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Linear Algebraic Groups These are the exercises (with solutions) and the exam for the course Algebra II, winter term 2014/2015 at Bonn. Students were not assumed to have heard algebraic geometry before, just algebra. In particular, basic affine and projective geometry were introduced from scratch. The course covered the standard material up to the structure results about max- imal tori and Borel subgroups. Root systems and data were covered, as well as how to associate a root system to a semisimple group. Contents 1 Some pointers to the literature 2 2 List of results proved in the course 3 3 Exercises 7 4 Solutions to Exercises 12 5 Exam and solutions 26
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Page 1: Linear Algebraic Groups Contents - Institut für Mathematikploog/WS2014/lag.pdf · homogenous spaces G=Hand the proof that morphisms of linear algebraic groups have closed image from

Linear Algebraic Groups

These are the exercises (with solutions) and the exam for the course Algebra II,winter term 2014/2015 at Bonn.Students were not assumed to have heard algebraic geometry before, just algebra.In particular, basic affine and projective geometry were introduced from scratch.The course covered the standard material up to the structure results about max-imal tori and Borel subgroups. Root systems and data were covered, as well ashow to associate a root system to a semisimple group.

Contents

1 Some pointers to the literature 2

2 List of results proved in the course 3

3 Exercises 7

4 Solutions to Exercises 12

5 Exam and solutions 26

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1 Some pointers to the literature

A. Borel: Linear Algebraic Groups Springer (1969, 1997)

J. Humphreys: Linear Algebraic Groups Springer (1975, 1981)

T. Springer: Linear Algebraic Groups Birkhauser (1981, 1998)

P. Tauvel, R. Yu: Lie Algebras and Algebraic Groups Springer (2005)

A. Onishchik, E. Vinberg: Lie Groups and Algebraic Groups Springer (1990)

R. Goodman, N.Wallach: Symmetry, Representations and Invariants Springer (2009)

The first three books (Borel, Humphreys, Springer) are the classical textbooks onthe subject. They share the following features: the ground field is algebraicallyclosed and of arbitrary characteristic; rationality questions (i.e. non-algebraicallyclosed fields of definition) are treated at the end; they cover the structure theory oflinear algebraic groups including the classification of reductive/semisimple groups;they avoid schemes but use ringed spaces to define quotients. Of these books, Ilike Humphrey’s the best: it does a good job of explaining why things happen.The other three books all work over the complex numbers, and have different aims.Tauvel/Yu is really concerned with Lie algebras. It covers basically all of the foun-dational results otherwise cited (including commutative algebra, sheaves, grouptheory, projective geometry, root systems) — at the price that Lie algebras areintroduced as late as in §19 and linear algebraic groups in §21. Moreover, thebook is extremely light on examples. It treats a lot of the finer structure theory,including reductive, Borel, parabolic, Cartan groups. In my opinion, this volume ismore a reference than a textbook to learn the subject from. The sections relevantto the course are §10,18,21–23,25–28.Goodman/Wallach starts with very explicit descriptions of the classical groups(SLn, SOn, Spn) and then develops the theories of Lie groups and of algebraicgroups in parallel. Thus it is very example-driven: for example, maximal toriand roots are all developed first for the classical groups. Later on, Chapter 11 isdevoted to a rapid development of linear algebraic groups, up to Borel subgroupsand maximal tori (but without the classification). I took the construction ofhomogenous spaces G/H and the proof that morphisms of linear algebraic groupshave closed image from Appendix A.Onishchik/Vinberg is written in a unique style: the book grew out of a 1967Moscow seminar; almost all of the theory is presented in a series of problems(generally with concise solutions). The book develops both (real and complex) Liegroups and complex linear algebraic groups, in a very quick and efficient fashion.They classify semisimple Lie algebras.

To a large extent, I have been following the lecture notes of Tamas Szamuely froma 2006 course at Budapast: http://www.renyi.hu/~szamuely/lag.pdf. It hasthe advantage of starting out with low technical demands (no schemes or sheavesare used; only quasi-projective varieties). The proof of conjugacy of maximal toriusing group cohomology is from these notes.

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2 List of results proved in the course

Affine algebraic geometry

Proposition 1: Connected components of affine algebraic groups coincide withirreducible components. The identity component is a normal subgroup.

Proposition 2: Mor(X, Y ) = Hom(A(Y ), A(X)) for affine varieties X, Y . Forany finitely generated, reduced K-algebra there is a unique affine variety X withA(X) ∼= A.

Proposition 3: A(X × Y ) ∼= A(X)⊗K A(Y ) for affine varieties X, Y .

Embedding theorem

Theorem: Any affine linear group G is a closed subgroup of some GLn.

Chevalley Lemma: If H ⊂ G is a closed subgroup of a linear algebraic group,then there exists a finite-dimensional representationG→ GL(V ) and a 1-dimensionallinear subspace L ⊆ V such that H = Stab(L).

Proposition 4: For a normal subgroup H ⊆ G of a linear algebraic group G,there is a finite-dimensional representation G→ GL(V ) with kernel H.

Jordan decomposition

Theorem: Jordan decomposition for a linear algebraic group G.

Theorem (Kolchin): If G ⊂ GLn is a unipotent group, then there exists anon-zero vector fixed by G.

Burnside Lemma: If V is a finite-dimensional K-vector space and A ⊆ End(V )a subalgebra without A-stable subspaces except 0 and V , then A = End(V ).

Theorem (structure of commutative linear algebraic groups G): Gs, Gu

are closed subgroups of G, and Gs×Gu∼−→ G is an isomorphism of affine algebraic

groups.

Actions and representations

Proposition 5: G diagonalisable ⇐⇒ G is isomorphic to a closed subgroup ofDn ⇐⇒ G = Gs is commutative

Lemma: X∗(Dn) = Zn and X∗(Dn) is a basis of A(Dn).

Theorem (structure of diagonalisable groups G): G diagonalisable ⇐⇒X∗(G) is a finitely generated abelian group (without p-torsion if char(K) = p) andX∗(G) is a basis of A(G) ⇐⇒ G ∼= µd1 × · · · × µdr ×Dm with p 6 | di.Proposition 6: G diagonalisable ⇐⇒ all finite-dimensional G-representationsdecompose into characters

Theorem (Weyl groups are finite): If G is a linear algebraic group and T ⊆ Ga torus, then the quotient W (G, T ) := NG(T )/CG(T ) is finite.

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Connected solvable groups

Theorem (Lie-Kolchin): If G ⊂ GLn is a connected solvable subgroup, thenthere exists a complete flag of G-invariant subspaces in V , i.e. G can be conjugatedinto Tn.

Theorem (structure of connected nilpotent groups): If G is a connectednilpotent linear algebraic group, then Gu, Gs are closed, normal subgroups of Gand Gs×Gu

∼−→ G is an isomorphism of affine algebraic groups, and Gs is a torus.

Tangent spaced and Lie algebras

Lemma: If X is an affine homogeneous space for a linear algebraic group G, thenX is smooth.

Theorem: If G is a connected, 1-dimensional linear algebraic group, then eitherG = Ga or G = Gm. [Only proved for char(K) = 0.]

Lemma: If G ⊆ GLn is a linear algebraic group, then the action of G on its Liealgebra g = {A ∈ gln | XAf ∈ I(G) ∀f ∈ I(G)} is given by g · A = gAg−1.

Proposition 7: For a linear algebraic group G, tangent space at the unit and Liealgebra coincide: L(G) ∼= TeG.

Quotients by normal subgroups

Theorem: If H ⊆ G is a closed normal subgroup of a linear algebraic group, thenG/H is a linear algebraic group.

Proposition 8: If A ⊂ B are K-algebras, with B finitely generated over A, thenthe extension property holds: ∀

b∈Bb 6=0

∃a∈Aa6=0

∀ϕ : A→Kϕ(a)6=0

∃Φ: B→KΦ(b)6=0

: Φ|A = ϕ .

Proposition 9: If f : X → Y is a morphism of affine varieties, then f(X) containsan open subset of its closure f(X).

Corollary: If f : G→ H is a morphism of affine algebraic groups, then the imagef(G) is a closed subgroup of H.

Quasi-projective varieties

Proposition 10: The Plucker map pd : Gr(d, V )→ P(ΛdV ) is a closed embedding.

Lemma: X proper, Z ⊆ X closed ⇒ Z proper.X1, X2 proper ⇒ X1 ×X2 proper.X proper, ϕ : X → Y morphism ⇒ ϕ(X) closed, proper subvariety of Y .X proper and affine ⇒ X is a finite set.

Theorem: Projective varieties are proper.

Nakamaya Lemma: If A is a commutative ring, M ⊂ A a maximal ideal and Na finitely generated A-module with N = MN , then there is an f ∈ A \M withfN = 0.

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Homogeneous spaces and quotients

Theorem: If H ⊂ G is a closed subgroup of a linear algebraic group, then G/His a quasi-projective with a morphism π : G → G/H and if ϕ : G → X is anyG-equivariant morphism of homogeneous G-spaces with ϕ(H) = ϕ(1), then thereexists a unique morphism ψ : G/H → X with ϕ = ψπ.Moreover, if G acts on a quasi-projective variety Y such that H ⊆ Gy for somey ∈ Y , then the natural map G/H → G · y is a morphism.

Proposition 11: Let B be a finitely generated K-algebra without zero divisorsand A ⊂ B is a subalgebra. If there is b ∈ B, b 6= 0 such that all Φ: B → K withΦ(b) 6= 0 are uniquely determined by Φ|A, then B ⊆ Quot(A).

Proposition 12: Let f : M → N and h : M → P be regular, dominant morphismsof affine varieties such that there exists a non-empty open set U ⊆M with f(m1) =f(m2) ⇒ h(m1) = h(m2) ∀m1,m2 ∈ U , then there is a rational map g : N 99K Pwith h = gf .

Borel and parabolic subgroups

Orbit lemma: If a linear algebraic group G acts on a quasi-projective varietyX, then (i) every orbit is open in its closure, (i) orbits of minimal dimension areclosed and, in particular, (iii) closed orbits exist.

Borel Fixed Point Theorem: Any action of a connected solvable group G ona projective variety X has a fixed point.

Theorem: Any two Borel subgroups of a linear algebraic group are conjugate.

Proposition 13: Borel subgroups are parabolic.

Proposition 14: If ϕ : G� G′ is a surjective morphism of affine algebraic groupsand H ⊆ G is a parabolic (or Borel) subgroup, then so is ϕ(H) ⊆ G′.

Maximal tori

Proposition 15: For a connected solvable group G, there exists a torus T ⊂ Gsuch that T ↪→ G� G/Gu is an isomorphism.

Corollary: A connected solvable group G is a semi-direct product G = Gu o T ,where T ∼= G/Gu is a maximal torus.

Theorem: Any maximal tori in a connected linear algebraic group are conjugate.

Structure results

Theorem: Let G be a connected linear algebraic group, B ⊆ G a Borel subgroupand T ⊆ G a maximal torus. Then G is covered by all Borel subgroups, and Gs iscovered by all maximal tori: G =

⋃g∈G gBg

−1 and Gs =⋃g∈G gTg

−1.

Proposition 16: Let G be a connected linear algebraic group and T ⊆ G a torus.Then CG(T ) is connected.

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Theorem: Given B ⊆ G, a Borel subgroup of a connected linear algebraic group,then NG(B) = B.

Proposition 17: There is a bijection between the generalised flag variety G/Band the set B of all Borel subgroups of G.

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3 Exercises (five problems each week)

1. Let N be the set of all matrices in GLn(K) with exactly one non-zero entry inevery row and every column. Show that N is a closed subgroup of GLn(K), thatits identity component N◦ = Dn is the subgroup of diagonal matrices, that N hasn! connected components and that N is the normaliser of Dn.

2. Give examples of non-closed subgroups of GL2(C) and compute their closures.

3. Describe the Hopf algebra structures on the coordinate rings of Ga and GLn.

4. Prove that a T0 topological group is already T2. Show that an infinite linearalgebraic group is always T0 but never T2. Explain the discrepancy!

5. Show that the product of irreducible affine K-varieties is again irreducible.This fails for non-algebraically closed fields K: exhibit zero divisors in C⊗R C.

6. Prove that the group Un of unipotent upper triangular matrices is nilpotent.

7. Prove that a group G is solvable if and only if it has a composition series withabelian factors, i.e. there is a chain of subgroups G = G0 ) G1 ) · · · ) Gm = {1}such that each Gi+1 is normal in Gi and all Gi/Gi+1 are abelian.

8. What is the Jordan decomposition for a finite group? For Ga?

9. Find a closed subgroup G of GL2 such that Gs is not a closed subset.

10. Compute the centre C of SL2(K), assuming char(K) 6= 2. Show that thequotient group PSL2(K) := SL2(K)/C is an affine algebraic group.

(Hint: embed SL2 ⊂ A4 as a Zariski-closed subset, then check that the action of C

on SL2 extends to an action of C on A4. Now map A4/C to some affine space as a

Zariski-closed subset.)

11. Prove that the group Tn of upper triangular matrices is solvable.

12. Show that Ga and Gm are not isomorphic as affine algebraic groups.

13. Let ϕ : X → Y be a morphism of affine varieties. Show that ϕ is dominant(i.e. the image of X is dense in Y ) if and only if ϕ∗ : A(Y )→ A(X) is injective.

14. Let H ⊂ GLn be an arbitrary subgroup. Show that the Zariski-closure His a linear algebraic group. Moreover, prove that closure preserves the followingproperties: H commutative; H normal; H solvable; H unipotent.

15. Show that none of the following implications among properties of linearalgebraic groups can be reversed:

unipotent

��torus +3 diagonalisable +3 abelian +3 nilpotent +3 solvable

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16. The group SL3 naturally acts on K3. Writing x1, x2, x3 for the standard basisof K3, this induces an action of SL3 on polynomials p(x1, x2, x3) ∈ K[x1, x2, x3] byg · p = pg. Compute the weight spaces for the torus T := D3 ∩ SL3 on the vectorspace K[x1, x2, x3]2 of polynomials of degree 2, and draw the weights.

17. Show that the normaliser of D2 in GL2 is solvable, but not conjugated to asubgroup of T2.

18. Compute the Weyl group of GL3 with respect to the torus D3.

19. For any linear algebraic group G, let H :=⋂χ∈X∗(G) ker(χ). Show that H

is a closed, normal subgroup of G and that G/H is diagonalisable. Also showX∗(G) ∼= X∗(G/H).

20. (Continuation of 19.) Compute H and G/H for G = GLn.

21. Assume char(K) 6= 2 and let Γ ∈ M(n,K) with associated bilinear formKn × Kn, (x, y) 7→ xtΓy, and O(Γ) its isometry group. Show that the tangentspace of O(Γ) at I = In is TIO(Γ) = {A ∈M(n,K) | AtΓ + ΓA = 0}.(Hint: use the Cayley transform c(A) = (I + A)(I − A)−1 for A ∈ M(n,K) with

det(I −A) 6= 0 and show that c(A) ∈ O(Γ) if and only if AtΓ + ΓA = 0.)

22. Compute the dimensions of SOn and Spn.

23. Assume char(K) = 0. Let G be a linear algebraic group, all of whose elementshave finite order. Show that G is finite.

24. For affine varieties X and Y , show that dim(X × Y ) = dim(X) + dim(Y ).

25. If X is an irreducible, smooth affine variety and Z ( X a closed subvariety,prove dim(Z) < dim(X).

26. For a linear algebraic groupG, use its comultiplication to define an associative,unital K-algebra structure on A(G)∗ = HomK(A(G), K) such that the induced Liealgebra structure coincides with the one from left invariant vector fields.

27. Exhibit Gm as a closed subgroup of SO2. Then find a two-dimensionaltorus T ⊂ SO4 and compute the weights for the action induced on the adjointrepresentation, T ↪→ SO4 → GL(so4).

28. Let A be a finite-dimensional, associative and unital K-algebra. Show thatthe group of units is a linear algebraic group. What is its Lie algebra?

29. Show that the differential of the adjoint representation of a linear algebraicgroup G is given by ad := d(Ad)e : g→ End(g), ad(A)(B) = [A,B].

30. Show that a morphism ϕ : G → H of linear algebraic groups induces ahomomorphism of Lie algebras dϕe : g→ h.

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31. Find a symmetric, non-degenerate bilinear form Γ on Kn such that T :=SO(Γ) ∩ Dn is a torus of dimension m if n = 2m or n = 2m + 1. Prove that T isa maximal torus in SO(Γ): if H ⊆ SO(Γ) is abelian with T ⊆ H, then T = H.

32. Compute the orbits of the natural actions of GLn,Tn,Un,Dn on An and drawthem for n = 2. Describe orbit closures as unions of orbits.

33. Show that the adjoint representation of SL2 has disconnected isotropy groups.

34. Let G be a unipotent linear algebraic group and X an affine G-variety. Showthat all orbits of G in X are closed.

35. For a homogeneous ideal I ⊆ K[x0, . . . , xn], geometrically compare thevarieties V (I) ⊆ An+1 and V (I) ⊆ Pn. Prove the homogeneous Nullstellensatz.

36. Compute the orbits of the actions of GL3 on P2, of GL4 on Gr(2, 4), and ofGL2 on P1 × P1 (diagonal action). Also compute isotropy groups for all orbits.

37. Show that any non-degenerate conic in P2 is isomorphic to P1. Deducethat homogeneous coordinate rings are not isomorphism invariants of projectivevarieties.

38. Show that A1 and P1 are homeomorphic, but A2 and P2 are not.

39. For an affine, irreducible variety X and a point p ∈ X, show that the localring of X at p is given by the localisation of A(X) at the maximal ideal Mp.

40. Show that Aut(P1) = PGL2.

41. Show that varieties are compact in the Zariski topology: any open cover hasa finite subcover.

42. Show that a locally compact Hausdorff space X is compact if and only if theprojection X × Y → Y is a closed map for all topological spaces Y .

43. Let X = V (y2 − x3) ⊂ A2 be the cuspidal cubic. Show that the mapA1 → X, t 7→ (t2, t3) is regular, birational and a homeomorphism, but not anisomorphism of varieties. Extend this to an example of a morphism of projectivevarieties with the same properties.

44. Let A be a commutative ring with unit and N a finitely generated A-module.Show that a surjective module endomorphism N → N is an isomorphism.

45. Find subgroups of SL2 such that the quotient is respectively projective, affineor strictly quasi-affine.

46. Compute the dimensions of Grassmannians Gr(d, n) and flag manifolds Fl(n).

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47. Show that two irreducible varieties X and Y are birational, i.e. their functionfields are isomorphic: K(X) ∼= K(Y ), if and only if there exist affine open subsetsUX ⊆ X and UY ⊆ Y which are isomorphic: UX ∼= UY .

48. Assume char(K) = p > 0. Show that the map Gm → Gm, x 7→ xp is abijective morphism of affine algebraic groups, but is not an isomorphism.

49. Let G be a linear algebraic group, acting on a quasi-projective variety X.Show that orbits of minimal dimension are closed; in particular, closed orbits exist.

(Hint: you can use the statement of exercise 25.)

50. LetG be a connected projective algebraic group. Show thatG is commutative.

51. Find Borel subgroups in SO4, Sp4, T4 and U4.

52. Find all parabolic subgroups P with T4 ⊆ P ⊆ GL4.

53. Let the connected linear algebraic group G act on a quasi-projective variety Xwith finitely many orbits. Show that every irreducible closed G-invariant subset inX is the closure of a G-orbit. Find a counterexample for an action with infinitelymany orbits.

54. Find a connected linear algebraic group G and a maximal solvable subgroupU ⊂ G such that U is disconnected.

55. Classify all root systems in the Euclidean plane E := R2.

(Note: A root system in an Euclidean space (E, (−,−)) is a subset Φ ⊂ E such that

(RS1) Φ is finite, spans E and 0 /∈ Φ;

(RS2) for any α ∈ Φ, Rα ∩ Φ = {α,−α};(RS3) for any α ∈ Φ, the reflection sα : E → E, x 7→ x− 2(x,α)

(α,α) x preserves Φ;

(RS4) for any α, β ∈ Φ: 〈β, α〉 := 2(β,α)(α,α) ∈ Z. )

56. Compute the radicals R(GLn), R(SLn), R(Un).

57. Let char(K) = 0 and U a commutative, unipotent group. Show that U ∼= Gra

for some r ∈ N.

58. Let G be a finite group and A a trivial G-module, i.e. an abelian group whichhas the trivial left G-action. Show that H1(G,A) = Hom(G,A).

59. Let ϕ : G → H be a surjective morphism of linear algebraic groups, andT ⊆ G a maximal torus (or a maximal connected normal unipotent subgroup,respectively). Show that ϕ(T ) ⊆ H has the same property.

60. Let (E,Φ) be a root system. A base of (E,Φ) is a subset S ⊆ Φ such thatS is a basis for E and every root β ∈ Φ can be written as β =

∑α∈Smαα with

either all mα ≥ 0 or all mα ≤ 0. The elements of S are also called simple roots,and their non-negative linear combinations positive roots and denoted Φ+.Draw bases and positive roots for the root systems in E = R2, of exercise 55.

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61. Let G be a connected algebraic group, B ⊆ G a Borel subgroup and V arational G-module. Show that the invariant subspaces coincide: V G = V B.

62. Let G be a linear algebraic group. Show that R(G)u = Ru(G), i.e. theunipotent part of the radical is the unipotent radical.

63. For a linear algebraic group G, show that G/R(G) is semisimple and thatG/Ru(G) is reductive.

64. Let G be a group with subgroups H,N ⊆ G. Show that the following notionsare equivalent — then G = N oH is called a semi-direct product of H by N :(1) There is a short exact sequence 1 → N

ι−→ Gπ−→ H → 1 admitting a section

σ : H → G, i.e. πσ = idH .(2) N is normal in G, and NH = G and N ∩H = 1.(3) There is a homomorphism α : H → Aut(N) and G is isomorphic to the group

N oα H defined by (n, h) · (n′, h′) := (nα(h)(n′), hh′) on the set N ×H.Also show that the existance of a retraction % : G→ N , i.e. %ι = idN , is equivalentto a splitting G ∼= N ×H of G as a direct product.

65. Show that the following subsets define root systems of rank n:{ej − ei | i, j ∈ {1, . . . , n+ 1}, i 6= j} ⊂ Qn+1 (called type An){±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Qn (called type Dn)

66. Show that centraliser of(i 00 −i

)∈ PGL2(C) is disconnected.

67. For B ⊆ G a Borel subgroup of a connected linear algebraic group G, showthat Z(B) = Z(G).

68. Prove directly for G = SLn and for G = SOn (with the bilinear form fromProblem 31) that G is covered by Borel subgroups and that maximal tori coincidewith their centralisers.

69. Let G be a connected linear algebraic group with a maximal torus and B bethe set of Borel subgroups of G with its natural T -action. Show that there is abijection between the fixed point set BT and the Weyl group W (G).

70. Show that the following subsets define root systems of rank n:{±ei | i ∈ {1, . . . , n}} ∪ {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Qn (type Bn){±2ei | i ∈ {1, . . . , n}} ∪ {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Qn (type Cn)

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4 Algebra II — solutions to exercise sheet 1

1. Let N be the set of all matrices in GLn(K) with exactly one non-zero entry inevery row and every column. Show that N is a closed subgroup of GLn(K), thatits identity component N◦ = Dn is the subgroup of diagonal matrices, that N hasn! connected components and that N is the normaliser of Dn.

Solution: It is clear that Dn ⊂ N and that σDn ⊂ N for any permutation σ ∈ Sn (letting permutations act on

rows, say). It is also clear that N =⋃σ σDn and that this union is disjoint. We observe that Dn is connected: it

is an open subset of An, hence irreducible (because An is irreducible — its affine coordinate ring is K[x1, . . . , xn]

and an integral domain), hence connected. This shows all statements except that N is the normaliser of Dn. This

is an easy result purely about groups.

2. Give examples of non-closed subgroups of GL2(C) and compute their closures.

Solution: All of the following examples are closed in the norm topology of GL2(C) but not Zariski-closed.

GL2(R). It is Zariski-dense in GL2(C): denoting G = GL2(R) and X = GL2(C), then G = {x ∈ X | x = c(x)}where c : X → X is complex conjugation (a bijection, but not a morphism of complex varieties). For any f ∈ A(X)

with f(G) = 0, we also have c(f) ∈ A(X), and hence the two polynomials with real coefficients f + c(f) and

(f − c(f))/i. However, G is the real affine variety in affine (n2 + 1)-sace with just one equation. Hence there canbe no further polynomials vanishing on all of G, hence I(G) = 0 and G = V (I(G)) = V (0) = X.

Z, embedded via n 7→(1 n0 1

). This subset is not closed, because it is discrete infinite. Its closure is U2

∼= Ga.

S1, the 1-dimensional real sphere, embedded via S1 ↪→ GL2(C), x 7→( x 00 1/x

). Its closure is Gm via S1 ⊂ C∗ = Gm

and the same embedding.

3. Describe the Hopf algebra structures on the coordinate rings of Ga and GLn.

Solution: For G = Ga is A(G) = K[x] with comultiplication ∆(x) = x ⊗ 1 + 1 ⊗ x from m(g, h) = g + h.

Moreover, coinverse ι(x) = −x and counit ε(x) = 0.For G = GLn is A(G) = K[x11, . . . , xnn, y]/(1 − y · det(x)). Comultiplication ∆(xij) =

∑l xil ⊗ xlj ; coinverse

ι(xij) = (−1)i+jy det(xij , where xij is the matrix obtained from x = (xij) by cutting row i and column j — this

is precisely the i, j-entry of the inverse matrix to x, computed using the adjoint matrix via X ·adj(X) = det(X)In,note that y = 1/det(x) ∈ A(G); counit ε(xij) = δij .

4. Prove that a T0 topological group is already T2. Show that an infinite linearalgebraic group is always T0 but never T2. Explain the discrepancy!

Solution: Let G be a T0 topological group. We first show that the singleton {e} is a closed subset, where e ∈ Gis the neutral element. Given any element x ∈ G, there is either an open neighbourhood Ux of x with e /∈ U or

an open neighbourhood V of e with x /∈ e. (This is the definition of T0, applied to the two points x, e.) In thelatter case, we may assume that V = V −1 (if not, replace V with V ∩ V −1, this is still an open neighbourhood

of e) and use the homeomorphism f : G → G, y 7→ xy, which maps e 7→ x. Let Ux := f(V ), this is an open

neighbourhood of x = f(e) with e /∈ Ux (otherwise e ∈ f(V ) = xV , hence x−1 ∈ V , contradicting x /∈ V andV = V −1). We now take the union U :=

⋃x 6=e Ux over all these neighbourhoods. By construction, this is an

open set with U = G \ {e}, so {e} is indeed closed.

The map G × G → G, (g, h) 7→ gh−1 is continuous, since G is a topological group. Its preimage of the closedsubset {e} is the diagonal ∆G = {(g, g) | g ∈ G}, which is therefore closed. It is a general (and easy) fact that a

topological space X is T2 if and only if the diagonal ∆X is a closed subset of X ×X.

The statements about linear algebraic groups follow from general properties of the Zariski topology: points ofaffine varieties are closed because they correspond on maximal ideals; hence varieties are T0 (and even T1 but

note that schemes have more points and are only T0 in general). Since any two non-empty open subsets of an

irreducible variety meet, varieties are never T2.Explanation: this exercise shows that while linear algebraic groups are groups with a topology, they are nottopological groups. The reason is that the topology on G×G for a variety is not the product topology.

5. Show that the product of irreducible affine K-varieties is again irreducible.This fails for non-algebraically closed fields K: exhibit zero divisors in C⊗R C.

Solution: Let A := A(X) and B := A(Y ) be the affine coordinate rings. We know that A(X × Y ) ∼= A ⊗K B.

Also, X irreducible corresponds to I(X) being a prime ideal or, equivalently, A(X) being an integral domain.Therefore, we have to show that if A and B are integral domains, then so is A⊗K B.

Let f =∑i ai⊗ bi ∈ A⊗B be a zero divisor. We can assume that the bi are linearly independent. If x ∈ X, then

f(x,−) =∑i ai(x)bj ∈ B is a zero divisor in B, hence zero. With the bj linearly independent, we find ai(x) = 0

for all i, thus ai = 0 (Nullstellensatz!) and so f = 0.

In C⊗R C, one computes (1⊗ i+ i⊗ 1)(1⊗ 1 + i⊗ i) = 0.

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Algebra II — solutions to exercise sheet 2

6. Prove that the group Un of unipotent upper triangular matrices is nilpotent.

Solution: This a straightforward computation with commutators of matrices. For example, C2(Un) = [Un,Un]

consists of matrices with zeros on the secondary diagonal. In the central series, Cn(Un) is trivial.

7. Prove that a group G is solvable if and only if it has a composition series withabelian factors, i.e. there is a chain of subgroups G = G0 ) G1 ) · · · ) Gm = {1}such that each Gi+1 is normal in Gi and all Gi/Gi+1 are abelian.

Solution: We use the fact that for a subgroup H ⊆ G holds: H ⊇ D(G) = [G,G] if and only if H is normal in

G and G/H is abelian.If G is solvable, then by definition the derived series Dn(G) trivialises. Put Gi := Di(G). Then by the fact, each

Gi+1 = D(Gi) is normal in Gi with abelian quotient. The series terminates after finitely many steps because Gis solvable.

If we are given the chain of subgroups, then invoking the fact in the reverse direction, we get D(Gi) ⊆ Gi+1.

Thus inductively, Di(G) ⊆ Gi and therefore the derived series trivialises.

8. What is the Jordan decomposition for a finite group? For Ga?

Solution: If char(K) = 0, then all elements of a finite group G (considered embedded in some GLn) are semisimple.

To see this, note that a unipotent matrix U has all eigenvalues 1. If U 6= In, then U has non-trivial Jordan blocksand Uk 6= In for k ≥ 1. Therefore U can never have finite order, i.e. belong to G. Note that this argument fails

in finite characteristic: for example, the matrix U =(1 10 1

)over a field K with char(K) = 2 is unipotent with

U 6= I2 but U2 = I2.Ga ∼= U2, so all elements of Ga are unipotent.

9. Find a closed subgroup G of GL2 such that Gs is not a closed subset.

Solution: The subgroup T2 of upper triangular matrices( ∗ ∗0 ∗). The subset of semisimple elements is

(T2)s ={(

a b0 d

)| a, d ∈ K∗, b ∈ K with a 6= d or a = d, b = 0

}.

It is the complement of matrices(a b0 d

)with a = d, b 6= 0, and this subset is not Zariski-open: for example, its

closure are the matrices(a b0 d

)with a = d (and arbitrary b), which is not all of T2.

10. Compute the centre C of SL2(K), assuming char(K) 6= 2. Show that thequotient group PSL2(K) := SL2(K)/C is an affine algebraic group.

(Hint: embed SL2 ⊂ A4 as a Zariski-closed subset, then check that the action of C

on SL2 extends to an action of C on A4. Now map A4/C to some affine space as a

Zariski-closed subset.)

Solution: By direct computation, C = {(1 00 1

),(−1 0

0 −1

)} is the group with two elements. With SL2 =

{(x1, x2, x3, x4) ∈ A4 | x1x4 − x2x3 = 1}, the group is a closed subset of A4. The group C acts by ±id,

which obviously extends to A4. An element of A4/C is a quadruple up to sign, and we can map

A4/C → A6, ±(x1, x2, x3, x4) 7→ (x1x2, x1x3, x1x4, x2x3, x2x4, x3x4).

It is straightforward (and necessary) to check that this map is (a) well-defined and (b) cut out by the three

polynomial equations y1y6 = y2y5 = y3y4, with coordinates A(A4) = K[x1, . . . , x4] and A(A6) = K[y1, . . . , y6].The map is not injective, but it is injective on the subset PSL2.

This exhibits PSL2 = SL2/C ⊂ A6 as a Zariski-closed subset. (The equation det = 1, i.e. x1x4 − x2x3 = 1,

for SL2 is invariant under C.) It remains to observe that multiplication PSL2 × PSL2 → PSL2 and inversionPSL2 → PSL2 are given by polynomial maps.

Alternative: the map A4/C → A10, ±(x1, x2, x3, x4) 7→ (x21, x22, x

23, x

24, x1x2, x1x3, x1x4, x2x3, x2x4, x3x4) is

injective and can therefore also be used to embed PSL2 in an affine space.

Note: This exercise shows that some linear algebraic groups do not come with a natural embedding into a lineargroup. For such groups, it is easier to check that they are affine algebraic groups.

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Algebra II — solutions to exercise sheet 3

11. Prove that the group Tn of upper triangular matrices is solvable.

Solution: Start by checking D(Tn) = Un and then use the method of exercise 6. See Tauvel/Yu §10.8.

12. Show that Ga and Gm are not isomorphic as affine algebraic groups.

Solution: In fact, the underlying varieties are not isomorphic: X := Ga = K and Y := Gm = K∗ have affine

coordinate rings A(X) = K[x] and A(Y ) = K[y, y−1]. From the equivalence between affine varieties and reduced,finitely generated algebras we know that X ∼= Y if and only if K[x] ∼= K[y, y−1]. One way to see that these rings

are not isomorphic is by their groups of units: A(X)∗ = K[x]∗ = K∗ whereas A(Y )∗ = K[y, y−1]∗ = K∗ × Z,

mapping (λ, n) 7→ λyn. Note that the field K is fixed from the outset: in A(X), every automorphism is a multipleof the identity, whereas this is not the case for A(Y ). (Abstractly, it may be hard to see that the groups K∗

and K∗ × Z are non-isomorphic. There are many bizarre isomorphisms, for example S1 ∼= C∗ as groups.) This

solution is equivalent to comparing the automorphism groups as affine varieties: Aut(X) = A(X)∗.

A different approach uses automorphisms of Ga and Gm as affine algebraic groups. The only automorphisms of

Gm as an affine algebraic group are λ 7→ λ and λ 7→ λ−1. By contrast, Ga has no such automorphisms at all.

Note that the Lie group analogs of Ga and Gm are isomorphic via R→ R>0, t 7→ et.

13. Let ϕ : X → Y be a morphism of affine varieties. Show that ϕ is dominant(i.e. the image of X is dense in Y ) if and only if ϕ∗ : A(Y )→ A(X) is injective.

Solution: Assume ϕ dominant and let f ∈ A(Y ) with 0 = ϕ∗(f) = fϕ ∈ A(X). As a continuous function

Y → K, the vanishing locus of f is a closed subset of Y . However, f is zero on the image of ϕ, hence on a denseset. So the smallest possible vanishing locus is Y , hence f = 0.

Now assume that ϕ∗ : A(Y ) → A(X) is injective. Put Z := im(ϕ), this is the Zariski-closure of the image of ϕ.

If ϕ is not dominant, i.e. Z ( Y , then I(Z) ) I(Y ). Hence there is f ∈ I(Z) with f /∈ I(Y ); this is a functionf : Y → K which is non-zero and vanishes on Z. Therefore, the composition fϕ = ϕ∗(f) = 0, contradicting the

injectivity of ϕ∗.

14. Let H ⊂ GLn be an arbitrary subgroup. Show that the Zariski-closure His a linear algebraic group. Moreover, prove that closure preserves the followingproperties: H commutative; H normal; H solvable; H unipotent.

Solution: (We repeatedly use: if f : X → Y is a continuous map with f(A) ⊆ B for subsets A ⊆ X,B ⊆ Y , then

f(A) ⊆ f(A). In fact, this property (for all A) is equivalent to continuity.)For any h ∈ H, left multiplication lh : GLn → GLn is a homeomorphism preserving the subset H. Therefore,

it induces a homeomorphism of closures, lh : H → H. As this works for all h ∈ H, we get H · H ⊆ H. Now

right multiplication by h ∈ H is a homeomorphism GLn → GLn restricting to H → H, hence mapping H → H.We get H · H = H. Analogously, inversion GLn → GLn, g 7→ g−1 is a homeomorphism preserving H, hence it

provides a homeomorphism of H. Therefore, H is a subgroup of GLn, and thus a linear algebraic group.

If H is commutative, then for any h ∈ H, the map [h,−] : H → H has trivial image {1}. As the set {1} is closed,

the induced map on closures is [h,−] : H → {1} and so [h, g] = 1 for all h ∈ H, g ∈ H. Now for g ∈ H, the map[−, g] : H → {1} induces [−, g] : H → {1}. Hence H is commutative as well.

For H normal, use the map H → H,h 7→ ghg−1 for g ∈ G. Similar reasoning works for H solvable. For H

unipotent, use that unipotency in GLn is given by the equations (In − g)n = 0.

15. Show that none of the following implications among properties of linearalgebraic groups can be reversed:

unipotent

��torus +3 diagonalisable +3 abelian +3 nilpotent +3 solvable

Solution: Any finite abelian subgroup of GL2 is diagonalisable (char(K) = 0) but not a torus.

Ga ∼= U2 is abelian but not diagonalisable (it is unipotent).U3 is nilpotent but not abelian.

T2 is solvable but not nilpotent.Any torus is nilpotent but not unipotent.Wanted: an example of a nilpotent group which is neither abelian nor unipotent!

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Algebra II — solutions to exercise sheet 4

16. The group SL3 naturally acts on K3. Writing x1, x2, x3 for the standard basisof K3, this induces an action of SL3 on polynomials p(x1, x2, x3) ∈ K[x1, x2, x3] byg · p = pg. Compute the weight spaces for the torus T := D3 ∩ SL3 on the vectorspace K[x1, x2, x3]2 of polynomials of degree 2, and draw the weights.

Solution: The vector space V := K[x1, x2, x3]2 has a basis x21, x1x2, x1x3, x22, x2x3, x

23. The torus T = D3 ∩ SL3

is isomorphic to D2, and the basis of V also turns out to give all simultaneous eigenvectors for the T -module:

t =

t1 0 00 t2 0

0 0 1/t1t2

, t · x21 = t21x21, t · x1x2 = t1t2x1x2, t · x1x3 = t−1

2 x1x3 etc.

We have X∗(T ) = X∗(D2) = Z2, and the character associated to (a1, a2) ∈ Z2 is χ(a1,a2) : T → Gm, t 7→ ta11 ta22 .

Therefore, we get the following six weight spaces (this is the full set due to dim(V ) = 6)

Vχ(2,0)= Kx21, Vχ(0,2)

= Kx22, Vχ(−1,−1)= Kx23, Vχ(1,1)

= Kx1x2, Vχ(0,−1)= Kx1x3, Vχ(−1,0)

= Kx2x3.

The set of weights is {(2, 0), (0, 2), (1, 1), (−1,−1), (0,−1), (−1, 0)} ⊂ Z2.

17. Show that the normaliser of D2 in GL2 is solvable, but not conjugated to asubgroup of T2.

Solution: The normaliser already appeared in Exercise 1: N2 = D2∐A2 (disjoint union), where A2 = {

(0 ∗∗ 0

)}

is the subset of matrices with zeroes on the diagonal. The group N2 is solvable because 1 ⊂ D2 ⊂ N2 is a

composition series with abelian factors (N2/D2∼= Z/2Z).

We show that there are no invariant subspaces for the action of N2 on K2: as a D2-module, K2 has the invariantsubspaces

(∗0

)and

(0∗)

but neither of these is A2-invariant. Therefore, N2 cannot embed into T2.

This exercise shows that both assumptions of connected, solvable are necessary in the Lie-Kolchin theorem.

18. Compute the Weyl group of GL3 with respect to the torus D3.

Solution: Put G := GL3 and T := D3. A direct computation shows that the centraliser CG(T ) = T .

Let t ∈ T be a diagonal matrix and g ∈ G. We want to see when gtg−1 is diagonal. Replacing g by (det(g))−1/3g,

if necessary, we can assume det(g) = 1. (This simplifies the formula for g−1 via the adjoint matrix a bit.) Alengthy computation by hand then shows that g has exactly one non-zero entry in each row, and each column.

Therefore g is obtained from t via a permutation matrix, and W (D3,GL3) = S3, the symmetric group on 3 letters.

19. For any linear algebraic group G, let H :=⋂χ∈X∗(G) ker(χ). Show that H

is a closed, normal subgroup of G and that G/H is diagonalisable. Also showX∗(G) ∼= X∗(G/H).

Solution: Each ker(χ) = χ−1(1) is a closed subgroup, hence their intersection H is closed. Also kernels are

normal, so again their intersection H is normal.

In order to show that G/H is diagonalisable, we will prove that it is commutative and consists of semisimpleelements. For any g1, g2 ∈ G and χ ∈ X∗(G), we have χ([g1, g2]) = χ(g1)χ(g2)χ(g1)−1χ(g2)−1 = 1, hence

[G,G] ⊆ ker(χ), and so [G,G] ⊆ H. This implies G/H commutative.Denoting by π : G→ G/H the projection, any element of G/H is of the form π(g). From the properties of Jordan

decomposition we know that π(g) = π(g)sπ(g)u = π(gs)π(gu). We claim that χ(u) = 1 for any u ∈ Gu. This will

imply u ∈ H, and thus π(g) = π(g)s. Now a character is an action on a 1-dimensional vector space, so it has tobe linearisable — but then the eigenvalues of a unipotent element are all 1, hence χ(u) = 1.The homomorphism X∗(π) = π∗ : X∗(G/H) → X∗(G) is injective, because π is surjective. Moreover, X∗(π) is

also surjective: if χ ∈ X∗(G), then χι = 1 by the definition of H, where ι : H → G denotes the embedding. Henceχ gives a well-defined character G/H → Gm.

20. (Continuation of 19.) Compute H and G/H for G = GLn.

Solution: Any character of G is trivial on D(G) = [G,G]. For GLn, the commutator subgroup is D(GLn) = SLn.(This requires a computation.) Hence H ⊇ SLn. Since SLn is normal in GLn with quotient GLn/SLn ∼= Gm and

there is the non-trivial character det : GLn → Gm, we find H = SLn.

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Algebra II — solutions to exercise sheet 5

21. Assume char(K) 6= 2 and let Γ ∈ M(n,K) with associated bilinear formKn × Kn, (x, y) 7→ xtΓy, and O(Γ) its isometry group. Show that the tangentspace of O(Γ) at I = In is TIO(Γ) = {A ∈M(n,K) | AtΓ + ΓA = 0}.(Hint: use the Cayley transform c(A) = (I + A)(I − A)−1 for A ∈ M(n,K) with

det(I −A) 6= 0 and show that c(A) ∈ O(Γ) if and only if AtΓ + ΓA = 0.)

Solution: [Goodman/Wallach, exercise 1.4.5.5] An element g ∈ GLn is a Γ-isometry if and only if (gx)tΓ(gy) for

all x, y ∈ Kn. This is equivalent to gtΓg = Γ. We compute (where the last step uses 2 6= 0 in K)

c(A) ∈ O(Γ) ⇐⇒ c(A)tΓc(A) = Γ ⇐⇒ (I −A)−t(I +A)tΓ(I +A)(I −A)−1 = Γ

⇐⇒ (I +A)tΓ(I +A) = (I −A)tΓ(I −A) ⇐⇒ Γ +AtΓA+AtΓ + Γ = Γ +AtΓA−AtΓ− Γ

⇐⇒ 2(AtΓ + ΓA) = 0 ⇐⇒ AtΓ + ΓA = 0.

Let now A ∈M(n,K) with AtΓ + ΓA = 0. The set of t ∈ A1 such that det(I − tA) = 0 is finite; let U ⊂ A1 be itsopen complement. (Note that U is also an affine variety, similarly to how Gm ⊂ A1 is.) We consider the morphism

of affine varieties, γ : U → GLn, t 7→ c(tA). Obviously γ(0) = I and by the above, γ(t) ∈ O(Γ) for all t ∈ U . Its

differential at 0 is a map dγ0 : T0U = K → TIO(Γ). We will show that dγ0(1) = dγ(t)/dt(0) = 2A; this will implyTIO(Γ) ⊇ {A ∈M(n,K) | AtΓ+ΓA = 0}. For the computation, apply the usual rules for differentiating products

and powers: d/dt|t=0((I+tA)·(I−tA)−1) = d/dt|t=0(I+tA)·I+I·d/dt|t=0(I−tA)−1 = A−(I−0·A)−2(−A) = 2A;

this is possible since A and I commute. (Over K = C, one can also use the geometric series for (I − tA)−1.)For B ∈ M(n,K), consider the regular function ψB : GLn → A1, g 7→ tr((gtΓg − Γ)B). Note gtΓg − Γ = 0 ⇐⇒tr((gtΓg − Γ = 0)B) = 0 ∀B; this follows from the non-degeneracy of the trace bilinear form on M(n,K). Via

ψB , we restrict to deriving scalar-valued functions. For A ∈ M(n,K), we have the directional derivative DA =∑ij Aij ∂/∂xij , and DAψB(I) = tr((AtΓ + ΓA)B). This can be seen in matrix coordinates: writing Γ = (Γij)

etc., we have tr((gtΓg − Γ = 0)B) =∑ijkl(gkiΓklglj − Γkl)bji. Thus TIO(Γ) ⊆ {A ∈M(n,K) | AtΓ + ΓA = 0}.

22. Compute the dimensions of SOn and Spn.

Solution: We apply the previous exercise to compute the tangent spaces TISOn and TISpn. The dimensions of

these tangent spaces coincide with the dimensions of the groups, since algebraic groups are smooth.

For the orthogonal group On, the bilinear form is given by Γ = In, hence TISOn = {A ∈M(n,K) | At +A = 0}.The condition At + A = 0 is equivalent to A skew-symmetric. Hence the diagonal entries are zero, and A is

determined by its upper triangular part. Therefore dim(TIOn) = (n − 1) + (n − 2) + · · · + 1 = 12n(n − 1).

Moreover, On and SOn have the same tangent space at I, because SOn is the identity component of On.For the symplectic group Spn, the bilinear form on K2n is given by Γ = J =

( 0 In−In 0

), where In is the identity

matrix in M(n,K). We get TISpn = {A ∈ M(2n,K) | AtJ + JA = 0}. (Here I = I2n.) Writing A =(B CD E

)as

a block matrix, we find the condition AtJ + JA = 0 turns out to be equivalent to D = Dt, C = Ct, B = −Et.Hence C and D are symmetric n × n-matrices, B is an arbitrary n × n-matrix and E is completely determined

by B. The dimension therefore is dim(TISpn) = 2 · 12

(n+ 1)n+ n2 = n(2n+ 1).

23. Assume char(K) = 0. Let G be a linear algebraic group, all of whose elementshave finite order. Show that G is finite.

Solution: We can assume that G is connected because for any linear algebraic group, G/G◦ is a finite group (G◦

connected component of the neutral element). Put Gt := {g ∈ G | gt = 1}. This is a closed subset of G and by

assumption G =⋃tGt. We deduce G = Gt for some fixed t from the following general fact:

A union of countably many subvarieties Zi ( X does not equal X: we can assume that Zi = V (fi) are hypersur-

faces and we do induction on dim(X). The case dim(X) = 1 is obvious (K algebraically closed and char(K) = 0implies K uncountable). Now take two non-proportional functions h1, h2 ∈ A(X) and consider the pencil ofhypersurfaces Lt := V (h1 + th2 with t ∈ A1. These are uncountably many pairwise different hypersurfaces, so

not all Zi can be among the Lt. This gives us points lying in X but not on any Zi.The map f : G → G, g 7→ gt is a morphism of affine algebraic groups; in particular, it is continuous. We have

f−1(1) = G, due to G = Gt. However, the fibre f−1(1) is finite, as can be seen from embedding G ⊆ GLn —

here we use char(K) = 0. Hence G is finite (and with the assumption G connected, even G = {1}).

24. For affine varieties X and Y , show that dim(X × Y ) = dim(X) + dim(Y ).

Solution: This boils down to showing T(x,y)(X × Y ) ∼= TxX ⊕ TyY which is easy in either of the definitions of

tangent space. Now let U ⊆ X and V ⊆ Y be the open, dense subsets where the dimensions of tangent spacesare minimal. It is then clear that dim(X) + dim(Y ) is the minimum attained on the open subset U ×V ⊆ X×Y .

25. If X is an irreducible, smooth affine variety and Z ( X a closed subvariety,prove dim(Z) < dim(X).

Solution: Denote the inclusion morphism ι : Z ↪→ X; its induced morphism ι∗ : A(X)� A(Z) is then surjective.We get induced surjections MX,P � MZ,P of maximal ideals and also M2

X,P � M2Z,P . Hence, we obtain a

surjection MX,P /M2X,P �MZ,P /M

2Z,P (of cotangent spaces at P ). There exist regular functions f ∈ MX,P

which don’t vanish on Z (as X is irreducible). Choose such an f of minimal degree. Then f cannot be in M2X,P ,

so MX,P /M2X,P �MZ,P /M

2Z,P is not an isomorphism, so dim(Z) ≤ dim(TpZ) < dim(TpX) = dim(X) for p ∈ Z.

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Algebra II — solutions to exercise sheet 6

26. For a linear algebraic groupG, use its comultiplication to define an associative,unital K-algebra structure on A(G)∗ = HomK(A(G), K) such that the induced Liealgebra structure coincides with the one from left invariant vector fields.

Solution: For two functionals ψ1, ψ2 ∈ A(G), we define their tensor product to be ψ1 ⊗ ψ2 ∈ (A(G) ⊗ A(G))∗

with (ψ1⊗ψ2)(f1⊗f2) := ψ1(f1)·ψ2(f2) ∈ K for any f1, f2 ∈ A(G). Using the comultiplication ∆ = µ∗ : A(G)→A(G)⊗A(G), we define ψ1 · ψ2 := (ψ1 ⊗ ψ2)∆ ∈ A(G)∗.This product is associative: coassociativity of ∆ means (id ⊗∆)∆ = (∆ ⊗ id)∆: A(G) → A(G)⊗3. Composingψ1 ⊗ ψ2 ⊗ ψ3 : A(G)⊗3 → K gives (ψ1 · ψ2) · ψ3 = ((ψ1 ⊗ ψ2)∆⊗ ψ3)∆ = (ψ1 ⊗ (ψ2 ⊗ ψ3)∆)∆ = ψ1 · (ψ2 · ψ3).

Now we use the isomorphisms θ : DerK(A(G),K(e)) //DerK(A(G), A(G))λ(G) : εoo where ε(d)(f) = d(f)(e) and

θ(v)(f) = Dv(f) with Dv(f)(g) = v(λgf) for f ∈ A(G) and g ∈ G. Let ψ1, ψ2 ∈ DerK(A(G),K(e)) ⊂ A(G)∗, and

f ∈ A(G) with ∆(f) = a1⊗b1+· · ·+ar⊗br. ThenDψ1(f) =

∑i ψ1(bi)·ai andDψ2

Dψ1(f)(e) =

∑i ψ1(bi)·ψ2(ai):

Dψ1(f)(g) = ψ1(λgf) = ψ1

(∑i ai(g) · bi

)=∑i ai(g) · ψ1(bi) and

Dψ2Dψ1

(f)(e) = ψ2(Dψ1(f)) = ψ2

(∑i ψ1(bi) · ai

)=∑i ψ1(bi) · ψ2(ai).

Hence ε[Dψ2, Dψ1

] = ψ2 · ψ1 − ψ1 · ψ2 = [ψ2, ψ1].

27. Exhibit Gm as a closed subgroup of SO2. Then find a two-dimensionaltorus T ⊂ SO4 and compute the weights for the action induced on the adjointrepresentation, T ↪→ SO4 → GL(so4).

Solution: First,(a bc d

)∈ O2 ⇐⇒ a = d, b = −c. Solving for a linear combination of Gm → Gm, x 7→ x and

x 7→ x−1 with determinant 1, we find a = x/2 + 1/2x and b = −ix/2 + i/2x, where i ∈ K is a fixed root of −1;hence that D(x) :=

( x/2+1/2x −ix/2+i/2xix/2−i/2x x/2+1/2x

)∈ SO2.

Using block matrices, we obtain T := D2 = (Gm)2 ↪→ SO4, (x, y) 7→(D(x) 0

0 D(y)

). As T is diagonalisable, the

representation T ↪→ SO4 → GL(so4), t ·M = tMt−1 for t = (x, y) ∈ T and M ∈ so4 = {M ∈M4(K) |Mt = −M}has exactly two non-trivial eigenspaces. The weights are (1, 0), (0, 1) ∈ X∗(T ) = Z2 and the weight space for χ(1,0)

is {(A B−B 0

)| A =

(0 a−a 0

)∈ so2, B =

( b1 b2ib1 ib2

)}, by explicit computation (note D(x)

(0 1−1 0

)D(x−1) =

(0 1−1 0

)):

(x, y) ·M =(D(x) 0

0 D(y)

)(A B−B C

)(D(x−1) 0

0 D(y−1)

) != χ1,0(x, y)M = xM

28. Let A be a finite-dimensional, associative and unital K-algebra. Show thatthe group of units is a linear algebraic group. What is its Lie algebra?

Solution: For any unit a ∈ A, the map La : A→ A given by left multiplication with a is an isomorphism. Hence

we get an injection A∗ ↪→ GL(A), a 7→ La. In order to see that the image is closed, we consider T ∈ End(A) and

show that TRb = RbT for all b ∈ A is equivalent to T = La for some a ∈ A. The implication (⇐=) is trivial, soassume that T commutes with all Rb and put a := T (1); then T (b) = TRb(1) = RbT (1) = ab = La(b). Hence

A∗ ⊂ GL(A) is the zero set of finitely many commutator equations [−, Rb] = 0, which are polynomial.The Lie algebra of A∗ is A with Lie bracket from the algebra commutator: [a1, a2] = a1a2 − a2a1, since for any

a ∈ A, because derivating the equation gRbg−1 = Rb gives GRb +RbG = 0 for G ∈ End(A).

29. Show that the differential of the adjoint representation of a linear algebraicgroup G is given by ad := d(Ad)e : g→ End(g), ad(A)(B) = [A,B].

Solution: Wallach/Goodman, Theorem 1.5.7

30. Show that a morphism ϕ : G → H of linear algebraic groups induces ahomomorphism of Lie algebras dϕe : g→ h.

Solution: Embedding H ↪→ GL(V ) for some finite-dimensional vector space V , we obtain a rational representation

ϕ : G→ GL(V ) and want to show that dϕe : g→ End(V ) has image in h. Let v ∈ g, h ∈ H and f ∈ I(H) ⊂ A(G).

Then (Ddϕ(v)f)(h) = dϕe(v)(λhf) = v(ϕ∗(λhf)) = 0, because (λhf)(ϕ(g)) = f(hϕ(g)) and f vanishes on H.Hence Ddϕ(v)(I(H)) = 0, which means that d%e(v) ∈ h.

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Algebra II — solutions to exercise sheet 7

31. Find a symmetric, non-degenerate bilinear form Γ on Kn such that T :=SO(Γ) ∩ Dn is a torus of dimension m if n = 2m or n = 2m + 1. Prove that T isa maximal torus in SO(Γ): if H ⊆ SO(Γ) is abelian with T ⊆ H, then T = H.

Solution: Let Γ have entries 1 on the skew diagonal and 0 else; e.g. Γ =(0 11 0

)for n = 2. Then Γ is

obviously symmetric and non-degenerate. For n = 2m even, SO(Γ) ∩ D2m contains the diagonal matricesdiag(x1, . . . , xm, x

−11 , . . . , x−1

m ). If n = 2m + 1 is odd, then SO(Γ) ∩ D2m+1 contains the diagonal matrices

diag(x1, . . . , xm, 1, x−11 , . . . , x−1

m ). In either case, these form a torus Dm in SO(Γ).Now let g ∈ SO(Γ) such that gh = hg for all h ∈ SO(Γ)∩Dn. Each h acts on the standard basis vectors ei ∈ Kn

by a character χi, i.e. h(ei) = χi(h) · ei. These characters are

n = 2m : χ1 = x1, . . . , χm = xm, χm+1 = x−11 , . . . , χ2m = x−1

m ,

n = 2m+ 1 : χ1 = x1, . . . , χm = xm, χm+1 = 1, χm+2 = x−11 , . . . , χ2m+1 = x−1

m .

In either case, the characters are pairwise distinct. Hence, all weight spaces are one-dimensional. As g centralises

SO(Γ) ∩Dn, it must preserves all weight spaces, thus acts diagonally itself. But then g ∈ Dn.

32. Compute the orbits of the natural actions of GLn,Tn,Un,Dn on An and drawthem for n = 2. Describe orbit closures as unions of orbits.

Solution: GLn acts on An with two orbits: 0 is a fixed point, i.e. a closed orbit; An \ {0} is an open orbit.

There are n+ 1 orbits for the Tn-action whose closures are nested: O0 = {0} and Oi = Ai×{0} \Oi−1 for i > 0;so that Oi = O0 ∪ . . . ∪Oi.The Un-orbit of a vector v = (v1, . . . , vn) is (∗, . . . , ∗, vi, 0, . . . , 0), where vi 6= 0 and vi+1 = . . . = vn = 0. In

particular, all orbits are of the form Ai−1 × {(x, 0, . . . , 0)} and hence closed.Dn acts on An with 2n orbits: any map t : {1, . . . , n} → {0, 1} determines a type of vectors v ∈ Nn by vi = 0 ⇐⇒t(i) = 0. These types are preserved by the Dn-action and classify orbits. The closed orbit is {0}, corresponding

to t = (0, . . . , 0); the orbit of type (1, . . . , 1), i.e. of vectors without zero components is open and dense. Giventwo orbits Ot and Os of types t and s, then Ot ⊆ Os if and only if s(i) = 0 ⇒ t(i) = 0 for all i = 1, . . . , n.

33. Show that the adjoint representation of SL2 has disconnected isotropy groups.

Solution: [The point is that some (not all) isotropy groups are disconnected. Note that both SL2 and sl2 are

irreducible; this exercise shows that these properties are not enough to ensure irreducible isotropy.]We consider the adjoint representation SL2 → GL(sl2), which is given by g · M = gMg−1 for g ∈ SL2 and

M ∈ sl2, i.e. a traceless matrix. The isotropy group of M =(0 10 0

)is F = {g ∈ SL2 | gM = Mg}. A quick matrix

computation gives(a bc d

)∈ F ⇐⇒ a = d, c = 0. Together with det(g) = 1, we get F = {

(a b0 a

)| a, b ∈ K, a2 = 1}.

This set decomposes into the two components(1 ∗0 1

)and

(−1 ∗0 −1

).

34. Let G be a unipotent linear algebraic group and X an affine G-variety. Showthat all orbits of G in X are closed.

Solution: This is the theorem of Kostant–Rosenlicht (1961). The property even characterises unipotent groups.

Let O = G · x be an orbit. By replacing X with O, we can assume that O is dense in X. Next, O is also open

in X: the image of the map G → X, g 7→ g · x contains an open subset U ⊆ X (this is a general property ofmorphisms between affine varieties) and thus G · x =

⋃g∈G g · U is open.

Denote by Z := X \ O the closed complement. Then G acts on Z, hence on the ideal I(Z) ⊆ A(X). The latter

action is locally finite, i.e. there exists a finite-dimensional, G-invariant subspace V ⊂ I(Z). As G is unipotent, therepresentation G→ GL(V ) has a fixed vector 0 6= f ∈ V G (Kolchin). This means λg(f) = f , i.e. f(g ·x) = f(x) for

all g ∈ G. Hence f is constant on O, and thus constant on O = X. However, f(Z) = 0, so f = 0, a contradicton.

35. For a homogeneous ideal I ⊆ K[x0, . . . , xn], geometrically compare thevarieties V (I) ⊆ An+1 and V (I) ⊆ Pn. Prove the homogeneous Nullstellensatz.

Solution: Denote π : An+1 → Pn, (a0, . . . , an) 7→ (a0 : . . . : an), also X := V (I) ⊆ An+1 and Y := V (I) ⊆ Pn.

The crucial observation: x ∈ An+1 satisfies f(x) = 0 for all f ∈ I (i.e. x ∈ X) if and only if f(π(x)) = f([x]) = 0 forall homogeneous f ∈ I, i.e. [x] ∈ Y . Hence X consists of lines through the origin, and all fibres of π : X \{0} → Y

are K∗. In particular, π(X \ {0}) = Y and π−1(Y ) ∪ {0} = X. [This is why X is called the affine cone of Y .]

For the Nullstellensatz, let I ⊂ K[x0, . . . , xn] be a radical ideal with (x0, . . . , xn) 6⊆ I. We first consider its affine

variety X = V (I) ⊆ An+1: the affine Nullstellensatz yields that X contains points different from the origin (I is

contained in maximal ideals, and it cannot coincide with the homogeneous maximal ideal by assumption). Fromthe above, we then see that the projective variety Y = V (I) ⊆ Pn contains points.

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Algebra II — solutions to exercise sheet 8

36. Compute the orbits of the actions of GL3 on P2, of GL4 on Gr(2, 4), and ofGL2 on P1 × P1 (diagonal action). Also compute isotropy groups for all orbits.

Solution: GL3 acts transitively on A3 \ {0}, hence it also acts transitively on P2. The isotropy group of the point(1 : 0 : 0) consists of the block matrices of type

(∗∗∗◦∗∗◦∗∗

).

For the action of GL4 on Gr(2, 4), we consider an arbitrary subspace V ⊂ K4 with dim(V ) = 2. If v1, v2 is a basisfor V , then we can apply an automorphism of K4, which maps v1 to e1 = (1, 0, 0, 0). Hence we can assume that

the basis is e1, v for some v ∈ K4. The subgroup T4 ⊂ GL4 preserves e1, up to scalars. Using an appropriate

triangular matrix, we can fix e1 and map v to a vector of the form (0, ∗, ∗, ∗). At this point we use the subgroupGL1×GL3 ⊂ GL4, and with GL3 acting transitively on A3 \{0}, we can map (0, ∗, ∗, ∗) 7→ (0, 1, 0, 0) = e2. Hence

GL4 acts transitively on Gr(2, 4). The isotropy group of the subspace V = (∗, ∗, 0, 0) consists of matrices whoselower left 2× 2-block is zero.

The diagonal action of GL2 on P1 × P2 has two orbits: one is the diagonal ∆ := {(p, p) | p ∈ P1} (this is a closed

orbit), the other is the open complement U := P1 × P1 \ ∆. It is obvious that ∆ is GL2-invariant. Moreover,∆ ∼= P1 and GL2 acts transitively on P1; hence ∆ is an orbit. For a point ((x0 : x1), (y0 : y1)) /∈ ∆, we have

x0y1 − x1y0 6= 0, and applying an appropriate matrix multiplication, we get( y1 −y0−x1 x0

)· ((x0 : x1), (y0 : y1)) = ((y1x0 − y0x1 : 0), (0 : y1x0 − y0x1)) = ((1 : 0), (0 : 1)) ∈ P1 × P1.

The isotropy group of any point in P1 × P1 is D2.

37. Show that any non-degenerate conic in P2 is isomorphic to P1. Deducethat homogeneous coordinate rings are not isomorphism invariants of projectivevarieties.

Solution: A conic in P2 is, by definition, the vanishing locus of some homogeneous polynomial of degree two:

a00x20 + a11x21 + a22x22 + a01x0x1 + a02x0x2 + a12x1x2 = 0 with aij ∈ K. This zero set can be also described in

matrix form xtMx = 0, where the symmetric bilinear form M can be diagonalised to

M =

a00 a01/2 a02/2a01/2 a11 a12/2

a02/2 a12/2 a22

c0 0 0

0 c1 0

0 0 c2

with ci either 0 or 1 (K is algebraically closed). This is given by a linear coordinate change, i.e. an automorphismof P2. However, if one of the ci = 0, then the quadric is degenerate. Hence every non-degenerate quadric in P2 is

isomorphic to V (x20 + x21 + x22). Moreover, P1 → P2, (y0 : y1) 7→ (y20 : y0y1 : y21) is a morphism which is injective

and whose image is cut out by V (x20 − x21 + x22). Hence P1 ∼= V (x20 − x21 + x22) ∼= V (x20 + x21 + x22) using the

coordinate change x1 7→ ix1, and all non-degenerate conics are isomorphic to P1.For the second claim, note that the homogeneous coordinate ring of P1 is K[x, y], the polynomial ring in

two variables (with its standard grading). However, the homogeneous coordinate ring of a quadric is e.g.S = K[x, y, z]/(x2 + y2 + z2). These rings are not isomorphic, for example because the associated affine va-

rieties aren’t (V (x2 + y2 + z2) ⊂ A3 is singular).

38. Show that A1 and P1 are homeomorphic, but A2 and P2 are not.

Solution: As topological spaces, both A1 and P1 = A1 ∪ {∞} are sets of the same cardinality as K, with the

cofinite topology. Hence they are homeomorphic.

When comparing A2 and P2, let a curve be a closed, irreducible subset which is neither a point nor the wholespace. We claim that for any curve C ⊂ A2, there exists another curve C′ ⊂ A2 with C ∩ C′ = ∅. (If C = V (f),

then C′ = V (f + 1) will work.) However, in P2 there are curves which intersect all other curves. (For example, aline V (x0) will work: if V (g) is a curve given by a homogeneous polynomial g ∈ K[x0, x1, x2], then either g ∈ (x0)and V (g)∩V (x0) is even infinite, or g(0, x1, x2) 6= 0 and defines a finite set of points in P1 — this set is non-emptybecause K is algebraically closed.) These are topological properties, so A2 and P2 cannot be homeomorphic.

39. For an affine, irreducible variety X and a point p ∈ X, show that the localring of X at p is given by the localisation of A(X) at the maximal ideal Mp.

40. Show that Aut(P1) = PGL2.

Solution: We already know that GL2 acts on P1 with kernel Z(GL2) ∼= Gm. Hence PGL2 acts faithfully on P1,

and we have to show that all automorphisms of the projective line come from linear maps.Let ϕ ∈ Aut(P1) be an arbitrary automorphism. Put (x0 : x1) := ϕ(1 : 0). If (x0 : x1) 6= (1 : 0), then x1 6= 0, andthen matrix ψ :=

( 0 1/x1x1 −x0

)satisfies ψϕ(1 : 0) = (1 : 0). Hence ψϕ induces an automorphism of P1\{(1 : 0)} = A1.

As A1 is an affine variety, we have End(A1) = End(A(A1)) = End(K[x]), so the automorphism must be of theform ψϕ = ax + b, i.e. a linear polynomial. Then θ :=

(a b0 1

)maps (x : 1) 7→ (ax + b : 1) = ψϕ(x : 1) and

(1 : 0) 7→ (a : 0) = (1 : 0). Hence θ = ψϕ and ϕ = ψ−1θ ∈ PGL2.[It is also true that Aut(Pn) = PGLn+1 but the proof requires considerably harder methods.]

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Algebra II — solutions to exercise sheet 9

41. Show that varieties are compact in the Zariski topology: any open cover hasa finite subcover.

Solution: This holds true for any noetherian topological space, i.e. a space X where descending chains of closed

subsets stabilise or, equivalently, ascending chains of open subsets stabilise. If X =⋃i∈I Ui is an open cover,

then choose a well-ordering on I and define Vj :=⋃i≤j Ui. Then the Vj form an ascending chain of open subsets.

[In fact, a topological space is noetherian if and only if every open subset is compact.]

Affine varieties are noetherian topological spaces because their affine coordinate rings are noetherian, so that

asscending chains of ideals stabilise or, equivalently, descending chains of Zariski-closed subsets stabilise. Thesame reasoning works for projective varieties. For quasi-affine or quasi-projective varieties, we observe that a

subspace of a noetherian topological space is still noetherian.

42. Show that a locally compact Hausdorff space X is compact if and only if theprojection X × Y → Y is a closed map for all topological spaces Y .

Solution: Let X be compact and take a closed subset Z ⊆ X × Y . We have to show that π(Z) ⊆ Y is closed,

where π : X × Y → Y is the projection. Let y ∈ Y \ π(Z). Then (x, y) /∈ Z for all x ∈ X. By the definitionof the product topology, for each x ∈ X, there exist open neighbourhoods x ∈ Ux ⊆ X and y ∈ Vx ⊆ Y such

that Ux × Vx ⊆ X × Y \ Z. In particular,⋃x∈X Ux = X. By compactness of X, there are finitely many points

x1, . . . , xr such that X = Ux1 ∪ · · · ∪ Uxr . Let V := Vx1 ∩ · · · ∩ Vxr , then y ∈ V ⊆ Y open. By construction,V ⊂ Y \ π(Z), so that Y \ π(Z) is open, hence π(Z) closed.

For the reverse implication, assume that X is not compact. Then we consider its one-point compactification

Y := X = X ∪ {∞} — this is a compact space; Y is Hausdorff space since X is locally compact and Hausdorff.Moreover, let Z := {(x, x) | x ∈ X} ⊂ X × X be the diagonal. As X is Hausdorff, Z ⊂ X ×X is closed; thus X is

also closed in X × X. Then π(Z) = X ⊂ X, a closed subset in a compact Hausdorff space, hence compact itself.

The statement actually holds for arbitrary topological spaces X: for the reverse implication, let X =⋃i∈I Ui be

an open covering without any finite subcovering. Let now Y := X∪{ω}, topologised with the following open sets:

all subsets of X; all sets containing {ω}∪ (X \U) where U is a finite union of the Ui. This is indeed closed under

arbitrary unions and finite intersections. Moreover, by our assumption on the covering Ui, the subset X ⊂ Y isnot closed, i.e. {ω} ⊂ Y is not open. Let Z be the closure of the diagonal ∆ := {(x, x) | x ∈ X} in X × Y . By

hypothesis, its closure π(Z) is closed in Y , hence π(Z) = Y . Hence (x, ω) ∈ Z for some x ∈ X. By the definitionsof closure, of the topology on Y and of the product topology,

(V × ({ω} ∪ (X \ Ui))

)∩∆ 6= ∅ for all i ∈ I and

open sets x ∈ V ⊆ X. Hence V ∩ (X \ Ui) 6= ∅, i.e. V 6⊆ Ui for all V and i. But this would imply x /∈ Ui for all

i ∈ I, contradicting that the Ui cover X.

43. Let X = V (y2 − x3) ⊂ A2 be the cuspidal cubic. Show that the mapA1 → X, t 7→ (t2, t3) is regular, birational and a homeomorphism, but not anisomorphism of varieties. Extend this to an example of a morphism of projectivevarieties with the same properties.

Solution: The map is given by polynomials, so it is regular. It is bijective, because for any point (x, y) ∈ X, wehave y2 = x3 and (with K algebraically closed), there are two square roots of x, call them t and −t. Moreover,

t6 = x3 = y2, regardless of sign. But exactly one of the two roots satisfies t3 = y. Next, A1 and X are infinitesets with the cofinite topology, hence every bijection between them is automatically a homeomorphism.

Finally, the two varieties are birational: K(A1) = Quot(K[t]) = K(t), i.e. the function field of the affine line is

the field of rational functions in one variable. But A(X) = K[x, y]/(y2 − x3) = K[t2, t3]) has the same quotientfield: Quot(K[t2, t3]) = K(t), as t3/t2 = t in Quot(A(X)).

The two curves are not isomorphic because their affine coordinate rings aren’t. For example, the tangent spaceT0X = (M0/M2

0 )∗ is 2-dimensional, whereas all tangent spaces of A2 are 1-dimensional.By homogenisation, we get an example of projective curves: P1 → V (x0x22 − x31), (t0 : t1) 7→ (t30 : t0t21 : t31).

44. Let A be a commutative ring with unit and N a finitely generated A-module.Show that a surjective module endomorphism N → N is an isomorphism.

Solution: The theorem of Vasconcelos.

Write f : N → N and use it to consider N as an A[X]-module, with X · n := f(n). Then XN = N by theassumption that f is surjective. Applying the Nakayama lemma (with I = (X) ⊂ A[X]) yields an elementY ∈ A[X] such that Y N = 0 and Y ∈ 1+(X), i.e. Y = 1+XZ for some Z ∈ A[X]. Let n ∈ ker(f), then X ·n = 0

and hence 0 = (1 +XZ)n = n+ ZXn = n, so that f is indeed injective.

45. Find subgroups of SL2 such that the quotient is respectively projective, affineor strictly quasi-affine.

Solution: The quotient SL2/(T2 ∩ SL2) ∼= P1 is a projective variety; SL2/(D2 ∩ SL2) ∼= PSL2 is an affine variety;

SL2/U2∼= A2 \ {0} is quasi-affine but not affine.

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Algebra II — solutions to exercise sheet 10

46. Compute the dimensions of Grassmannians Gr(d, n) and flag manifolds Fl(n).

Solution: We already know that Fl(n) = GLn/Tn is a homogeneous space. This implies that tangent spaces at the

flag manifold are given by quotients TeGLn/TeTn, hence dim(Fl(n)) = dim(gln)− dim(tn) = n2 − n(n+ 1)/2 =n(n− 1)/2, where tn ⊂ gln is the subset of all upper triangular matrices.

The Grassmannian Gr(d, n) has a transitive action by GLn, so Gr(d, n) = GLn/H where H is the stabiliser

of some d-dimensional subspace U ⊂ Kn. Taking U to be the span of the first d standard basis vectors ofKn, we find H = {

(A B0 D

)| A ∈ GLd, B ∈ M(d × n − d), D ∈ GLn−d}. Again, we compute the dimension as

dim(Gr(d, n)) = dim(gln)− dim(TeH) = n2 − (d2 + d(n− d) + (n− d)2) = nd− d2 = d(n− d).

47. Show that two irreducible varieties X and Y are birational, i.e. their functionfields are isomorphic: K(X) ∼= K(Y ), if and only if there exist affine open subsetsUX ⊆ X and UY ⊆ Y which are isomorphic: UX ∼= UY .

Solution: For any open affine subset ∅ 6= U ⊆ X, we have Quot(A(U)) = K(X). Hence UX ⊆ X and UY ⊆ Y

with UX ∼= UY implies K(X) ∼= Quot(A(UX)) ∼= Quot(A(UY )) ∼= K(Y ).Now, assumeK(X) ∼= K(Y ). Pick an open affine subset V ⊆ X; this givesK(X) ∼= K(V ). Choose an isomorphism

ϕ : K(V ) → K(Y ). Let f1, . . . , fr ∈ A(V ) be generators of the affine coordinate ring of Y . Then the ϕ(fi) are

rational functions on Y , and there is an open subset U ⊆ Y on which all ϕ(fi) are regular. Shrink U , if necessary,to ensure it is affine. Then ϕ induces a morphism ϕ : A(V )→ A(U). This map of K-algebras must be injective,

since the induced map of quotient fields is. Hence we obtain a dominant morphism α : U → V of affine varieties.

Repeating this argument for ϕ−1 : K(Y )→ K(X) yields a dominant morphism β : V → U ′ for some open, affinesubset U ′ ⊆ Y . The morphisms α and β are inverse on an open subset.

48. Assume char(K) = p > 0. Show that the map Gm → Gm, x 7→ xp is abijective morphism of affine algebraic groups, but is not an isomorphism.

Solution: The map is obviously a group homomorphism, due to characteristic p. It is a morphism of varieties,

since it is evidently given by a polynomial. It is surjective, because K is algebraically closed (every element of

the field has roots of any order). It is injective because of xp = yp ⇐⇒ (x− y)p = 0 ⇐⇒ x− y = 0.However, it is not an isomorphism: variety morphisms Gm → Gm are the same as endomorphisms of the K-algebra

A(Gm) = K[T, T−1). The map F : Gm → Gm, F (x) = xp induces F ∗ : K[T, T−1] → K[T, T−1], f(T ) 7→ f(T p).

This ring homomorphism is not surjective!

49. Let G be a linear algebraic group, acting on a quasi-projective variety X.Show that orbits of minimal dimension are closed; in particular, closed orbits exist.

(Hint: you can use the statement of exercise 25.)

Solution: Let x ∈ X and Z := G · x be the closure of the orbit of x. Then Z is a G-invariant, closed subset of X.

We already know that O := G · x ⊆ Z is open. Assume the orbit is not closed. Then Z \ O is closed in Z andG-invariant, hence quasi-projective and a union of orbits. Therefore, dim(Z \O) < dim(Z) and since dimensions

are non-negative and finite, orbits of minimal dimension must be closed.

50. LetG be a connected projective algebraic group. Show thatG is commutative.

Solution: We consider the two morphisms f, p : G × G → G with f(g, h) = ghg−1h−1 and p(g, h) = g. Let

U ⊂ G be an open affine subset of G such that e ∈ U . Then p(G × G \ f−1(U)) ⊂ G is a closed subset not

containing e. Thus there is an open subset V ⊂ G such that p−1(V ) ⊆ f−1(U), hence f(p−1(V )) ⊆ U . Letg ∈ V ; then f({g} × G) = f(p−1(g)) ⊆ U . However, with G proper and connected, this set is the singleton {e}since f(g, e) = e. Thus f(p−1(V )) = {e}, but p−1(V ) ⊂ G×G is a dense subset, and so f(G×G) = {e}, i.e. G

is commutative.

Sketch of an alternative proof if char(K) = 0: for any g ∈ G, let cg : G→ G, h 7→ ghg−1. Its derivative at e gives

an automorphism Ad(g) := dce : TeG → TeG. In particular, we get a morphism Ad: G → GL(TeG). With Gproper and connected, its image must be the identity of TeG. Like in the affine case, TeG is a Lie algebra and we

now know that it is abelian. G connected and char(K) = 0 imply that G is commutative.

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Algebra II — solutions to exercise sheet 11

51. Find Borel subgroups in SO4, Sp4, T4 and U4.

Solution: Recall that GL4/TT4 ∼= Fl(K4), the variety of full flags in K4. Given the subgroup SO4 ⊂ GL4,

we consider the subset of isotropic flags Fl0(Kn, β) = {V 0 ( V 1 ( V 2 | dim(V i) = i, β|V i×V i = 0} whereβ : K4 ×K4 → K is the bilinear form. This is a closed subset, so that Fl0(Kn, β) is a projective variety. SO4

acts transitively on isotropic flags (similar proof as with GLn and full flags). Using the bilinear form β from

Exercise 31, a particular isotropic flag is given by V 0 = 0 ⊂ V 1 = Ke1 ⊂ V 2 = Ke1 +Ke3. From g ∈ GL4 withge1 = e1, ge3 = e3 and gtβg = β, we get

β =

0 0 0 1

0 0 1 0

0 1 0 01 0 0 0

, g =

1 ∗ 1 ∗0 1 0 ∗0 0 1 ∗0 0 0 1

and in particular the stabiliser group in SO4 is solvable.

T4 and U4 are solvable and connected, hence their Borel groups are the full group in each case.

52. Find all parabolic subgroups P with T4 ⊆ P ⊆ GL4.

Solution: There are 8 such parabolic subgroups, there are given by block matrices of the formats

GL4 =

(∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗

),

(∗∗∗∗∗∗∗∗∗∗∗

),

(∗∗∗∗∗∗∗∗∗∗∗

),

(∗∗∗∗∗∗∗∗∗∗∗

),

(∗∗∗∗∗∗∗∗∗∗∗∗∗

),

(∗∗∗∗∗∗∗∗∗∗∗∗

),

(∗∗∗∗∗∗∗∗∗∗∗∗

),

(∗∗∗∗∗∗∗∗∗∗

)= T4.

53. Let the connected linear algebraic group G act on a quasi-projective varietyX with finitely many orbits. Show that every irreducible G-invariant subset in Xis the closure of a G-orbit. Find a counterexample for an action with infinitelymany orbits.

Solution: We know that orbits in general are quasi-projective and open in their closure. An irreducible G-

invariant subset Z ⊆ X is a union of orbits. Since there are only finitely many orbits, Z contains an orbit O as

an open subset; the closure O must then be Z.Counterexamples abound. (For a trivial one, take G = 1, Z = X = A1.)

54. Find a connected linear algebraic group G and a maximal solvable subgroupU ⊂ G such that U is disconnected.

Solution: This is problem 17 in disguise: we take G = GL2 and U := N(D2) =( ∗ 00 ∗)∪(0 ∗∗ 0

). Then U is

obviously disconnected, it is solvable by problem 17. It is a matrix calculation to show that the only subgroup of

G strictly containing U is G.

55. Classify all root systems in the Euclidean plane E := R2.

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Algebra II — solutions to exercise sheet 12

56. Compute the radicals R(GLn), R(SLn), R(Un).

Solution: On the one hand, R(GLn) is the identity component of the intersection over all Borel subgroups. With

the Borel subgroups B := Tn and B− of upper/lower triangular matrices, we find R(GLn) ⊆ B ∩B− = Dn. Onthe other hand, R(GLn) is the maximal normal, solvable, connected subgroup: we get R(GLn) = K∗ · In from

R(GLn) normal: for n = 2, we have(1 11 0

)(a 00 b

)(0 11 −1

)=(b a−b0 a

)∈ R(GL2) ⊆ D2, hence a = b.

The same computation for SLn shows R(SLn) = 1, i.e. SLn is semisimple.R(Un) = Un because Un is unipotent, hence solvable; connected; normal in itself.

57. Let char(K) = 0 and U a commutative, unipotent group. Show that U ∼= Gra

for some r ∈ N.

Solution: One way is to use exponential and logarithm for matrices (introduced in the classification of connected,1-dimensional groups). Consider U ⊆ Un for some n (Kolchin). The map log: U → n is then polynomial, i.e. a

morphism of affine varieties, where n is an additive subgroup of nilpotent matrices. As U is commutative, the

logaritm is even a group homomorphism, and its inverse is exponential. Now n is a K-vector space, so in particularisomorphic to Gra for some r ∈ N.

58. Let G be a finite group and A a trivial G-module, i.e. an abelian group whichhas the trivial left G-action. Show that H1(G,A) = Hom(G,A).

Solution: This is an obvious application of the definitions: note that trivial G-action implies B1(G,A) = 0, so

that H1(G,A) = Z1(G,A) = Hom(G,A), since the 1-cocycle condition collapses ϕ(g1g2) = ϕ(g1) + g1(ϕ(g2)) =

ϕ(g1) + ϕ(g2).

59. Let ϕ : G → H be a surjective morphism of linear algebraic groups, andT ⊆ G a maximal torus (or a maximal connected normal unipotent subgroup,respectively). Show that ϕ(T ) ⊆ H has the same property.

Solution: Humphreys §21.3

60. Let (E,Φ) be a root system. A base of (E,Φ) is a subset S ⊆ Φ such thatS is a basis for E and every root β ∈ Φ can be written as β =

∑α∈Smαα with

either all mα ≥ 0 or all mα ≤ 0. The elements of S are also called simple roots,and their non-negative linear combinations positive roots and denoted Φ+.Draw bases and positive roots for the root systems in E = R2, of exercise 55.

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Algebra II — solutions to exercise sheet 13

61. Let G be a connected algebraic group, B ⊆ G a Borel subgroup and V arational G-module. Show that the invariant subspaces coincide: V G = V B.

Solution: Trivially, V G ⊆ V B . Let v ∈ V B . We look at the morphism G → G · v, g 7→ gv. Because of v ∈ V B ,this map factors through G/B → G · v. However, G/B is projective and the orbit G · v is affine; both varieties

are connected. Hence, G/B → G · v is constant, and v ∈ V G.

62. Let G be a linear algebraic group. Show that R(G)u = Ru(G), i.e. theunipotent part of the radical is the unipotent radical.

Solution: Springer 7.6.3

63. For a linear algebraic group G, show that G/R(G) is semisimple and thatG/Ru(G) is reductive.

Solution: We know that the surjective group homomorphism π : G → G/R(G) maps Borel subgroups to Borelsubgroups. Then R(G) = (

⋂B B)◦, by a characterisation of the radical. Hence G/R(G) has trivial radical by

( ⋂π(B)

π(B))◦⊂ π(R(G)).

For the statement about G/Ru(G), show that if an extension 1 → G′ → G → G′′ → 1 the outer groups G′ and

G′′ are unipotent, then so is the middle one G. (This follows from Jordan decomposition.) Now the preimage

of unipotent and normal subgroup U ⊆ G/Ru(G) in G is Ru(G) ⊆ U ⊆ G. Then U is normal in G and alsounipotent (from the extension 1→ Ru(G)→ U → U/Ru(G)→ 1, hence Ru(G) = U .

64. Let G be a group with subgroups H,N ⊆ G. Show that the following notionsare equivalent — then G = N oH is called a semi-direct product of H by N :(1) There is a short exact sequence 1 → N

ι−→ Gπ−→ H → 1 admitting a section

σ : H → G, i.e. πσ = idH .(2) N is normal in G, and NH = G and N ∩H = 1.(3) There is a homomorphism α : H → Aut(N) and G is isomorphic to the group

N oα H defined by (n, h) · (n′, h′) := (nα(h)(n′), hh′) on the set N ×H.Also show that the existance of a retraction % : G→ N , i.e. %ι = idN , is equivalentto a splitting G ∼= N ×H of G as a direct product.

Solution: (1) ⇐⇒ (2): Given the split extension, then N = ker(ι, hence is normal in G. Given g ∈ G, thenπσπ(g) = π(g), so σπ(g) = gn for some n ∈ N ; altogether g = n−1σπ(g) ∈ NH. Finally, if g ∈ N ∩ σ(H), theng = σ(h), hence 1 = π(g) = π(σ(h)) = h; so N ∩ σ(H) = 1. The reverse implication is proved similarly.

(1) ⇒ (3): The homomorphism α : H → Aut(N) is defined as follows: for h ∈ H and n ∈ N , let α(h)(n) =

σ(h)nσ(h)−1. Then π(α(h)(n)) = πσ(h) · π(n) · πσ(h)−1 = h · 1 · h−1 = 1, hence α(h)(n) ∈ N .

65. Show that the following subsets define root systems of rank n:{ej − ei | i, j ∈ {1, . . . , n+ 1}, i 6= j} ⊂ Qn+1 (called type An){±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Qn (called type Dn)

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Algebra II — solutions to exercise sheet 14

66. Show that centraliser of(i 00 −i

)∈ PGL2(C) is disconnected.

Solution: In fact, the centraliser of A :=(1 00 −1

)in SL2 is disconnected (th exercise follows from this).

Now M :=(a bc d

)(1 00 −1

)(d −b−c a

)=( ad+bc −2ab

2cd −(ad+bc)

), so M = A ⇐⇒ ad+ bc = 1, ab = cd = 0, leading to

CA(SL2) ={( a 0

0 1/a

)}∪{( 0 b−1/b 0

)},

which is a disjoint union.

67. For B ⊆ G a Borel subgroup of a connected linear algebraic group G, showthat Z(B) = Z(G).

Solution: Trivially, Z(B) = CB(B) ⊆ CG(B) ⊆ CG(G) = Z(G). Let g ∈ Z(G). Then g ∈ B′ for some Borel

subgroup B′ ⊆ G (because Borel groups cover G). Since Borel subgroups are conjugated, there is an h ∈ G with

B′ = hBh−1. As g is central, we have g = h−1gh ∈ B.

68. Prove directly for G = SLn and for G = SOn (with the bilinear form fromProblem 31) that G is covered by Borel subgroups and that maximal tori coincidewith their centralisers.

Solution: G = SLn with B = Tn ∩ SLn (upper triangular matrices of determinant 1). If M ∈ SLn is anymatrix, then there is a base change matrix g ∈ GLn such that gMg−1 is upper triangular (this is a weak form

of the Gauss algorithm). Moreover, we can assume that det(B) = 1 (because K is algebraically closed). Hence

gMg−1 ∈ B or, equivalently, M ∈ g−1Bg, a Borel subgroup of G.T = Dn ∩ SLn (diagonal matrices of determinant 1) is a maximal torus in G. Let d = diag(d1, . . . , dn) ∈ T and

g ∈ G. Then gd is obtained from g by multiplying the i-th row of g by di, whereas in dg the i-th column of g is

multiplied by di. If gij 6= 0, then we get digij = djgij , hence gij = 0 unless di = dj . If g ∈ CG(T ), then we havedg = gd for all d ∈ T , hence g is a diagonal matrix.

G = SOn with B = Tn ∩ SOn a Borel subgroup. Here, any matrix can be brought into upper triangular formusing the Gram-Schmidt algorithm. Hence Borel subgroups again cover G.

The argument for CG(T ) = T proceeds as above (we can work with diagonal tori by choosing an appropriate

symmetric bilinear form).

69. Let G be a connected linear algebraic group with a maximal torus and B bethe set of Borel subgroups of G with its natural T -action. Show that there is abijection between the fixed point set BT and the Weyl group W (G).

Solution: We let W = NG(T )/CG(T ) act on BT by n · B := nBn−1. In order to show that this is well-defined, we first prove that CG(T ) ⊆

⋂B∈BT B. As CG(T ) is connected (all torus centralisers are) and nilpotent

(because it has the unique maximal torus T ), it is solvable and hence contained in some B ∈ BT . For another

B′ = gBg−1 ∈ BT , we have maximal tori T, gTg−1 ⊆ B′. Hence these are conjugated to each other under someb′ ∈ B′. Then CG(T ) = C(b′gTg−1b′−1) = b′gCG(T )g−1b′−1 ⊆ B′.Let B,B′ ∈ BT , i.e. T ⊆ B,B′ ⊆ G. By conjugacy of Borel subgroups, there is g ∈ G with B = gB′g−1.

Then T ⊆ B and gTg−1 ⊆ B are maximal tori, so by conjugacy of maximal tori in B, there is b ∈ B withbgTg−1b−1 = T . Hence n := bg ∈ NG(T ). Then also nB′n−1 = B, so n ∈ W maps B′ 7→ B. Hence W acts

transitively on BT .

Assume now that some B ∈ BT is fixed under W . So we want to show that nBn−1 = B implies n ∈ CG(T ). ButnBn−1 = B implies n ∈ NG(B) = B by the theorem about normalisers of Borel subgroups, so n ∈ B ∩NG(T ) =

NB(T ) = CB(T ) ⊆ CG(T ), where the last equality follows from the semi-direct decomposition B = Bu o T andJordan decomposition.

70. Show that the following subsets define root systems of rank n:{±ei | i ∈ {1, . . . , n}} ∪ {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Qn (type Bn){±2ei | i ∈ {1, . . . , n}} ∪ {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Qn (type Cn)

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5 About the exam

Topics in descending importance.

Calculations with matrices and matrix groups

For example, for any explicitly defined matrix group G, you could check that itis a linear algebraic group, compute is centre, commutator subgroup, Lie algebraetc. Or check for properties such as solvable, unipotent, Borel etc.

Exercises along these lines have been 6, 11, 16, 17, 18, 32, 36, 51, 52, 68.

Properties of linear algebraic groups, orbits and homogeneous spaces

There have been some recurring arguments that you should be able to apply as well:embedding a group in GLn; Chevalley lemma; Jordan decomposition; the struc-ture of diagonalisable and connected solvable groups; Borel fixed point theorem(and a map from a connected, projective variety to an affine variety is constant);conjugacy of Borel subgroups and of maximal tori.

Exercises 14, 15, 46, 49, 53, 61, 67, 69.

Basic algebraic geometry

Affine geometry: Zariski topology, singular points, tangent spaces.Projective geometry: projective space, Grassmannians, proper morphisms.

Exercises: 13, 35, 43, 47.

Lie algebras and adjoint representation

You should know the definitions of the classical groups (GLn, SLn, SOn, Spn) andof their Lie groups. If the non-standard bilinear forms are needed, I will list them;you don’t have to memorise these.

Exercises 22, 27, 30.

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1 2 3 4 5 6 7 8∑ Matriculation number

Exam Algebra II

K is an algebraically closed field. All algebras and varieties are defined over K.

1. An affine transformation of A2 is a bijection A2 ∼−→ A2 mapping affine linesto affine lines.a) Show that the set G of affine transformations of A2 is a linear algebraic group.b) Describe a maximal torus in G.

2. Prove that a connected linear algebraic group consisting of semisimple elementsis a torus. Show by example that the connectedness assumption is necessary.

3. Let G be a connected solvable linear algebraic group. Does G have a uniquemaximal torus? (Proof or counterexample.)

4. Compute the Lie algebra sl2 of SL2 from first principles, choose a basis andcompute the matrix of ad(g) : sl2 → sl2 for arbitrary g ∈ SL2 in the chosen basis.

5. Let B be a non-degenerate symmetric bilinear form on K2n. Show that theset of full isotropic flags V 0 ⊂ V 1 ⊂ V 2 ⊂ . . . ⊂ K2n (i.e. dim(V i) = i andB|V i×V i = 0 for all i) is a projective variety.

6. Compute the roots of the group Sp4. You can use that this group has rank 2.

7. Let B ⊆ G be a Borel subgroup of a connected linear algebraic group G. Showthat the centres coincide: Z(B) = Z(G).

8. Let G = GLn and B ⊂ G a Borel subgroup. Let W be the Weyl group of G.a) Show G =

⋃w∈W BwB. (Explain first that BwB is a well-defined subset of G.)

b) Show that the union is disjoint.

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Solutions for exam

1. An affine transformation of A2 is a bijection A2 ∼−→ A2 mapping affine linesto affine lines.a) Show that the set G of affine transformations of A2 is a linear algebraic group.b) Describe a maximal torus in G.

Solution: a) First solution: It is clear that G is a group. Let us check first that G ⊂ Aut(A2) (automorphisms

as an affine variety). For this, we have to check g∗(f) = fg ∈ A(A2) = K[x, y]. If deg(f) = 1, i.e. f = ax+ by+ cfor some a, b, c ∈ K, then we know g∗(f) is again a polynomial of degree 1 — this is condition that G preserves

affines lines (i.e. zero sets V (ax + by + c)). Since g∗ is obviously a ring automorphism (of the ring of all K-

valued functions on A2, for example), we see that g∗ preserves A(A2), as every polynomial is generated fromlinear forms. In particular, we have g∗ ∈ Aut(A2) = Aut(K[x, y]), but the above argument shows even more:

g∗ ∈ Aut(K +Kx+Ky) = GL3.

Second solution: If ϕ ∈ G is arbitrary with ϕ(O) 6= O (where O = (0, 0) ∈ A2 is a fixed origin), then we considerthe translation v : A2 → A2, x 7→ x− ϕ(O). Translations are affine transformations, so vϕ ∈ G with vϕ(O) = O.

Hence vϕ is an affine transformation of A2 fixing O, hence it is a linear automorphism of K2, i.e. vϕ ∈ GL2.

There are two natural subgroups in G: translations G2a∼= V ⊂ G and linear transformations GL2 = GO ⊂ G

(the subgroup fixing O). Then V ∩GL2 = 1 and V GO = G by the above argument. An immediate computation

shows that GO is a normal subgroup of G: for g ∈ G and v(x) = x + v the translation along v ∈ A2, we have

g−1vg(x) = g−1(g(x) + v) = x+ g−1(v), which is again a translation.Altogether we see that the abstract group G has a semi-direct product decomposition G = G0 o V . Since

GO ∼= GL2 and v ∼= G2a are affine varieties,so is their product (which is the underlying set of G). The composition

is given by linear functions, so in particular is polynomial.

b) Here we use the semi-direct decomposition from above. Since Ga is unipotent, so is V ∼= G2a. Tori consist of

semisimple elements, hence any torus in G has to be disjoint from V . We find that a maximal torus in G is a

maximal torus in GL2, for example D2.

2. Prove that a connected linear algebraic group consisting of semisimple elementsis a torus. Show by example that the connectedness assumption is necessary.

Solution: Let G = Gs be connected. We have to show that G is commutative (because a torus is, by definition,

a connected commutative group all of whose elements are semisimple).

Let B ⊆ G be a Borel subgroup. Then B is connected solvable, so has a semi-direct product decompositionB = Bu n T where T ⊆ B ⊆ G is a maximal torus. From the assumption G = Gs we know Bu = 1, so B = T .

In particular, B is commutative and hence nilpotent. This forces G = B. (In the course, we have seen that if a

Borel subgroup is nilpotent (or normal), then it is the whole group.)Connectedness is necessary: any finite group consists of semisimple elements (this uses char(K) = 0 and Jordan

decomposition).

3. Let G be a connected solvable linear algebraic group. Does G have a uniquemaximal torus? (Proof or counterexample.)

Solution: G does not necessarily have only one maximal torus. (In fact, we have shown that connected nilpotentgroups are characterised by having a unique maximal torus.)

A counterexample can be found in B := T2 ∩SL2: this is a Borel subgroup of SL2, so is conncted solvable. It has

the standard maximal torus T := D2 ∩ SL2 ={( x 0

0 1/x

)}but another torus is

(1 10 1

)T(1 −10 1

)={( x 1/x−x

0 1/x

)}.

4. Compute the Lie algebra sl2 of SL2 from first principles, choose a basis andcompute the matrix of ad(g) : sl2 → sl2 for arbitrary g ∈ SL2 in the chosen basis.

Solution: By definition, SL2 = V (x1x4 − x2x3 − 1) ⊂ A4. We want to compute the tangent space at the pointe = (1, 0, 0, 1). For a variety cut out by a single polynomial f and a point p, the tangent space at p is given bythe vanishing locus of

∑i ∂f/∂xi(p) xi = 0. In our situation, we get the linear equation x1 + x4 = 0, i.e. the Lie

algebra TeSL2 = sl2 = {M ∈M(2,K) | tr(A) = M} is given by traceless matrices.

A (very typical) basis for sl2 consists of the three matrices H =(1 00 −1

), X =

(0 10 0

), Y =

(0 01 0

).

A group element g =(a bc d

)acts on a tangent vector M ∈ sl2 by ad(g)(M) = gMg−1. Hence we get

ad(g)(H) =(a bc d

)(1 00 −1

)(d −b−c a

)=

( ad+bc −2ab−2cd −bc−ad

)= (ad+ bc)H − 2abX − 2cdY ;

ad(g)(X) =(a bc d

)(0 10 0

)(d −b−c a

)=

(−ac a2−c2 ac

)= −acH + a2X − c2Y ;

ad(g)(Y ) =(a bc d

)(0 01 0

)(d −b−c a

)=

( bd −b2d2 −bd

)= bdH − b2X + d2Y.

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Hence the matrix of ad(g) in the basis (H,X, Y ) isad+ bc −ac bd

−2ab a2 −b2−2cd −c2 d2

.

5. Let B be a non-degenerate symmetric bilinear form on K2n.Show that the set of full isotropic flags V 0 ⊂ V 1 ⊂ V 2 ⊂ . . . ⊂ K2n (i.e. dim(V i) =i and B|V i×V i = 0 for all i) is a projective variety.

Solution: We write Fn for the set of isotropic flags in K2n. Because B is non-degenerate, the maximal dimension

of an isotropic subspace in K2n is n. Now a flag V · ∈ Fn is partial flag (of ordinary subspaces) in K2n; the latteris given by the partial flag manifold Xn := GL2n/P where the parabolic subgroup has block form P =

(Tn ∗0 ∗

).

In particular, Xn is a projective variety. The subset Fn ⊂ Xn is closed: B is a bilinear form, so isotropicity is a

polynomial condition. Hence, Fn is projective.

6. Compute the roots of the group Sp4. You can use that this group has rank 2.

Solution: With I =(1 00 1

)and J =

(0 I−I 0

), the symplectic group is defined by Sp4 = {g ∈ GL4 | gtJg = J}.

It is an easy computation that diag(x, y, x−1, y−1) ∈ Sp4, hence T := D4 ∩ Sp4 is a 2-dimensional torus, so is a

maximal torus (because we can use that Sp4 has rank 2). As Sp4 is given by a bilinear form, its Lie algebra is

sp4 = {X ∈M(4,K) | XtJ + JX = 0} = {(A BC D

)| Bt = B,Ct = C,D = −At}.

In particular, A =( a1 a2a3 a4

), D =

(0 d−d 0

), B =

( b1 b2b2 b3

), C =

( c1 c2c2 c3

).

The adjoint action of Sp4 (and hence that of T ) on sp4 is given by ad(g)(X) = gXg−1. We explicitly compute

ad(t)(X) =

x 0 0 00 y 0 0

0 0 x−1 00 0 0 y−1

a1 a2 b1 b2a3 a4 b2 b3c1 c2 −a1 −a3c2 c3 −a2 −a4

x−1 0 0 0

0 y−1 0 0

0 0 x 00 0 0 y

=

a1 xy−1a2 x2b1 xyb2

x−1ya3 a4 xyb2 y2b3x−2c1 x−1y−1c2 −a1 −x−1ya3

x−1y−1c2 y−2c3 −xy−1a2 −a4

and we are now looking for 8 eigenvectors relative to characters (i.e. functions xmyn). They are easy to see:

character weight eigenvector (all other matrix entries 0)

1 0 a1 = 0 or a4 = 0

xy−1 e1 − e2 a2 = 1x−1y e2 − e1 a3 = 1

x2 2e1 b1 = 1

xy e1 + e2 b2 = 1y2 2e2 b3 = 1

x−2 −2e1 c1 = 1

x−1y−1 −e1 − e2 c2 = 1y−2 −2e2 c3 = 1

Therefore, the roots (i.e. the non-zero weights) are ±2e1,±2e2,±(e1 − e2),±(e1 + e2).

7. Let B ⊆ G be a Borel subgroup of a connected linear algebraic group G. Showthat the centres coincide: Z(B) = Z(G).

Solution: This is exercise 67.

8. Let G = GLn and B ⊂ G a Borel subgroup. Let W be the Weyl group of G.a) Show G =

⋃w∈W BwB. (Explain first that BwB is a well-defined subset of G.)

b) Show that the union is disjoint.

Solution: a) We choose the standard torus and Borel subgroup: T = Dn ⊂ B = Tn ⊂ GLn. The Weyl group

of GLn is the permutation group W = NG(T )/CG(T ) = NG(T )/T ∼= Sn. In particular, from CG(T ) = T (which

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has been an exercise for T = Dn ⊂ GLn and is easy to check again) and T ⊂ B, we see that nB = ntB forn ∈ NG(T ) and t ∈ T . Note that W is the subgroup of permutation matrices in GLn. (This is a special feature

of GLn — for other reductive groups, the Weyl group is not necessarily a subgroup in a canonical way!)

One way to see GLn = BWB is via the Gauss algorithm: bringing an arbitrary matrix g ∈ GLn in row echelon(hence upper triangular) form starts with (g | In) and uses row transformations to end in (U | L) with Lg = U

where U is an upper triangular matrix and L is a lower triangular matrix, i.e. U,Lt ∈ B. However, in general

one has to apply a permutation (e.g. if the top left entry of g is zero); i.e. there is a permutation matrix w1 ∈Wsuch that the row algorithm works on w1g (using w1g means that we permute rows of g). We record that for any

g ∈ G, there are w1 ∈W,L ∈ Bt, U ∈ B with wg = L−1U .

We arrive at g = w−11 L−1U . Now there is a unique permutation matrix w2 ∈ W such that w−1

1 L−1w2 ∈ B— i.e. we undo the row permutations of w−1

1 by appropriate column permutations of w2. Eventually, we get

g = w−11 L−1w2w

−12 U = U ′wU ∈ BwB where U ′ := w−1

1 L−1w2 ∈ B by construction and w := w−12 .

b) If BwB ∩ Bw′B 6= ∅, then there exist b1, b2, b3, b4 ∈ B such that b1wb2 = b3w′b4, hence b−13 b1wb2b

−14 = w′

and w′ ∈ BwB. This immediately implies Bw′B ⊆ BwB. The other inclusion follows by symmetry.