Last update: 4 July 2013
Our object is to prove that if and are as in §6, then is a universal central extension of in a sense to be defined. The reader is referred to the lecturer's paper in Colloque sur la théorie des groupes algébriques, Bruxelles, 1962 and for generalities to Schur's papers in J. Reine Angew. Math. 1904, 1907, 1911.
Definition: A central extension of a group is a couple where is a group, is a homomorphism of onto and center of
|(a)||as in §6.|
|(b)||the natural homomorphism of one Chevalley group onto another constructed from a smaller weight lattice. E.g., and|
|(c)||a topological covering of a connected topological group; i.e., is a local isomorphism, carrying a neighborhood of isomorphically onto one of We note that is central since a discrete normal subgroup of a connected group is necessarily central. To see this, let be a discrete normal subgroup of a connected group and let Since the map given by has a discrete and connected image, for all|
Definition: A central extension of a group is universal if for any central extension of there exists a unique homomorphism such that i.e., the following diagram is commutative:
We abbreviate universal central extension by u.c.e. We develop this property in a sequence of statements.
(i) If a u.c.e. exists, it is unique up to isomorphism.
If and are u.c.e. of let and be such that and Now and Hence is the identity on by the uniqueness of in the definition of a u.c.e. Similarly is the identity on
(ii) If is a u.c.e. of then and hence where is the derived group of
Consider the central extension where and Now if and then and hence Thus, and
(iii) If and is a central extension of then where is a central subgroup of on which is trivial. Moreover,
We have Hence, where Also,
(iv) If then possesses a u.c.e.
For each we introduce a symbol Let be the group generated by subject to the condition that commutes with for all If then by using induction on the length of an expression in we see that extends to a central homomorphism of onto
(a) covers all central extensions of To see this, let be any central extension of Choose such that Since is central, the satisfy the condition on the Hence, there is a homomorphism such that and thus
(b) If and also denotes the restriction of to then covers all central extensions uniquely. By (iii), we have that covers all central extensions. If is a central extension of and if then center of Thus, is a homomorphism of into an Abelian group. Since is trivial and
Remark: Part (a) shows that if is any group then there is a central extension covering all others.
(iv') If is a central extension of which covers all others and if then is a u.c.e.
(v) If and are central extensions, then so is provided
If let be the map Now center of since because center of Now is a homomorphism, so is trivial, and center of
(vi) Exercise: In (v), is a u.c.e. of if and only if is a u.c.e. of
Definition: A group is said to be centrally closed if is a u.c.e. of
(vii) Corollary: If is a u.c.e. of then is centrally closed.
(viii) If is centrally closed, then every central extension of splits; i.e, there exists a homomorphism such that
(ix) is a u.c.e. of if and only if every diagram of the form
can be uniquely completed, where is a central extension of and is a homomorphism.
One direction is immediate by taking and Conversely, suppose the diagram is given and is a u.c.e. Let be the subgroup of If is given by then is central. Since is a u.c.e. of there is a unique homomorphism such that Now the homomorphism where is given by satisfies If then let be given by Since we have and proving the uniqueness of
Definition: A linear (respectively projective) representation of a group is a homomorphism of into some (respectively
Since is a central extension of we have the following result:
(x) Corollary: If is a u.c.e. of then every projective representation of can be lifted uniquely to a linear representation of
(xi) Topological situation: If is a topological group one can replace the condition by is connected, the condition is a u.c.e. by is a universal covering group in the topological sense, and the condition is centrally closed by is simply connected in the above discussion and obtain similar results.
Definition: If is a u.c.e of the group then we call the Schur multiplier of
If we write to indicate the dependence on then a homomorphism leads to a corresponding one by (ix). Thus is a functor from the category of groups such that to the category of Abelian groups with the following property: if is onto, then so is
Remark: Schur used different definitions (and terminology) since he considered only finite groups but did not require that If our definitions are equivalent to his. One of Schur's results, which we shall not use, is that if is finite then so is
Theorem 10: Let be an indecomposable root system and a field such that and, if then If is the corresponding universal Chevalley group (abstractly defined by the relations (A),(B),(B'),(C) of §6), if is the group defined by the relations (A),(B),(B') (we use (B') only if and if is the natural homomorphism from to then is a u.c.e. of
Remark: There are exceptions to the conclusion. E.g. and are such. Indeed and is a central extension of For see Schur. It can be shown that the number of couples for which the conclusion fails is finite.
Since and u.c.e. exist for both and The conclusion becomes is centrally closed, by the above remarks. We need only show that every central extension of splits, i.e., there exists so that i.e., the relations defining can be lifted to
We may assume but then is a central extension of by (v). We need only show
(1) If is a central extension of then the relations (A),(B),(B') can be lifted to
Let a central subgroup of We have:
(2) A commutator with depends only on the classes mod to which and belong.
Choose so that Then in for all We define so that and so that
(3) and then (and later in terms of the by the same formulas which define the and the in terms of the Note that this choice is not circular because of (2). We shall show that the relations (A),(B),(B') hold with in place of
(4) for all
Set with Conjugating (3) by we get and the left side equals by (3). Similarly we have
(4') for all
(5) If and are roots, and is not a root, then and commute for all Set We must show Clearly, from the definitions we have
(6) is additive in both positions.
(5a) Assume If then by Lemma 20(c) and by (4) with If then where by Lemma 20(c) and by (4) with In both cases, by (6) for some or Choose so Then we get
(5b) Assume and If there is a root so that set in the preceding argument and obtain Choose so that and By and (6), whence and If there is no such then is of type and is a long root. In this case, however, with and roots. Thus, with by Lemma 33. Since commutes with all factors but the last by (5a), it also commutes with the last.
(5c) Assume and At least we have using in the argument above. We may also assume that is not a prime. If it were, then and would be powers of and (5) would be immediate. Referring to the proof of (5b), we see it will suffice to be able to choose so that are squares and If is finite of characteristic 2, this is possible since all elements of are squares. Otherwise, set Then and we need only choose so that and Since at most values of are to be avoided and in the present case, this too is possible. T his completes the proof of (5).
(7) preserves the relations (A). The element is in and hence the transform of by is itself. However, by (3), (4), (5) this transform is also
(8) preserves the relations (B). We have where One proves by induction on the number of roots of the form If this is just (5). If the inductive hypothesis and (7) imply satisfies (6), and then the argument in (5a) may be used.
(9) preserves the relations (B'). This follows from (4').
This completes the proof of the theorem.
Exercise: Assume is the original Lie algebra with coefficients transferred by means of a Chevalley basis to a field whose characteristic does not divide any Also assume is indecomposable of Prove:
|(a)||The relations form a defining set for Hint: define and show that the relations of Theorem 1 hold.|
|(b)||the derived algebra of|
|(c)||Every central extension of splits. Hint: parallel the proof of Theorem 10.|
Corollary 1: The relations (A),(B),(B') can be lifted to any central extension of
|(a)||is centrally closed. Each of its central extensions splits. Its Schur multiplier is trivial. It yields the u.c.e. of all the Chevalley groups of the given type, and covers linearly all of the projective representations of these groups.|
|(b)||If is finite or more generally an algebraic extension of a finite field, then (a) holds with replaced by|
This follows from various of the generalities at the beginning of this section.
E.g., if is finite, and is excluded, then and all have trivial Schur multipliers, and the natural central extensions are all universal.
Corollary 3: Assume and are as above. If is infinite and divisible implies there exists with then the Schur multiplier of i.e., is also divisible.
Elements of the form in generate We have by Lemma 39(a). By induction, we get for arbitrary Since for we can find such that the proof is complete.
Corollary 3a: If is infinite and divisible by a set of primes including 2, then is also divisible by these primes.
Corollary 3b: If is infinite and divisible, then any central extension of by a kernel which is a reduced group (no divisible subgroups other than 1) is trivial; i.e. it splits.
Let be a central extension of with reduced. Since is u.c.e. we have so that Since is divisible, so is Hence and Thus, there is a homomorphism so that Therefore, on and on
Corollary 3c: If is infinite and divisible, then any finite dimensional projective representation of can be lifted uniquely to a linear representation.
Assume Since we have Let be the natural projection. Since is finite, we have is finite and thus is reduced. Consider the central extension of where and Now is reduced, so by Corollary 3b, we have with on If then with
Example: Corollary 3c says, for example, that every finite dimensional representation of can be lifted to a linear one. (The novelty is that the representation is not assumed to be continuous.)
Theorem 11: If is an indecomposable root system, if and if (i.e. we exclude of type or and of type then uniquely covers all central extensions of for which the kernel has no
By Theorem 10, we could assume or However, the proof does not use this assumption or Theorem 10. If is a central extension of such that has no then we wish to show (A), (B), and (B') can be lifted to
(1) Assume is divisible by Choose so that and We claim relations (A), (B) and (B') hold on the
(1a) If are roots, not a root, and then and commute, We have with Taking powers, we get which implies since has no
(1b) The relations (A) hold. Taking powers of we get as before.
(1c) Exercise: Relations (B) and (B') also hold.
(2) General case.
(2a) can be embedded in a group which is divisible by and has no We have a homomorphism of a free Abelian group onto Now is a divisible group, and we can identify with Hence say, and is a divisible group. Moreover, since has no where is the of Thus, projects faithfully into which is divisible and has no
(2b) Conclusion of proof. Form the direct product of and with amalgamated, and define by
Now satisfies the assumptions of (1) so the relations (A), (B), (B') can be lifted to However, by Lemma 32', the lifted group is its own derived group and hence contained in
Corollary 1: Every projective representation of in a field of characteristic can be lifted to a linear one.
Corollary 2: The Schur multiplier of is a
These are easy exercises.
Since the kernel of the map above turns out to be the Schur multiplier of its structure for arbitrary is of some interest. The result is:
Theorem 12: (Moore, Matsumoto) Assume is an indecomposable root system and a field with If is the universal Chevalley group based on and if is the group defined by (A), (B), (B'), and if is the natural map from to with the Schur multiplier of then is isomorphic to the abstract group generated by the symbols subject to the relations:
Remark: These relations are satisfied by the norm residue symbol in class field theory, which is a significant aspect of Moore's work.
(1) If is the group generated (in G') by all a fixed long root, then We know that form a generating set for Using the Weyl group we can narrow the situation to at most two roots with long, short, and Hence, and by Lemma 37(f). This shows a will suffice.
(2) is a mapping onto This follows from (1).
(3) is a homomorphism. We must show that the relations hold if is replaced by The relations (a) are obvious. A special case of (e) has been shown in Lemma 39(d). The other relations (b), (c), (d) follow from the commutator relations connecting the and the
(3') Assume is not of type In this case there is a root so that Thus or By relation (c),
(4) is an isomorphism. This is done by constructing an explicit model for
Now let be a connected topological group. A covering of is a couple such that is a connected topological group and is a homomorphism of onto which maps a neighborhood of 1 in homeomorphically onto a neighborhood of 1 in i.e., which is a local isomorphism. A covering is universal if it covers all other covering groups. If is a universal covering, we say that is simply connected.
|(a)||A covering of a connected group is necessarily central as was noted at the beginning of this section.|
|(b)||If a universal covering exists, then it is unique and each of its coverings of other covering groups is unique. This follows from the fact that a connected group is generated by any neighborhood of 1.|
|(c)||If is a Lie group, then a universal covering for exists and simple connectedness is equivalent to the property that every continuous loop can be shrunk to a point (See Chevalley, Lie Groups or Cohn, Lie Groups.)|
Theorem 13: If is a universal Chevalley group over viewed as a Lie group, then is simply connected.
Before proving Theorem 13, we shall first state a lemma whose proof we leave as an exercise.
Lemma 41: If are complex numbers such that and there exist and such that
|Proof of Theorem 13.|
Let be a covering of Locally is invertible so we may set on some neighborhood of 1 in We shall show that can be extended to a homomorphism of onto i.e., covers It suffices to show that can be extended to all of so that the relations (A), (B), (B'), (C) hold on the
Consider the relations
Since is locally an isomorphism, there is such that (A) holds for If then set Using induction and Lemma 41, we see that is well defined. Clearly, (A) then holds for all Alternatively, we could note that is topologically equivalent to and hence simply connected. Thus, extends to a homomorphism of into and (A) holds. Clearly the extension of to is unique.
To obtain the relations (B), let be roots let be the set of roots of the form and let be the corresponding unipotent subgroup of Topologically, is equivalent to for some and is hence simply connected. As before can be extended to a homomorphism of into and the relations (B) hold. This extension is consistent with those above, by the uniqueness of the latter.
We now consider where Hence, if is near 1 in then is near 1 in Thus, is multiplicative near 1 and hence Abelian everywhere (recall that is central). We then have by Lemma 37(f). Since has square roots, we have is multiplicative, i.e., (C) holds.
Examples: and are simply connected. These cases can also be proved by induction on (See Chevalley, Lie Groups, Chapter II.)
(a) If is replaced by in the preceding discussion, then relations (A), (B) can be lifted exactly as before. Also is still multiplicative if one of the two arguments is positive. Further is generated by a fixed long root, and, if type is excluded, then or Prove all of this.
(b) Moore has constructed a universal covering of and has determined the fundamental group in case is a field, using appropriate modifications of the definitions (here is totally disconnected.)
(c) Let be a Chevalley group over the corresponding universal group, and the natural homomorphism. If is algebraically closed and if only appropriate coverings are allowed, then is a universal covering of in the sense of algebraic groups.
We close this section with a result in which the coefficients may come from any ring (associative with 1). The development is based in part on a letter from J. Milnor. Let be the ring, and let be the group of infinite matrices which are equal to the identity everywhere except for a finite invertible block in the upper left hand corner. Thus, Let be the subgroup of generated by the elementary matrices where is the usual matrix unit. For example, if is a field, then a simple group whose double coset decomposition involves the infinite symmetric group. Indeed, if is a Euclidean domain, then
The relation shows (b) and hence also If then because in we have
We call the Whitehead group of This concept is used in topology. The case in which is of particular interest.
Example: If is a Euclidean domain, then the group of units. (See Milnor, Whitehead Torsion.)
By Lemma 42 and (iv), has a u.c.e. Set This notation is partly motived by the following exact sequence.
is a functor from rings to Abelian groups with the following property: if is onto, then so is the associated map
Remark: is known to the lecturer in the following cases:
|(a)||If is a finite field (or an algebraic extension of a finite field), then|
|(b)||If is any field, see Theorem 12.|
Theorem 14: Let be the abstract group generated by the symbols subject to the relations
|(A)||is additive in|
(a) is central. If choose large enough so that is a product of with Let be the subgroup of generated by the Now by (A) and (B), any element of can be expressed as Since in this form is unique, is an isomorphism. Also by (A) and (B), if Thus, If then and since is an isomorphism we have In particular, commutes with all Similarly, commutes with all and hence with all Thus, is in the center of
(b) is universal. From (B), it follows that Hence it suffices to show it covers all central extensions. Let be a central extension of and let be the center of We must show that we can lift the relations (A) and (B) to Fix and choose Choose so that We will prove that the satisfy the equations (A) and (B).
(b1) If then and commute. Choose and write Since commutes up to an element of with and it commutes with Hence
(b2) fixed, is Abelian.
(b3) The relations (A) hold. The proof is exactly the same as that of statement (7) in the proof of Theorem 10.
(b4) in is independent of the choice of If set Transforming by and using (b1) we get with in place of
(b5) The relations (B) hold. We will use:
If, are elements of a group such that commutes with and such that commutes with and then Since a commutes with The other conditions insure Now assume are distinct. Choose so that by and (b4).
This completes the proof of the theorem.
Let denote the subgroup of generated by with
Corollary 1: If then is centrally closed.
Corollary 2: If is a finite field and then is centrally closed.
This follows from Corollary 1 and the equations
Remarks: (a) It follows that if is a finite field and if is not centrally closed, then either or and The exact set of exceptions is:
Exercise: Prove this.
(b) The argument above can be phrased in terms of roots, etc. As such, it carries over very easily to the case in which all roots have one length. The only other exception is
(c) By a more complicated extension of the argument, it can also be shown that the universal Chevalley group of type or over a finite field (or an algebraic extension of a finite field) is centrally closed if is large enough. Hence, only a finite number of universal Chevalley groups with indecomposable and finite fail to be centrally closed.
Now we sketch a proof that is a group of order The notation above will be used. The proof depends on the following result:
(1) For is generated by symbols subject to the relations
Identifying with the usual and using we see that the relations (B) here imply those of Theorem 14. Since the last relation may be written and are the only units of we have defined by the usual relations (A), (B), (C) of §6. Perhaps there are other rings, e.g., the integers, for which this result holds. For the proof of (1) see W. Magnus, Acta Math. 64 (1934), which gives the reference to Nielsen, who proved the key case (it takes some work to cast Nielsen's result into the above form). The case with (B) replaced by (B') is simpler and is proved in an appendix to Kurosh, Theory of Groups.
Now let refer to elements of
(2) If is the kernel of and then is generated by and
As usual, we only require (C) when and Setting amounts to dividing by which thus equals The relation which may be deduced from (B) as in the proof of Lemma 37, then yields
Assume not. There is a natural map This maps onto which (see Remark (a) after the proof of Theorem 13) generates the kernel of Thus is centrally closed, hence simply connected. Since can be contracted to (by the polar decomposition, which will be proved in the next section) which is not simply connected since is a nontrivial covering, we have a contradiction.
It now follows from (2), (3) and Theorem 14 that
By Corollary 1 above the same conclusion holds with replaced by any with
Exercise: Let be translations of the underlying space, with again a field. I.e., is the group of all matrices of the form where is generated by Prove:
If the relation
|(2)||If is finite, (C) may be omitted.|
|(3)||If is large enough, the group defined by (A) and (B) is a u.c.e. for|
|(4)||Other analogues of results for|
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.