Download 3.5. Separable morphisms. Recall that a morphism φ : X → Y of irre

ALGEBRAIC GROUPS
43
3.5. Separable morphisms. Recall that a morphism φ : X → Y of irreducible varieties is called dominant if its image is dense in Y . In the case
of affine varieties, that is equivalent to φ∗ : k[Y ] → k[X] being injective.
Hence, even for arbitrary irreducible X, Y , the map φ∗ : k(Y ) → k(X) induced by a dominant morphism is injective. So we can view k(X) as an
extension field of k(Y ). The morphism φ is called separable if k(X) is a
separable extension of k(Y ).
Go back to the case that X, Y are affine. The composite of φ∗ and dX :
k[X] → ΩX is a derivation k[Y ] → ΩX . So by the universal property of
differentials, we get induced a k[Y ]-module map
φ̄∗ : ΩY → ΩX
such that dX ◦ φ∗ = φ̄∗ ◦ dY .
Let x ∈ X, y = φ(x). The k[X]-module kx viewed as a k[Y ]-module via
φ∗ is ky . We used this before to see that φ∗ induced the linear map
dφx : Tx (X) = Derk (k[X], kx ) → Ty (Y ) = Derk (k[Y ], ky ), D $→ D ◦ φ∗ .
Equivalently, we can view dφx as the map
dφx : Homk[X] (ΩX , kx ) → Homk[Y ] (ΩY , ky ), θ $→ θ ◦ φ̄∗ .
Applying adjointness of tensor and hom, we can even view dφx as a linear
map
Homk (ΩX (x), k) → Homk (ΩY (y), k).
Theorem 3.15. Let φ : X → Y be a morphism of irreducible varieties.
(i) Assume that x ∈ X and y = φ(x) ∈ Y are simple points and that
dφx is surjective. Then, φ is a dominant separable morphism.
(ii) Assume φ is a dominant separable morphism. Then the points x ∈ X
with the property of (i) form a non-empty open subset of X.
Proof. We may assume X and Y are affine and ΩX , ΩY are free k[X]- resp.
k[Y ]-modules of rank d = dim X resp. e = dim Y . In particular, X and Y
are smooth.
The map φ̄∗ : ΩY → ΩX of k[Y ]-modules induces a homomorphism of
free k[X]-modules
ψ : k[X] ⊗k[Y ] ΩY → ΩX .
We can represent ψ as a d × e matrix A with entries in k[X], fixing bases
for ΩX , ΩY . Suppose that dφx is surjective. Then, A(x), which represents
the dual map (dφx )∗ : ΩY (y) → Ωx (x), is injective hence a matrix of rank
e. Hence the rank of A is at least e, hence equal to e since rank cannot be
more than the number of columns. This shows that ψ is injective.
Hence φ̄∗ is injective too. Since ΩX and ΩY are free modules, this implies
that φ∗ : k[Y ] → k[X] must be injective, so φ is dominant. Moreover,
injectivity of ψ implies injectivity of
k(X) ⊗k[Y ] ΩY → k(X) ⊗k(X) ΩX .
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ALGEBRAIC GROUPS
This is the map α in the exact sequence
β
α
k(X) ⊗k(Y ) Ωk(Y )/k −→ Ωk(X)/k −→ Ωk(X)/k(Y ) −→ 0.
Hence k(X) is a separable extension of k(Y ) by the differential criterion for
separability.
The proof of (ii) is similar to the proof of (ii) in 3.12.
!
Corollary 3.16. Let G be a connected algebraic group.
(i) If X is a variety on which G acts transitively, then X is irreducible
and smooth. In particular G is smooth.
(ii) Let φ : X → Y be a G-equivariant morphism between two varieties
on which G acts transitively. Then φ is separable if and only if dφx
is surjective for some x ∈ X which is if and only if dφx is surjective
for all x ∈ X.
(iii) Let φ : G → H be a surjective homomorphism of algebraic groups.
Then, φ is separable if and only if dφe is surjective.
Proof. (i) Take x ∈ X. The orbit map g $→ gx is onto, so X is irreducible
as G is. Since at least one point of x is simple, and the action is transitive,
we see that all points of x are simple, so X is smooth.
(ii) This follows at once from the theorem.
(iii) Apply (ii) to X = G, Y = G" .
!
3.6. Lie algebras. Notation: if G is an algebraic group, let g denote the
tangent space Te (G) to G at the identity. At the moment, g is just a vector
space of dimension dim G (we will see in a while that it has additional
structure as a Lie algebra). For example, the tangent spaces to GLn , SLn ,
Sp2n , SOn , . . . at the identity will be denoted gln , sln , sp2n , son , . . . . At the
moment these are all just vector spaces!
Example 3.17.
(1) First, GLn . Then:
gln = Derk (k[GLn ], ke )
is the n2 dimensional vector space on basis {ei,j | 1 ≤ i, j ≤ n} where
ei,j is the point derivation
f $→
∂f
(e).
∂Ti,j
Here, Ti,j is the ij-coordinate function, and k[GLn ] is the localization
of the polynomial ring k[Ti,j ] at determinant. I will always identify
the vector space gln with the vector space of n × n matrices over k,
so that ei,j is identified with the ij matrix unit.
(2) SLn . Since SLn = V (det −1) inside of GLn , sln is a canonically
embedded subspace of gln . Indeed, it will be all matrices
!
X=
ai,j ei,j
i,j
ALGEBRAIC GROUPS
45
such that
X(det −1) = 0.
But (calculate!)
!
!
!
(
ai,j ei,j )(
sgn(w)T1,w1 . . . Tn,wn − 1) =
ai,i .
i,j
w∈Sn
i
So sln is all matrices in gln of trace zero.
(3) Sp2n . Recall that Sp2n = {x ∈ GLn | xt Jx = J} where
"
#
0
In
J=
−In 0
as a closed subgroup of GL2n . Let T be the 2n × 2n matrix with
ij entry Ti,j , the ij-coordinate function on GL2n . Then, Sp2n =
V (T t KT −J) (4n2 polynomial equations written as one matrix equation!). Then sp2n is the X ∈ gl2n such that X(T t JT − J) = 0. You
calculate: this is exactly the condition that X t J + JX = 0. Now
you can calculate dim Sp2n by linear algebra...
(4) Similarly, son (in characteristic )= 2) is all X ∈ gln satisfying X T +
X = 0.
(5) Also recall that Tn is all upper triangular invertible matrices. The
tangent space tn will be all upper triangular matrices embedded into
gln . Similarly you can compute the tangent spaces of Dn , Un , . . . .
Okay, now let’s introduce on g the structure of a Lie algebra. It is convenient to go via an intermediate: L(G). This is defined to be the vector
space
L(G) = {D ∈ Derk (k[G], k[G]) | D(λx f ) = λx D(f ) for all f ∈ k[G], x ∈ G.
(Recall: (λx f )(g) = f (x−1 g)). We call L(G) the left invariant derivations
because they are the derivations commuting with the left regular action of
G on k[G].
Now, Derk (k[G], k[G]) is a Lie algebra with operation being the commutator: [D, D" ] = D ◦ D" − D" ◦ D. Obviously, if D and D" are left
invariant derivations, so is [D, D" ]. Therefore L(G) is a Lie subalgebra of
Derk (k[G], k[G]).
Lemma 3.18. Let G be a connected algebraic group, g = Te (G). The maps
L(G) → g, D $→ eve ◦ D
and
g → L(G), X $→ X̃,
where X̃ : k[G] → k[G] is the derivation with (X̃f )(g) = X(λg−1 f ) for all
g ∈ G, are mutually inverse isomorphisms.
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Proof. We’d better first make sure the maps make sense, i.e. that eve ◦ D
is a point derivation and that X̃ is a left invariant derivation. Only the last
thing is tricky:
(X̃(λg f ))(h) = X(λh−1 λg f ) = X(λh−1 g f ) = (X̃f )(g −1 h) = (λg (X̃f ))(h).
Now let’s compute eve ◦ X̃:
(eve ◦ X̃)(f ) = (X̃f )(e) = X(f )
for all f ∈ k[G], hence eve ◦ X̃ = X.
!
Finally, let’s compute ev
e ◦ D for a left invariant derivation D:
!
(ev
e ◦ D(f ))(g) = (eve ◦D)(λg −1 f ) = (D(λg −1 f ))(e) = (λg −1 (Df ))(e) = (Df )(g).
!
Hence ev
e ◦ D = D.
!
Because of the lemma, we get on g an induced structure as a Lie algebra,
induced by the Lie algebra structure on L(G):
[X, Y ] := eve ◦ [X̃, Ỹ ].
Example 3.19. Let’s compute the Lie algebra structure on gln . We need
to work out ẽi,j . Well, for g ∈ G,
∂ !
(
Tp,r (g)Tr,q )(e) = δq,j Tp,i (g).
(ẽi,j Tp,q )(g) = ei,j (λg−1 Tp,q ) =
∂Ti,j r
Therefore ẽi,j Tp,q = δj,q Tp,i .
Now you can compute the commutator [ẽi,j , ẽk,l ]: it acts on each Tp,q in
the same way as δj,k ẽi,l − δi,l ẽk,j . Hence we’ve worked out
[ei,j , ek,l ] = δj,k ei,l − δi,l ek,j .
The right hand side is just the commutator ei,j ek,l − ek,l ei,j of the matrix
units. So in general:
[X, Y ] = XY − Y X
for X, Y ∈ gln – the usual commutator as matrices!
Lemma 3.20. Let H be a closed subgroup of G and set I = I(H). Then,
h = {X ∈ g | XI = 0} = {X ∈ g | X̃I ⊆ I}.
In particular, h is a Lie subalgebra of g.
Proof. That
h = {X ∈ g | XI = 0}
is just the definition of tangent space of a closed subvariety... If XI = 0,
consider X̃f for f ∈ I. Evaluating at h ∈ H,
(X̃f )(h) = X(λh−1 f )
which is zero as λh−1 f is still in I (as H is a subgroup!). So X̃f vanishes on
H, i.e. is containing in I. This shows X̃I ⊆ I.
The opposite, namely that if X̃I ⊆ I then XI = {0} is easy.
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Finally, observe that if D, D" ∈ L(G) satisfy DI ⊆ I, D" I ⊆ I, then the
commutator [D, D" ] also does. This implies that h is a subalgebra of g. !
Now we get the Lie algebra structure in all the above examples: sln , son ,
. . . , since they are all just Lie subalgebras of gln , where the Lie bracket is
just the commutator as matrices.
The next lemma I’m going to leave to you to supply the proof (hint: its
easier to rephase things in terms of L(G) and L(H))...
Lemma 3.21. Let φ : G → H be a morphism of connected algebraic groups.
Then dφe : g → h is a Lie algebra homomorphism.
3.7. Some differential calculations. I now want to compute differentials
of various natural morphisms. In all cases, since an arbitrary algebraic group
can be embedded as a closed subgroup of some GLn , it is enough to make
the computation in the case of GLn , when we can be very explicit.
Example 3.22.
(1) Let µ : G × G → G be multiplication. Then,
dµ(e,e) : g ⊕ g → g is addition.
Proof: Suppose G = GLn . Recall g⊕g ∼
= Derk (k[G]⊗k[G], k(e,e) ),
the isomorphism mapping (X, Y ) to the map (f ⊗ g $→ X(f )g(e) +
f (e)Y (g). By definition,
!
(dµ(e,e) (ei,j , ek,l ))(Tr,s ) = (ei,j , ek,l )(
Tr,t ⊗ Tt,s ) = δi,r δj,s + δk,s δr,k
t
= (ei,j + ek,l )(Tr,s ).
Hence dµ(e,e) (ei,j , ek,l ) = ei,j + ek,l , i.e. dµ(e,e) is addition.
(2) Let i : G → G be the inverse map. Then, die : g → g is the map
X $→ −X.
Proof: Consider the composite G → G × G → G, g $→ (g, i(g)) $→
gi(g) = e. The composite is a constant function, so its differential
is zero. But the differential of a composite is the composite of the
differentials, so appying (1), 0 = d ide +die . The differential of the
identity map is the identity map, so we are done.
(3) Fix x ∈ G. Let Int x : G → G be the automorphism of algebraic
groups g $→ xgx−1 . The differential d(Int x)e : g → g is a Lie algebra
automorphism. It is usually denoted Ad x : g → g.
For example, suppose G = GLn . Then, for a matrix X ∈ gln ,
(Ad x)(X) = xXx−1 , i.e Ad x is just the Lie algebra automorphism
given by conjugation by x. Hence, for H and closed subgroup of G
and x ∈ H, Ad x : h → h is just conjugation by x too – it leaves the
subspace h of gln invariant.
Proof: Let us compute (Int x)∗ Ti,j :
((Int x)∗ Ti,j )(g) = Ti,j (xgx−1 ) = (µ∗ (µ∗ Ti,j ))(x, g, x−1 )
!
=
Ti,k (x)Tk,l (g)Tl,j (x−1 ).
k,l
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Hence:
(Int x)∗ Ti,j =
!
xi,k Tk,l (x−1 )l,j .
k,l
The ij-entry of (Ad x)(X) is
!
(Ad x)(X)(Ti,j ) =
xi,k X(Tk,l )(x−1 )i,j
k,l
which is the ij-entry of xXx−1 .
(4) Here is a consequence of (3). For each x ∈ G, Ad x : g → g is
an invertible linear map (even a Lie algebra automorphism) So you
can think of Ad as a group homomorphism from G to GL(g) (or
even to Aut(g) which is a closed subgroup of GL(g)). I claim that
Ad : G → GL(g) is a morphism of algebraic groups.
Proof: Embed G as a closed subgroup of some GLn . Then by (3),
Ad x is conjugation by x, which is clearly a morphism of varieties
– since it is given by matrix multiplication and inversion which are
polynomial operations.
(5) The image of Ad : G → GL(g) is a closed connected subgroup of
Aut(g), denoted Ad G. It is interesting to consider the kernel of Ad,
a closed normal subgroup of G. Obviously, every element of Z(G)
belongs to ker Ad, i.e. Z(G) ⊆ ker Ad. I warn you that equality need
not hold here. However it usually does, for example it always does
in characteristic 0.
For example, if G = GLn , ker Ad = Z(G) (check directly: ker Ad
is the invertible matrices which commute with all other matrices),
the scalar matrices. In this case Ad G is the group known as P GLn .
As an abstract group, P GLn ∼
= GLn /{scalars}.
Similarly, if G = SLn , ker Ad is the scalar matrices of determinant
one, i.e. the matrices


ω 0
0


 0 ... 0 
0
0
ω
where ω runs over all nth roots of unity. The group Ad SLn is known
as P SLn . As an abstract group P SLn ∼
= SLn /{scalars}.
Warning. In characteristic p|n, Z(SLn ) is trivial. So P SLn ∼
=
SLn as an abstract group, the isomorphism being the map g $→
Ad g. However, SLn is not isomorphic to P SLn as an algebraic
group: the problem is that Ad is a bijective morphism that is NOT
an isomorphism of varieties, i.e. its inverse is not a morphism of
varieties.
(6) Because of (4), we have a morphism Ad : G → GL(g). So we can
consider its differential again,
d Ad : g → gl(g).
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49
The map d Ad is usually denoted ad. I claim that ad X ∈ gl(g) is
the map Y $→ [X, Y ] (in particular, ad X is a derivation of g, so ad
is a Lie algebra homomorphism g → Der(g)). In other words,
The differential of Ad is ad.
The proof is so nasty I’m not going to type it in. . .
Exercise 3.23.
(8) Here is an example to show that ker Ad may be
larger than Z(G) if chark = p > 0. Let G be the two dimensional
closed subgroup of GL3 consisting of all matrices of the form


a 0 0
 0 ap b 
0 0 1
where a )= 0 and b are arbitrary elements of k. Describe the Lie
algebra g explicitly as a subspace of gl3 . Now compute Z(G) and
ker Ad and show that they are not equal.
(9) Consider the morphism Ad : SL2 → P SL2 . Both SL2 and P SL2 are
three dimensional algebraic groups, so their Lie algebras are three
dimensional too. Consider the differential ad : sl2 → psl2 . Using
Example (8) above, compute the kernel of ad, an ideal in the Lie
algebra sl2 . Deduce by 3.16 that the morphism Ad is separable if
and only if chark )= 2. (So in characteristic 2 – when Ad is a bijective
morphism – it is inseparable so it cannot be an isomorphism).
(10) Let chark = p > 0. Take X ∈ g = Derk (k[G], ke ). Recall g is
isomorphic to L(G), the left invariant derivations of k[G], via the
map X $→ X̃. Show that (X̃)p is a left invariant derivation of k[G].
Hence, there is a unique element X [p] ∈ g with
*
[p] = (X)
+ p.
X
This gives an extra operation on g in characteristic p, the map X $→
X [p] , which makes g into what is known as a restricted Lie algebra.
Describe this map explicitly in the one dimensional cases G = Ga
and G = Gm .