. Hence, if h is any monomial appearing in ⟨ f, g ⟩∈ P, we get ⟨ hf, g ⟩≠ 0. But hf ∈ I_d(λ)⊆ (I_λ)_d(λ), a contradiction. This shows that the Tanisaki ideal is the unique largest homogeneous symmetric ideal of P intersecting S^λ⊆ P_d(λ) trivially, as mentioned in <cit.>.

From the inclusion I ⊆ I_λ it follows that _(P/I_λ) ≤_(M^λ) = nλ. The final point is now that actually _(P/I_λ) = nλ (proved in <cit.>, recalled in <cit.>), hence I = I_λ. This shows that I_λ is the only homogeneous ideal in ℋ_M^λ which proves at once connectedness of ℋ_M^λ and that the Tanisaki ideal is its only torus-fixed point. We can even strengthen this result as follows.


    Let I ⊆ P be a not necessarily homogeneous symmetric ideal such that P/I ≅ M^λ as S_n-modules. If P/I is supported only at the origin, i.e., V(I) = {0}, then I = I_λ.



    It is enough to prove that I ⊆ I_λ. Assume this is not true and there exists f ∈ I ∖ I_λ. Write f = f_1 + f_2 + ⋯ as a sum of its homogeneous parts. Then there is a smallest degree d such that there exists some g ∈ S^λ⊆ P_d(λ) with ⟨ f_d, g ⟩≠ 0. Let h be an arbitrary monomial appearing in ⟨ f_d, g ⟩, then ⟨ hf, g ⟩ = ⟨ hf_d, g ⟩≠ 0. Replacing f by hf ∈ I ∖ I_λ we are now given a possibly inhomogeneous element f ∈ I such that all homogeneous components of f except for f_d(λ) annihilate S^λ⊆ P_d(λ). Since S^λ occurs in P only in degrees ≥ d(λ), the S_n-submodule of I generated by all permutations of f in particular contains a polynomial p = p_d(λ) + p_d(λ)+1 + ⋯ with 0 ≠ p_d(λ)∈ S^λ⊆ P_d(λ) such that all permutations of p span an S_n-module isomorphic to S^λ. By Lemma <ref>, then, all p_d for d > d(λ) are divisible by p_d(λ). Therefore, we can eliminate all degree >d(λ) terms of p with multiples of p itself, creating only new terms in strictly higher degrees. Since V(I) = {0}, there exists some N with ^N ⊆ I. Therefore, eventually, p_d(λ)∈ I. But then, again by Lemma <ref>, I contains the entire Specht ideal of type λ, hence P/I cannot contain any copy of S^λ which contradicts P/I ≅ M^λ.



    The Hilbert–Chow morphism γ_M^λ⟶^n/S_n is finite.



    The Hilbert–Chow morphism is projective by Proposition <ref>. It is also equivariant with respect to the torus action. Since the Tanisaki ideal I_λ is in the closure of the torus-orbit of every point, the upper semicontinuity of fiber dimensions reduces the finiteness of γ to the finiteness of the fiber over the origin in ^n/S_n. Set-theoretically, this fiber is a singleton by Proposition <ref>.



    Let q ∈O(λ). Then the fiber of the Hilbert–Chow morphism γ_M^λ⟶^n/S_n over the point corresponding to the orbit S_n · q is a singleton.


For the proof, we need a few facts from the theory of higher Specht polynomials in the case of products of symmetric groups G  S_l_1×⋯× S_l_k, ∑_i=1^k l_i = n, acting on the polynomial ring P via G ⊆ S_n, see <cit.>. Fundamentally, the irreducible representations of G are exactly the tensor products of Specht modules , ν^i ⊢ l_i. This tensor product can be realized concretely as a G-submodule of P_d(ν^1) + … + d(ν^k) spanned by the products of the Specht polynomials for the ν^i. In fact, the smallest degree d such that P_d contains a copy of  is precisely d(ν^1) + … + d(ν^k), and in this degree the multiplicity of  is 1. Denote by 𝐓 (T_1, …, T_k) a k-tuple of tableaux with T_i of shape ν^i, filled in such a way that the integers l_1 + … + l_i-1 + 1, …, l_1 + … + l_i all appear exactly once in T_i. We write F_𝐓∏_i=1^k F_T_i. Lemma <ref> generalizes to this setting, i.e., every higher Specht polynomial F_𝐓^𝐒 (see <cit.> for the notation) is divisible by F_𝐓. This can be proven similarly as Lemma <ref>.


    We extend the proof of Proposition <ref>. Let q ∈O(λ)⊆^n. Up to a permutation, the first l_1 entries of q are equal to a_1 ∈, the following l_2 entries are equal to a_2 ∈, and so on, where a_1, …, a_k are all distinct. Each multiplicity l_i is a sum of some parts of λ in such a way that every part of λ shows up for exactly one of the l_i. In other words, there is a surjective map l:[m] → [k] such that l_i = ∑_j ∈ l^-1(i)λ_j. Denote by μ^i ⊢ l_i the partition given by those parts of λ indexed by l^-1(i). We now denote by q = q_1, q_2, …, q_s the distinct points of the S_n-orbit of q. For every ideal I ⊆ P in the fiber of the Hilbert–Chow morphism over this orbit, the Chinese remainder theorem gives a canonical S_n-equivariant isomorphism
    
    P/I ≅ P/I_q_1⊕⋯⊕ P/I_q_s,    f ↦ (f, …, f),

    where the S_n-module structure on the right side is defined similarly as in the proof of Proposition <ref>. For the uniqueness of I it will be enough to prove the uniqueness of the I_q_i because I = ⋂_i=1^s I_q_i. Without loss of generality we only show the uniqueness of I_q.
    
    Let G = S_l_1×⋯× S_l_k denote the stabilizer of q, so that P/I_q is a G-module. We define I' ⊆ P to be the apolar ideal to the G-module
    
    V ( S^μ^1⊗ 1 ⊗⋯⊗ 1 ) ⊕⋯⊕( 1 ⊗ 1 ⊗⋯⊗ S^μ^k) ⊆ P,

    all direct summands embedded in the smallest possible degrees d(μ^1), …, d(μ^k), i.e., V is spanned by all Specht polynomials for the partitions μ^1, …, μ^k in the respective sets of variables.
    (If some of the μ^i = (l_i) are trivial, we do not include the corresponding direct summands into V.)
    For any f ∈ P we can write f = f^1 + … + f^k + g, where f^i is the sum of all terms of f which only involve the i-th set of variables, i.e., those indexed by l_1 + … + l_i-1 + 1, …, l_1 + … + l_i, and g is the sum of all remaining terms. Certainly, differentiating a polynomial with respect to a variable it does not contain gives zero. Therefore, since the i-th direct summand of V only involves the i-th set of variables, it is clear that ⟨ f, V ⟩ = 0 if and only if ⟨ f^i, S^μ^i⟩ = 0 for all i = 1, …, k, where S^μ^i is understood again as the span of all Specht polynomials of type μ^i in the i-th set of variables.
    This proves
    
    P/I' ≅( [x_1,…,x_l_1]/I'_1 ) ⊗⋯⊗( [x_n-l_k+1,…, x_n]/I'_k ),

    where I'_1 is the apolar ideal to S^μ^1⊆[x_1,…,x_l_1]_d(μ^1) etc., i.e., the I'_i = I_μ^i are the Tanisaki ideals corresponding to the partition μ^i ⊢ l_i. In particular, P/I' ≅ M^μ^1⊗⋯⊗ M^μ^k as G-modules.
    
    Let now I'_q be the translate of I' whose only support point is q; this translation is G-equivariant. We claim I_q ⊆ I'_q. For this, we first observe that the canonical map V → P/I_q is injective, i.e., no direct summand of V is contained in I_q. The reason is that the restriction of the Chinese remainder isomorphism above to S^λ⊆ P_d(λ) is injective. Now, S^λ⊆ P_d(λ) is generated by the Specht polynomials of shape λ, each of which is a multiple of the product of the corresponding Specht polynomials of the μ^i. In particular, any Specht polynomial for μ^i is non-zero in P/I_q_i, hence indeed V injects into P/I_q.
    To see that I_q ⊆ I'_q, we can first translate both ideals G-equivariantly into the origin. Note that V still injects into P/I_0 because the Specht polynomials generating V are not affected by this translation. Now assume that there exists f ∈ I_0 ∖ I'. Writing f = f^1 + … + f^k + g as above, we can assume without loss of generality that ⟨ f^1, S^μ^1⟩≠ 0. Multiplying f by some monomial h appearing in ⟨ f^1, S^μ^1⟩, we may replace f by hf, so that we can assume that the degree d(μ^1) part of f^1 does not annihilate S^μ^1. Hence the G-submodule of I_0 generated by f contains an element f' which spans a G-submodule isomorphic to S^μ^1⊗ 1 ⊗⋯⊗ 1 such that the degree d(μ^1) part of f' is a Specht polynomial in S^μ^1. Moreover, f' does not have any non-zero homogeneous components of smaller degrees because P_<d(μ^1) does not contain any copy of S^μ^1⊗ 1 ⊗⋯⊗ 1. Therefore, we may assume f' = a + b with a ∈ S^μ^1⊆ P_d(μ^1) a Specht polynomial and b ∈ P_>d(μ^1). As f' spans a G-module isomorphic to S^μ^1⊗ 1 ⊗⋯⊗ 1, the polynomial b lies in the corresponding isotypic component of P and is divisible by a by Lemma <ref>. Thus inductively, since ^N ⊆ I_0 for some N, we obtain a ∈ I_0. Therefore, the canonical map V → P/I_0 has a non-trivial kernel, contradicting what we said above. So indeed, I_q ⊆ I'_q. This implies
    
    l_1μ^1⋯l_kμ^k = _(P/I'_q) ≤_(P/I_q)    = _(P/I)/r
       = nλ/|S_n|/|G| = nλ/nl_1,…,l_k = l_1μ^1⋯l_kμ^k,

    so for dimension reasons actually I_q = I'_q. This proves the uniqueness of I_q.



    Connectedness follows from the fact that the Tanisaki ideal I_λ is the only torus-fixed point. Since the smoothable component has dimension m, it suffices to show that every other potential irreducible component has dimension ≤ m. Since the Hilbert–Chow morphism is finite by Corollary <ref>, the dimension of every irreducible component agrees with the dimension of its image. So it remains to prove that all partitions μ for which the preimage of O(μ)/S_n under the Hilbert–Chow morphism is non-empty have at most m parts. For this, let I ⊆ P be a symmetric ideal with P/I ≅ M^λ as S_n-modules and such that the S_n-orbit given by V(I) lies in O(μ)/S_n. Then P/√(I)≅ M^μ and the canonical surjection P/I ↠ P/√(I) is S_n-equivariant. Therefore, μ⊵λ. This in particular implies that the number of non-zero parts of μ is at most m, as desired.
    The remaining statement is precisely Proposition <ref>.


Remarkably, it is possible to give explicit generators for the Tanisaki ideal I_λ, as follows.


    Let λ⊢ n and let λ' be its transpose. For S ⊆ [n] we write e_r(x_S) for the elementary symmetric polynomial of degree r in the variables indexed by S. Then the Tanisaki ideal I_λ is generated by the polynomials e_r(x_S) with |S| ≥ r ≥ r_λ(|S|)  |S| - n + 1 + ∑_i=1^n-|S|λ'_i.


A short computation shows that |S| ≥ r_λ(|S|) is equivalent to |S| ≥ n - λ_1 + 1. Under this assumption, if S ⊆ S', then r_λ(|S|) ≥ r_λ(|S'|). This together with the relation e_r(x_S ∖ i) = e_r(x_S) - x_i e_r-1(x_S ∖ i) for i ∈ S shows that I_λ is in fact generated already by the full elementary symmetric polynomials e_r(x_[n]) for 1 ≤ r ≤ n and by the e_r_λ(|S|)(x_S). In fact, among the full elementary symmetric polynomials, it is enough to require e_r(x_[n]) to lie in the ideal for 1 ≤ r ≤ m, where m is the number of non-zero parts of λ.


    Let λ = (λ_1,λ_2) ⊢ n be a partition with only 2 non-zero parts. Then _M^λ≅^2.



It suffices to show that the Tanisaki ideal I_λ is a smooth point of _M^λ since it is the only torus-fixed point. The tangent space at [I_λ] is isomorphic to _P^S_n(I_λ,P/I_λ), the vector space of all S_n-equivariant P-module homomorphisms I_λ→ P/I_λ. Since I_λ lies in the smoothable component of _M^λ which has dimension 2, it is enough to show that the dimension of the tangent space is at most 2.
		
	For this, note first that r_λ(n-l) = l+1 for all 0 ≤ l ≤λ_2 and that λ_2 + 1 = r_λ(n-λ_2) = r_λ(n-λ_2 - 1) = ⋯ = r_λ(n-λ_1 + 1). We abbreviate e_r = e_r(x_[n]). We will now repeatedly use the relation e_r(x_S ∖ i) = e_r(x_S) - x_i e_r-1(x_S ∖ i) for all i ∈ S. First we can observe that
	
    e_2(x_[n-1]) = e_2 - x_n e_1(x_[n-1]) = e_2 - x_n e_1 + x_n^2,

	hence x_n^2 ∈ I_λ. Let I  (p_1,x_i^2) ⊆ I_λ. Using the Newton identities and the last equation we can see that e_2(x_[n-1]) ∈ I. Similarly, e_r(x_[n-1]) ∈ I for all r ≥ 2 = r_λ(n-1). Inductively, using the same idea, for all l ≤λ_2 we have
	
    e_r(x_[n-l])    = e_r(x_[n-l]∪ n) - x_n e_r-1(x_[n-l]) 
       = e_r(x_[n-l]∪ n) - x_n e_r-1(x_[n-l]∪ n) + x_n^2 e_r-2(x_[n-l]),

	and for r ≥ r_λ(n-l) = l+1 = r_λ(n-l+1) + 1 all three terms lie in I. If λ_1 ≤λ_2 + 1, this implies I_λ = I. If λ_1 ≥λ_2 + 2, we continue as follows: For all l in the range λ_1 - 1 ≥ l ≥λ_2 + 1, the partial elementary symmetric polynomials e_r_λ(|S|)(x_S), |S| = n-l, all have the same degree λ_2 + 1, and for |S| = n-λ_1+1 = λ_2 + 1 we get the square-free monomial of degree λ_2 + 1, so all other generators in this range are redundant. In the end we obtain I_λ = (p_1,x_i^2, S_n(x_1 x_2 ⋯ x_λ_2+1)). On the other hand, P/I_λ≅ M^λ = ⊕_0 ≤ k ≤λ_2 S^(n-k,k) as follows from the combinatorial formula for the Kostka numbers. These two facts together give even more: If λ_1 ≥λ_2 + 2, what we have proven so far applied to the partition μ (λ_1 - 1, λ_2 + 1) shows further that I_λ is already generated by I together with only the Specht polynomials of shape μ = (λ_1 - 1,λ_2 + 1). Indeed, the canonical surjection
	
    ⊕_0 ≤ k ≤λ_2+1 S^(n-k,k)≅ P/I_μ↠ P/I_λ≅⊕_0 ≤ k ≤λ_2 S^(n-k,k)

	has kernel S^μ, embedded in degree d(μ) = λ_2 + 1. Moreover, each Specht module S^(n-k,k) in P/I_λ is embedded in degree k, hence in degree < λ_2 + 1. Therefore, if I' denotes the ideal generated by I and all Specht polynomials of shape μ, we have a canonical surjection P/I' ↠ P/I_λ which is an isomorphism in degrees ≤λ_2 + 1. But I_λ is generated in degrees ≤λ_2 + 1, so I' = I_λ.
		
	Since M^λ does not contain any copy of S^μ, we deduce that a homomorphism f ∈_P^S_n(I_λ,P/I_λ) is determined already by its values f(p_1) = α· 1, f(p_2) = β· 1 and f(x_i^2-x_j^2) = γ· (x_i - x_j) for some α, β, γ∈. Hence it suffices to show that α, β, γ must satisfy some non-trivial linear relation. If λ_1 > λ_2 we will show that necessarily β = 0 and if λ_1 = λ_2 = n/2 we will show that α = n/2γ, finishing the proof. We start with λ_1 > λ_2. Since x_1 x_2 ⋯ x_λ_2+1∈ I_λ, we can consider its image under f. Write f(x_1 x_2 ⋯ x_λ_2+1) = f_0 + f_1 + … + f_λ_2 as a sum of its graded pieces. Then x_1^2 x_2 ⋯ x_λ_2 + 1∈ I_λ is both a multiple of x_1^2 and of x_1 x_2 ⋯ x_λ_2+1 and therefore maps to both β/n x_2 ⋯ x_λ_2 + 1 and x_1 f(x_1 x_2 ⋯ x_λ_2+1) at the same time. Therefore, all graded pieces of the two must be equal in P/I_λ, in particular
	
    β x_2 ⋯ x_λ_2 + 1 - n x_1 f_λ_2 - 1∈ I_λ.

	Hence, this polynomial must annihilate every Specht polynomial of shape λ. However, since λ_1 > λ_2, there exists a Specht polynomial not involving the variable x_1 at all (putting the 1 in the upper right corner of the tableau) and containing the monomial x_2 ⋯ x_λ_2 + 1 (putting these indices successively in the second row, for example). This Specht polynomial is annihilated by the above polynomial if and only if β = 0.
		
	If λ_1 = λ_2 = n/2, we have the following relation: Define g  (x_1-x_2) (x_3-x_4) ⋯ (x_n-1 - x_n). Then
	
    g p_1 = ∑_i  oddg/x_i - x_i+1 (x_i^2 - x_i+1^2).

	This implies that (α - n/2γ) · g ∈ I_λ which, since g is a Specht polynomial of shape λ, is only possible if α = n/2γ.




 §.§ Two ideal inclusions

While exploring Tanisaki ideals we encountered the containments of a certain ideal generated by Specht polynomials, a simple monomial ideal and the Tanisaki ideal. Although these inclusions are not used in our understanding of the invariant Hilbert schemes, we do include them as we have not found them in the literature.

For a partition μ⊢ n of length m and 1 ≤ k ≤ m-1 we define R_k(μ):=μ_m+μ_m-1+…+μ_k+1+1, and R_m(μ) := 1. We define an ideal I_μ associated with μ as 
    I_μ = (p_1,…,p_m-1,S_n · x_1^m,S_n · (x_1x_2⋯ x_R_k(μ))^k : 1 ≤ k ≤ m-1).
 

We claim I_μ⊆ I_μ and (p_1,…,p_m-1,_T |  (T) μ) ⊆ I_μ. The Tanisaki ideal of μ is the associated graded ideal to the S_n-orbit of any point a ∈^n whose stabilizer in S_n is conjugate to the Young subgroup S_μ <cit.>. To see the first inclusion let a_1,…,a_m ∈ be the distinct coordinates of a such that a_i occurs μ_i times in a. Then for any 1 ≤ k ≤ m and any subset I⊂ [n] of cardinality R_k(μ) we have {a_i | i ∈ I}∩{a_1…,a_k}≠∅ by the pigeonhole principle. Thus the polynomial ∏_i=1^R_k(μ)∏_j=1^k(x_i-a_j) vanishes on the entire orbit S_n ·{a}. Also ∏_i=1^m(x_1-a_i) vanishes on the orbit. However, the top degree terms of those polynomials are precisely the monomial generators of the ideal I_μ. This proves I_μ⊂ I_μ.
For the other inclusion, we note that for a ∈^n we know that p_i-p_i(a) is contained in the vanishing ideal of S_n ·{a} and the containment of _T with (T) μ follows from <cit.>.


    We have (p_1,…,p_n,_T :  (T) μ) ⊆I_μ⊆ I_μ. 




It remains to prove the first inclusion. For this, we show the stronger containment

    (_T |  (T) μ) ⊆ (S_n · x_1^m, S_n · (x_1 x_2 ⋯ x_R_k(μ))^k : 1 ≤ k ≤ m-1) .

Let T be a Young tableau of shape λμ with l columns C_1,…,C_l. We have 
    _T = ±∏_k=1^l∏_i < j 
     i,j ∈ C_k (x_i-x_j).
 Any monomial occurring in _T is a permutation of the monomial

    x_1^|C_1|-1x_2^|C_1|-2⋯ x^1_|C_1|-1· x_|C_1|^|C_2|-1 x_|C_1|+1^|C_2|-2⋯ x^1_|C_1|+|C_2|-2⋯ x_|C_1|+…+|C_l-1|-l+2^|C_l|-1⋯ x_|C_1|+…+|C_l|-l^1.

This monomial lies in the ideal (S_n · x_1^m, S_n · (x_1 x_2 ⋯ x_R_k(μ))^k | 1 ≤ k ≤ m-1) if and only if there exists an integer 1 ≤ k ≤ m such that for at least R_k(μ) distinct indices j, x_j^k divides the monomial. For each column C_i we have max{|C_i|-k,0} distinct variables occurring with exponent at least k in the monomial. Now, 

    ∑_i=1^lmax{|C_i|-k,0}   = ∑_i=1, |C_i| ≥ k+1^l |C_i|-k 
       = ∑_t ≥ k+1∑_i=1, 
     |C_i|=t^l t-k 
       = ∑_t ≥ k+1 (λ_t-λ_t+1)(t-k) 
       = ∑_r ≥ 1 (λ_k+r-λ_k+r+1)r 
       = λ_k+1-λ_k+2+2λ_k+2-2λ_k+3+…
       = λ_k+1 + λ_k+2 + λ_k+3 + …
       = R_k(λ)-1.

Since #{ j : (λ')_j ≥ k } = λ_k and because μλ, there is an integer s with μ_1+…+μ_s > λ_1 + … +λ_s. However, this is equivalent to n - (μ_1+…+μ_s) < n - (λ_1+…+λ_s) which in turn is equivalent to μ_m + … + μ_s+1 + 1 < λ_ (λ)+…+λ_s+1+1. The left hand side equals R_s(μ) and the right hand side equals R_s(λ). Thus, we obtain R_s(μ) ≤ R_s(λ)-1, as desired.


In general, both ideal inclusions are strict. The first inclusion is strict already for the partition (2,1), and the second inclusion is strict, for example, for (3,3,1,1).




§ CLASSIFICATION FOR R ≤ 2N




 §.§ Homogeneous symmetric ideals

Our classification is based on the theory of higher Specht polynomials <cit.>. These polynomials can be used to produce an explicit isotypic decomposition of the degree d part P_d of the polynomial ring P. This decomposition becomes uniform if the number of variables n is at least 2d <cit.>.


    Let n ≥ 3. Below we give explicit decompositions of P_d into irreducible S_n-representations for d ≤ 3 using slightly modified higher Specht polynomials. In every direct summand, all indices are meant to run through all of [n] but such that different letters are assigned different values.
    
    P_1 = ⟨ p_1 ⟩⊕⟨ x_i - x_j ⟩ = S^(n)⊕ S^(n-1,1).

    For n ≥ 4 we have
    
    P_2    = ⟨ p_1^2 ⟩⊕⟨ p_2 ⟩
       ⊕⟨ p_1(x_i - x_j) ⟩⊕⟨ x_i^2 - x_j^2 ⟩
       ⊕⟨ (x_i - x_j)(x_k - x_l) ⟩
       = 2S^(n)⊕ 2S^(n-1,1)⊕ S^(n-2,2),

    while for n=3 instead
    
    P_2    = ⟨ p_1^2 ⟩⊕⟨ p_2 ⟩
       ⊕⟨ p_1(x_i - x_j) ⟩⊕⟨ x_i^2 - x_j^2 ⟩
       = 2S^(3)⊕ 2S^(2,1).

    For n ≥ 6 we have
    
    P_3    = ⟨ p_1^3 ⟩⊕⟨ p_1 p_2 ⟩⊕⟨ p_3 ⟩
       ⊕⟨ p_1^2(x_i - x_j) ⟩⊕⟨ p_2(x_i - x_j) ⟩⊕⟨ p_1(x_i^2 - x_j^2) ⟩⊕⟨ x_i^3 - x_j^3 ⟩
       ⊕⟨ p_1(x_i-x_j)(x_k-x_l) ⟩⊕⟨ (x_i+x_j+x_k+x_l)(x_i-x_j)(x_k-x_l) ⟩
       ⊕⟨ (x_i-x_j)(x_i-x_k)(x_j-x_k) ⟩
       ⊕⟨ (x_i-x_j)(x_k-x_l)(x_s-x_t) ⟩
       = 3S^(n)⊕ 4S^(n-1,1)⊕ 2S^(n-2,2)⊕ S^(n-2,1,1)⊕ S^(n-3,3),

    while for n=5,
    
    P_3    = ⟨ p_1^3 ⟩⊕⟨ p_1 p_2 ⟩⊕⟨ p_3 ⟩
       ⊕⟨ p_1^2(x_i - x_j) ⟩⊕⟨ p_2(x_i - x_j) ⟩⊕⟨ p_1(x_i^2 - x_j^2) ⟩⊕⟨ x_i^3 - x_j^3 ⟩
       ⊕⟨ p_1(x_i-x_j)(x_k-x_l) ⟩⊕⟨ (x_i+x_j+x_k+x_l)(x_i-x_j)(x_k-x_l) ⟩
       ⊕⟨ (x_i-x_j)(x_i-x_k)(x_j-x_k) ⟩
       = 3S^(5)⊕ 4S^(4,1)⊕ 2S^(3,2)⊕ S^(3,1,1),

    for n=4,
    
    P_3    = ⟨ p_1^3 ⟩⊕⟨ p_1 p_2 ⟩⊕⟨ p_3 ⟩
       ⊕⟨ p_1^2(x_i - x_j) ⟩⊕⟨ p_2(x_i - x_j) ⟩⊕⟨ p_1(x_i^2 - x_j^2) ⟩⊕⟨ x_i^3 - x_j^3 ⟩
       ⊕⟨ p_1(x_i-x_j)(x_k-x_l) ⟩
       ⊕⟨ (x_i-x_j)(x_i-x_k)(x_j-x_k) ⟩
       = 3S^(4)⊕ 4S^(3,1)⊕ S^(2,2)⊕ S^(2,1,1),

    and for n=3,
    
    P_3    = ⟨ p_1^3 ⟩⊕⟨ p_1 p_2 ⟩⊕⟨ p_3 ⟩
       ⊕⟨ p_1^2(x_i - x_j) ⟩⊕⟨ p_2(x_i - x_j) ⟩⊕⟨ p_1(x_i^2 - x_j^2) ⟩
       ⊕⟨ (x_1-x_2)(x_1-x_3)(x_2-x_3) ⟩
       = 3S^(3)⊕ 3S^(2,1)⊕ S^(1,1,1).





Table <ref> provides the first main result of this section. It provides all homogeneous symmetric ideals I ⊆ P = [x_1,…,x_n] for n ≥ 3 and such that r = (P/I) ≤ 2n. Note that all unspecified indices in each appearing generator are meant to be distinct and run through all of [n]. The generators of the form (x_i-x_j)(x_k-x_l) must therefore be interpreted as only occurring for n ≥ 4 but not for n=3. To avoid redundancy in the table, for row 2 one should take r ≥ n+3. For example, for r = n+2 the first ideal of row 2 is the same as that of row 5 for [a:b] = [-1:n]. Finally, the second ideal of row 7 has colength 7>2n for n=3, so we only consider it for n ≥ 4.


    Let n ≥ 3. Table <ref> lists all homogeneous symmetric ideals I with _ P/I = r ≤ 2n together with the isomorphism class of the S_n-module P/I. For all occurring representations ρ, the invariant Hilbert scheme _ρ is irreducible. Moreover, each point of _ρ corresponding to a homogeneous symmetric ideal is smooth or singular according to the last column of Table <ref>, where a point [a:b] of ^1 means that the corresponding ideal is the only singular point in this row.


Let us point out that the representation of Row 6 in Table <ref> is not a direct sum of permutation modules. In particular, _ρ can be non-empty even for such ρ. In fact, for ρ = S^(n)⊕ 2 S^(n-1,1), we have _ρ≅^1 as the proof of Theorem <ref> shows, so the only embedded symmetric deformations of the homogeneous ideal in Row 6 are those obtained from it via the action of 𝔾_a.

Recently, Griffin studied the associated graded ideals of vanishing ideals of two distinct S_n-orbits as analogs to Tanisaki ideals <cit.>. Interestingly, the unique singular point in Row 7(c) for [a:b] = [1:0] is a homogeneous ideal which lies in the smoothable component of _ρ (since _ρ is irreducible) with ρ = 2S^(n)⊕ 2 S^(n-1,1) but the proof of Theorem <ref> shows that it cannot be obtained as the associated graded of any radical ideal.


    The fact that ^2 and (x_i-x_j)_≥ 2 + ^d for all d ≥ 3 are singular points of the invariant Hilbert scheme for any n ≥ 3 in particular recovers the well-known fact that Hilb_r(^3) is singular for all r ≥ 4. 



    This is a case distinction using Lemma <ref>. If I_1 = P_1, then I =, and if I_1 = ⟨ x_i-x_j ⟩, then I = (x_i-x_j) + ^r for some r, so P/I = r S^(n), corresponding to row 1 in Table <ref>. Hence, we can assume that either I_1 = ⟨ p_1 ⟩ or I_1 = 0.

    
    We will begin with I_1 = ⟨ p_1 ⟩.
    In this case, (P/I)_≤ 1 = n, so (P/I)_≥ 2 = r-n ≤ n. Now, (S^(n-2,2)) = n(n-3)/2. For n ≥ 6, this is always >n. Hence, the dimension constraint r ≤ 2n implies (x_i-x_j)(x_k-x_l) ∈ I for all n ≥ 6. Let us, therefore, first assume (x_i-x_j)(x_k-x_l) ∈ I for any n ≥ 3 (for n=3 this is a vacuous condition).
    
    If both of p_2 and x_i^2-x_j^2 do not lie in I, then necessarily I = (p_1,(x_i-x_j)(x_k-x_l)) + ^3 since already (P/I)_≤ 2 = 2n. This ideal is the first ideal in row 7 of Table <ref>. If p_2 ∈ I but x_i^2 - x_j^2 ∉I, we get either the ideal of row 6 or, for n ≥ 4, the second ideal of row 7. This follows using Lemma <ref> and observing that (x_i-x_j)(x_i-x_k)(x_j-x_k) ∈ ((x_i-x_j)(x_k-x_l)) ⊆ I for all n ≥ 4, as the following equation shows:
    
    (x_1-x_2)(x_1-x_3)(x_2-x_3)    = (x_3 - x_1) · (x_1-x_2)(x_3-x_4) 
       + (x_1-x_2) · (x_1-x_3)(x_2-x_4).

    For n=3 there are two more possibilities here, namely rows 12 and 13.
    If both p_2 ∈ I and x_i^2 - x_j^2 ∈ I, then I = (p_1) + ^2, row 3. Otherwise, x_i^2 - x_j^2 ∈ I but p_2 ∉I. Applying the Reynolds operator to (x_1^2 - x_2^2)x_1 gives a scalar multiple of n p_3 - p_2p_1, hence p_3 ∈ I. Therefore, (P/I)_3 does not contain any trivial representation. Moreover, (P/I)_≤ 2 = n+1. However, (x_1-x_2)(x_1-x_3)(x_2-x_3) ∈ (x_i^2-x_j^2) for all n ≥ 3. Indeed, the formula
    
    (x_1-x_2)(x_1-x_3)(x_2-x_3) = x_1(x_3^2-x_2^2) + x_2(x_1^2-x_3^2) + x_3(x_2^2-x_1^2)

    can easily be verified. Moreover, by Lemma <ref>, for all n ≥ 3 we have p_2(x_i-x_j), x_i^3-x_j^3 ∈ (p_1(x_i-x_j),x_i^2-x_j^2,(x_i-x_j)(x_k-x_l)) ⊆ I, implying I_3 = P_3, so we obtain the second ideal of row 4.
    
    Otherwise, still assuming I_1 = ⟨ p_1 ⟩, we have n ∈{4,5} and (x_i-x_j)(x_k-x_l) ∉I.
    If n=5, then necessarily I = (p_1,p_2,x_i^2-x_j^2) + ^3 for dimension reasons, which is row 8.
    If n=4, then we have (P/I)_≤ 2≥ 6, so necessarily x_i^2-x_j^2 ∈ I for dimension reasons. Applying again the Reynolds operator to (x_1^2-x_2^2)x_1, we get p_3 ∈ I. Hence, if p_2 ∈ I, we get row 9. If instead p_2 ∉I, then we get the first ideal of row 10. This finishes the case I_1 = ⟨ p_1 ⟩.

    
    Secondly, we assume I_1 = 0, so (P/I)_≤ 1 = n+1. For n ≥ 5 and n=3, all irreducible S_n-representations except for the trivial and the alternating one have dimension ≥ n-1. Moreover, the least degree k such that P_k contains an alternating representation is k = n2 which is >n for all n ≥ 4. Hence, for all n ≥ 5, this forces that (P/I)_≥ 2 is either 0, a direct sum of only trivial representations or S^(n-1,1), and in the last case (even for n=3,4) this forces I_3 = P_3 for dimension reasons, so we obtain the third ideal of row 7 for some (a,b) ≠ (0,0). If (P/I)_2 = 0, then I = ^2, so we get the first ideal of row 4 (even for n=3,4). Otherwise, assume we have any n ≥ 3 and that (P/I)_≥ 2 only consists of trivial representations, necessarily of exactly (r-n-1) many. We claim that we can only get the two ideals of row 2 or the ideal of row 5 for some (a,b) ≠ (0,0) (the case (a,b) = λ (-1,n) gives the same as the first ideal of row 2). The reasoning is as follows. Let J be the ideal generated by all non-trivial representations in all degrees 2 ≤ k ≤ n-1 and by ^n. Then, clearly, J ⊆ I. But we also know J ⊆ (x_i-x_j)_≥ 2 + ^n. Next, we claim that (P/J)_k has dimension at most 1 for all k ≥ 3. For this, observe that p_1(x_i-x_j) ∈ I and x_i^k-x_j^k ∈ I for all k ≥ 2 for dimension reasons. Applying again the Reynolds operator to p_d(x_1-x_2)x_1, d ∈{1,2}, and to (x_1^k-x_2^k)x_1 gives scalar multiples of n p_d p_2 - p_d p_1^2 and np_k+1 - p_kp_1, respectively. The claim (P/J)_k ≤ 1 for all k ≥ 3 easily follows from this. For dimension reasons, then, J_k = (x_i-x_j)_k or I_k = P_k for all k ≥ 3. Hence, the same is true for I_k for all k ≥ 3. This shows that we indeed get one of the three ideals in rows 2 and 5.
    
    The only remaining cases are n ∈{3,4} and such that (P/I)_≥ 2 contains some non-trivial representation which is also different from S^(n-1,1). For n=4, then, we must have S^(2,2) in (P/I)_2, i.e., (x_i-x_j)(x_k-x_l) ∉I. Hence, (P/I)_≤ 2≥ 7. Moreover, both p_1(x_i-x_j) and x_i^2-x_j^2 are necessarily in I for dimension reasons, and some linear combination of p_1^2 and p_2 must also be contained in I, again for dimension reasons. Hence, we get either the second ideal of row 10 or the ideal of row 11 for some (a,b) ≠ (0,0). Finally, for n=3, the only possibility for a non-trivial representation in (P/I)_≥ 2 not equal to S^(2,1) is the alternating representation S^(1,1,1) in degree 3, generated by (x_1-x_2)(x_1-x_3)(x_2-x_3). Then I contains (p_1(x_i-x_j),x_i^2-x_j^2) for dimension reasons. But we have already seen that then automatically (x_1-x_2)(x_1-x_3)(x_2-x_3) ∈ I as well, so this is impossible.



    For all n ≥ 3 we have
    
    x_1^3-x_2^3 ∈ (x_n p_1(x_1-x_2), p_1(x_1^2-x_2^2), p_2(x_1-x_2), (x_i-x_j)(x_k-x_l)), 
    
            p_2(x_1-x_2), x_1^3-x_2^3 ∈ (x_n p_1(x_1-x_2),x_i^2-x_j^2,(x_i-x_j)(x_k-x_l)).




    All claims follow from the containments f,g ∈ ((x_i-x_j)(x_k-x_l)), where
    
    f     n(n-1)(x_1^3-x_2^3) - (n^2-3n+3)p_2(x_1-x_2) 
       - (2n-3)p_1(x_1^2-x_2^2) + n(n-2)x_n p_1(x_1-x_2), 
    
        g     (n-2)p_2(x_1-x_2) - n x_n p_1(x_1-x_2) + n x_n (x_1^2-x_2^2) 
       - n(n-2)(x_1-x_2)(x_1^2-x_n^2) + (n-2)(x_1-x_2) ∑_i=2^n (x_1^2-x_i^2).

    The ideal ((x_i-x_j)(x_k-x_l)) is the vanishing ideal of the set Z {S_n.(b,a,a,…,a) : a,b ∈}, in particular it is radical. The radicality follows from the fact that ((x_i-x_j)(x_k-x_l)) is the Specht ideal for the partition (n-2,2), and all Specht ideals are radical by <cit.>.[This was first proved by Haiman and Woo in an unpublished manuscript titled “Garnir modules, Springer fibers, and Ellingsrud–Strømme cells on the Hilbert Scheme of points.”] Hence, it is enough to show that f and g vanish on  Z. Since both are divisible by x_1-x_2, it is enough to prove their vanishing on (b,a,a,…,a) and (a,b,a,…,a). These are simple computations.




 §.§ Singular points in Table <ref>

We now indicate how to obtain the last column of Table <ref>, i.e., how to decide which ideals correspond to smooth points of _ρ. For all ideals in rows 8–13 we have checked smoothness via , implementing the algorithm of Lehn and Terpereau <cit.> for computing the tangent space. For smoothness one only has to check that the tangent space dimension agrees with the dimension of the smoothable component and argue that it contains the respective ideal which is the case for all ideals in Table <ref>. The theoretical background of the alogrithm of Lehn and Terpereau also provides the foundation for our proofs for rows 1–7, so we recall some basic results of loc. cit. here.

Let I ⊆ P = [x_1, …, x_n] be a homogeneous symmetric ideal. Let N_1 ⊆ P be a minimal graded S_n-submodule generating I as a P-module, and let n_1 _(N_1). Note that each graded piece of N_1 is an S_n-representation and that n_1 is usually larger than β_0(I), the minimal number of generators of I.

Let K (P ⊗_ N_1 → I). Observe that K is an S_n-module via the inclusion into P ⊗_ N_1, where S_n acts diagonally on P ⊗_ N_1. Let now N_2 ⊆ P ⊗_ N_1 be any graded subvector space generating K as a P-module, giving a surjection P ⊗ N_2 → K. Let n_2 _(N_2). 


    Lehn and Terpereau require N_2 ⊆ P ⊗_ N_1 to be a minimal graded S_n-submodule generating K as a P-module. This is indeed important for the rest of their algorithm but both minimality and the S_n-module structure of N_2 are actually not needed for the tangent space computation alone. In fact, even the minimality of N_1 is not strictly necessary, only the grading and the S_n-module structure are.


From the short exact sequence 0 → K → P ⊗_ N_1 α→ I → 0 and the definition of N_2 we get a commutative diagram

    0 [r]
               _P(I,P/I) [r,"α^∗"]
                   _P(P ⊗ N_1, P/I) [r] [dr,"β^∗"']
                       _P(K,P/I) [d,hook] 
             _P(P ⊗ N_2, P/I),


where the upper row is exact and β: P ⊗ N_2 → P ⊗ N_1 is the canonical map. Now, the presence of the Reynolds operator yields that taking S_n-invariants is an exact functor. Therefore, α^∗ induces an isomorphism 

    ^S_n_P(I,P/I) ≅(β^∗) ∩^S_n_P(P ⊗ N_1, P/I).

If P/I ≅ρ as S_n-modules, the left hand side is naturally isomorphic to T_[I]_ρ, and the right hand side is isomorphic to

    (β^∗) ∩_S_n(N_1,ρ),

giving an effective method to compute the dimension of T_[I]_ρ, since β^∗ can be represented by an n_2 × n_1 matrix with entries in P/I, after choosing bases of N_1 and N_2.



  §.§.§ Row 1
 I = (x_i - x_j) + ^r. Here, N_1 = ⟨ x_i - x_j ⟩⊕⟨ p_1^r ⟩ and P/I = r S^(n) = ⟨ 1 ⟩⊕⟨ p_1 ⟩⊕⋯⊕⟨ p_1^r-1⟩, hence any f ∈_S_n(N_1,P/I) satisfies f(x_i - x_j) = 0 and f(p_1^d) = ∑_i=0^r-1α_i p_1^i for some α_i ∈. In particular, _(T_[I]) ≤ r. On the other hand, I lies in the smoothable component of , whose dimension is r. The last claim follows by noticing that the vanishing ideal of r distinct diagonal points (a_1, …, a_1), …, (a_r, …, a_r) contains (x_i - x_j) and the product of the r polynomials p_1 - n a_i whose top degree part is precisely p_1^r. 



  §.§.§ Row 2(a)+4(a)+5
 I = (x_i - x_j)_≥ 2 + ^d for d ≥ 2. Note that for d=2 this is the first ideal of row 4 while for d=3 we get the ideal of row 5 corresponding to [a:b] = [-1:n].
Next, we observe that P/(x_i-x_j)_≥ 2 has constant Hilbert function 1 in all degrees ≥ 2 and that a homogeneous polynomial of degree ≥ 2 belongs to (x_i-x_j)_≥ 2 if and only if it vanishes at (1,1,…,1).
A minimal S_n-stable generating set of I is then given by

    N_1 = ⟨ x_i^2 - x_j^2 ⟩⊕⟨ p_1(x_i - x_j)  ⟩⊕⟨ (x_i - x_j)(x_k - x_l) ⟩⊕⟨ p_1^2 - n p_2 ⟩⊕⟨ p_1^d ⟩.

Moreover, P/I = d S^(n)⊕ S^(n-1,1) = ⟨ 1 ⟩⊕⟨ p_1 ⟩⊕⋯⊕⟨ p_1^d-1⟩⊕⟨ x_i - x_j ⟩.
Hence, any f ∈_S_n(N_1,P/I) is of the form

    f((x_i - x_j)(x_k - x_l))    = 0, 
    
        f(x_i^2 - x_j^2)    = α (x_i - x_j), 
    
        f(p_1 (x_i - x_j))    = β (x_i - x_j), 
    
        f(p_1^2 - n p_2)    = ∑_i=0^d-1γ_i p_1^i, 
    
        f(p_1^d)    = ∑_i=0^d-1δ_i p_1^i.


If this f lies in ^S_n_P(I,P/I), then necessarily γ_0 = γ_1 = ⋯ = γ_d-2 = 0. Indeed, applying the Reynolds operator to p_1(x_1-x_2)x_1 ∈ I gives a non-zero rational multiple of p_1(p_1^2 - n p_2), hence f maps the latter to both a non-zero rational multiple of ∑_i=0^d-1γ_i p_1^i+1 and also to a multiple of the Reynolds operator applied to β (x_1 - x_2) x_1, which in turn is a multiple of p_1^2 - n p_2 and hence 0 in P/I.

We claim that also δ_0 = 0. For this, observe that p_1(x_1-x_2) · p_1^d-1 maps to both β(x_1-x_2) · p_1^d-1 = 0  I and to (x_1 - x_2) ∑_i=0^d-1δ_i p_1^i = δ_0(x_1-x_2)  I, hence δ_0 = 0.

So far we have proved _(^S_n_P(I,P/I)) ≤ d+2. We claim that equality holds. For this, observe that there is no linear dependence among the degree 2 generators of N_1 and that every homogeneous multiple of (x_i-x_j) and p_1^d-1 with strictly larger degree already lies in I. Therefore, there is no linear relation among the remaining α,β,γ_d-1,δ_1,…,δ_d-1 which actually involves at least one of α, β, γ_d-1. Therefore, the only possible linear relations are among δ_1, …, δ_d-1. By what we have explained above, such a relation arises from a syzygy of the generators of N_1 such that the coefficient g of p_1^d is a homogeneous polynomial which is non-zero in P/I. If (g) ≥ 2, then after rescaling we have g = p_1^k  I for some 2 ≤ k ≤ d-1. But such a syzygy does not exist since no power p_1^k (including k = 0 and k = 1) is contained in (x_i - x_j)_≥ 2. Therefore, the only case left is g ∈⟨ x_i - x_j ⟩. But this gives precisely the relation we have already used above to show δ_0 = 0. This proves _(^S_n_P(I,P/I)) = d+2.

Finally, in order to show that I is actually a singular point of _ρ, where ρ = d S^(n)⊕ S^(n-1,1), we must show that I lies in the smoothable component which is easily seen to have dimension d+1. For this, let a, b, c_2, c_3, …, c_d ∈ such that a ≠ b and c_2, c_3, …, c_d are all distinct. Define J_(a,b,c_2,c_3,…,c_d) to be the vanishing ideal of the points given by the S_n-orbit of (b,a,a,…,a) and the d-1 diagonal points (c_2,…, c_2), …, (c_d, …, c_d). The product of the d polynomials p_1 - ((n-1)a + b) and p_1 - n c_i clearly lies in J, and its top degree part is precisely p_1^d. Also, for all degree 2 generators of N_1 except for p_1^2 - np_2 there is a polynomial in J whose top degree part is precisely this generator. This is evidenced by the polynomials (x_i - x_j)(x_i + x_j - (a+b)), (x_i - x_j)(p_1 - ((n-1)a + b)) and (x_i - x_j)(x_k - x_l). Fixing c_2, …, c_d and letting (a,b) tend to (0,0) linearly, i.e., along the family (ta,tb), we obtain the ideal J_(c_2,…,c_d) = ((p_1) + ^2) ∩_(c_2,…,c_2)∩⋯∩_(c_d,…,c_d). For all distinct c_2, …, c_d, this ideal therefore lies in the smoothable component. Moreover, it still contains the above polynomials with (a,b) replaced by (0,0). We claim that J_(c_2,…,c_d) also contains p_1^2 - n p_2. But this is clear since this polynomial is obviously contained in (p_1) + ^2 and vanishes on all diagonal points. Hence, letting (c_2, …, c_d) tend to (0,…, 0) linearly, i.e., along the family (tc_2,…, tc_d) which lies entirely in the smoothable component, the limiting ideal contains p_1^2 - np_2 and all the above polynomials with all parameters replaced by 0, which is a generating set for I. Therefore, this limiting ideal must be I itself, and hence I lies in the smoothable component.

Row 2(b). I = (p_1(x_i-x_j),x_i^2-x_j^2,(x_i-x_j)(x_k-x_l)) + ^d with d ≥ 3. We first note that (x_i-x_j)_≥ 3⊆ I, hence a minimal S_n-stable generating set of I is given by

    N_1 = ⟨ x_i^2 - x_j^2 ⟩⊕⟨ p_1(x_i - x_j)  ⟩⊕⟨ (x_i - x_j)(x_k - x_l) ⟩⊕⟨ p_1^d ⟩.

Observe that this is the same as for Row 2(a) omitting the generator p_1^2 - n p_2. A simpler version of the previous argument then shows that I lies in the smoothable component; in fact, I is the associated graded ideal of every radical ideal in _ρ. Moreover, the same tangent space computations as for Row 2(a) show _(^S_n_P(I,P/I)) ≤ d+1. Indeed, the argument for δ_0 = 0 still works, and this time the γ_i do not appear. This proves that I is a smooth point of _ρ.



  §.§.§ Row 3
 I = (p_1) + ^2 is the Tanisaki ideal for the partition λ = (n-1,1) of length 2, so I is a smooth point by Proposition <ref>.



  §.§.§ Row 4(b)
 I = (p_1,x_i^2-x_j^2,(x_i-x_j)(x_k-x_l)) + ^3. For all n ≥ 3, ^3 is already contained in (p_1,x_i^2-x_j^2,(x_i-x_j)(x_k-x_l)) by the proof of the classification above. As ρ = P/I = ⟨ 1 ⟩⊕⟨ p_2 ⟩⊕⟨ x_i - x_j ⟩ = 2S^(n)⊕ S^(n-1,1), picking N_1 = ⟨ p_1 ⟩⊕⟨ x_i^2 - x_j^2 ⟩⊕⟨ (x_i - x_j)(x_k - x_l) ⟩, any f ∈^S_n_P(I,P/I) has the form

    f(p_1)    = α· 1 + β p_2, 
    
        f(x_i^2 - x_j^2)    = γ (x_i - x_j), 
    
        f((x_i-x_j)(x_k-x_l))    = 0,

so _(^S_n_P(I,P/I)) ≤ 3. Since the dimension of the smoothable component of _ρ is 3 as well, it suffices to show that I lies in the latter. For this, pick a radical ideal as in Row 4(a) but this time such that (n-1)a+b = nc. The associated graded is then I.



  §.§.§ Row 5
 I = (ap_1^2+bp_2,p_1(x_i-x_j),x_i^2-x_j^2,(x_i-x_j)(x_k-x_l)) + ^3. In case [a:b] = [-1:n], I is the ideal of Row 2(a) with d = 3, so we know that this point is singular. We claim that I is a smooth point for all different [a:b]. The point here is that the 2-dimensional torus T fixes this 1-dimensional set of ideals, and there are precisely two T-fixed points, namely [a:b] = [-1:n] and [a:b] = [1:0]. It is therefore enough to prove that the ideal obtained for [a:b] = [1:0] corresponds to a smooth point of _ρ.
First, we note that (p_1^2,p_1(x_i - x_j)) = (p_1)_≥ 2, hence ^3 is already contained in I = (p_1^2,p_1(x_i-x_j),x_i^2-x_j^2,(x_i-x_j)(x_k-x_l)), as was the case in Row 4(b).
Hence, we can choose the generators of I as an S_n-stable basis for N_1. Moreover, ρ = P/I = ⟨ 1 ⟩⊕⟨ p_1 ⟩⊕⟨ p_2 ⟩⊕⟨ x_i-x_j ⟩ = 3S^(n)⊕ S^(n-1,1). Hence, any f ∈^S_n_P(I,P/I) has the form

    f(p_1^2)    = α· 1 + β p_1 + γ p_2, 
    
        f(p_1(x_i - x_j))    = δ (x_i - x_j), 
    
        f(x_i^2 - x_j^2)    = ϵ (x_i - x_j), 
    
        f((x_i-x_j)(x_k-x_l))    = 0.

But necessarily α = 0 since f maps p_1^2(x_i - x_j) to both α (x_i - x_j) and δ p_1 (x_i - x_j) = 0  I. Hence, _(^S_n_P(I,P/I)) ≤ 4, and the dimension of the smoothable component of _ρ is 4 as well.
So finally we prove that I lies in the smoothable component. For this we can pick the vanishing ideal J of two distinct diagonal points (c,…,c), (d, …, d) and the S_n-orbit of (b,a,a,…,a) for a ≠ b where we assume (n-1)a+b = nc. Then, clearly, (p_1 - nc)(p_1 - nd) ∈ J with top degree part p_1^2. Hence, the associated graded of J must be I.



  §.§.§ Row 6
 I = (p_1,p_2,(x_i-x_j)(x_k-x_l)) + ^3. We assume n ≥ 4; the case n=3 is similar if one replaces all occurrences of (x_i-x_j)(x_k-x_l) by (x_1-x_2)(x_1-x_3)(x_2-x_3). Here, N_1 = ⟨ p_1, p_2, p_3, (x_i-x_j)(x_k-x_l)⟩ and ρ = P/I = ⟨ 1 ⟩⊕⟨ x_i - x_j ⟩⊕⟨ x_i^2 - x_j^2 ⟩ = S^(n)⊕ 2S^(n-1,1). Hence, any f ∈^S_n_P(I,P/I) has the form

    f(p_1)    = α· 1, 
    
        f(p_2)    = β· 1, 
    
        f(p_3)    = γ· 1, 
    
        f((x_i-x_j)(x_k-x_l))    = 0.

We use the containment

    p  x_1 x_2 (x_1-x_2) p_1 - (x_1^2 - x_2^2) p_2 + (x_1 - x_2) p_3 ∈ ((x_i-x_j)(x_k-x_l)),

which can be proven as before using the radicality of ((x_i-x_j)(x_k-x_l)). Hence,

    0 = f(p) = - β (x_1^2 - x_2^2) + γ (x_1 - x_2).

This implies β = γ = 0, hence _(^S_n_P(I,P/I)) ≤ 1. On the other hand, the additive group (,+) acts freely on _ρ by translation of the support along a diagonal vector in ^n, so every irreducible component of _ρ has dimension ≥ 1. Therefore, I must be a smooth point even though there is no radical ideal in _ρ by Proposition <ref>. In fact, since I is the only homogeneous ideal in _ρ, this proves that _ρ≅^1 is precisely the (,+)-orbit of I. So in some sense, I has no meaningful embedded symmetric deformations. 



  §.§.§ Row 7(a)
 I = (p_1,(x_i-x_j)(x_k-x_l)) + ^3. The tangent space dimension for n=3 can be computed to be 4 in  via the algorithm of Lehn–Terpereau, so we may assume n ≥ 4. In this case a minimal S_n-stable set of generators is given by N_1 = ⟨ p_1, p_3, p_2(x_i-x_j), (x_i-x_j)(x_k-x_l) ⟩. Moreover, ρ = P/I = ⟨ 1 ⟩⊕⟨ p_2 ⟩⊕⟨ x_i - x_j ⟩⊕⟨ x_i^2 - x_j^2 ⟩ = 2S^(n)⊕ 2S^(n-1,1).
Hence, any f ∈^S_n_P(I,P/I) has the form

    f(p_1)    = α· 1 + β p_2, 
    
        f(p_3)    = γ· 1 + δ p_2, 
    
        f(p_2(x_i-x_j))    = ϵ (x_i-x_j), 
    
        f((x_i-x_j)(x_k-x_l))    = 0.

We claim α = 0. For this, we again look at the first polynomial in the proof of Lemma <ref>:

    p     n(n-1)(x_1^3-x_2^3) - (n^2-3n+3)p_2(x_1-x_2) 
       - (2n-3)p_1(x_1^2-x_2^2) + n(n-2)x_n p_1(x_1-x_2) ∈ ((x_i-x_j)(x_k-x_l)).

Therefore, f(p) = 0. At the same time, x_1^3 - x_2^3 ∈ I spans a copy of S^(n-1,1), so it must map to ϵ' (x_1 - x_2) for some ϵ'. Hence,

    0 = f(p)    = (n(n-1) ϵ' - (n^2-3n+3) ϵ) (x_1-x_2) 
       + n(n-2) α x_n(x_1-x_2) - (2n-3)(x_1^2 - x_2^2) α.

Hence, the coefficient of x_1-x_2 must vanish. Moreover, if α was non-zero, then necessarily

    g  n(n-2) x_n(x_1-x_2) - (2n-3)(x_1^2 - x_2^2) ∈ I.

This is equivalent to the existence of a linear form h with g - p_1 h ∈ ((x_i-x_j)(x_k-x_l)). Since the latter ideal is the vanishing ideal of the set {S_n.(b,a,a,…,a) : a,b ∈}, we only need to show that there is no linear form h such that g - p_1 h vanishes on this entire set. The polynomial g vanishes on all the points with b not occurring as the first or second coordinate. A simple computation shows that then necessarily h = λ (x_1-x_2) for some λ∈. Another computation shows that there is no λ such that g - λ p_1 (x_1-x_2) vanishes on (b,a,a,…,a) for all a,b ∈, a contradiction. This shows α = 0 and hence _(^S_n_P(I,P/I)) ≤ 4. 


Since the smoothable component of _ρ has dimension 4 as well, we only need to show that I is smoothable. Indeed, let J be the vanishing ideal of the two S_n-orbits of (b,a,a,…,a) and (d,c,c,…,c) where a ≠ b, c ≠ d and p_1(b,a,a,…,a) = p_1(d,c,c,…,c). The associated graded of J then clearly contains p_1 and (x_i - x_j)(x_k - x_l), hence must be I.

Row 7(b). I = (p_1,p_2,(x_i-x_j)(x_k-x_l)) + ^4 for n ≥ 4. Here, ^4 is already contained in (p_1,p_2,(x_i-x_j)(x_k-x_l)), so these generators form a minimal S_n-stable basis of N_1. Moreover, ρ = P/I = ⟨ 1 ⟩⊕⟨ p_3 ⟩⊕⟨ x_i - x_j ⟩⊕⟨ x_i^2 - x_j^2 ⟩ = 2S^(n)⊕ 2 S^(n-1,1).
Hence, any f ∈^S_n_P(I,P/I) has the form

    f(p_1)    = α· 1 + β p_3, 
    
        f(p_2)    = γ· 1 + δ p_3, 
    
        f((x_i-x_j)(x_k-x_l))    = 0,

yielding _(^S_n_P(I,P/I)) ≤ 4. Therefore, it suffices again to prove that I is smoothable. One may pick any radical ideal J as we did for Row 7(a) but such that additionally p_2(b,a,a,…,a) = p_2(d,c,c,…,c) holds. Then, the associated graded of J contains p_1, p_2 and (x_i-x_j)(x_k-x_l), hence must be I.

Row 7(c). I = (p_1^2,p_2,ap_1(x_i-x_j)+b(x_i^2-x_j^2),(x_i-x_j)(x_k-x_l)) + ^3. As for Row 5 above, the 2-dimensional torus T fixes this 1-dimensional set of ideals and there are precisely two T-fixed points, namely [a:b] ∈{[1:0],[-2:n]}. It suffices to prove that the first choice yields a singular point while the second choice yields a smooth point. For n=3 this can again be checked in , so we may assume n ≥ 4.
We first assume [a:b] = [1:0]. We let N_1 = ⟨ p_1^2, p_2, p_3, p_1(x_i-x_j), (x_i-x_j)(x_k-x_l) ⟩ and ρ = P/I = ⟨ 1 ⟩⊕⟨ p_1 ⟩⊕⟨ x_i - x_j ⟩⊕⟨ x_i^2 - x_j^2 ⟩ = 2S^(n)⊕ 2S^(n-1,1).
Hence, any f ∈^S_n_P(I,P/I) has the form

    f(p_1^2)    = α_0 · 1 + α_1 p_1, 
    
        f(p_2)    = β_0 · 1 + β_1 p_1, 
    
        f(p_3)    = γ_0 · 1 + γ_1 p_1, 
    
        f(p_1(x_i-x_j))    = δ_0 (x_i-x_j) + δ_1 (x_i^2-x_j^2), 
    
        f((x_i-x_j)(x_k-x_l))    = 0.

We claim α_0 = β_0 = γ_0 = 0. This follows again from the containment

    p_3(x_1-x_2) - p_2(x_1^2-x_2^2) + p_1 x_1 x_2 (x_1-x_2) ∈ ((x_i-x_j)(x_k-x_l)).

Applying f, we obtain

    0 = γ_0 (x_1 - x_2) - β_0 (x_1^2 - x_2^2) + α_0/n x_2(x_1-x_2).

Here, we have used n p_1 x_1 = p_1^2 + ∑_i=2^n (x_1 - x_i) p_1 for the last term. Note moreover that p_1 x_1 x_2 (x_1-x_2). must also map to δ_0 (x_1-x_2)x_1x_2 + δ_1 (x_1^2-x_2^2)x_1x_2 = 0  I, because ^3 ⊆ I. Therefore, α_0 = 0 since x_2(x_1-x_2) ∉I because otherwise x_1^2 - x_2^2 = x_1(x_1-x_2) + x_2(x_1-x_2) ∈ I which is not the case. Finally, we also get β_0 = γ_0 = 0, as claimed. 


We claim next that no more relations among α_1,β_1,γ_1,δ_0,δ_1 exist. Clearly, there is no syzygy in degree 2. For a syzygy in degree 3, the coefficient of p_3 must be zero, otherwise p_3 would be contained in the ideal generated by the remaining generators of N_1. Moreover, the only linear relation among α_1,β_1,γ_1,δ_0,δ_1 which could follow from a syzygy in degree 3 would be δ_0 = 0 because ^3 ⊆ I and (p_1)_≥ 2⊆ I. In fact, δ_0 = 0 is equivalent to the existence of a linear form g such that p_1(x_1-x_2)g ∈ (p_1^2, p_2, (x_i-x_j)(x_k-x_l))_3 and g(x_1-x_2) ∉I. We claim that such a g does not exist.

To see this, we may assume g ∈⟨ x_i - x_j ⟩. Note that if g does not involve the variables x_1 and x_2, then (x_1-x_2)g ∈ ((x_i-x_j)(x_k-x_l)) ⊆ I, so we may assume that g involves only x_1 and x_2. The containment p_1(x_1-x_2)g ∈ (p_1^2, p_2, (x_i-x_j)(x_k-x_l))_3 is equivalent to the existence of two linear forms g',g” such that

    p_1(x_1-x_2)g + p_1^2g' + p_2 g”∈ ((x_i-x_j)(x_k-x_l)),

which in turn is equivalent to the vanishing of this polynomial on the vanishing set of ((x_i-x_j)(x_k-x_l)). Picking a point (a,a,b,a,…,a) where b is not in position 1 or 2, the first summand vanishes and we get

    ((n-1)a+b)^2g'(a,a,b,a,…,a) + ((n-1)a^2+b^2)g”(a,a,b,a,…,a) = 0

for all a,b ∈, and all analogous equations where b is at the i-th position with 3 ≤ i ≤ n. Hence, in all these n-2 equations all coefficients of a^3,a^2b,ab^2,b^3, which are linear forms in the coefficients of g' and g”, must vanish. A computation shows that the only way in which this is possible is that g' = μ (x_1-x_2) and g” = ν (x_1-x_2) for some μ, ν∈. Coming back to p_1(x_1-x_2)g + p_1^2g' + p_2 g”, applying the transposition (1,2) to this polynomial, adding the result to it and evaluating on (b,a,a,…,a) shows that g = λ (x_1 + x_2) for some λ∈. Hence, the polynomial

    λ p_1 (x_1+x_2) + μ p_1^2 + ν p_2

must vanish on (b,a,a,…,a) for all a,b ∈. It can be checked that this is only possible if λ = μ = ν = 0. In particular, we obtain g = 0, so a g with the desired properties does not exist.

Finally, any syzygy in degree ≥ 4 does not yield any linear relation among the remaining α_1,β_1,γ_1,δ_0,δ_1, again because ^3 ⊆ I and (p_1)_≥ 2⊆ I. Thus, we have proved _(^S_n_P(I,P/I)) = 5.

We now consider the ideal I corresponding to [-2:n]. Note that it is enough to prove _(^S_n_P(I,P/I)) ≤ 4 and that I lies in the smoothable component in order to see that I is a smooth point of _ρ. Then necessarily all ideals in the one-dimensional family lie in the smoothable component. Moreover, all of them must be smooth points of _ρ except for the other T-fixed point corresponding to [1:0] above, which then must be singular.

First, ^3 is already contained in the ideal generated by

    N_1 = ⟨ p_1^2, p_2, -2p_1(x_i - x_j) + n(x_i^2 - x_j^2), (x_i-x_j)(x_k-x_l) ⟩.

Moreover, ρ = P/I = ⟨ 1 ⟩⊕⟨ p_1 ⟩⊕⟨ x_i-x_j ⟩⊕⟨ p_1(x_i-x_j) ⟩ = 2S^(n)⊕ 2S^(n-1,1).
Hence, any f ∈^S_n_P(I,P/I) has the form

    f(p_1^2)    = α_0 · 1 + α_1 p_1, 
    
        f(p_2)    = β_0 · 1 + β_1 p_1, 
    
        f(-2p_1(x_i-x_j) + n(x_i^2 - x_j^2))    = γ_0 (x_i-x_j) + γ_1 p_1(x_i-x_j), 
    
        f((x_i-x_j)(x_k-x_l))    = 0.

We use the containment

    p     (n-2) p_1^2(x_1-x_2) - n(n-2)p_2(x_1-x_2) 
       - (n-1)(-2p_1(x_1-x_2)+n(x_1^2-x_2^2))(x_1-x_2) 
       + 2(n-1)(-2p_1(x_1-x_3)+n(x_1^2-x_3^2))(x_1-x_2) ∈ ((x_i-x_j)(x_k-x_l)).

Hence,

    0 = f(p)    = ((n-2)α_0- n(n-2)β_0)(x_1-x_2) + ((n-2)α_1 -n(n-2)β_1) p_1(x_1-x_2) 
       - (n-1) γ_0 (x_1-x_2)^2 + 2(n-1) γ_0 (x_1-x_2)(x_1-x_3).

We obtain α_0 = n β_0 and at least one more non-trivial linear relation among α_1, β_1, γ_0, hence _(^S_n_P(I,P/I)) ≤ 4.

Finally, we need to show that I lies in the smoothable component of _ρ. For this, consider the vanishing ideal J of the two S_n-orbits of (b,a,a,…,a) and (d,c,c,…,c) with a,b,c,d ∈ and a ≠ b, c ≠ d, b ≠ d but such that b-a = d-c. It can be checked that then the polynomial

    -2 p_1(x_1-x_2) + n(x_1^2 - x_2^2) - (n-2)(b-a)(x_1-x_2)

lies in J. With this it is easy to see that all generators of I lie in the associated graded of J, hence the latter must agree with I. 




 §.§ Irreducibility in Table <ref>



    Let ρ be any of the S_n-representations in Table <ref>. Then _ρ is irreducible. 



For rows 1, 3, 6 and 8–13 this follows from smoothness and connectedness in each case.

For the remaining rows 2, 4, 5 and 7, we have seen that all homogeneous symmetric ideals lie in the respective smoothable components, proving connectedness of _ρ in each case. It is therefore enough to consider the homogeneous symmetric ideals I which are singular points of _ρ and to prove, in each case, that all ideals whose associated graded equals I lie in the smoothable component again.

The three singular ideals in rows 2, 4 and 5 can all be written as I = (x_i-x_j)_≥ 2 + ^d, where d=2 for row 4, d=3 for row 5, and d>3 for row 2.
Let now J ⊆ P be any symmetric ideal whose associated graded is I. For d ≥ 3, we use

    N_1 = ⟨ x_i^2 - x_j^2 ⟩⊕⟨ p_1(x_i - x_j)  ⟩⊕⟨ (x_i - x_j)(x_k - x_l) ⟩⊕⟨ p_1^2 - n p_2 ⟩⊕⟨ p_1^d ⟩

as generators for I. We claim that J has generators of the form

    (x_i - x_j)(x_k - x_l), 
    
        p_1(x_i - x_j) + a_1(x_i - x_j), 
    
        x_i^2 - x_j^2 + a_2(x_i - x_j), 
    
        p_1^2 - np_2 + b_1 p_1 + b_2, 
    
        p_1^d + ∑_i=0^d-1 c_i p_1^i,

for some a_1,a_2,b_1,b_2,c_i ∈. For instance, since (x_i - x_j)(x_k - x_l) ∈ I and I is the associated graded of J, there is an element (x_i - x_j)(x_k - x_l) + ℓ + c ∈ J for some linear form ℓ and some constant c. Applying the transposition (i,j) to this polynomial yields -(x_i - x_j)(x_k - x_l) + (i,j)(ℓ) + c ∈ J, hence the sum must lie in J as well, and equals ℓ + (i,j)(ℓ) + 2c. It follows that ℓ = λ (x_i - x_j) for some λ∈ and that c = 0. For the same reason, we also get ℓ = μ (x_k - x_l) for some μ∈, hence ℓ = 0. The reasoning for the other generators is similar.

Note that for all k ≥ 1, we have p_1^k(x_1 - x_2) ≡ -a_1 p_1^k-1(x_1-x_2)  J and hence p_1^k(x_1-x_2) ≡ (-a_1)^k(x_1-x_2)  J for all k ≥ 0.
Let c_d  1. Then we have

    J    ∋( ∑_k=0^d c_k p_1^k )(x_1-x_2) - (p_1(x_1-x_2) + a_1(x_1-x_2)) ( ∑_k=0^d-1 c_k+1p_1^k ) 
       = c_0 (x_1-x_2) - a_1 ∑_k=0^d-1 c_k+1 p_1^k (x_1-x_2) 
       ≡( (-a_1)^d + c_d-1(-a_1)^d-1 + … + c_0 ) (x_1-x_2)  J .

Hence, we must have (-a_1)^d + c_d-1(-a_1)^d-1 + … + c_0 = 0.
Moreover,

    J    ∋-1/(n-2)!∑_σ∈ S_nσ (x_1(p_1(x_1-x_2) + a_1(x_1-x_2))) - p_1(p_1^2 - np_2 + b_1 p_1 + b_2) 
       = a_1 (p_1^2 - n p_2) - b_1 p_1^2 - b_2 p_1 
       ≡ -b_1 p_1^2 - (a_1 b_1 + b_2)p_1 - a_1 b_2  J .

Since d ≥ 3, we obtain b_1 = 0 and hence also b_2 = 0. Together with the previous equation we obtain a flat family over an irreducible base isomorphic to an affine space of dimension d+1 (since we can eliminate the three variables b_1,b_2,c_0). Since the subset of the smoothable component consisting of all ideals whose associated graded is I has dimension d+1 as well, the general member of this family must be a smoothable ideal, hence every member of the family is a smoothable ideal, proving our claim.

For d=2 we use instead the monomial generators of I = ^2 and obtain the following generators for J:

    x_i x_j + a_1 (x_i + x_j) + a_2 p_1 + a_3, 
    
        x_i^2 + b_1 x_i + b_2 p_1 + b_3.

Since _(P/J) = n+1, the vanishing set of J must contain at least one point on the diagonal. Hence, using the 𝔾_a-action, we may assume that the origin is contained in the vanishing set of J, i.e., a_3 = b_3 = 0. But which parameters a_1,a_2,b_1,b_2 do then actually give us an ideal J of the correct colength? Clearly, the associated graded of J contains ^2 for every choice of parameters. But equality need not hold in general; the obstruction are linear forms contained in J. These linear forms, however, can only arise from syzygies of the monomial generators of ^2 in degree 3, i.e., from the identities x_i^2 x_j = x_i (x_i x_j) and (x_i x_j) x_k = x_i (x_j x_k). Using these, we obtain all possible linear forms contained in J. Their coefficients yield 3 linearly independent homogeneous polynomials F_1,F_2,F_3 of degree 2 in a_1,a_2,b_1,b_2. The coefficients of the F_i, in turn, are integer constants or linear polynomials in n. Treating n as a variable, we consider the ideal (F_1,F_2,F_3) in the polynomial ring [a_1,a_2,b_1,b_2,n]. A computation in  then provides the minimal primes of this ideal (over ). They are 𝔭_1  (a_1,b_1,a_2-b_2) and 𝔭_2  (b_2+a_1+(n-1)a_2, a_1^2 + (n-2)a_1 a_2 + a_2 b_1). Both are geometrically prime, i.e., prime over , even if n is regarded as an integer again. The first prime ideal 𝔭_1 only parametrizes ideals defining a subscheme of ^n for which the origin is not reduced. Precisely, apart from ^2 itself, the members of the family defined by 𝔭_1 have the form ((p_1) + ^2) ∩_(c,c,…,c) for some c ∈∖{0}. We have already seen that these ideals all lie in the smoothable component.
The flat family defined by 𝔭_2 must then contain, in particular, the set of all symmetric ideals whose associated graded is ^2 and which define a subscheme of ^n for which the origin is a reduced point. Now, the general radical ideal in _ρ has associated graded equal to ^2, and the subset whose vanishing set contains the origin has dimension 2. On the other hand, ([a_1,a_2,b_1,b_2]/𝔭_2) = 2 as well. Therefore, by irreducibility, every member of the family defined by 𝔭_2 is smoothable, concluding the argument.

For the last remaining singular ideal I = (p_1^2,p_2,p_1(x_i-x_j),(x_i-x_j)(x_k-x_l)) + ^3 of Row 7(c) the reasoning is as follows. We assume first n ≥ 4.
Let J ⊆ P be any symmetric ideal with (J) = I. We use

    N_1 = ⟨ p_1^2 ⟩⊕⟨ p_2 ⟩⊕⟨ p_1(x_i-x_j) ⟩⊕⟨ (x_i-x_j)(x_k-x_l) ⟩⊕⟨ p_3 ⟩

as generators for I.
Then, clearly, J has generators of the form 

    (x_i-x_j)(x_k-x_l), 
    
        p_1^2 + A p_1 + B, 
    
        p_2 + C p_1 + D, 
    
        p_1(x_i-x_j) + E (x_i-x_j), 
    
        p_3 + F p_1 + G.

First, we note that I is not the associated graded ideal of any radical ideal. The reason is that the only possible radical ideals are the vanishing ideals of two distinct orbits represented by (b,a,a,…,a) and (d,c,c,…,c) for a ≠ b, c ≠ d. If p_1(b,a,a,…,a) = p_1(d,c,c,…,c), then clearly p_1 is contained in the associated graded but not in I. Otherwise the associated graded contains some linear combination of p_1(x_1-x_2) and x_1^2 - x_2^2 in which the latter occurs with a non-zero coefficient while I only contains the former.

Next, we claim that the vanishing set of J contains a diagonal point. To see this, note that the vanishing set of a symmetric ideal J whose associated graded is I, which is not radical and whose vanishing set does not contain a diagonal point must consist of precisely one orbit generated by (b,a,a,…,a) for a ≠ b. Every point must have multiplicity 2 by symmetry and since _(P/J) = 2n. The stabilizer of the subscheme of (P/J) supported at a single point, say (b,a,a,…,a) is precisely S_n-1, embedded into S_n as the permutations fixing the first index. Since its multiplicity is 2, the subscheme supported at (b,a,a,…,a) is defined by the ideal ((x_1-b)^2, x_2 - a, …, x_n - a), so that J is the intersection of all permutations of this ideal. However, this is impossible since the associated graded then would not contain p_1(x_1-x_2) as can be checked after reducing to a=0 via the additive group action.

Hence, we may assume V(J) contains the origin, so that all constant terms of the above generators are zero. If now E ≠ 0, then localizing at p_1 + E shows that (x_i - x_j) is contained in the localized ideal, so this ideal corresponds to Row 1, contradicting the fact that P/J ≅ρ. Hence, E = 0. If A ≠ 0, we can localize at p_1 + A and obtain the ideal generated by (x_i-x_j)(x_k-x_l) and p_1,p_2,p_3 which is contained in the ideal of row 6; this is again impossible. Hence, A = 0 as well. The only parameters we are left with are C and F. This proves that the subset of _ρ of ideals J with (J) = I and whose vanishing set contains the origin is either irreducible of dimension 2 or has dimension ≤ 1. Therefore, taking the additive group action back in, the subset of _ρ of ideals J with (J) = I (without conditions on V(J)) is either irreducible of dimension 3 or has dimension ≤ 2.

Let now H be the locally closed subset of _ρ of all ideals having the same Hilbert function h as I. We endow H with the reduced scheme structure. By Subsection <ref>, there is a natural map φ H ⟶^S_n ×^∗_ρ,h(^n), sending an ideal I to its associated graded (I). Note that the target of φ consists precisely of the ideals in Row 7(c), so it is irreducible of dimension 1. Moreover, all fibers of φ over closed points corresponding to homogeneous ideals I' have a component of dimension at least 3 containing I'. Indeed, for any such ideal in Row 7(c) other than I itself consider the vanishing ideal J of two distinct orbits represented by (b,a,a,…,a) and (d,c,c,…,c), a ≠ b, c ≠ d. If p_1(b,a,a,…,a) ≠ p_1(d,c,c,…,c), then there is a unique α∈ such that α p_1(x_1 - x_2) + (x_1^2 - x_2^2) + β (x_1 - x_2) ∈ J for some β∈. This shows that J belongs to Row 7(c).
Conversely, given any α∈, the existence of β such that the above polynomial belongs to J defines exactly one linear condition on a,b,c,d. Observe that this means that the fiber of φ over (J) contains a 3-dimensional subset of the smoothable component containing (J).

Let Z be the smoothable component of _ρ. Consider the composition

    (Z ∩ H)_red↪ H φ⟶^S_n ×^∗_ρ,h(^n).

By the upper semicontinuity of fiber dimensions (on the source!), the intersection φ^-1([I]) ∩ Z has dimension at least 3. This proves, first, that φ^-1([I]) is actually irreducible of dimension 3 and, second, that this entire fiber φ^-1([I]) lies in the smoothable component Z. This in turn proves the irreducibility of _ρ for n ≥ 4.

For n = 3, the role of (x_i-x_j)(x_k-x_l) is played by (x_1-x_2)(x_1-x_3)(x_2-x_3). With this replacement all arguments stay the same.



































§ CONCLUSION




 §.§ Connectedness, irreducibility, singularities

The last two properties are in general not well-understood even for usual Hilbert schemes of points, and so the same is true in the invariant setting. For general reductive groups, invariant Hilbert schemes may be disconnected, see <cit.> and the references therein. On the other hand, all examples we have seen in this article are connected.


    Is _ρ connected whenever it is non-empty, at least if ρ is a direct sum of permutation modules?



    What is a minimal example of ρ such that _ρ is reducible?


As a consequence of the next proposition, reducible examples exist, for example, in the case where ρ = [S_n]^l is a direct sum of l regular representations if n and l are large enough.


    If ρ = [S_n]^⊕ l, the Hilbert–Chow morphism
    
    γ_ρ⟶_l(^n/S_n)

is an isomorphism.



    There are inverse natural transformations between the S_n-invariant Hilbert functor of ^n for ρ = [S_n]^⊕ l and the usual Hilbert functor of ^n/S_n for length l subschemes. By <cit.>, for a finite type -scheme T, the natural transformation corresponding to the Hilbert–Chow morphism
    
    ℋilb^S_n_[S_n]^⊕ l(^n)(T) →ℋilb_l(^n/S_n)(T)

    sends an S_n-stable quasi-coherent ideal sheaf ℐ⊆𝒪_T[x_1,…,x_n] to its invariant subsheaf
    
    ℐ^S_n⊆𝒪_T[p_1,…,p_n].

    The inverse map sends a quasi-coherent ideal sheaf 𝒥⊆𝒪_T[p_1,…,p_n] to
    
    𝒥⊗_𝒪_T[p_1,…,p_n]𝒪_T[x_1,…,x_n] ↪𝒪_T[x_1,…,x_n].

    This map is well-defined because 𝒪_T[x_1,…,x_n] is a free module over 𝒪_T[p_1,…,p_n] of rank n!. More precisely, the quotient
    
    ( 𝒪_T[p_1,…,p_n] / 𝒥) ⊗_𝒪_T[p_1,…,p_n]𝒪_T[x_1,…,x_n]

    has the correct isotypic decomposition since 𝒪_T[p_1,…,p_n] / 𝒥 is locally free of rank l.
    In order to prove that the two maps are inverse to each other, observe that
    
    𝒥⊆(𝒥⊗_𝒪_T[p_1,…,p_n]𝒪_T[x_1,…,x_n])^S_n

    and
    
    ℐ^S_n⊗_𝒪_T[p_1,…,p_n]𝒪_T[x_1,…,x_n] ⊆ℐ.

    The two natural surjections induced by these inclusions are surjections of locally free 𝒪_T-modules of the same finite rank and hence isomorphisms, proving equality.
    

In Proposition <ref>, the target is reducible for every n ≥ 3 if l is sufficiently large by <cit.>, hence so is the source. Another series of examples with precisely two irreducible components is given by n ≥ 4 and l = 8 by <cit.>.




 §.§ Stabilization

Given an S_n-representation ρ = ⊕_i w_i S^λ^i, and k ≥ 0, we let ρ(k) be the S_n+k-representation obtained from replacing each λ^i = (λ^i_1,…,λ^i_m_i) ⊢ n by λ^i(k)  (λ^i_1 + k, λ^i_2,…,λ^i_m_i) ⊢ n+k.


    Is it true that for every S_n-representation ρ there exists k_0 such that for all k ≥ k_0 we have
    
    _ρ(k)≅_ρ(k+1)?



Our results seem to suggest an affirmative answer but we have not succeeded in defining natural maps.




alpha

