By Biasi C., de Mattos D.

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1) Lemma. If Λ is a discrete subgroup of G, then there is a Borel fundamental domain for Λ in G. That is, there is a Borel subset F of G, such that the natural map F → Λ\G defined by g Λg is bijective. Proof. Because Λ is discrete, there is a nonempty, open subset U of G, such that (U U −1 ) ∩ Λ = {e}. Since G is second countable (or, if you prefer, since G is σ -compact), there is a sequence {gn } of elements of G, such that ∪∞ n=1 U gn = G. Let ∞ Ugn F= n=1 ΛUgi . i

4, pp. 519–520]. 2, pp. 447–451] proves that all of the classical simple groups except SO(p, q) are connected. 2, p. 521]. 40 CHAPTER 3. INTRODUCTION TO SEMISIMPLE LIE GROUPS References [Bor4] A. , Springer, New York, 1991. [Car] É. Cartan: Les groupes réels simples finis et continus, Ann. Sci. École Norm. Sup. 31 (1914) 263–355. [Hel2] S. Helgason: Differential Geometry, Lie Groups, and Symmetric Spaces. Academic Press, New York, 1978. ) [Hum1] J. E. Humphreys: Introduction to Lie Algebras and Representation Theory.

X , ] implies the descending chain condition on Zariski-closed subsets, so A can be written as a finite union of irreducibles. ] #7. Let H be a connected subgroup of SL( , R). Show that if H ⊂ A1 ∪ A2 , where A1 and A2 are Zariski-closed subsets of SL( , R), then either H ⊂ A1 or H ⊂ A2 . [Hint: The Zariski closure H = B1 ∪ · · · ∪ Br is an irredundant union of irreducible, Zariski-closed subsets (see Exercise 6). For h ∈ H, we have H = hB1 ∪ · · · ∪ hBr , so uniqueness implies that h acts as a permutation of {Bj }.