Last update: 1 August 2013
The significance of the results so far to the theory of semisimple algebraic groups will now be indicated.
Let be an algebraically closed field. A subset is said to be algebraic if there exists a subset such that for all The algebraic subsets of are the closed sets of the Zariski topology on For set for all
Let Define by Then is an algebraic subset of is a matric algebraic group if is a subgroup of for some and some algebraically closed field and is an algebraic subset of If is a subfield of is defined over if has a basis of polynomials with coefficients in
|(h)||any finite subgroup.|
A map of algebraic groups is a homomorphism if it is a group homomorphism and each of the matric coefficients is a rational function of the A homomorphism is an isomorphism if there exists a homomorphism such that and A homomorphism is defined over if each of the rational functions above has its coefficients in
Except for the last assertion, the following results are proved in Séminaire Chevalley (1956-8), Exposé 3.
Let be a matric algebraic group. Then the following are equivalent:
|(ii)||The image of an algebraic group under a rational homomorphism is algebraic.|
|(iii)||A group generated by connected algebraic subgroups is algebraic and connected (e.g. (a) - (g) are connected). It is defined over the perfect field if each of the subgroups is.|
If is an algebraic group, the radical of is the maximal connected solvable normal subgroup. is semisimple if (1) and (2) is connected.
Example: has radical
For the remainder of this section we assume that is algebraically closed, is the prime field, is a Chevalley group based on and the lattice. (Since a change of basis in is given by polynomials with integral coefficients we may speak of a basis over
Theorem 6: With the preceding notations:
|(a)||is a semisimple algebraic group relative to|
|(b)||is a maximal connected solvable subgroup (Borel subgroup).|
|(c)||is a maximal connected diagonalizable subgroup (maximal torus).|
|(d)||is the normalizer of and|
|(e)||and are all defined over relative to|
Remark: and are determined by the abstract group
|(a)||is maximal solvable and has no subgroups of finite index.|
|(b)||is maximal nilpotent and every subgroup of finite index is of finite index in its normalizer.|
|Proof of Theorem 6.|
(a) by This is a rational homomorphism. So since is a connected algebraic group so is Hence is algebraic and connected. Let Since is solvable and normal it is finite by the Corollary to Theorem 5. Since is also connected and hence is semisimple.
(b and c) is the image of under and hence is algebraic and connected; so is connected, algebraic, and solvable. Let Then (some simple root so and hence by Corollary 6 of Theorem 4' is not solvable and hence (b) holds. is a maximal connected diagonalizable subgroup of (for any larger subgroup must intersect nontrivially). Hence is a maximal connected diagonalizable subgroup of (by a theorem in Chevalley's Séminaire); so (c) holds.
(d) is clear. To prove (e) it suffices by (iii) to prove:
Lemma 34: Let and Then:
|(a)||is defined over and is an isomorphism over|
|(b)||is defined over and is a homomorphism over|
Let be a basis of formed of weight vectors. Choose so that then write and choose so that If is of weight then is of weight Since it follows that if is the matric coordinate function then All other coefficients of are polynomials over in hence also in This set of polynomial relations defines as a group over Now is an inverse of so the map is an isomorphism over The proof of (b) is left as an excercise.
We can recover the lattices and from the group as follows. Let Define by This is a character defined over generates a lattice the character group of The are determined by as the unique minimal unipotent subgroups normalized by If then where is called a global root. Define the lattice generated by all Then
Exercise: There exists a such that and (The action of on is given by the action of on the character group).
We summarize our results in:
Existence Theorem: Given a root system a lattice with (where and are the root and weight lattices, respectively), and an algebraically closed field then there exists a semisimple algebraic group defined over such that and are realized as the lattices of global roots and characters, respectively, relative to a maximal torus. Furthermore can be taken over the prime field.
The classification theorem, that up to every semisimple algebraic group over has been obtained above, is much more difficult, (See Séminair Chevalley, 1956-8).
We recall that
Lemma 35: Let be algebraically closed, a Chevalley group over a basis for Define by for Then the map given by is an isomorphism over of algebraic groups.
Write Given we can find such that (for and is divisible). Then acts on as multiplication by i.e. as This shows that maps onto Clearly is a rational mapping defined over Let be the basis of dual to (i.e. Write Then so exists and is defined over
Theorem 7: Let be an algebraically closed field and the prime subfield. Let be a Chevalley group parametrized by and viewed as an algebraic group defined over as above. Then:
|(a)||is an open subvariety of defined over|
|(b)||If is the number of positive roots, then the map defined by is an isomorphism of of varieties over|
(a) We consider the natural action of on relative to a basis over made up of products of and such that For we set and then a function on over We claim that if and only if Assume Since fixes up to a nonzero multiple and if then it follows that If with the same considerations show that If makes all positive roots negative then by the equation and Theorem 4' the two cases above are exclusive and exhaustive, whence (a).
The map is composed of the two maps and We will show that these are isomorphisms over For this follows from Lemma 35. Consider Let be a basis for the underlying vector space, made up of weight vectors in the lattice and the corresponding coordinate functions on For each root choose as in the proof of Lemma 34. Set Choosing an ordering of the positive roots consistent with addition^,we see at once that an integral polynomial in the earlier and that is an integral polynomial in the for all Thus is an isomorphism over and similarly for To prove is an isomorphism we order the so that consist respectively of subdiagonal unipotent, diagonal, super-diagonal unipotent matrices (see Lemma 18, Cor. 3), and then we may assume that they consist of all of the invertible matrices of these types. Let be in and let the subdiagonal entries of the diagonal entries of the superdiagonal entries of be labelled with respectively. We order the indices so that precedes in case and Then in the three cases above resp. resp. increased by an integral polynomial in preceeding We may now inductively solve for the as rational forms over in the the division by the forms representing the being justified by the fact that they are nonzero on Thus is an isomorphism over and (b) follows.
Example: In consists of all such that the minors are nonsingular.
Remark: It easily follows that the Lie algebra of is
We can now easily prove the following important fact (but will refer the reader to Séminaire Bourbaki, Exp. 219 instead). Let be a Chevalley group over viewed as above as an algebraic matric group over the prime field, and the corresponding ideal over (consisting of all polynomials over which vanish on Then the set of zeros of in any algebraically closed field is just the Chevalley group over of the same type (same root system and same weight lattice) as Thus we have a functorial definition in terms of equations of all of the semisimple algebraic groups of any given type.
Corollary 1: Let be as above. Let be a Chevalley group constructed using instead of but with the same Assume that Then the homomorphism taking for all and is a homomorphism of algebraic groups over
Consider first By Theorem 7 we need only show that is rational over The nonzero coordinates of are The nonzero coordinates of are Each of the former is a monomial in the latter (because and hence is rational over Now for (resp. can be chosen with coefficients in (for so that is rational over Since we conclude that is rational over
Corollary 2: The homomorphism (of Corollary 6 to Theorem 4') is a homomorphism of algebraic groups over
This is a special case of Corollary 1.
Corollary 3: Assume and are fixed, that is universal, are fields and and are the corresponding Chevalley groups. Then
Clearly Suppose Then (see Theorem 4') with defined over the prime field. We must show that i.e. where Write with Applying of Theorem 7, we get Since is defined over and is defined over all Hence
Remark: Suppose and is a Chevalley group over Then has the structure of a complex Lie group, and all the preceding statements have obvious modifications in the language of Lie groups, all of which are true. For example, all complex semisimple Lie groups are included in the construction, and in Theorem 7 is an isomorphism of complex analytic manifolds.
This is a typed excerpt of Lectures on Chevalley groups by Robert Steinberg, Yale University, 1967. Notes prepared by John Faulkner and Robert Wilson. This work was partially supported by Contract ARO-D-336-8230-31-43033.