THE BOREL-DE SIEBENTHAL’S THEOREM

THE BOREL-DE SIEBENTHAL’S THEOREM
P. GILLE, NOVEMBER 2010
This is the following.
0.1. Theorem. Let F be a field of characteristic 6= 2, 3. Let G/F be a
reductive group and let H/F be a reductive subgroup of maximal rank. Then
H = ZG(Z(H))0.
The original theorem [2] is about compact Lie groups and the specialists
know for a long time that is generalizes as stated. Our goal is to present
here a proof of that result. The following is well known in the case of subtori
[3, 15.3.2].
0.2. Lemma. Let G/F be a reductive group and let T/F be a maximal torus.
Let S 1/2 T be a subgroup.
(1) The F–group ZG(S) is smooth and ZG(S)0 is reductive.
(2) If T is split, let U® be the root groups associated to ©(G, T). Then
ZG(S)0 is generated by T and the root subgroups U® for the ® 2
©(G, T) 1/2 T°Ë mapping to 0 in S°Ë.
Proof. (1) Since S is of multiplicative type, the group ZG(S) is smooth [4,
XI.5.2]. We show that G/k is reductive at the end of the proof.
(2) We can assume that F is algebraically closed. The F-group ZG(S)0
is smooth so is a linear algebraic group. The group ZG(S)0 is generated
by T and root subgroups U® of G [1, 13.20]. for ® running over R :=
©(G, T) ker(T°Ë ! S°Ë). Similarly, the radical M of ZG(S)0 is normalized
by T, hence it is generated by (M T)0 and the root subgroups U® of M.
We claim that M = (M T)0. For sake of contradiction, assume that U®
is a root subgroup of M. Its conjugate U−® in ZG(S)0 is then as well a
root subgroup of ZG(S)0, hence M contain a semisimple group of rank one,
which contradicts the solvability of M. Thus M is a torus and we conclude
that ZG(S)0 is reductive. °Ë
We first look at the behaviour of Theorem 0.1 under central extensions.
0.3. Lemma. Under the hypothesis of the theorem, let S be a central subgroup
(of multiplicative type) of G and denote by f : G ! G/S the quotient
map.
(1) S 1/2 H and H/S is a reductive subgroup of maximal rank of G/S.
(2) If H/S = ZG/S(Z(H/S))0, then H = ZG(Z(H))0.
Proof. (1) follows from the fact that the center of G is included in all maximal
tori of G.
1
2 P. GILLE, NOVEMBER 2010
(2) We have an exact sequence of algebraic groups
1 ! S ! ZG(Z(H)) ! ZG/S(Z(H/S))
Since S 1/2 H 1/2 ZG(Z(H))0, we have the following exact diagram
1 −−−−! S −−−−! ZG(Z(H))0 −−−−! ZG/S(Z(H/S))0
|| [ [
1 −−−−! S −−−−! H −−−−! H/S −−−−! 1.
If H/S = ZG/S(Z(H/S))0, it follows by diagram chase that H = ZG(Z(H))0.
°Ë
We can now proceed to the proof of Theorem 0.1.
Proof. Reduction to the case H semisimple: Let S = Z(H)0 be the connected
center of H Then H 1/2 ZG(S) and H/S is a semisimple subgroup
of G/S. If the result is known in the semisimple case, we have H/S =
ZZG(S)/S(Z(H/S))0. Lemma 0.3 shows that H = ZZG(S)(Z(H))0, hence
H = ZG(Z(H))0.
Furthermore Lemma 0.3 applied to Z(G) permits to assume that G is
semisimple adjoint. We can assume moreover that k is algebraically closed.
We consider a maximal (split) torus T of H.
Case H maximal proper semisimple group of G: We choose compatible orderings
on the roots systems ©(H, T) ( ©(G, T) = T°Ë. Since F is of characteristic
6= 2, 6= 3, ©(H, T) is a closed subsystem of ©(G, T) [4, XXIII.6.6].
Let A be the root lattice of H, i.e. the sublattice of T°Ë generated by ©(H, T).
The center Z(H) of H is a diagonalisable group whose character group is
Z(H)°Ë = T°Ë/A.
To show that H = ZG(Z(H))0, we note first that ZG(Z(H))0 is reductive
by Lemma 0.2. Since ZG(Z(H))0 contains the semisimple group H,
ZG(Z(H))0 is necessarily semisimple. We claim that Z(H) 6= 1. For sake of
contradiction, assume that A = T°Ë, i.e. that H is adjoint. Then our basis
for ©(H, T) would be a basis for ©(G, T), contradiction. So Z(H) 6= 1 and
ZG(Z(H))0 is a proper subgroup of G. Since H is a maximal semisimple
subgroup, we conclude that H = ZG(Z(H))0.
General case: By dimension reasons, there is a chain of semisimple groups
H = H0 ( H1 · · · ( Hn−1 ( Hn = G
such that Hi is maximal in Hi+1 for i = 0, …, n − 1. By induction on n, we
can assume that
THE BOREL-DE SIEBENTHAL’S THEOREM 3
H = ZHn−1(Z(H))0
=
3
ZG(Z(H)) Hn−1
´0
=
3
ZG(Z(H)) ZG
¡
Z(Hn−1)
¢´0
[maximal case]
= ZG(Z(H))0 [ Z(Hn−1) 1/2 Z(H) ].
Thus H = ZG(Z(H))0 as desired. °Ë
Acknowledgements. We thank Skip Garibaldi, Simon P´epin-Lehalleur
and Gopal Prasad for their useful comments.
References
[1] A. Borel, Linear Algebraic Groups (Second enlarged edition), Graduate text in Mathematics
126 (1991), Springer.
[2] A. Borel et J. de Siebenthal, Les sous-groupes fermées de rang maximum des groupes
de Lie clos, Commentarii Math. Helv. 22 (1949), 200-221.
[3] T.A. Springer, Linear Algebraic Groups, second edition (1998), Birk¨auser.
[4] Séminaire de G´eom´etrie alg´ebrique de l’I.H.E.S., 1963-1964, sch´emas en groupes,
dirig´e par M. Demazure et A. Grothendieck, Lecture Notes in Math. 151-153. Springer
(1970).
UMR 8552 du CNRS, DMA, Ecole Normale Sup´erieure, F-75005 Paris, France
E-mail address

Philippe.Gille@ens.fr

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