Table Of ContentLinearly reductive algebraic groups and Hilbert’s theorem
Matei Ionita, Nilay Kumar
May 15, 2014
Throughout this note, we fix an algebraically closed field k of characteristic zero. All algebraic
varieties are taken to be affine and irreducible.
1 Basic Definitions
We begin with the definition of an algebraic group:
Definition 1.1. Let G be an (affine) variety equipped with a morphism µ : G×G → G such that
the set of points of G with the operation given by µ is a group, and the operation of inversion
i : G → G is a morphism. In this case we say that (G,µ,i) is an algebraic group. We take
morphismsbetweentwoalgebraicgroupstobemapsthataresimultaneouslymorphismsofvarieties
andgrouphomomorphisms. Inparticular,wedefinealinearalgebraicgrouptobeaclosedsubgroup
of some GL(n).
The basic object of study is, of course, a representation:
Definition 1.2. Let V be a k-vector space. We say that ρ : G → GL(V) is an (algebraic)
representation of G if ρ is a morphism of algebraic groups. By abuse of notation we will denote
therepresentationbyV andρ(g)v byg·v. AsubrepresentationofV isavectorsubspaceW ⊂ V
such that ρ(g)W = W for all g ∈ G, i.e. W is G-invariant as a subspace of V. We say that V
is reducible if it contains a non-trivial subrepresentation and irreducible otherwise. Finally, we
denote by VG = {v ∈ V | g ·v = v for all g ∈ G} the invariant subspace of V under G. We
take morphisms between two representations of G to be G-equivariant linear maps, i.e. linear maps
commuting with ρ.
It will be convenient for our purposes to consider instead the category dual to the category of
G-representations above.
Definition 1.3. Let G be an affine algebraic group over k, and A be the coordinate ring of G. An
algebraic corepresentation of the group G is a vector space V over k together with a linear map
µ : V → V ⊗ A, which satisfies the following two conditions:
V k
1. Let (cid:15) : A → k be the coidentity, i.e. the map that evaluates f ∈ A on the identity of G. The
composition
V µV V ⊗ A 1⊗(cid:15) V
k
is the identity in V.
1
2. Let µ : A → A⊗ A be the comultiplication. Then the following diagram commutes:
A k
V µV V ⊗ A
k
µV µV⊗1A
V ⊗ A 1V⊗µA V ⊗ A⊗ A
k k k
Remark. There are several remarks to be made here. First let us make explicit the duality between
corepresentations and representations. We regard the vector space in Definition 1.3 as the dual V∗
ofthevectorspaceV inDefinition1.2,whichisallowedbecauseV ∼= V∗. Thenacorepresentationis
a linear map µ : V∗ → V∗⊗ A. This can be extended to a ring homomorphism on the polynomial
V k
algebra of V∗ as µ : SymV∗ → SymV∗⊗ A.
V k
Note that SymV∗ and A are finitely generated k-algebras, which are the affine coordinate rings
of V and G respectively. Then µ induces a morphism ρ of affine varieties: ρ : G×V → V. The
V
duals of properties (1) and (2), stated below, make this morphism into a group action:
1. Let e ∈ G be the identity. Then the following composition is the identity in V:
(e,v) ρ
V → G×V → V
i.e. ρ(e,v) = v.
2. Let m : G×G → G be the group operation. Then the following diagram commutes:
G×G×V (m,1V) G×V
(e,ρ) ρ
ρ
G×V V
In other words, ρ(g g ,v) = ρ(g ,ρ(g ,v)).
1 2 1 2
The fact that µ is linear shows that ρ is linear on the V factor. Moreover, using (1) and (2) we see
V
that ρ(g−1,ρ(g,v)) = v. These facts allow us to think of ρ as a homomorphism from G to GL(V),
so we obtain a representation of G on V. This construction is clearly invertible, i.e. starting from a
representation ρ and dualizing will give a ring homomorphism ρ∗ : SymV∗ → SymV∗⊗ A, where
k
ρ∗ is the pullback by ρ. Since ρ is linear on the V factor, ρ∗ is linear, so in particular it takes V∗
to V∗⊗ A. The map µ in the definition of a corepresentation is precisely the restriction of ρ∗ to
k V
V∗.
Definition 1.4. Let µ : V → V ⊗A be a corepresentation of a group G. We say that a vector
x ∈ V is G-invariant if µ(x) = x⊗1 and we denote the G-invariant subspace of all such vectors
by VG. Moreover, a subspace U ⊂ V is called a subrepresentation if µ(U) ⊂ U ⊗A and V is
called irreducible if it has no non-trivial subrepresentations.
The following property demonstrates the utility of corepresentations.
Proposition 1.5. Every corepresentation V of G is locally finite-dimensional, i.e. every x ∈ V is
contained in a finite-dimensional corepresentation of G.
Proof. See [2], Proposition 4.6.
2
We illustrate the concepts just introduced for two simple examples: the multiplicative group
G and the additive group G .
m a
Definition 1.6. Given a vector space V and an integer m ∈ Z, it’s easy to see that the map
V → V ⊗k[t,t−1]
v (cid:55)→ v⊗tm
is a corepresentation of G . It is called the corepresentation of weight m, and we denote it by
m
V .
(m)
Proposition 1.7. Every corepresentation V of G is a direct sum
m
(cid:77)
V = V
(m)
m∈Z
Proof. See [2], Proposition 4.7.
Proposition 1.8. Every corepresentation V of G is given by
a
(cid:88)∞ sn
µ(v) = fn(v)⊗
n!
n=0
for some endomorphism f ∈ End(V) which is locally nilpotent (that is, for every v, there exists
some n such that fn(v) = 0).
Proof. See [2], Proposition 4.8.
2 Linear Reductivity
We now restrict ourselves to linearly reductive algebraic groups, whose representations and corep-
resentations, as we shall see, decompose quite simply.
Definition 2.1. An algebraic group G is said to be linearly reductive if for every epimorphism
φ : V → W of G-corepresentations the induced map on G-invariants φG : VG → WG is surjective.
There are a few equivalent definitions that are useful.
Proposition 2.2. The following are equivalent:
1. G is linearly reductive.
2. For every epimorphism φ : V → W of finite-dimensional representations the induced map on
G-invariants φG : VG → WG is surjective.
3. If V is any finite-dimensional representation and v ∈ V is G-invariant modulo a proper
subrepresentation U ⊂ V, then the coset v+U contains a nontrivial G-invariant vector.
Proof. See [2], Proposition 4.37.
As a first example, we note the following:
Proposition 2.3. Every finite group G is linearly reductive.
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Proof. Let V be a finite-dimensional corepresentation and v ∈ V be a vector invariant modulo a
subrepresentation U ⊂ V. Defining
1 (cid:88)
v(cid:48) = g·v,
|G|
g∈G
we find that v(cid:48) is G-invariant and
1 (cid:88)
v(cid:48)−v = (g·v−v),
|G|
g∈G
which is contained in U. This satisfies condition (3) above, and hence G is linearly reductive.
Example2.4. TheadditivegroupG isnotlinearlyreductive. Indeed,considerthetwo-dimensional
a
representation given by
(cid:18) (cid:19)
1 t
t (cid:55)→ .
1
Then the restriction to the x-axis k[x,y] → k[x,y]/(y) = k[x] is a surjective homomorphism of
Ga-representations. But k[x,y]Ga = k[y] and k[x]Ga = k[x] and hence the corresponding map on
invariants is not surjective.
Theorem 2.5. G is linearly reductive if and only if every finite-dimensional corepresentation V of
G is completely reducible, i.e. V splits as a direct sum of irreducible corepresentations.
Proof. Itsufficestoshowthatanyfinite-dimensionalcorepresentationV ofGwithasubrepresenta-
tion W decomposes as a direct sum of corepresentations V = W ⊕W(cid:48). Consider first Hom (V,W)
k
and Hom (W,W), as spaces of linear maps, as G-corepresentations. Linear reductivity of G im-
k
plies that the surjective restriction map Hom (V,W) → Hom (W,W) induces a surjective map
k k
Hom (V,W)G → Hom (W,W)G. Note now that Hom (V,W)G is precisely those morphisms that
k k k
are G-equivariant, i.e. morphisms of G-corepresentations. Consider now the identity morphism
Id ∈ Hom (W,W)G, which lifts via surjectivity to a φ ∈ Hom (V,W). Denoting W(cid:48) = kerφ, we
W k k
obtain a short exact sequence of G-corepresentations
0 W(cid:48) V φ W 0.
Treating the identity Id as the inclusion ι : W → V, we find that φ◦ι = Id : W → W, and
W W
hence we obtain a splitting V = W(cid:48)⊕W.
Consider a surjective map φ : V → W of G-corepresentations. By hypothesis, the above
sequence splits. W and W(cid:48) are each stable under G, therefore VG = W(cid:48)G ⊕WG, and hence the
induced map VG → WG is surjective.
This characterization of linearly reductive groups shows their importance; one can hope to
classify the irreducible (co)representations and express arbitrary (co)representations as direct sums
of these. Moreover, the linearly reductive groups are the ones Hilbert’s theorem holds for. For
the rest of this section, we focus on discussing examples of linearly reductive groups, so that,
when we prove Hilbert’s theorem, we already know what kind of groups it applies to. As a first
example, Proposition 1.8 shows that G is linearly reductive. Furthermore, it is a basic fact in
m
the representation theory of Lie groups that compact Lie groups are linearly reductive (where
compactness is taken in the sense of the Euclidean topology). The proof of this is analytic, and we
do not give it here. The idea is to generalize the proof of Proposition 2.3, by replacing the sum over
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elements of g with an integral over the compact group G. The technical difficulty lies in defining
an invariant measure on G, known as the Haar measure.
We are interested in proving that certain noncompact linear algebraic groups are linearly re-
ductive. An important family of groups for which this holds are the complex special linear groups
SL(n,C). Our proof is based on Weyl’s unitary trick, which is a way of using the representation
theoryofcompactsubgroups, knowntobelinearlyreductive, togetinformationaboutnoncompact
groups, such as SL(n,C). We begin by proving the following lemma about linear algebraic groups.
For the remainder of the section, we assume that k = C.
Lemma 2.6. Let G be a linear algebraic group, and K a subgroup of G that is Zariski-dense in G
and is compact in the Euclidean topology. Then G is linearly reductive.
Proof. Let V → W be a surjective morphism of G-corepresentations. Clearly V and W are also
corepresentations of K. But K is compact, and we assumed that all compact groups are linearly
reductive. This gives a surjection VK → WK. We claim that, in fact, VK = VG and WK = WG.
Note that this proves that VG → WG is surjective, which is enough to conclude that G is linearly
reductive.
Clearly VG ⊂ VK as K is a subgroup of G. To see that VK ⊂ VG, we fix v ∈ VK and, using
the equivalence of representations and corepresentations, we have a morphism of affine varieties
φ : G×{v} → V
φ is continuous, therefore φ−1v is Zariski closed in G×{v}. But, since v ∈ VK, K ×{v} ⊂ φ−1v,
so K ×{v} ⊂ φ−1v. Since K is Zariski dense in G, this means G×{v} ⊂ φ−1v, so v ∈ VG.
This lemma gives an easy way to prove that a linear algebraic group G is linearly reductive:
we simply need to exhibit a compact, Zariski dense subgroup K. For example, an easy proof that
G = GL(1) is linearly reductive uses the compact subgroup U(1) of GL(1). Since GL(1) is 1-
m
dimensional, any closed subset is either finite or the whole of GL(1). U(1) is infinite, therefore
U(1) = GL(1). In the next proposition, we apply Lemma 2.6 to show that SL(n) is linearly
reductive by taking K = SU(n). The proof of density of SU(n) in SL(n) follows [1]. The following
elementary lemma will be useful.
Lemma 2.7. Let V be an R-vector space and let f : V ⊗R C → C be a holomorphic function such
that f(V ⊗RR) = 0. Then f is identically zero.
Proof. Suppose dimV = 1. Then it is a basic complex-analytic fact that if f : C → C holomorphic
vanishesontherealline,f mustbeidenticallyzero. IfdimV = n,wehavef holomorphicsatisfying
f(x ,...,x ) for all x ∈ R. Viewing f as a function of the first variable by fixing x ,...,x , we
1 n i 2 n
find using the reasoning of the n = 1 case that f(z ,x ,...,x ) = 0 for all z ∈ C. Repeating this
1 2 n 1
argument, we find that f(z ,...,z ) = 0 for all z ∈ C.
1 n i
Proposition 2.8. SL(n,C) is linearly reductive.
Proof. Using Lemma 2.6, it suffices to show that SU(n) is Zariski-dense in SL(n). Suppose not –
then there exists an f ∈ k[SL(n)] not identically zero but satisfying f(SU(n)) = 0. Consider now
the Lie algebra su(n), which is a real subspace of sl(n) with the property that su(n)⊗RC = sl(n).
Note now that the exponential map exp : sl(n) → SL(n) is holomorphic and maps su(n) ⊂ sl(n)
to SU(n). Thus the composition f ◦exp : sl(n) → C is holomorphic and satisfies f(expsu(n)) = 0.
By the previous lemma, f ◦exp must be identically zero on sl(n), whence f must be identically
zero, contradicting our original assumption.
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Remark. The proof of Proposition 2.8 did not use any particular features of SL(n), apart from the
fact that its Lie algebra sl(n) has a compact real form su(n). Therefore the same proof generalizes
to other linear algebraic groups G, as long as we can find a compact real form for Lie(G). In
particular, for G = GL(n), we have u(n)⊗R C = gl(n), so we can take K = U(n), which proves
that GL(n) is linearly reductive. Similarly, for the symplectic group SL(2n,C), let U(n,H) be the
quaternionic unitary group, which is compact; we have u(n,H)⊗RC = sp(2n,C), which proves that
Sp(2n,C) is linearly reductive.
3 Hilbert’s Theorem
In this section we obtain an impressive payoff for the somewhat nontransparent definitions of
corepresentations and linear reductivity. Under this formalism, the proof of Hilbert’s theorem on
the finite generation of rings of invariants is almost trivial. Before we study the theorem itself, we
give some motivation for our interest in this result.
Suppose we have an action of a linear algebraic group G on an affine variety X. We look
at the quotient X/G, i.e. the set of orbits in X under G, and ask whether it is also an affine
variety. In most cases it will not: if there exists an orbit which is not Zariski closed in X, then
the corresponding point in X/G is not closed, so we need the formalism of schemes in order to talk
about the quotient. However, in the case that X/G is an affine variety, the easiest way to describe
it is through its coordinate ring. An obvious candidate for this is (A[X])G, the ring of invariants
under G. We would like, then, for this ring to be finitely generated over k, since the equivalence
of categories between finitely generated k-algebras and affine varieties will give an affine variety
structure on X/G.
That being said, we turn to Hilbert’s theorem. We follow the proof in [2], Theorem 4.51.
Theorem 3.1 (Hilbert). Let G be a linearly reductive algebraic group, which acts on a polynomial
ring S, preserving the grading. Then the ring of invariant polynomials SG is finitely generated.
Proof. The invariant ring inherits a grading from S:
(cid:77)
SG = SG∩S
e
e≥0
It’s useful to consider SG, the span of the invariants of positive degree, and the ideal J that SG
+ +
generatesinS. SinceS isNoetherian, J isgeneratedbyfinitelymanypolynomialsf ,...,f ∈ SG.
1 N +
This means that the following S-module homomorphism is surjective.
φ : S ⊕···⊕S → J
N
(cid:88)
(h ,...,h ) (cid:55)→ h f
1 N i i
i=1
Weclaimthatf ,...,f generateSG. Toseethis, wepickahomogenousinvarianth ∈ SG. Weuse
1 N
induction on degh to show that h ∈ k[f ,...,f ]. If degh = 0, then this is clear. For degh > 0,
1 N
h ∈ SG, so in paticular h ∈ JG. The map φ above is a surjective morphism of G corepresentations,
+
so since G is linearly reductive the induced map
SG⊕···⊕SG → JG
6
is also surjective. Therefore there exist h(cid:48),...,h(cid:48) ∈ SG such that
1 N
N
(cid:88)
h = h(cid:48)f
i i
i=1
The f have positive degree, therefore degh(cid:48) < degh for all i. By the inductive hypothesis, this
i i
means h(cid:48) ∈ k[f ,...,f ], so h ∈ k[f ,...,f ].
i 1 N 1 N
We are now interested in generalizing Hilbert’s theorem by replacing the polynomial ring S
with an arbitrary finitely generated k-algebra R. The following lemma will be useful in the proof.
Lemma 3.2. Let G be a linear algebraic group that acts on a finitely generated k-algebra R. Then
there exists a set of generators r ,...,r for R such that the linear span of r ,...,r is a G-
1 N 1 N
invariant vector subspace of R.
Proof. Let f ,...,f be any set of generators. By Proposition 1.6, each f is contained in a finite
1 M i
(cid:76)
dimensional corepresentation V ⊂ R. Consider V = V ; it is G-invariant because each of its
i i i
factors is, and it is finite dimensional. Therefore we can take r ,...,r to be any basis of V.
1 N
Apartfrombeingusefulinthefollowingproof,thislemmahasausefulgeometricinterpretation.
Let X be the affine algebraic variety corresponding to R. The lemma gives a surjective morphism
of corepresentations Sym(V∗) → R. This corresponds to an injective, G-equivariant morphism of
affine varieties
X → V = AN
In other words, every affine variety X can be equivariantly embedded in an affine space on which
G acts linearly.
Theorem 3.3. Let G be a linearly reductive algebraic group, which acts on a finitely generated
k-algebra R. Then the invariant ring RG is finitely generated.
Proof. As in the remark above, the choice of generators r ,...,r gives a surjective k-algebra
1 N
homomorphism:
φ
S = k[x ,...,x ] → R
1 N
x (cid:55)→ r
i i
This map induces a corepresentation of G on S, by composition with the corepresentation µ of G
on R:
φ µ
S → R → R⊗A → S ⊗A
Note that R⊗A → S⊗A is generally not a homomorphism, since R is a quotient of S, and not a
subalgebra. However, thedefinitionofacorepresentationonlyrequiresalinearmapbetweenvector
spaces, which is clearly satisfied here. By Hilbert’s theorem, SG is finitely generated; let f ,...,f
1 k
be generators. Now, since G is linearly reductive, the map
SG → RG
induced by φ is surjective, so RG is generated by φ(f ),...,φ(f ).
1 k
7
References
[1] Darij Grinberg, Symmetric subspace of linear operators (answer), MathOverflow, URL: http:
//mathoverflow.net/a/43474 (visited on 2014-05-11).
[2] Shigeru Mukai, An introduction to invariants and moduli, Cambridge University Press, 2003.
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