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ALGEBRAIC GROUPS
2.4. Orbits of algebraic groups. The reason that algebraic groups are so
important is that they act on algebraic varieties!!
Let G be an algebraic group and X be a variety (not necessarily affine!).
We say that G acts on X, or that X is a G-variety, if we are given a morphism
ρ:G×X →X
of varieties that makes X into a G-set in the usual sense. I will use all
the usual language of group actions: in particular I will write gx instead of
ρ(g, x) whenever it is convenient. For example, G acts on itself in various
ways: by conjugation, by left multiplication, . . . .
The orbit of x ∈ X is the set Gx = {gx|g ∈ G}. The action is transitive if
Gx = X for x ∈ X. The set of fixed points of g resp. G on X is denoted X g
resp. X G . The stabilizer of x ∈ X in G is denoted Gx , clearly a subgroup
of G. For x ∈ X, the orbit map
fx : G → Gx, g $→ gx
induces a bijection f¯x between the set G/Gx of cosets of Gx in G and the
orbit Gx. Of course it would be nice if both G/Gx and the orbit Gx were varieties in some natural way so that this map f¯x was a morphism of varieties...
we will see that is the case eventually.
More generally, for a subset Y ⊆ X, let CG (Y ) = {g ∈ G | gy =
y for all y ∈ Y } and NG (Y ) = {g ∈ G | gY ⊆ Y }. If you take Y to be
a subgroup of G and the action of G on itself by conjugation, these are the
usual notions of centralizer and normalizer of a subgroup of G. Finally, the
center of G is the normal subgroup Z(G) = {z ∈ G | gz = zg for all g ∈ G}
which is just the set of fixed points of G on itself for the conjugation action.
Lemma 2.11. Let G act on X. Let Y, Z be subsets of X with Z closed.
(i) The set {g ∈ G | gY ⊆ Z} is closed; in particular NG (Z) is closed.
(ii) For each x ∈ X, the stabilizer Gx is a closed subgroup of G, in
particular CG (Y ) is closed.
(iii) The fixed point set X g of g ∈ G is closed in X; in particular X G is
closed.
Proof. (i) For each y ∈ X, the orbit map fy : G !
→ X, g $→ gy is a morphism.
So fx−1 (Z) is closed in G. So the intersection y∈Y fy−1 (Z) is closed in G,
which is {g ∈ G | gY ⊆ Z}.
!
(ii) Observe Gx = {g ∈ G | g{x} ⊆ {x}, CG (Y ) = y∈Y Gy and apply (i).
(iii) To see that X g is closed, consider ψ : X → X × X defined by
x $→ (x, gx). The fixed point set X g is the inverse image under ψ of the
diagonal, which is closed since X is a variety.
!
The lemma shows things like centralizers of subsets, normalizers of closed
subsets, fixed point sets, . . . are closed. However: orbits themselves will not
be closed in general. In fact the structure of orbits of an algebraic group on
a variety can be extremely interesting...
ALGEBRAIC GROUPS
25
Theorem 2.12. Let G be a connected algebraic group acting on X. Each
orbit Gx is irreducible, open in its closure (hence is an irreducible variety
in its own right), and its boundary Gx − Gx is a union of orbits of strictly
smaller dimension. In particular, orbits of minimal dimension are closed
(so closed orbits exist).
Proof. Let O = Gx. It is the image of G under the orbit map fx , hence
contains a non-empty
open subset U of its closure. But G acts transitively
"
on O, so O = g∈G gU is open in its closure too.
Therefore O − O is closed so all of its irreducible components have strictly
smaller dimension than O itself. It is G-stable, so it is a union of orbits. !
Let us now pause to give some examples of algebraic group actions and
their orbits.
Example 2.13.
(1) Let G = GLn = GL(V ) where V = k n (viewed as
affine n-space). Note the coordinate ring k[V ] of V is the polynomial
algebra k[T1 , . . . , Tn ] where Ti is the ith coordinate function. Incidentally, if you like the basis free approach then k[V ] is the algebra
of all functions on V , which is just S(V ∗ ), the symmetric algebra of
the dual space.
What are the orbits of G on V ? There are just two, the point {0},
a closed orbit of dimension 0, and the rest V − {0}, an open orbit
of dimension (n + 1). The stabilizer of a point in the open orbit is
conjugate to the closed subgroup


1
∗
 0



H= .

 ..
∗ 
0
Note dim G = n2 , dim H = n(n − 1) and dim G − dim H = n which is
the dimension of the open orbit. This is true in general: dim Gx =
dim G − dim Gx for all x ∈ X, though we are not quite ready to
prove this yet.
(2) Note V $ = V − {0} is an open subset of V , hence a variety in its
own right. Let P(V ) ∼
= Pn−1 be the projective variety consisting of
all 1-subspaces of V . There is an obvious function
φ : V $ → P(V ), x $→ [x].
This is actually a morphism. To prove this, we will use the following
simple lemma that reduces the check to affine open subsets:
Lemma 2.14. Let φ : X → Y be a function. Suppose there is an
affine open covering Y = U1 ∪ · · · ∪ Un and open subsets V1 , . . . , Vn
of X such that
(i) φ(Vi ) ⊆ Ui ;
(ii) f ◦ φ ∈ OX (Vi ) whenever f ∈ OY (Ui ).
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ALGEBRAIC GROUPS
Then φ is a morphism of varieties.
In our case, we take Ui = Pni−1 , the affine open subset of Pn . We
take Vi to be φ−1 (Ui ), i.e. the vectors in V that have non-zero i-th
coordinate. So Vi is the principal open subset D(Tj ) in the original
affine space V . So we have that k[Vi ] = k[T1 , . . . , Tn , Ti−1 ]. Now you
can easily see that φ∗ (k[Ui ]) is the subalgebra k[T1 /Ti , . . . , Tn /Ti ] of
k[Vi ]. This checks the condition in the lemma, so φ is a morphism.
(3) By a similar argument to (2), the map
ρ : G × P(V ) → P(V ), (g, +v,) $→ +gv,
is a morphism of varieties. This is just the natural action of G on
1-spaces in V . This action is transitive, and the stabilizer of a point
is conjugate to


∗
∗
 0



H= .

.
 .
∗ 
0
This time dim G = n2 , dim H = n(n + 1) + 1 and dim G − dim H =
n − 1 = dim Pn−1 as we expected.
(4) Now I want to refine this example. It is time to introduce the Grassmann varieties Gd (V ). Here, V is an n-dimensional vector space,
and 0 ≤ d ≤ n. As a set, Gd (V ) is the set of all d-dimensional
subspaces of V . (For example, G1 (V ) = P(V ) ∼
= Pn−1 ). We want to
give Gd (V ) the structure of a variety.
There is an injective map
d
)
ψ : Gd (V ) → P( (V )), +v1 , . . . , vd , $→ v1 ∧ · · · ∧ vd .
So to endow Gd (V ) with the structure of a (projective) variety, it
suffices to show that the image of ψ is closed. We will do this by
*
showing its intersection with the affine open subsets of P( d (V )) are
all closed.
*
Fix an ordered basis v1 , . . . , vn for V . Then d (V ) has basis
{vi1 ∧ · · · ∧ vid | 1 ≤ i1 < · · · < id /= n}.
Without loss of generality, lets consider the affine open subset U ⊂
*
P( d (V )) consisting of the span of all vectors whose v1 ∧ · · · ∧ vd coordinate is non-zero. Let V0 = +v1 , . . . , vd ,. Clearly, for a ddimensional subspace W of V , ψ(W ) ∈ U if and only if the projection
π of V onto V0 along the basis maps W isomorphically onto V0 . Then,
wi = π −1 vi = vi + xi where
n
+
xi =
ai,j vj
j=d+1
ALGEBRAIC GROUPS
27
for some a$i,j s. Conversely, given any choice of the d(n − d) scalars
ai,j ’s there is a unique subspace W with ψ(W ) ∈ U . So we have
parametrized ψ −1 (U ) by d × (n − d)-matrices. Now let us think
about the coordinates of ψ(W ) in U . The image of W under ψ is
+
ai,j v1 ∧ · · · ∧ vj ∧ · · · ∧ vd + (∗)
v1 ∧ · · · ∧ vd +
j>d
(vj in the ith position) where (∗) consists of wedges with two or more
from v1 , . . . , vd replaced by higher vk ’s. Obviously the coefficients in
(∗) will look like some polynomial functions in the ai,j ’s. Hence, the
coordinates of ψ(W ) with respect to the standard coordinates of U
is of the form
(. . . , ai,j , . . . ; . . . , fk (ai,j ), . . . ),
where the fk are some polynomial functions in the ai,j ’s. As W
varies, we get all such points for arbitrary ai,j ∈ Ad(n−d) . This
shows that
ψ(Gd (V )) ∩ U = {(x, f (x) | x ∈ Ad(n−d) }
where f is a morphism from Ad(n−d) to some other affine space. But
the graph of a morphism is always closed 1.17(i).
Hence we can identify Gd (V ) with its image under ψ to give it the
structure of a projective algebraic variety (as a closed subvariety of
some Pn ).
(5) Now consider the natural action of G = GL(V ) on Gd (V ). By
*
(3), it is just the restriction of the action of G on P( d (V )) to
Gd (V ), hence the action is a morphism. Moreover, it is obviously
transitive. This shows that Gd (V ) is an irreducible variety, and you
can compute its dimension by computing the stabilizer of a point to
get dim Gd (V ) = d(n − d).
(6) As a final example, we have the flag variety F (V ). This is the
following closed subvariety of G1 (V ) × G2 (V ) × · · · × Gn (V ):
F (V ) = {(f1 , f2 , . . . , fn ) | fi ∈ Gi (V ), fi ⊂ fi+1 }.
Clearly GL(V ) acts morphically on F (V ), and the action is transitive. Hence F (V ) is an irreducible projective variety. Consider the
standard flag f + = (f1 , . . . , fn ) where fi = +v1 , . . . , vi ,. Its stabilizer
on G is the subgroup Tn of all upper triangular matrices. Hence:
dim F (V ) = dim GLn − dim Tn = n(n − 1)/2.
Exercise 2.15.
(1) Prove the lemma in Example 3 above.
(2) Use the lemma to verify that the map ρ in Example 4 above really
is a morphism of varieties, as was claimed.
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ALGEBRAIC GROUPS
2.5. The Jordan decomposition. Let us review some linear algebra. Let
V be a finite dimensional vector space and a ∈ End(V ). Then, a is called
semisimple if there is a basis for V with respect to which a is a diagonal
matrix, a is called nilpotent if as = 0 for some s ≥ 1 (equivalently, all
eigenvalues of a are zero) and a is called unipotent if a − 1 is nilpotent
(equivalently, all eigenvalues of a are 1).
Given a set S of pairwise commuting elements of End(V ), you can always
find a basis for V with respect to which all elements of S are upper triangular matrices. If in addition all elements of S are semisimple, you can find a
basis for V with respect to which all elements of S are diagonal. It follows
easily from this that the product of two commuting semisimple, resp. nilpotent, resp. unipotent matrices is again semisimple, resp. nilpotent, resp.
unipotent. This is definitely false if you remove the word “commuting”.
Lemma 2.16. Let a ∈ End(V ).
(i) There are unique elements as , an ∈ End(V ) such that as is semisimple, an is nilpotent, as an = an as and a = as + an .
(ii) There are polynomials P, Q without constant term so that as =
P (a), an = Q(a). Hence, as and an leave invariant any subspace
W of V that a stabilizes, and (a|W )s = (as )|W , (a|W )n = (an )|W .
Similarly for the induced maps on V /W .
(iii) Let φ : V → W be a linear map and b ∈ End(W ) be such that
φ ◦ a = b ◦ φ. Then, φ ◦ as = bs ◦ φ, φ ◦ an = bn ◦ φ.
,
Proof. (i) Let (T − λi )ni be the characteristic polynomial of a, where the
λi are the distinct eigenvalues of a. Let
Vi = {v ∈ V | (a − λi )ni v = 0},
a non-zero a-stable subspace of V . By CRT, there exists P ∈ k[T ] such that
P (T ) ≡ 0
(mod T ),
P (T ) ≡ λi
(mod (T − λi )ni ) for all i.
Set as = P (a). Then, as stabilizes all Vi and the restriction of as to Vi
is scalar multiplication by λi . Hence, as is semisimple and a − as has all
eigenvalues zero, so is nilpotent. Everthing in (i) and (ii) follow pretty easily
once we have checked the uniqueness...
For uniqueness, suppose that a = bs + bn is another such decomposition.
Since as and an are polynomials in a, bs and bn commute with as and an .
Hence, as − bs = bn − an is both semisimple and nilpotent, hence zero. !
The lemma is known as the additive Jordan decomposition. As a corollary,
we get the multiplicative Jordan decomposition:
Corollary 2.17. Let a ∈ GL(V ). There exist unique as , au ∈ GL(V ) such
that as is semisimple, au is unipotent and a = as au = au as . They have
properties similar to (ii) and (iii) in the theorem too.
Proof. Let a = as + an be the additive Jordan decomposition. Since a is
invertible, none of its eigenvalues are 0. Hence, taking a basis with respect
ALGEBRAIC GROUPS
29
to which both as and an are upper triangular matrices, you see that none of
the eigenvalues of as are 0. Hence, as is invertible. Now set au = 1 + a−1
s an ,
and check as and au do the job.
!
We can also think about a ⊕ b and a ⊗ b, if a ∈ End(V ), b ∈ End(W ). So,
a⊕b is the endomorphism of V ⊕W such that (a⊕b)(v+w) = a(v)+b(w), and
a⊗b is the endomorphism of V ⊗W such that (a⊗b)(v⊗w) = a(v)⊗b(w). You
can check easily that if a, b are semisimple (resp. nilpotent, resp. unipotent)
then so are a⊕b and a⊗b. It then follows from the uniqueness in the Jordan
decompositions that:
Lemma 2.18. Let a = as au and b = bs bu be the Jordan decompositions of
a ∈ GL(V ), b ∈ GL(W ). Then, a ⊕ b = (as ⊕ bs ) + (au ⊕ bu ) is the Jordan
decomposition of a ⊕ b ∈ GL(V ⊕ W ), and a ⊗ b = (as ⊗ bs )(au ⊗ bu ) is the
Jordan decompostion of a ⊗ b ∈ GL(V ⊗ W ).
The theory just developed generalizes to infinite dimensional vector spaces
V providing we restrict our attention to locally finite endomorphisms a, i.e.
endomorphisms such that any v ∈ V belongs to a finite dimensional ainvariant subspace. We call a locally finite endomorphism a of V semisimple
if its restriction to every finite dimensional a-invariant subspace is semisimple, and define nilpotent, unipotent similarly.
For a general locally finite a ∈ End(V ), we have its Jordan decompositions: a = as + an and a = as au , with all the properties of the finite
dimensional case holding. To define as : take v ∈ V , find a finite dimensional a-invariant subspace W containing v, and define
as (v) = (a|W )s (v).
The fact that this is well-defined independent of the choice of W follows from
the uniqueness statement in the finite dimensional Jordan decomposition.
The elements an and au are defined similarly... and properties like (ii) and
(iii) of the lemma hold.
Now let G be an algebraic group. Let g ∈ G. We know by 2.4 that the
right translation of functions ρ(g) : k[G] → k[G] is a locally finite endomorphism of the vector space k[G]. Therefore, we have a Jordan decomposition
ρ(g) = ρ(g)s ρ(g)u .
Theorem 2.19. For any g ∈ G, there are unique elements gs , gu ∈ G such
that ρ(g)s = ρ(gs ), ρ(g)u = ρ(gu ) and g = gs gu = gu gs . Moreover, if
φ : G → H is a morphism of algebraic groups, φ(gs ) = φ(g)s , φ(gu ) = φ(g)u .
Proof. Let m : k[G] ⊗ k[G] → k[G] be the algebra multiplication. We have
that
m ◦ (ρ(g) ⊗ ρ(g)) = ρ(g) ◦ m.
Hence, by 2.16(iii) and 2.18,
m ◦ (ρ(g)s ⊗ ρ(g)s ) = ρ(g)s ◦ m.
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ALGEBRAIC GROUPS
Hence, ρ(g)s is an automorphism of k[G], so ξ : f $→ (ρ(g)s f )(e) is an
algebra homomorphism k[G] → k, i.e. there is a point gs ∈ G such that ξ is
evaluation at gs .
Now we compare ρ(g)s and ρ(gs ): note λ(x) and ρ(g) commute for all
x ∈ G. Hence, λ(x) commutes with ρ(g)s too by 2.16(iii).
(ρ(g)s f )(x) = (λ(x−1 )ρ(g)s f )(e) = (ρ(g)s λ(x−1 )f )(e)
= (λ(x−1 )f )(gs ) = f (xgs ) = (ρ(gs )f )(x).
Hence ρ(g)s = ρ(gs ).
In a similar way, you get a gu ∈ G such that ρ(g)u = ρ(gu ). But the
representation ρ : G → GL(k[G]) is faithful, so now we see that gs gu =
gu gs = g since we know ρ(g)s ρ(g)u = ρ(g)u ρ(g)s = ρ(g).
Now let φ : G → H be a homomorphism of algebraic groups. Then
we have φ∗ : k[H] → k[G]. Take g ∈ G and set h = φ(g). You check
that φ∗ ◦ ρH (h) = ρG (g) ◦ φ∗ . Hence, φ∗ ◦ ρH (h)s = ρG (g)s ◦ φ∗ . Hence
φ∗ ◦ ρH (hs ) = ρG (gs ) ◦ φ∗ . For any a ∈ k[H],
(φ∗ (ρH (hs )(a))(e) = ρH (hs )(a)(e) = a(hs ).
This equals
ρG (gs )(φ∗ (a))(e) = φ∗ (a)(gs ) = a(φ(gs )).
Since this is true for all a, we get that hs = φ(gs ). The argument for
unipotent parts is similar.
!
In the case G = GLn , gs is just the semisimple part of g viewed as an
automorphism of V = k m , and gu is its unipotent part. To see this, let
0 /= f ∈ V ∗ . For v ∈ V , define f˜(v) ∈ k[G] by
f˜(v)(g) = f˜(gv).
This gives an injective map f˜ : V → k[G], and satisfies
f˜(gv) = ρ(g)f˜(v).
Hence:
f˜(gs v) = ρ(g)s f˜(v) = ρ(gs )f˜(v) = f˜(gs v)
where the gs on the left is the semisimple part of g ∈ GL(V ) in the old
sense and the gs ’s on the right hand side are the semisimple part of g in the
abstract Jordan decomposition.
For an arbitrary G, we can embed G as a closed subgroup of some GLn .
Again, the Jordan decomposition g = gs gu of g as an element of the abstract
group G and as an automorphism of V coincide.
Exercise 2.20.
(3) Show that the set Gu = {gu |g ∈ G} of all unipotent
elements of G is closed.
ALGEBRAIC GROUPS
31
2.6. Unipotent algebraic groups. An algebraic group G is called unipotent if all of its elements are unipotent.
Theorem 2.21. Let G be a unipotent closed subgroup of GLn . Then there
is g ∈ GLn such that gGg −1 ⊆ Un .
Proof. Let V = k n . We need to show that G fixes some flag in V . Proceed
by induction on n. If there is a proper subspace W of V stablized by G,
then by induction G fixes a flag in W and in V /W and we can glue them
together...
So we need to consider the case that G does not stabilize and subspace of
V , i.e. G acts irreducibly on the vector space V . Then by Wedderburn’s theorem, the elements of G span the vector space End(V ) of all endomorphisms
of G.
Since G is unipotent, all elements of G have trace n. Hence,
trV (h − gh) = trV ((1 − g)h) = 0
for all g, h ∈ G, hence for all g ∈ G and h ∈ End V . Taking h to be
the various matrix units, you now get that 1 − g = 0, i.e. g = 1. Hence
G = {e}...
!
Corollary 2.22. Unipotent algebraic groups are nilpotent.
Here is a nice application of the theory so far:
Theorem 2.23 (Rosenlicht). Let G be a unipotent algebraic group acting
on an affine variety X. Then all orbits of G on X are closed.
Proof. Let O be an orbit. Replacing X by O, we may assume that O is open
dense in X. Let Y be its complement. Consider the action of G on k[X]
by translation of functions. This action is locally finite, by the same proof
as 2.4. Moreover, G stabilizes Y hence it leaves I(Y ) < k[X] invariant. By
the above theorem, we can find a non-zero function f ∈ I(Y ) fixed by G.
But then f is constant on O, so since O is dense, f is constant on all of
X. This shows f is a non-zero scalar, hence I(Y ) = k[X] and Y = ∅. Thus
O = X.
!
Let me finally make an important definition.
Lemma 2.24. Let X and Y be closed subgroups of G. If X normalizes Y ,
then XY is a closed subgroup of G.
Proof. It is immediate that XY is a subgroup. It is the image of X × Y
under a morphism of varieties, so it contains a non-empty open subset of its
closure by 1.28. Therefore XY is closed by 2.7(ii).
!
Now let G be an arbitrary connected algebraic group. Suppose that X
and Y are two closed connected normal solvable subgroups. Then XY is
again a closed normal subgroup of G by the lemma, and it is connected and
solvable (exercise). Hence XY is another closed connected normal solvable
32
ALGEBRAIC GROUPS
subgroup. It follows that G contains a unique closed connected normal
solvable subgroup of maximal dimension. This is called the radical of G,
denoted R(G). An algebraic group is called semisimple if its radical R(G) =
{e}.
Similarly you define the unipotent radical of G, denoted Ru (G). This is the
unique closed connected normal unipotent subgroup of maximal dimension.
An algebraic group is called reductive if is unipotent radical Ru (G) = {e}.
Since Ru (G) is unipotent, it is nilpotent, hence solvable, hence Ru (G) ≤
R(G). So semisimple groups are reductive.
For example, SLn is a semisimple group but GLn has a one dimensional
radical consisting of the scalar matrices. Thus GLn is not semisimple, but
since scalar matrices are semisimple, its unipotent radical is trivial, so GLn
is reductive.
Later on, we shall restrict our attention to the study of reductive algebraic
groups, for which there is a beautiful structure theory and classification.