An epimorphic subgroup arising from Roberts' counterexample

Abstract

In 1994, based on Roberts' counterexample to Hilbert's fourteenth problem,
A'Campo-Neuen constructed an example of a linear action of a 12-dimensional
commutative unipotent group H_0 on a 19-dimensional vector space V such that
the algebra of invariants k[V]^{H_0} is not finitely generated. We consider a
certain extension H of H_0 by a one-dimensional torus and prove that H is
epimorphic in SL(V). In particular, the homogeneous space SL(V)/H provides a
new example of a homogeneous space with epimorphic stabilizer that admits no
projective embeddings with small boundary.

Full text

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arXiv:1102.3543v1 [math.AG] 17 Feb 2011
AN EPIMORPHIC SUBGROUP ARISING FROM
ROBERTS’ COUNTEREXAMPLE
ROMAN AVDEEV
Abstract. In 1994, based on Roberts’ counterexample to Hilbert’s fourteenth problem,
A’Campo-Neuen constructed an example of a linear action of a 12-dimensional commu-
tative unipotent group H0on a 19-dimensional vector space V such that the algebra of
invariants k[V ]H0is not finitely generated. We consider a certain extension H of H0
by a one-dimensional torus and prove that H is epimorphic in SL(V ). In particular,
the homogeneous space SL(V )/H provides a new example of a homogeneous space with
epimorphic stabilizer that admits no projective embeddings with small boundary.
1. Introduction
We work over an algebraically closed field k of characteristic zero. Throughout the
paper all topological terms relate to the Zarisky topology, all groups are supposed to be
algebraic and their subgroups closed.
Let G be a connected affine algebraic group and H a subgroup of it. We recall that
a projective embedding with small boundary of the homogeneous space G/H is an open
G-equivariant embedding ρ: G/H ֒→ X, where X is an irreducible normal projective G-
variety and codimX(X\ρ(G/H)) ? 2. For a given homogeneous space G/H, the existence
of such embedding implies that the algebra k[G/H] of regular functions on G/H consists
of constants, that is, k[G/H] = k. A subgroup H ⊂ G with k[G/H] = k is said to be
epimorphic. Various characterizations, properties, and examples of epimorphic subgroups,
as well as several conjectures and open problems concerning them, can be found in [BBI],
[BBII], [BBK], and [Gr, §23B].
It turns out that not every homogeneous space G/H with epimorphic H admits a
projective embedding with small boundary. A criterion for this is given by Theorem 1
below. To formulate this theorem, we need to recall some additional notions. A subgroup
H ⊂ G is said to be observable if G/H is a quasi-affine variety. An observable subgroup
H ⊂ G is said to be a Grosshans subgroup if the algebra k[G/H] is finitely generated
over k.
Theorem 1. Let H ⊂ G be an epimorphic subgroup. Then the following conditions are
equivalent:
(a) G/H admits a projective embedding with small boundary;
(b) there is a character χ of H such that Kerχ is a Grosshans subgroup in G.
A complete proof of this theorem can be found in [Ar, Theorem 1.1].
Under a certain assumption on H, a combinatorial classification of all projective em-
beddings with small boundary for a given homogeneous space G/H is obtained in [Ar,
§3]. Another problem arising in connection with Theorem 1 is to construct examples
Date: February 18, 2011.
2010 Mathematics Subject Classification. 14E25, 14M99, 14M17.
Key words and phrases. Algebraic group, homogeneous space, epimorphic subgroup, embedding,
Hilbert’s fourteenth problem.
Partially supported by Russian Foundation for Basic Research, grant no. 09-01-00648.
1

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2ROMAN AVDEEV
of epimorphic subgroups H ⊂ G such that the corresponding homogeneous spaces G/H
admit no projective embeddings with small boundary. In view of condition (b) of Theo-
rem 1, such examples should be based on examples of observable subgroups H0⊂ G such
that the algebra k[G/H0] is not finitely generated over k. In their turn, examples of this
kind are provided by linear counterexamples to Hilbert’s fourteenth problem.
We recall that, in general, Hilbert’s fourteenth problem asks whether for a subfield L
of the field K(x1,...,xn) of rational functions in n variables over a field K such that
L ⊃ K the algebra L ∩ K[x1,...,xn] is finitely generated over K. (Here K[x1,...,xn]
is the algebra of polynomials in x1,...,xn.) An important special case of this problem
asks whether the algebra K[x1,...,xn]Gof invariants of a linear action of a group G on
an n-dimensional vector space is finitely generated over K. (This special case is obtained
from the general one with L being the quotient field of K[x1,...,xn]G.) Counterexamples
to this special case of Hilbert’s fourteenth problem are called linear counterexamples.
By this moment, counterexamples to Hilbert’s fourteenth problem are not so many in
number. The first one, which appears also to be a linear counterexample, was discov-
ered by Nagata in 1958 [N]. In Nagata’s example, a 13-dimensional unipotent group acts
on a 32-dimensional vector space. Much later, in 1997, Nagata’s counterexample was
considerably simplified by Steinberg [St] whose counterexample is now known as Nagata-
Steinberg’s counterexample. In this example, the dimension of the subgroup is reduced
to 6 and that of the vector space to 18. The second counterexample, which is not linear,
was constructed in 1990 by Roberts [R] who used an approach completely different from
that of Nagata. In 1994, based on Roberts’ counterexample, A’Campo-Neuen [A] con-
structed a linear counterexample involving an action of a 12-dimensional unipotent group
on a 19-dimensional vector space. Subsequently, this counterexample was followed by a
series of linear counterexamples (see, for instance, [Fr] and references therein). We note
that among linear counterexamples to Hilbert’s fourteenth problem the key role is played
by counterexamples involving linear actions of unipotent groups.
A natural way of obtaining examples of homogeneous spaces with epimorphic stabilizer
that admit no projective embeddings with small boundary is as follows. First, one takes a
linear counterexample to Hilbert’s fourteenth problem involving an action of a unipotent
group H0on a vector space V over k. Second, one fixes a connected reductive group G ⊂
GL(V ) containing H0. Third, one chooses an appropriate one-dimensional torus S ⊂ G
normalizing H0such that the subgroup H = SH0is epimorphic in G. In this situation,
the homogeneous space G/H admits no projective embeddings with small boundary, see
Proposition 1 in Section 2.
The first example of a homogeneous space with epimorphic stabilizer that admits no
projective embeddings with small boundary was mentioned in [BBII, 7(b)]. In this ex-
ample, G = SL2×... × SL2 (16 copies) and H is obtained by extending the group in
Nagata’s counterexample by a one-dimensional torus. An analogous example based on
Nagata-Steinberg’s counterexample was considered (with complete proofs) in [Ar, §2]. In
the present paper, we construct a new example of that kind based on A’Campo-Neuen’s
counterexample. The precise formulations and construction are given in Section 2.
In connection with these results the following question may be of interest.
Question. Let a unipotent group H0act linearly on a finite-dimensional vector space V
and let G be a connected reductive subgroup of GL(V ) containing H0. Suppose that the
algebra k[V ]H0is not finitely generated over k. Is there a one-dimensional torus S ⊂ G
normalizing H0such that the group H = SH0is epimorphic in G?

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AN EPIMORPHIC SUBGROUP ARISING FROM ROBERTS’ COUNTEREXAMPLE3
2. Construction of the subgroup
We put G = SL19and denote by V the space of the tautological representation of G. We
fix a basis e1,e2,...,e19in V . Further on, for any element of G, its matrix is considered
with respect to this basis.
Let Gabe the additive group of k.
We consider a subgroup H0≃ (Ga)12embedded in G as follows:
µ = (µ0,µ1,...,µ11) ?→ h(µ) =
where E4, E15are the identity matrices of order 4, 15, respectively, and
?
E4
0
M(µ) E15
?
,
(1)M(µ) =

























µ1
µ2
0
µ3
µ4
0
µ5
µ6
0
µ7
µ8
µ9
µ10
µ11
0
µ0
µ1
µ2
0
0
0
0
0
0
µ0
µ7
0
0
0
0
0
0
0
0
0
0
0
0
0
µ0
µ5
µ6
0
0
0
0
µ10
µ11
µ0
µ3
µ4
0
0
0
0
0
µ8
µ9
0
0

























.
The result of A’Campo-Neuen is as follows.
Theorem 2 ([A]). The algebra k[V ]H0is not finitely generated over k.
Remark 1. In [A] this theorem is proved for any ground field of characteristic zero.
To construct our example, we consider a one-dimensional torus
S = {diag(s15,s15,s15,s15,s−4,...,s−4
(2)
?
???
15
) | s ∈ k×} ⊂ G,
where k×= k\{0} is the multiplicative group of k. Clearly, S normalizes H0. We now
put H = SH0. The main result of this paper is given by the following theorem.
Theorem 3. The subgroup H is epimorphic in G.
This theorem will be proved in Section 3.
Corollary 1. The homogeneous space G/H admits no projective embeddings with small
boundary.
This corollary follows from Theorem 3 and the following proposition.
Proposition 1. Suppose we are given a linear action of a unipotent group H0 on a
vector space V over k such that the algebra k[V ]H0is not finitely generated. Suppose that
a connected reductive subgroup G ⊂ GL(V ) containing H0 is fixed and there is a one-
dimensional torus S ⊂ G normalizing H0 such that the group H = SH0is epimorphic
in G. Then the homogeneous space G/H admits no projective embeddings with small
boundary.

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4ROMAN AVDEEV
Proof. It suffices to show that condition (b) of Theorem 1 does not hold. This is done by
the same argument as in [Ar, Lemma 2.2].
?
3. Proof of Theorem 3
Before we prove Theorem 3, let us fix some notation.
We recall that G = SL19. Let B (resp. U, T) be the Borel subgroup (resp. the
maximal unipotent subgroup, the maximal torus) in G consisting of all upper triangular
(resp. upper unitriangular, diagonal) matrices contained in G. We denote by X(B) the
weight lattice of B. The semigroup of dominant weights of B is denoted by X+(B),
X+(B) ⊂ X(B). For i = 1,2,...,18 we denote by πithe i-th fundamental weight of B,
which takes every upper triangular matrix to the product of its first i diagonal entries.
The simple G-module with highest weight λ ∈ X+(B) is denoted by V (λ), and its
highest weight vector with respect to B is denoted by vλ. Let Pλ⊂ G be the subgroup
that stabilizes the line ?vλ? ⊂ V (λ). This subgroup is a parabolic subgroup containing
the Borel subgroup B. We identify the weight lattice X(Pλ) of Pλwith a sublattice of
X(B) by means of the natural embedding B ֒→ Pλ.
Every dominant weight λ of B has the form λ = a1π1+ a2π2+ ... + a18π18for some
non-negative integers a1,a2,...,a18. If ai> 0 for some i ∈ {1,2,...,18} then Pλstabilizes
the line ?vπi? ⊂ V (πi). At that, Pλacts on vπiby multiplication by the weight πi. This
weight takes every matrix A ∈ Pλto the minor corresponding to the first i and last i rows
and columns of A. (The lower left (19 − i) × i block of A consists of zero entries.)
In this section, we identify elements s ∈ k×and their images in S, see (2).
We now proceed to prove Theorem 3. By [Gr, Lemma 23.5] it suffices to show that
there are no proper observable subgroups of G containing H. In view of [Gr, Lemma 7.7]
the proof will be completed if we check the following two conditions:
(1) for every non-trivial simple G-module V (λ) and every Borel subgroup?B ⊂ G a
(2) there are no proper reductive subgroups of G containing H.
Condition (1) follows from Lemma 1. Condition (2) will be checked using Lemma 2.
We now turn to formulate and prove lemmas.
highest weight vector of V (λ) with respect to?B is not invariant under H;
Lemma 1. Let?B ⊂ G be an arbitrary Borel subgroup and V (λ), λ ?= 0, an arbitrary
h ∈ H such that h · ? vλ?= ? vλ.
Proof. Assume that h·? vλ= ? vλfor all h ∈ H. Since λ ?= 0, we have λ = a1πi1+a2πi2+...+
The subsequent argument is divided into several steps.
Step 1. Since all Borel subgroups in G are conjugated, there exists an element g0∈ G
such that?B = g0Bg−1
(3)g−1
for all h ∈ H. Consider the Bruhat decomposition of g0:
(4)g0= uσb,
simple G-module with highest weight vector ? vλwith respect to?B. Then there is an element
amπim, where 1 ? m ? 18, 1 ? i1< i2< ... < im? 18, and ai> 0 for all i = 1,...,m.
0. Then ? vλ= αg0· vλfor some α ?= 0, whence
0hg0· vλ= vλ
where u ∈ U, σ ∈ NG(T), b ∈ B are some fixed elements. We may assume that σ = εσ0,
where σ0is a permutation matrix, ε = 1 for detσ0= 1, and ε = eπ√−1/19for detσ0= −1.
We now substitute expression (4) for g0in (3). Since b multiplies vλby a scalar, we have
σ−1u−1huσ · vλ= vλ
(5)

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AN EPIMORPHIC SUBGROUP ARISING FROM ROBERTS’ COUNTEREXAMPLE5
for all h ∈ H. Let τ : G → G be the map given by the formula τ(g) = σ−1u−1guσ. Taking
into account (5) we obtain τ(H) ⊂ Pλ.
Step 2. Let ν be the permutation of the set {1,2,...,19} that corresponds to σ. Then
σ(ej) = εeν(j)for j = 1,...,19. For each pair of matrices g = (gij) ∈ G, g = σ−1gσ we
have gij= gν(i),ν(j)for i,j = {1,...,19}. In particular, gjj= gν(j),ν(j)for j = 1,...,19.
We note that under the map g ?→ g entries of g lying in the same row (resp. column) are
transformed into elements of g that also lie in the same row (resp. column).
Step 3. Suppose s ∈ S. Then u−1su is an upper triangular matrix whose diagonal
entries are s15, s15, s15, s15, s−4,...,s−4. Further, the diagonal entries of the matrix
τ(s) = σ−1u−1suσ are again s15, s15, s15, s15, s−4,...,s−4, perhaps in another order. At
that, for every i = 1,...,18 the determinant of the upper left i × i block of τ(s) is equal
to the product of all diagonal entries of this block. Therefore, for every j = 1,...,m we
have πij(τ(s)) = sbjfor some bj∈ Z. Moreover, bj= 15kj− 4lj, where
kj= #{k ∈ {1,2,...,ij} | ν(k) ∈ {1,2,3,4}}
and
lj= #{k ∈ {1,2,...,ij} | ν(k) / ∈ {1,2,3,4}}.
Clearly, 0 ? kj ? 4, 0 ? lj ? 15, and kj+ lj = ij. The last equality implies that
(kj,lj) / ∈ {(0,0),(4,15)}, whence bj ?= 0 for all j = 1,...,m. Further, the condition
τ(s) · vλ= vλimplies that λ(τ(s)) = 1 for all s ∈ S and a1b1+ a2b2+ ... + ambm= 0.
We conclude that there exists j0∈ {1,...,m} with bj0> 0. Put i∗= ij0, k∗= kj0, and
l∗= lj0. Since bj0> 0, we have l∗< 15k∗/4, in particular, k∗> 0. Obviously, for every
matrix in τ(H) its lower left (19 − i∗) × i∗block consists of zero entries.
Step 4. Suppose that u =
0Q
unitriangular matrices of order 4 and 15, respectively, R and R′are 4×15 matrices, R′=
−P−1RQ−1. Let h = h(µ) ∈ H0be an arbitrary element. Recall that h =
for some µ ∈ k12, where M(µ) is the matrix in (1). Then
?PR
?
and u−1=
?P−1
R′
Q−1
0
?
, where P and Q are upper
?
E4
0
M(µ) E15
?
u−1hu =
?E4+ R′M(µ)P
Q−1M(µ)P
P−1R + R′M(µ)R + R′Q
E15+ Q−1M(µ)R
?
.
We consider the 15× 4 matrix D = D(h) = Q−1M(µ)P. Note that for every entry dpqof
D we have dpq= mpq+?cijmijwhere the sum is taken over all pairs (i,j) with i ? p,
j ? q, and (i,j) ?= (p,q), the coefficients cijbeing uniquely determined by the matrix u.
Now, using the latter observation and the explicit form (1) of the matrix M(µ), we can
successively choose elements µ11, µ10, µ9, µ8, µ7, µ0, µ6, µ5, µ4, µ3, µ2, µ1∈ k in such a
way that, for the corresponding element h0∈ H0, the submatrix D(h0) of u−1h0u has the