Keywords

1 Introduction

Let Q be a connected complex algebraic group. Set \(\mathfrak{q} =\mathop{ \mathrm{Lie}}\nolimits Q\). Then and . We will call the latter object the algebra of symmetric invariants of \(\mathfrak{q}\). An important property of is that it is isomorphic to \(Z\mathbf{U}(\mathfrak{q})\) as an algebra by a classical result of M. Duflo (here \(Z\mathbf{U}(\mathfrak{q})\) is the centre of the universal enveloping algebra of \(\mathfrak{q}\)).

Let \(\mathfrak{g}\) be a reductive Lie algebra. Then by the Chevalley restriction theorem is a polynomial ring (in \(\mathrm{rk\,}\mathfrak{g}\) variables). A quest for non-reductive Lie algebras with a similar property has recently become a trend in invariant theory. Here we consider finite-dimensional representations \(\rho: \mathfrak{g} \rightarrow \mathfrak{g}\mathfrak{l}(V )\) of \(\mathfrak{g}\) and the corresponding semi-direct products \(\mathfrak{q} = \mathfrak{g}\ltimes V\). The Lie bracket on \(\mathfrak{q}\) is defined by

$$\displaystyle{ [\xi +v,\eta +u] = [\xi,\eta ] +\rho (\xi )u -\rho (\eta )v }$$
(1)

for all \(\xi,\eta \in \mathfrak{g}\), v, u ∈ V. Let G be a connected simply connected Lie group with \(\mathop{\mathrm{Lie}}\nolimits G = \mathfrak{g}\). Then \(\mathfrak{q} =\mathop{ \mathrm{Lie}}\nolimits Q\) with \(Q = G\ltimes \exp (V )\).

It is easy to see that \(\mathbb{C}[V ^{{\ast}}]^{G} \subset \mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}}\) and therefore \(\mathbb{C}[V ^{{\ast}}]^{G}\) must be a polynomial ring if \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}}\) is, see [10, Section 3]. Classification of the representations of complex simple algebraic groups with free algebras of invariants was carried out by Schwarz [7] and independently by Adamovich and Golovina [1]. One such representation is the spin-representation of \(\mathop{\mathrm{Spin}}\nolimits _{7}\), which leads to \(Q =\mathop{ \mathrm{Spin}}\nolimits _{7}\ltimes \mathbb{C}^{8}\). Here \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}}\) is a polynomial ring in three variables generated by invariants of bi-degrees (0, 2), (2, 2), (6, 4) with respect to the decomposition \(\mathfrak{q} = \mathfrak{s}\mathfrak{o}_{7}\oplus \mathbb{C}^{8}\), see [10, Proposition 3.10].

In this paper, we treat another example, G = SL n , \(V = m(\mathbb{C}^{n})^{{\ast}}\oplus k\mathbb{C}^{n}\) with \(n\geqslant 2\), \(m\geqslant 1\), \(m\geqslant k\). Here \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}}\) is a polynomial ring in exactly the following three cases:

  • k = 0, \(m\leqslant n + 1\), and \(n \equiv t\pmod m\) with t ∈ {−1, 0, 1};

  • m = k, k ∈ {n − 2, n − 1};

  • \(n\geqslant m> k> 0\) and m − k divides n − m.

We also briefly discuss semi-direct products arising as \(\mathbb{Z}_{2}\)-contractions of reductive Lie algebras.

2 Symmetric Invariants and Generic Stabilisers

Let \(\mathfrak{q} =\mathop{ \mathrm{Lie}}\nolimits Q\) be an algebraic Lie algebra, Q a connected algebraic group. The index of \(\mathfrak{q}\) is defined as

$$\displaystyle{ \mathrm{ind}\mathfrak{q} =\min _{\gamma \in \mathfrak{q}^{{\ast}}}\mathrm{dim}\mathfrak{q}_{\gamma }, }$$

where \(\mathfrak{q}_{\gamma }\) is the stabiliser of γ in \(\mathfrak{q}\). In view of Rosenlicht’s theorem, \(\mathrm{ind}\mathfrak{q} =\mathrm{ tr.deg}\,\mathbb{C}(\mathfrak{q}^{{\ast}})^{Q}\). In case \(\mathrm{ind}\mathfrak{q} = 0\), we have \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}} = \mathbb{C}\). For a reductive \(\mathfrak{g}\), \(\mathrm{ind}\mathfrak{g} =\mathrm{ rk\,}\mathfrak{g}\). Recall that \((\mathrm{dim}\mathfrak{g} +\mathrm{ rk\,}\mathfrak{g})/2\) is the dimension of a Borel subalgebra of \(\mathfrak{g}\). For \(\mathfrak{q}\), set \(\mathbf{b}(\mathfrak{q}):= (\mathrm{ind}\mathfrak{q} + \mathrm{dim}\mathfrak{q})/2\).

Let {ξ i } be a basis of \(\mathfrak{q}\) and the structural matrix with entries in \(\mathfrak{q}\). This is a skew-symmetric matrix of rank \(\mathrm{dim}\mathfrak{q} -\mathrm{ ind}\mathfrak{q}\). Let us take Pfaffians of the principal minors of of size and let \(\mathbf{p} = \mathbf{p}_{\mathfrak{q}}\) be their greatest common divisor. Then p is called the fundamental semi-invariant of \(\mathfrak{q}\). The zero set of p is the maximal divisor in the so called singular set

$$\displaystyle{ \mathfrak{q}_{\mathrm{sing}}^{{\ast}} =\{\gamma \in \mathfrak{q}^{{\ast}}\mid \mathrm{dim}\mathfrak{q}_{\gamma }>\mathrm{ ind}\mathfrak{q}\} }$$

of \(\mathfrak{q}\). Since \(\mathfrak{q}_{\mathrm{sing}}^{{\ast}}\) is clearly a Q-stable subset, p is indeed a semi-invariant, \(Q\cdot \mathbf{p} \subset \mathbb{C}\mathbf{p}\). One says that \(\mathfrak{q}\) has the “codim-2” property (satisfies the “codim-2” condition), if \(\mathrm{dim}\mathfrak{q}_{\mathrm{sing}}^{{\ast}}\leqslant \mathrm{dim}\mathfrak{q} - 2\) or equivalently if p = 1.

Suppose that are homogenous algebraically independent polynomials. The Jacobian locus of these polynomials consists of all \(\gamma \in \mathfrak{q}^{{\ast}}\) such that the differentials d γ F 1, …, d γ F r are linearly dependent. In other words, if and only if (dF 1 ∧ … ∧ dF r ) γ = 0. The set is a proper Zariski closed subset of \(\mathfrak{q}^{{\ast}}\).

Suppose that does not contain divisors. Then by the characteristic zero version of a result of Skryabin, see [5, Theorem 1.1], \(\mathbb{C}[F_{1},\ldots,F_{r}]\) is an algebraically closed subalgebra of , each that is algebraic over \(\mathbb{C}(F_{1},\ldots,F_{r})\) is contained in \(\mathbb{C}[F_{1},\ldots,F_{r}]\).

Theorem 1 (cf. [3, Section 5.8])

Suppose that \(\mathbf{p}_{\mathfrak{q}} = 1\) and suppose that are homogeneous algebraically independent polynomials such that \(r =\mathrm{ ind}\mathfrak{q}\) and \(\sum _{i=1}^{r}\mathrm{deg}H_{i} = \mathbf{b}(\mathfrak{q})\) . Then is a polynomial ring in r generators.

Proof

Under our assumptions , see [5, Theorem 1.2] and [9, Section 2]. Therefore \(\mathbb{C}[H_{1},\ldots,H_{r}]\) is an algebraically closed subalgebra of by [5, Theorem 1.1]. Since , each symmetric \(\mathfrak{q}\)-invariant is algebraic over \(\mathbb{C}[H_{1},\ldots,H_{r}]\) and hence is contained in it. □

For semi-direct products, we have some specific approaches to the symmetric invariants. Suppose now that \(\mathfrak{g} =\mathop{ \mathrm{Lie}}\nolimits G\) is a reductive Lie algebra, no non-zero ideal of \(\mathfrak{g}\) acts on V trivially, G is connected, and \(\mathfrak{q} = \mathfrak{g}\ltimes V\), where V is a finite-dimensional G-module.

The vector space decomposition \(\mathfrak{q} = \mathfrak{g}\oplus V\) leads to \(\mathfrak{q}^{{\ast}} = \mathfrak{g}\oplus V ^{{\ast}}\), where we identify \(\mathfrak{g}\) with \(\mathfrak{g}^{{\ast}}\). Each element x ∈ V ∗ is considered as a point of \(\mathfrak{q}^{{\ast}}\) that is zero on \(\mathfrak{g}\). We have exp(V )⋅ x = ad∗(V )⋅ x + x, where each element of ad∗(V )⋅ x is zero on V. Note that \(\mathrm{ad}^{{\ast}}(V )\cdot x \subset \mathrm{ Ann}(\mathfrak{g}_{x}) \subset \mathfrak{g}\) and dim(ad∗(V )⋅ x) is equal to \(\mathrm{dim}(\mathrm{ad}^{{\ast}}(\mathfrak{g})\cdot x) = \mathrm{dim}\mathfrak{g} -\mathrm{dim}\mathfrak{g}_{x}\). Therefore \(\mathrm{ad}^{{\ast}}(V )\cdot x =\mathrm{ Ann}(\mathfrak{g}_{x})\).

The decomposition \(\mathfrak{q} = \mathfrak{g}\oplus V\) defines also a bi-grading on and clearly is a bi-homogeneous subalgebra, cf. [10, Lemma 2.12].

A statement is true for a “generic x” if and only if this statement is true for all points of a non-empty open subset.

Lemma 1

A function \(F \in \mathbb{C}[\mathfrak{q}^{{\ast}}]\) is a V -invariant if and only if F(ξ + ad∗(V )⋅ x, x) = F(ξ, x) for generic x ∈ V ∗ and any \(\xi \in \mathfrak{g}\) .

Proof

Condition of the lemma guaranties that for each v ∈ V, exp(v)⋅ F = F on a non-empty open subset of \(\mathfrak{q}^{{\ast}}\). Hence F is a V -invariant. □

For x ∈ V ∗, let \(\varphi _{x}\!: \mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q} \rightarrow \mathbb{C}[\mathfrak{g} + x]^{G_{x}\ltimes \exp (V )}\) be the restriction map. By [10, Lemma 2.5] . Moreover, if we identify \(\mathfrak{g} + x\) with \(\mathfrak{g}\) choosing x as the origin, then for any \(\mathfrak{q}\)-invariant F [10, Section 2]. Under certain assumptions on G and V the restriction map φ x is surjective, more details will be given shortly.

There is a non-empty open subset U ⊂ V ∗ such that the stabilisers G x and G y are conjugate in G for any pair of points x, y ∈ U see e.g. [8, Theorem 7.2]. Any representative of the conjugacy class {hG x h −1∣h ∈ G, x ∈ U} is said to be a a generic stabiliser of the G-action on V ∗.

There is one easy to handle case, \(\mathfrak{g}_{x} = 0\) for a generic x ∈ V ∗. Here \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q} = \mathbb{C}[V ^{{\ast}}]^{G}\), see e.g. [10, Example 3.1], and \(\xi +y \in \mathfrak{q}_{\mathrm{sing}}^{{\ast}}\) only if \(\mathfrak{g}_{y}\neq 0\), where \(\xi \in \mathfrak{g}\), y ∈ V ∗. The case \(\mathrm{ind}\mathfrak{g}_{x} = 1\) is more involved.

Lemma 2

Assume that G has no proper semi-invariants in \(\mathbb{C}[V ^{{\ast}}]\) . Suppose that \(\mathrm{ind}\mathfrak{g}_{x} = 1\), , and the map φ x is surjective for generic x ∈ V ∗ . Then \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}} = \mathbb{C}[V ^{{\ast}}]^{G}[F]\) , where F is a bi-homogeneous preimage of a generator of that is not divisible by any non-constant G-invariant in \(\mathbb{C}[V ^{{\ast}}]\) .

Proof

If we have a Lie algebra of index 1, in our case \(\mathfrak{g}_{x}\), then the algebra of its symmetric invariants is a polynomial ring. There are many possible explanations of this fact. One of them is the following. Suppose that two non-zero homogeneous polynomials f 1, f 2 are algebraically dependent. Then f 1 a = cf 2 b for some coprime integers a, b > 0 and some \(c \in \mathbb{C}^{^{\times } }\). If f 1 is an invariant, then so is a polynomial function \(\root{b}\of{f_{1}} = \root{ab}\of{c}\root{a}\of{f_{2}}\).

Since , it is generated by some homogeneous f. The group G x has finitely many connected components, hence is generated by a suitable power of f, say f = f d.

Let \(F \in \mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\) be a preimage of f. Each its bi-homogeneous component is again a \(\mathfrak{q}\)-invariant. Without loss of generality we may assume that F is bi-homogenous. Also if F is divisible by some non-scalar \(H \in \mathbb{C}[V ^{{\ast}}]^{G}\), then we replace F with F∕H and repeat the process as long as possible.

Whenever G y (with y ∈ V ∗) is conjugate to G x and φ y (F) ≠ 0, φ y (F) is a G y -invariant of the same degree as f and therefore is a generator of . Clearly and . If contains a homogeneous in \(\mathfrak{g}\) polynomial T that is not proportional (over \(\mathbb{C}(V ^{{\ast}})^{G}\)) to a power of F, then φ u (T) is not proportional to a power of φ u (F) for generic u ∈ V ∗. But . This implies that . It remains to notice that \(\mathbb{C}(V ^{{\ast}})^{G} =\mathrm{ Quot}\,\mathbb{C}[V ^{{\ast}}]^{G}\), since G has no proper semi-invariants in \(\mathbb{C}[V ^{{\ast}}]\), and by the same reason \(\mathbb{C}(V ^{{\ast}})^{G}[F] \cap \mathbb{C}[\mathfrak{q}] = \mathbb{C}[V ^{{\ast}}]^{G}[F]\) in case F is not divisible by any non-constant G-invariant in \(\mathbb{C}[V ^{{\ast}}]\). □

It is time to recall the Raïs’ formula [6] for the index of a semi-direct product:

$$\displaystyle{ \mathrm{ind}\mathfrak{q} = \mathrm{dim}V - (\mathrm{dim}\mathfrak{g} -\mathrm{dim}\mathfrak{g}_{x}) +\mathrm{ ind}\mathfrak{g}_{x}\ \text{ with}\ x \in V ^{{\ast}}\ \text{generic. } }$$
(2)

Lemma 3

Suppose that are homogenous polynomials such that φ x (H i ) with \(i\leqslant \mathrm{ind}\mathfrak{g}_{x}\) freely generate for generic x ∈ V ∗ and \(H_{j} \in \mathbb{C}[V ^{{\ast}}]^{G}\) for \(j>\mathrm{ ind}\mathfrak{g}_{x}\) ; and suppose that \(\sum \limits _{i=1}^{\mathrm{ind}\mathfrak{g}_{x}}\mathrm{deg}_{\mathfrak{g}}H_{i} = \mathbf{b}(\mathfrak{g}_{x})\) . Then \(\sum \limits _{i=1}^{r}\mathrm{deg}H_{i} = \mathbf{b}(\mathfrak{q})\) if and only if \(\sum \limits _{i=1}^{r}\mathrm{deg}_{V }H_{i} = \mathrm{dim}V\) .

Proof

In view of the assumptions, we have \(\sum \limits _{i=1}^{r}\mathrm{deg}H_{i} = \mathbf{b}(\mathfrak{g}_{x}) +\sum \limits _{ i=1}^{r}\mathrm{deg}_{V }H_{i}\). Further, by Eq. (2)

$$\displaystyle{ \begin{array}{l} \mathbf{b}(\mathfrak{q}) = (\mathrm{dim}\mathfrak{q} + \mathrm{dim}V - (\mathrm{dim}\mathfrak{g} -\mathrm{dim}\mathfrak{g}_{x}) +\mathrm{ ind}\mathfrak{g}_{x})/2 = \\ \quad = \mathrm{dim}V + (\mathrm{dim}\mathfrak{g}_{x} +\mathrm{ ind}\mathfrak{g}_{x})/2 = \mathbf{b}(\mathfrak{g}_{x}) + \mathrm{dim}V. \end{array} }$$

The result follows. □

From now on suppose that G is semisimple. Then both G and Q have only trivial characters and hence cannot have proper semi-invariants. In particular, the fundamental semi-invariant is an invariant. We also have . Set \(r =\mathrm{ ind}\mathfrak{q}\) and let x ∈ V ∗ be generic. If \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\) is a polynomial ring, then there are bi-homogenous generators H 1, …, H r such that H i with \(i>\mathrm{ ind}\mathfrak{g}_{x}\) freely generate \(\mathbb{C}[V ^{{\ast}}]^{G}\) and the invariants H i with \(i\leqslant \mathrm{ind}\mathfrak{g}_{x}\) are mixed, they have positive degrees in \(\mathfrak{g}\) and V.

Theorem 2 ([3, Theorem 5.7] and [10, Proposition 3.11])

Suppose that G is semisimple and \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}}\) is a polynomial ring with homogeneous generators H 1, …, H r . Then

  1. (i)

     \(\sum _{i=1}^{r}\mathrm{deg}H_{i} = \mathbf{b}(\mathfrak{q}) + \mathrm{deg}\mathbf{p}_{\mathfrak{q}}\) ;

  2. (ii)

     for generic x ∈ V ∗ , the restriction map is surjective, , and is a polynomial ring in \(\mathrm{ind}\mathfrak{g}_{x}\) variables.

It is worth mentioning that φ x is also surjective for stable actions. An action of G on V is called stable if generic G-orbits in V are closed, for more details see [8, Sections 2.4 and 7.5]. By [10, Theorem 2.8] φ x is surjective for generic x ∈ V ∗ if the G-action on V ∗ is stable.

3 \(\mathbb{Z}/2\mathbb{Z}\)-contractions

The initial motivation for studying symmetric invariants of semi-direct products was related to a conjecture of D. Panyushev on \(\mathbb{Z}_{2}\)-contractions of reductive Lie algebras. The results of [10], briefly outlined in Sect. 2, have settled the problem.

Let \(\mathfrak{g} = \mathfrak{g}_{0}\oplus \mathfrak{g}_{1}\) be a symmetric decomposition, i.e., a \(\mathbb{Z}/2\mathbb{Z}\)-grading of \(\mathfrak{g}\). A semi-direct product, \(\tilde{\mathfrak{g}} = \mathfrak{g}_{0}\ltimes \mathfrak{g}_{1}\), where \(\mathfrak{g}_{1}\) is an Abelian ideal, can be seen as a contraction, in this case a \(\mathbb{Z}_{2}\) -contraction, of \(\mathfrak{g}\). For example, starting with a symmetric pair \((\mathfrak{s}\mathfrak{o}_{n+1},\mathfrak{s}\mathfrak{o}_{n})\), one arrives at \(\tilde{\mathfrak{g}} = \mathfrak{s}\mathfrak{o}_{n}\ltimes \mathbb{C}^{n}\). In [4], it was conjectured that is a polynomial ring (in \(\mathrm{rk\,}\mathfrak{g}\) variables).

Theorem 3 ([4, 9, 10])

Let \(\tilde{\mathfrak{g}}\) be a \(\mathbb{Z}_{2}\) -contraction of a reductive Lie algebra \(\mathfrak{g}\) . Then is a polynomial ring (in \(\mathrm{rk\,}\mathfrak{g}\) variables) if and only if the restriction homomorphism \(\mathbb{C}[\mathfrak{g}]^{\mathfrak{g}} \rightarrow \mathbb{C}[\mathfrak{g}_{1}]^{\mathfrak{g}_{0}}\) is surjective.

If we are in one of the “surjective” cases, then one can describe the generators of . Let H 1, …, H r be suitably chosen homogeneous generators of and let H i • be the bi-homogeneous (w.r.t. \(\mathfrak{g} = \mathfrak{g}_{0}\oplus \mathfrak{g}_{1}\)) component of H i of the highest \(\mathfrak{g}_{1}\)-degree. Then is freely generated by the polynomials H i • (of course, providing the restriction homomorphism \(\mathbb{C}[\mathfrak{g}]^{\mathfrak{g}} \rightarrow \mathbb{C}[\mathfrak{g}_{1}]^{\mathfrak{g}_{0}}\) is surjective) [4, 9].

Unfortunately, this construction of generators cannot work if the restriction homomorphism is not surjective, see [4, Remark 4.3]. As was found out by Helgason [2], there are four “non-surjective” irreducible symmetric pairs, namely, (E 6, F 4), \((E_{7},E_{6}\oplus \mathbb{C})\), \((E_{8},E_{7}\oplus \mathfrak{s}\mathfrak{l}_{2})\), and \((E_{6},\mathfrak{s}\mathfrak{o}_{10}\oplus \mathfrak{s}\mathfrak{o}_{2})\). The approach to semi-directproducts developed in [10] showed that Panyushev’s conjecture does not hold for them. Next we outline some ideas of the proof.

Let G 0 ⊂ G be a connected subgroup with \(\mathop{\mathrm{Lie}}\nolimits G_{0} = \mathfrak{g}_{0}\). Then G 0 is reductive, it acts on \(\mathfrak{g}_{1}\cong \mathfrak{g}_{1}^{{\ast}}\), and this action is stable. Let \(x \in \mathfrak{g}_{1}\) be a generic element and G 0,x be its stabiliser in G 0. The groups G 0,x are reductive and they are known for all symmetric pairs. In particular, is a polynomial ring. It is also known that \(\mathbb{C}[\mathfrak{g}_{1}]^{G_{0}}\) is a polynomial ring. By [4] \(\tilde{\mathfrak{g}}\) has the “codim-2” property and \(\mathrm{ind}\tilde{\mathfrak{g}} =\mathrm{ rk\,}\mathfrak{g}\).

Making use of the surjectivity of φ x one can show that if \(\mathbb{C}[\tilde{\mathfrak{g}}^{{\ast}}]^{\tilde{\mathfrak{g}}}\) is freely generated by some H 1, …, H r , then necessary \(\sum \limits _{i=1}^{r}\mathrm{deg}H_{i}> \mathbf{b}(\tilde{\mathfrak{g}})\) for \(\tilde{\mathfrak{g}}\) coming from one of the “non-surjective” pairs [10]. In view of some results from [3] this leads to a contradiction.

Note that in case of \((\mathfrak{g},\mathfrak{g}_{0}) = (E_{6},F_{4})\), \(\mathfrak{g}_{0} = F_{4}\) is simple and \(\tilde{\mathfrak{g}}\) is a semi-direct product of F 4 and \(\mathbb{C}^{26}\), which, of course, comes from one of the representations in Schwarz’s list [7].

4 Examples Related to the Defining Representation of \(\mathfrak{s}\mathfrak{l}_{n}\)

Form now assume that \(\mathfrak{g} = \mathfrak{s}\mathfrak{l}_{n}\) and \(V = m(\mathbb{C}^{n})^{{\ast}}\oplus k\mathbb{C}^{n}\) with \(n\geqslant 2\), \(m\geqslant 1\), \(m\geqslant k\). According to [7] \(\mathbb{C}[V ]^{G}\) is a polynomial ring if either k = 0 and \(m\leqslant n + 1\) or \(m\leqslant n\), \(k\leqslant n - 1\). One finds also the description of the generators of \(\mathbb{C}[V ^{{\ast}}]^{G}\) and their degrees in [7]. In this section, we classify all cases, where \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}}\) is a polynomial ring and for each of them give the fundamental semi-invariant.

Example 1

Suppose that either \(m\geqslant n\) or m = k = n − 1. Then \(\mathfrak{g}_{x} = 0\) for generic x ∈ V ∗ and therefore \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q} = \mathbb{C}[V ^{{\ast}}]^{G}\), i.e., \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\) is a polynomial ring if and only if \(\mathbb{C}[V ^{{\ast}}]^{G}\) is. The latter takes place for (m, k) = (n + 1, 0), for m = n and any k < n, as well as for m = k = n − 1. Non-scalar fundamental semi-invariants appear here only for

  • m = n, where p is given by det(v)n−1−k with \(v \in n\mathbb{C}^{n}\);

  • m = k = n − 1, where p is the sum of the principal 2k×2k-minors of

In the rest of the section, we assume that \(\mathfrak{g}_{x}\neq 0\) for generic x ∈ V ∗.

4.1 The Case k = 0

Here the ring of G-invariants on V ∗ is generated by

$$\displaystyle{ \{\varDelta _{I}\mid I \subset \{ 1,\ldots,m\},\vert I\vert = n\}\ \ \text{[8, Section 9]}, }$$

where each Δ I (v) is the determinant of the corresponding submatrix of v ∈ V ∗. The generators are algebraically independent if and only if \(m\leqslant n + 1\), see also [7].

We are interested only in m that are smaller than n. Let n = qm + r, where \(0 <r\leqslant m\), and let I ⊂ {1, …, m} be a subset of cardinality r. By choosing the corresponding r columns of v we get a matrix w = v I . Set

$$\displaystyle{ F_{I}(A,v):= \mathrm{det}\left (v\vert Av\vert \ldots \vert A^{q-1}v\vert A^{q}w\right ),\ \text{ where }A \in \mathfrak{g},v \in V ^{{\ast}}. }$$
(3)

Clearly each F I is an SL n -invariant. Below we will see that they are also V -invariants. If r = m, then there is just one invariant, F = F {1,…, m}. If r is either 1 or m − 1, we get m invariants.

Lemma 4

Each F I defined by Eq.  (3) is a V -invariant.

Proof

According to Lemma 1 we have to show that F I (ξ + ad∗(V )⋅ x, x) = F(ξ, x) for generic x ∈ V ∗ and any \(\xi \in \mathfrak{s}\mathfrak{l}_{n}\). Since m < n, there is an open SL n -orbit in V ∗ and we can take x as E m . Let \(\mathfrak{p} \subset \mathfrak{g}\mathfrak{l}_{n}\) be the standard parabolic subalgebra corresponding to the composition (m, n − m) and let \(\mathfrak{n}_{-}\) be the nilpotent radical of the opposite parabolic. Each element (matrix) \(\xi \in \mathfrak{g}\mathfrak{l}_{n}\) is a sum ξ = ξ − + ξ p with \(\xi _{-}\in \mathfrak{n}_{-}\), \(\xi _{p} \in \mathfrak{p}\). In this notation \(F_{I}(A,E_{m}) = \mathrm{det}\left (A_{-}\vert (A^{2})_{-}\vert \ldots \vert (A^{q-1})_{-}\vert (A^{q})_{-,I}\right )\).

Let α = α A and β = β A be m×m and (n − m)×(n − m)-submatrices of A standing in the upper left and lower right corner, respectively. Then (A s+1)− = ∑ t = 0 s β t A − α s−t. Each column of A − α is a linear combination of columns of A − and each column of β t A − α j+1 is a linear combination of columns of β t A − α j. Therefore

$$\displaystyle{ F_{I}(A,E_{m}) = \text{det}\left (A_{-}\vert \ldots \vert (A^{q-1})_{ -}\vert (A^{q})_{ -,I}\right ) = }$$
$$\displaystyle{ \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \text{det}\left (A_{-}\vert \beta A_{-}\vert \ldots \vert \beta ^{q-2}A_{ -}\vert \beta ^{q-1}A_{ -,I}\right ). }$$
(4)

Notice that \(\mathfrak{g}_{x} \subset \mathfrak{p}\) and the nilpotent radical of \(\mathfrak{p}\) is contained in \(\mathfrak{g}_{x}\) (with x = E m ). Since \(\mathrm{ad}^{{\ast}}(V )\cdot x =\mathrm{ Ann}(\mathfrak{g}_{x}) = \mathfrak{g}_{x}^{\perp }\subset \mathfrak{g}\) (after the identification \(\mathfrak{g}\cong \mathfrak{g}^{{\ast}}\)), A − = 0 for any \(A \in \mathfrak{g}_{x}^{\perp }\); and we have β A = cE n−m with \(c \in \mathbb{C}\) for this A. An easy observation is that

$$\displaystyle\begin{array}{rcl} & & \text{det}\ \left (\xi _{-}\vert (\beta _{\xi } + cE_{n-m})\xi _{-}\vert \ldots \vert (\beta _{\xi } + cE_{n-m})^{q-1}\xi _{ -,I}\right ) = {}\\ & & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \text{det}\ \left (\xi _{-}\vert \beta _{\xi }\xi _{-}\vert \ldots \vert \beta _{\xi }^{q-1}\xi _{ -,I}\right ). {}\\ \end{array}$$

Hence F I (ξ + A, E m ) = F I (ξ, E m ) for all A ∈ ad∗(V )⋅ E m and all \(\xi \in \mathfrak{s}\mathfrak{l}_{n}\). □

Theorem 4

Suppose that \(\mathfrak{q} = \mathfrak{s}\mathfrak{l}_{n}\ltimes m(\mathbb{C}^{n})^{{\ast}}\) . Then \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\) is a polynomial ring if and only if \(m\leqslant n + 1\) and m divides either n − 1, n or n + 1. Under these assumptions on m, \(\mathbf{p}_{\mathfrak{q}} = 1\) exactly then, when m divides either n − 1 or n + 1.

Proof

Note that the statement is true for \(m\geqslant n\) by Example 1. Assume that \(m\leqslant n - 1\). Suppose that n = mq + r as above. A generic stabiliser in \(\mathfrak{g}\) is \(\mathfrak{g}_{x} = \mathfrak{s}\mathfrak{l}_{n-m}\ltimes m\mathbb{C}^{n-m}\). On the group level it is connected. Notice that , since G x has no non-trivial characters. Note also that \(\mathbb{C}[V ^{{\ast}}]^{G} = \mathbb{C}\), since m < n. If \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\) is a polynomial ring, then so is \(\mathbb{C}[\mathfrak{g}_{x}^{{\ast}}]^{G_{x}}\) by Theorem 2(ii) and either n − m = 1 or, arguing by induction, \(n - m \equiv t\pmod m\) with t ∈ {−1, 0, 1}.

Next we show that the ring of symmetric invariants is freely generated by the polynomials F I for the indicated m. Each element \(\gamma \in \mathfrak{g}_{x}^{{\ast}}\) can be presented as γ = β 0 + A −, where \(\beta _{0} \in \mathfrak{s}\mathfrak{l}_{n-m}\). Each restriction φ x (F I ) can be regarded as an element of . Equation (4) combined with Lemma 4 and the observation that \(\mathfrak{g}_{x}^{{\ast}}\cong \mathfrak{g}/\mathrm{Ann}(\mathfrak{g}_{x})\) shows that φ x (F I ) is either Δ I of \(\mathfrak{g}_{x}\) (in case q = 1, where F I (A, E m ) = detA −,I ) or F I of \(\mathfrak{g}_{x}\). Arguing by induction on n, we prove that the restrictions φ x (F I ) freely generate for x = E m (i.e., for a generic point in V ∗). Notice that n − m = (q − 1)m + r.

The group SL n acts on V ∗ with an open orbit SL n ⋅ E m . Therefore the restriction map φ x is injective. By the inductive hypothesis it is also surjective and therefore is an isomorphism. This proves that the polynomials F I freely generate \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\).

If m divides n, then \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q} = \mathbb{C}[F]\) and the fundamental semi-invariant is a power of F. As follows from the equality in Theorem 2(i), p = F m−1.

Suppose that m divides either n − 1 or n + 1. Then we have m different invariants F I . By induction on n, \(\mathfrak{g}_{x}\) has the “codim-2” property, therefore the sum of degφ x (F I ) is equal to \(\mathbf{b}(\mathfrak{g}_{x})\) by Theorem 2(i). The sum of V -degrees is m×n = dimV and hence by Lemma 3 \(\sum \mathrm{deg}F_{I} = \mathbf{b}(\mathfrak{q})\). Thus, \(\mathfrak{q}\) has the “codim-2” property. □

Remark 1

Using induction on n one can show that the restriction map φ x is an isomorphism for all m < n. Therefore the polynomials F I generate \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\) for all m < n.

4.2 The Case m = k

Here \(\mathbb{C}[V ^{{\ast}}]^{G}\) is a polynomial ring if and only if \(k\leqslant n - 1\); a generic stabiliser is \(\mathfrak{s}\mathfrak{l}_{n-k}\), and the G-action on \(V \cong V ^{{\ast}}\) is stable. We assume that \(\mathfrak{g}_{x}\neq 0\) for generic x ∈ V ∗ and therefore \(k\leqslant n - 2\).

For an N×N-matrix C, let Δ i (C) with \(1\leqslant i\leqslant N\) be coefficients of its characteristic polynomial, each Δ i being a homogeneous polynomial of degree i. Let \(\gamma = A + v + w \in \mathfrak{q}^{{\ast}}\) with \(A \in \mathfrak{g}\), \(v \in k\mathbb{C}^{n}\), \(w \in k(\mathbb{C}^{n})^{{\ast}}\). Having these objects we form an (n + k)×(n + k)-matrix

and set F i (γ) = Δ i (Y γ ) for each i ∈ {2k + 1, 2k + 2, 2k + 3, …, n + k}. Each F i is an \(\mathrm{SL}_{n}\times \mathop{\mathrm{GL}}\nolimits _{k}\)-invariant. Unfortunately, these polynomials are not V -invariants.

Remark 2

If we repeat the same construction for \(\tilde{\mathfrak{q}} = \mathfrak{g}\mathfrak{l}_{n}\ltimes V\) with \(k\leqslant n - 1\), then \(\mathbb{C}[\tilde{\mathfrak{q}}^{{\ast}}]^{\tilde{Q}} = \mathbb{C}[V ^{{\ast}}]^{\mathop{\mathrm{GL}}\nolimits _{n}}[\{F_{i}\mid 2k + 1\leqslant i\leqslant n + k\}]\) and it is a polynomial ring in \(\mathrm{ind}\tilde{\mathfrak{q}} = n - k + k^{2}\) generators.

Theorem 5

Suppose that \(m = k\leqslant n - 1\) . Then \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}}\) is a polynomial ring if and only if k ∈ {n − 2, n − 1}. In case k = n − 2, \(\mathfrak{q}\) has the “codim-2” property.

Proof

Suppose that k = n − 2. Then a generic stabiliser \(\mathfrak{g}_{x} = \mathfrak{s}\mathfrak{l}_{2}\) is of index 1 and since the G-action on V is stable, \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}}\) has to be a polynomial ring by [10, Example 3.6]. One can show that the unique mixed generator is of the form F 2k+2 H 2k − F 2k+1 2, where H 2k is a certain \(\mathrm{SL}_{n}\times \mathop{\mathrm{GL}}\nolimits _{k}\)-invariant on V of degree 2k and then see that the sum of degrees is \(\mathbf{b}(\mathfrak{q})\).

More generally, \(\mathfrak{q}\) has the “codim-2” property for all \(k\leqslant n - 2\). Here each G-invariant divisor in V ∗ contains a G-orbit of maximal dimension, say Gy. Set u = n − k − 1. If G y is not SL n−k , then \(\mathfrak{g}_{y} = \mathfrak{s}\mathfrak{l}_{u} \ltimes (\mathbb{C}^{u}\oplus (\mathbb{C}^{u})^{{\ast}}\oplus \mathbb{C})\) is a semi-direct product with a Heisenberg Lie algebra. Following the proof of [4, Theorem 3.3], one has to show that \(\mathrm{ind}\mathfrak{g}_{y} = u\) in order to prove that \(\mathfrak{q}\) has the “codim-2” property. This is indeed the case, \(\mathrm{ind}\mathfrak{g}_{y} = 1 +\mathrm{ ind}\mathfrak{s}\mathfrak{l}_{u}\).

Suppose that 0 < k < n − 2 and assume that is a polynomial ring. Then there are bi-homogeneous generators h 2, …, h n−k of \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\) over \(\mathbb{C}[V ^{{\ast}}]^{G}\) such that their restrictions to \(\mathfrak{g} + x\) form a generating set of for a generic x (with \(\mathfrak{g}_{x}\cong \mathfrak{s}\mathfrak{l}_{n-k}\)), see Theorem 2(ii). In particular, \(\mathrm{deg}_{\mathfrak{g}}\mathbf{h}_{t} = t\).

Take \(\tilde{\mathfrak{g}} = (\mathfrak{s}\mathfrak{l}_{n}\oplus \mathfrak{g}\mathfrak{l}_{k}) \ltimes V\), which is a \(\mathbb{Z}_{2}\)-contraction of \(\mathfrak{s}\mathfrak{l}_{n+k}\). Then \(\mathfrak{q}\) is a Lie subalgebra of \(\tilde{\mathfrak{g}}\). Note that \(\mathop{\mathrm{GL}}\nolimits _{k}\) acts on \(\mathfrak{q}\) via automorphisms and therefore we may assume that the \(\mathbb{C}\)-linear span of {h t } is \(\mathop{\mathrm{GL}}\nolimits _{k}\)-stable. By degree considerations, each h t is an SL k -invariant as well. The Weyl involution of SL n acts on V and has to preserve each line \(\mathbb{C}\mathbf{h}_{t}\). Since this involution interchanges \(\mathbb{C}^{n}\) and \((\mathbb{C}^{n})^{{\ast}}\), each h t is also a \(\mathop{\mathrm{GL}}\nolimits _{k}\)-invariant. Thus,

Since \(\tilde{\mathfrak{g}}\) is a “surjective” \(\mathbb{Z}_{2}\)-contraction, its symmetric invariants are known [4, Theorem 4.5]. The generators of are Δ j • with \(2\leqslant j\leqslant n + k\). Here degΔ j • = j and the generators of (\(\mathfrak{s}\mathfrak{l}_{n}\oplus \mathfrak{g}\mathfrak{l}_{k}\))-degrees 2, 3, …, n − k are Δ 2k+2 •, Δ 2k+3 •, …, Δ n+k •. As the restriction to \(\mathfrak{s}\mathfrak{l}_{n}\oplus \mathfrak{g}\mathfrak{l}_{k} + x\) shows, none of the generators Δ j • with \(j\geqslant 2k + 2\) lies in . This means that h t cannot be equal or even proportional over \(\mathbb{C}[V ^{{\ast}}]^{G}\) to Δ 2k+t • and hence has a more complicated expression. More precisely, a product Δ 2k+1 • Δ 2k+t−1 • necessary appears in h t with a non-zero coefficient from \(\mathbb{C}[V ^{{\ast}}]^{G}\) for \(t\geqslant 2\). Since deg V Δ 2k+1 • = 2k, we have \(\mathrm{deg}_{V }\mathbf{h}_{t}\geqslant 4k\) for every \(t\geqslant 2\). The ring \(\mathbb{C}[V ^{{\ast}}]^{G}\) is freely generated by k 2 polynomials of degree two. Therefore, the total sum of degrees over all generators of is greater than or equal to

$$\displaystyle{ \mathbf{b}(\mathfrak{s}\mathfrak{l}_{n-k}) + 4k(n - k - 1) + 2k^{2} = \mathbf{b}(\mathfrak{q}) + 2k(n - k - 2). }$$

This contradicts Theorem 2(i) in view of the fact that \(\mathbf{p}_{\mathfrak{q}} = 1\). □

4.3 The Case 0 < k < m

Here \(\mathbb{C}[V ^{{\ast}}]^{G}\) is a polynomial ring if and only if \(m\leqslant n\), [7]. If n = m, then \(\mathfrak{g}_{x} = 0\) for generic x ∈ V ∗. For m < n, our construction of invariants is rather intricate.

Let π 1, …, π n−1 be the fundamental weights of \(\mathfrak{s}\mathfrak{l}_{n}\). We use the standard convention, π i = ɛ 1 + … + ɛ i , \(\varepsilon _{n} = -\sum \limits _{i=1}^{n-1}\varepsilon _{i}\). Recall that for any t, \(1\leqslant t <n\), \(\varLambda ^{t}\mathbb{C}^{n}\) is irreducible with the highest weight π t . Let {e 1, …, e n } be a basis of \(\mathbb{C}^{n}\) such that each e i is a weight vector and ℓ t : = e 1 ∧ … ∧ e t is a highest weight vector of \(\varLambda ^{t}\mathbb{C}^{n}\). Clearly . Write n − k = d(m − k) + r with \(0 <r\leqslant (m - k)\). Let \(\varphi \!: m\mathbb{C}^{n} \rightarrow \varLambda ^{m}\mathbb{C}^{n}\) be a non-zero m-linear G-equivariant map. Such a map is unique up to a scalar and one can take φ with φ(v 1 + … + v m ) = v 1 ∧ … ∧ v m . In case r ≠ m − k, for any subset I ⊂ {1, …, m} with | I | = k + r, let \(\varphi _{I}\!: m\mathbb{C}^{n} \rightarrow (k + r)\mathbb{C}^{n} \rightarrow \varLambda ^{k+r}\mathbb{C}^{n}\) be the corresponding (almost) canonical map. By the same principle we construct \(\tilde{\varphi }\!: k(\mathbb{C}^{n})^{{\ast}}\rightarrow \varLambda ^{k}(\mathbb{C}^{n})^{{\ast}}\).

Let us consider the tensor product \(\mathbb{W}:= (\varLambda ^{m}\mathbb{C}^{n})^{\otimes d}\otimes \varLambda ^{k+r}\mathbb{C}^{n}\) and its weight subspace \(\mathbb{W}_{d\pi _{k}}\). One can easily see that \(\mathbb{W}_{d\pi _{k}}\) contains a unique up to a scalar non-zero highest weight vector, namely

$$\displaystyle{ w_{d\pi _{k}} =\sum _{\sigma \in S_{n-k}}\mathrm{sgn}(\sigma )(\ell_{k} \wedge e_{\sigma (k+1)} \wedge \ldots \wedge e_{\sigma (m)}) \otimes \ldots \otimes (\ell_{k} \wedge e_{\sigma (n-r+1)}\ldots \wedge e_{\sigma (n)}). }$$

This means that \(\mathbb{W}\) contains a unique copy of \(V _{d\pi _{k}}\), where \(V _{d\pi _{k}}\) is an irreducible \(\mathfrak{s}\mathfrak{l}_{n}\)-module with the highest weight dπ k . We let ρ denote the representation of \(\mathfrak{g}\mathfrak{l}_{n}\) on \(\varLambda ^{m}\mathbb{C}^{n}\) and ρ r the representation of \(\mathfrak{g}\mathfrak{l}_{n}\) on \(\varLambda ^{k+r}\mathbb{C}^{n}\). Let ξ = A + v + w be a point in \(\mathfrak{q}^{{\ast}}\). (It is assumed that \(A \in \mathfrak{s}\mathfrak{l}_{n}\).) Finally let ( , ) denote a non-zero \(\mathfrak{s}\mathfrak{l}_{n}\)-invariant scalar product between \(\mathbb{W}\) and that is zero on the \(\mathfrak{s}\mathfrak{l}_{n}\)-invariant complement of \(V _{d\pi _{k}}\) in \(\mathbb{W}\). Depending on r, set

$$\displaystyle\begin{array}{rcl} & \mathbf{F}(\xi ):= (\varphi (v)\otimes \rho (A)^{m-k}\varphi (v)\otimes \rho (A^{2})^{m-k}\varphi (v)\otimes \ldots \otimes \rho (A^{d})^{m-k}\varphi (v),\tilde{\varphi }(w)^{d}) & {}\\ & \text{ for }r = m - k; & {}\\ & \mathbf{F}_{I}(\xi ):= (\varphi (v)\otimes \rho (A)^{m-k}\varphi (v)\otimes \ldots \otimes \rho (A^{d-1})^{m-k}\varphi (v)\otimes \rho _{r}(A^{d})^{r}\varphi _{I}(v),\tilde{\varphi }(w)^{d})& {}\\ \end{array}$$

for each I as above in case r < m − k. By the constructions the polynomials F and F I are SL n -invariants.

Lemma 5

The polynomials F and F I are V -invariants.

Proof

We restrict F and F I to \(\mathfrak{g}^{{\ast}} + x\) with x ∈ V ∗ generic. Changing a basis in V if necessary, we may assume that x = E m + E k . If r < m − k, some of the invariants F I may become linear combinations of such polynomials under the change of basis, but this does not interfere with V -invariance. Now φ(v) is a vector of weight π m and \(\tilde{\varphi }(w)^{d}\) of weight − dπ k . Notice that dm + (k + r) = n + kd. If \(\sum \limits _{i=1}^{n+kd}\lambda _{i} = d\sum \limits _{i=1}^{k}\varepsilon _{i}\) and each λ i is one of the ɛ j , \(1\leqslant j\leqslant n\), then in the sequence (λ 1, …, λ n+kd ) we must have exactly one ɛ j for each \(k <j\leqslant n\) and d + 1 copies of each ɛ i with \(1\leqslant i\leqslant k\). Hence the only summand of ρ(A s)m−k φ(E m ) that plays any rôle in F or F I is ℓ k ∧A s e k+1∧…∧A s e m . Moreover, in A s e k+1∧…∧A s e m we are interested only in vectors lying in Λ m−kspan(e k+1, …, e n ).

Let us choose blocks α, U, β of A as shown in Fig. 1.Then up to a non-zero scalar F(A, E m + E k ) is the determinant of

$$\displaystyle\begin{array}{rcl} & & \left (U\vert \beta U + U\alpha \vert P_{2}(\alpha,U,\beta )\vert \ldots \vert P_{d-1}(\alpha,U,\beta )\right ), {}\\ & & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{ where }P_{s}(\alpha,U,\beta ) =\sum _{ t=0}^{s}\beta ^{t}U\alpha ^{s-t}. {}\\ \end{array}$$

Each column of Uα is a linear combination of the columns of U, a similar relation exists between β t Uα s+1 and β t Uα s. Therefore

$$\displaystyle{ \mathbf{F}(A,E_{m} + E_{k}) = \mathrm{det}\left (U\vert \beta U\vert \beta ^{2}U\vert \ldots \vert \beta ^{d-1}U\right ). }$$
(5)

We have to check that F(ξ + A, x) = F(ξ, x) for any A ∈ ad∗(V )⋅ x and any \(\xi \in \mathfrak{g}\), see Lemma 1. Recall that \(\mathrm{ad}^{{\ast}}(V )\cdot x =\mathrm{ Ann}(\mathfrak{g}_{x}) = \mathfrak{g}_{x}^{\perp }\subset \mathfrak{g}\). In case x = E m + E k , U is zero in each \(A \in \mathfrak{g}_{x}^{\perp }\) and β corresponding to such A is a scalar matrix. Therefore F(ξ + ad∗(V )⋅ x, x) = F(ξ, x).

Fig. 1
figure 1

Submatrices of \(A \in \mathfrak{s}\mathfrak{l}_{n}\)

Full size image

The case r < m − k is more complicated. If {1, …, k} ⊂ I, then \(I = \tilde{I}\sqcup \{ 1,\ldots,k\}\). Let \(U_{\tilde{I} }\) be the corresponding submatrix of U and \(\alpha _{\tilde{I} \times \tilde{I}}\) of α. One just has to replace U by \(U_{\tilde{I} }\) and α by \(\alpha _{\tilde{I}\times \tilde{I}}\) in the last polynomial P d−1(α, U, β) obtaining

$$\displaystyle{ \mathbf{F}_{I}(A,x) = \mathrm{det}\left (U\vert \beta U\vert \beta ^{2}U\vert \ldots \vert \beta ^{d-2}U\vert \beta ^{d-1}U_{\tilde{ I} } \right ). }$$

These are \(\binom{m - k}{r}\) linearly independent invariants in .

Suppose that {1, …, k} ⊄ I. Then ρ I (A d)r has to move more than r vectors e i with \(k + 1\leqslant i\leqslant m\), which is impossible. Thus, F I (A, x) = 0 for such I. □

Theorem 6

Suppose that 0 < k < m < n and m − k divides n − m, then \(\mathrm{ind}\mathfrak{g}_{x} = 1\) for generic x ∈ V ∗ and \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q} = \mathbb{C}[V ^{{\ast}}]^{G}[\mathbf{F}]\) is a polynomial ring, the fundamental semi-invariant is equal to F m−k−1 .

Proof

A generic stabiliser \(\mathfrak{g}_{x}\) is \(\mathfrak{s}\mathfrak{l}_{n-m}\ltimes (m - k)\mathbb{C}^{n-m}\). Its ring of symmetric invariants is generated by F = φ x (F), see Theorem 4 and Eq. (5). We also have \(\mathrm{ind}\mathfrak{g}_{x} = 1\). It remains to see that F is not divisible by a non-constant G-invariant polynomial on V ∗. By the construction, F is also invariant with respect to the action of SL m ×SL k . The group L = SL n ×SL m ×SL k act on V ∗ with an open orbit. As long as rk w = k, rk v = m, the L-orbit of y = v + w contains a point v′ + E k , where also rk v′ = m. If in addition the upper k×m-part of v has rank k, then L⋅ y contains x = E m + E k . Here F is non-zero on \(\mathfrak{g} + y\). Since the group L is semisimple, the complement of L⋅ (E m + E k ) contains no divisors and F is not divisible by any non-constant G-invariant in \(\mathbb{C}[V ^{{\ast}}]\). This is enough to conclude that \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q} = \mathbb{C}[V ^{{\ast}}]^{G}[\mathbf{F}]\), see Theorem 2.

The singular set \(\mathfrak{q}_{\mathrm{sing}}^{{\ast}}\) is L-stable. And therefore \(\mathbf{p}_{\mathfrak{q}}\) is also an SL m ×SL k -invariant. Hence p is a power of F. In view of Theorem 2(i), p = F m−k−1. □

Theorem 7

Suppose that 0 < k < m < n and m − k does not divide n − m, then \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\) is not a polynomial ring.

Proof

The reason for this misfortune is that \(\binom{m}{k + r}> \binom{m - k}{r}\) for r < m − k. One could prove that each F I must be in the set of generators and thereby show that \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{Q}\) is not a polynomial ring. But we present a different argument.

Assume that the ring of symmetric invariants is polynomial. It is bi-graded and SL m acts on it preserving the bi-grading. Since SL m is reductive, we can assume that there is a set {H 1, …, H s } of bi-homogeneous mixed generators such that and the \(\mathbb{C}\)-linear span is SL m -stable. The polynomiality implies that a generic stabiliser \(\mathfrak{g}_{x} = \mathfrak{s}\mathfrak{l}_{n-m}\ltimes (m - k)\mathbb{C}^{n-m}\) has a free algebra of symmetric invariants, see Theorem 2(ii), and by the same statement φ x is surjective. This means that r is either 1 or m − k − 1, see Theorem 4, s = m − k, and φ x is injective on . Taking our favourite (generic) x = E m + E k , we see that there is SL m−k embedded diagonally into G×SL m , which acts on as on \(\varLambda ^{r}\mathbb{C}^{m-k}\). The group SL m−k acts on in the same way. Since m − k does not divide n − m, we have \(m - k\geqslant 2\). The group SL m cannot act on an irreducible module \(\varLambda ^{r}\mathbb{C}^{m-k}\) of its non-trivial subgroup SL m−k , this is especially obvious in our two cases of interest, r = 1 and r = m − k − 1. A contradiction. □

Conjecture 1

It is very probable that \(\mathbb{C}[\mathfrak{q}^{{\ast}}]^{\mathfrak{q}} = \mathbb{C}[V ^{{\ast}}]^{G}[\{\mathbf{F}_{I}\}]\) for all \(n> m> k\geqslant \ 1\).