Lecture 7 - Zariski Topology and Regular Elements Prof

Home , Zariski topology

18.745 Introduction to Lie Algebras September 30, 2010 Lecture 7 - Zariski Topology and Regular Elements Prof. Victor Kac Scribe: Daniel Ketover

Deﬁnition 7.1. A topological space is a set 푋 with a collection of subsets 퐹 (the closed sets) satisfying the following axioms:

1) 푋 ∈ 퐹 and ∅ ∈ 퐹

2) The union of a ﬁnite collection of closed sets is closed

3) The intersection of arbitrary collection of closed sets is closed

4) (weak separation axiom) For any two points 푥, 푦 ∈ 푋 there exists an 푆 ∈ 퐹 such that 푥 ∈ 푆 but 푦 ∕∈ 푆

A subset 푂 ⊂ 푋 which is the complement in 푋 of a set in 퐹 is called open. The axioms for a topological space can also be phrased in terms of open sets.

Definition 7.2. Let X = 픽푛 where 픽 is a field. We define the Zariski topology on 푋 as follows. A set in 푋 is closed if and only if if is the set of common zeroes of a collection (possibly infinite) of 푛 polynomials 푃훼(푥) on 픽 Exercise 7.1. Prove that the Zariski topology is indeed a topology.

Proof. We check the axioms. For 1), let 푝(푥) = 1, so that 핍(푝) = ∅, so ∅ ∈ 퐹 . If 푝(푥) = 0, then 푛 푛 핍(푝) = 픽 , so 픽 ∈ 퐹 . For 2), ﬁrst take two closed sets 핍(푝훼) with 훼 ∈ 퐼 and 핍(푞훽) with 훽 ∈ 퐽 , and consider the family of polynomials 푓훼훽 = {푝훼푞훽 ∣ 훼 ∈ 퐼, 훽 ∈ 퐽}. Then 핍(푓훼훽) = 핍(푝훼) ∪ 핍(푞훽). Therefore 핍(푝훼) ∪ 핍(푞훽) is closed. This can be iterated for any n-fold union. For 3) closure 푛 under arbitrary intersections is obvious. For 4), suppose 푥 ∕= 푦 ∈ 픽 and 푥 = (푥1, 푥2, ..., 푥푛). Set 푋 = 핍(푝1, 푝2, ..., 푝푛) where 푝푖(푧) = 푧푖 − 푥푖. So 푥 ∈ 푋 but y is not. Example 7.1. If 푋 = 픽 the closed sets in the Zariski topology are precisely the ∅, 픽 and ﬁnite subsets of 픽.

Notation 7.1. Given a collection of polynomials 푆 on 픽푛 we denote by 핍(푆) ∈ 픽푛 the set of common zeroes of all polynomials in 푆. By deﬁnition, 핍(푆) are all closed subsets in 픽푛 in the Zariski topology. If 푆 contains precisely one non-constant polynomial, then 핍(푆) is called a hypersurface.

Proposition 7.1. Suppose the ﬁeld 픽 is inﬁnite and 푛 ≥ 1 then

1) The complement to a hypersurface in 픽푛 is an inﬁnite set. Consequently, the complement of 핍(푆), where 푆 contains a nonzero polynomial, is an inﬁnite set.

2) Every two non-empty Zariski open subsets of 픽푛 have a nonempty intersection

3) If a polynomial 푝(푥) vanishes on a nonempty Zariski open set then it is identically zero.

1 Example 7.2. The condition that 픽 be inﬁnite is crucial. In 3), for instance, the polynomial 2 푝(푥) = 푥 + 푥 on 픽2 vanishes at 1 and 0 but is not the zero polynomial.

Proof. For 1), We induct on 푛. When 푛 = 1, since any nonzero polynomial has at most deg 푃 roots, the complement of this set is infinite (since 픽 is). If 푛 ≥ 1 then any nonzero polynomial 푝(푥1, 푥2, ..., 푥푛) can be written as a polynomial in one variable with coefficients polynomi- 푑 푑−1 als in the other variables. So we can write 푝(푥) = 푝푑(푥2, 푥3, ..., 푥푛)푥1 + 푝푑−1(푥2, 푥3, ...푥푛)푥1 + ... + 푝0(푥2, 푥3, ..., 푥푛) where 푝푑(푥2, 푥3, ...푥푛) is a nonzero polynomial. By the inductive hypothesis, 표 표 표 표 표 표 we can find 푥2, 푥3, ..., 푥푛 ∈ 픽 such that 푝푑(푥2, 푥3, ..., 푥푛) ∕= 0 Now fixing these values, 푝(푥) = 표 표 표 푑 표 표 표 푑−1 표 표 표 푝푑(푥2, 푥3, ..., 푥푛)푥1 + 푝푑−1(푥2, 푥3, ...푥푛)푥1 + ... + 푝0(푥2, 푥3, ..., 푥푛) we are back in the 푛 = 1 표 표 표 표 표 푑 case. We can find an infinite number of values 푥1 such that 푝(푥) = 푝푑(푥2, 푥3, ..., 푥푛)(푥1) + 표 표 표 표 푑−1 표 표 표 푝푑−1(푥2, 푥3, ...푥푛)(푥1) + ... + 푝0(푥2, 푥3, ..., 푥푛) ∕= 0. Thus we have found an infinite number of points in 픽푛 where 푝 does not vanish. To prove the second claim of 1), just observe if 푆 a collection of polynomials containing the nonzero polynomial 푝(푥), we have 핍(푆)퐶 ⊃ 핍(푝)퐶 so by the first claim, we have that 핍(푆)퐶 contains an infinite set. 2) Consider 핍(푆1) and 핍(푆2) where 푆1 and 푆2 are nonzero sets of polynomials. It suffices to prove 퐶 퐶 that 핍(푝1) ∩ 핍(푝2) is nonempty for any fixed 푝1 ∈ 푆1 and 푝2 ∈ 푆2 nonzero polynomials. But 퐶 퐶 퐶 observe that 핍(푝1푝2) = 핍(푝1) ∪ 핍(푝2), so 핍(푝1푝2) = 핍(푝1) ∩ 핍(푝2) but by 1) we know that the term on the left is infinite, hence nonempty. 3) If 푝 vanishes on 핍(푆)퐶 for S containing a nonzero polynomial 푞 then 푝 vanishes on 핍(푞)퐶 . If 푝 were nonzero, then by 2), 핍(푞)퐶 ∩ 핍(푝)퐶 is nonempty. This is a contradiction, so 푝 is identically zero.

Let 픤 be a ﬁnite dimensional Lie algebra of dimension 푑 and consider the characteristic polynomial 푑 푑−1 of an endomorphism ad 푎 for some 푎 ∈ 픤: det픤(ad 푎 − 휆) = (−휆) + 푐푑−1(−휆) + ... + det(ad 푎).

This is a polynomial of degree 푑. Because (ad 푎)(푎) = [푎, 푎] = 0, we know det(ad 푎) = 0 and the constant term in the polynomial vanishes.

Exercise 7.2. Show that 푐푗 is a homogeneous polynomial on 픤 of degree 푑 − 푗.

Proof. By ”polynomial∑ in 픤” we mean that if we ﬁx a basis 푋1, 푋2, ..., 푋푑 of 픤 and write a general element 푎 = 푖 푎푖푋푖, then 푐푘(푎) is a polynomial in the 푎푖. Observe that ad 푎 is a 푑 by 푑 matrix whose entries 푏푖푗 are linear combinations of the 푎푖. Now consider det픤(ad 푎 − 휆). This is a polynomial in the entries of the matrix ad 푎 − 휆. A general element in the expansion of this 푘 determinant is 푏∗∗푏∗∗...(푏푖푖 − 휆)...(푏푗푗 − 휆). In order to get a term with 휆 we have to chose 푘 휆’s and 푑 − 푘 of the 푏∗∗ in the expansion of this product. Therefore the coeﬃcient of the polynomial 푘 휆 is homogeneous of degree 푑 − 푘 in the 푏∗∗, so it is also homogeneous of degree 푑 − 푘 in the 푎푖 since the 푏푖푗 are linear in the 푎푖.

Deﬁnition 7.3. The smallest positive integer such that 푐푟(푎) is not the zero polynomial on 픤 is called the rank of 픤. Note 1 ≤ 푟 ≤ 푛. An element 푎 ∈ 픤 is called regular if 푐푟(푎) ∕= 0. The nonzero polynomial 푐푟(푎) of degree 푑 − 푟 is called the discriminant of 픤. Proposition 7.2. Let 픤 be a Lie algebra of dimension 푑, rank 푟 over 픽

1) 푟 = 푑 if and only if 픤 is nilpotent.

2 2) If 픤 is nilpotent, the set of regular elements is 픤.

3) If 픤 is not nilpotent, the set of regular elements is inﬁnite if 픽 is.

Proof. 1) 푟 = 푑 means that det(ad 푎 − 휆) = (−휆)푑 for all 푎. This is true if and only if all the eigenvalues of ad 푎 are 0, which happens if and only if ad 푎 is nilpotent for all 푎. By Engel’s theorem, this is equivalent to 픤 being nilpotent. 푑 2) If 픤 is nilpotent, we know det(ad 푎 − 휆) = (−휆) and 푐푟 = 푐푑 = 1, which does not vanish anywhere, so all elements are regular. 퐶 3) If 픤 is not nilpotent, the set of regular elements is by definition 핍(푐푟) , which is the complement of a hypersurface, defined by a homogeneous polynomial of degree 푑 − 푟 > 0 by Ex 7.2. By part 1 of the previous theorem the complement of a hypersurface is an infinite set if 픽 is.

Example 7.3. For 픤 = 픤픩푛(픽), the rank is 푛 and we will ﬁnd its discriminant. Also, the set of regular elements consists of all matrices with distinct eigenvalues.

Exercise 7.3. 1) Show that the Jordan decomposition of ad 푎 is given by ad 푎 = (ad 푎푠)+(ad 푎푛) in 픤픩푛 2) If 휆1, ..., 휆푛 are eigenvalues of 푎푠 then 휆푖 − 휆푗 are eigenvalues of ad 푎푠 3) ad 푎푠 has the same eigenvalues as ad 푎.

Proof. For 2), If 휆1, ..., 휆푛 are eigenvalues of 푎푠 observe that if 퐸푖푗 is the matrix with 1 in the 푖푗 slot and zero elsewhere, then [푎푠, 퐸푖푗] = (푎푠푖푖 − 푎푠푗푗)퐸푖푗 so ad 푎푠 is diagonalizable since the 퐸푖푗 form a basis of 픤픩푛(픽). The eigenvalues are thus 휆푖 − 휆푗. As for 1), we already showed that ad 푎푠 is diagonalizable, also ad 푎푛 is nilpotent because 푎푛 is. Finally, ad 푎푠 and ad 푎푛 commute: [ad 푎푠, ad 푎푛]푣 = [푎푠, [푎푛, 푣]] − [푎푛, [푎푠, 푣]] = −[[푎푠, 푎푛], 푣] = 0 (since [푎푠, 푎푛] = 0). Thus by uniqueness of Jordan form, ad 푎 = ad 푎푠 + ad 푎푛. As for 3), we know from Jordan canonical form that the semisimple part of an operator and the operator itself have the same eigenvalues – and since ad 푎푠 is the semisimple part of the operator ad 푎, the claim follows.

Exercise 7.4. Deduce the statement∏ of example 1.3 by showing that the rank of 픤픩푛 is 푛 and that the discriminant is given by 푐푛(푎) = 푖∕=푗(휆푖 − 휆푗). Also compute 푐2(푎) for 픤픩2(픽) in terms of the matrix coeﬃcients of 푎. ∏ Proof. We must compute det(ad 푎 − 휆). This is equal to det(ad 푎 − 휆) = (휆 − 휆 − 휆). Note ∏ 푠 ∏ 푖,푗 푖 푗 when 푖 = 푗, we can pull out the (−휆), so we have (휆 −휆 −휆) = (−휆)푛 (휆 −휆 −휆). Thus 푖,푗 푖 푗 ∏ 푖∕=푗 푖 푗 푛 we see that the rank of 픤픩푛(픽 ) = 푛, and the discriminant is 푖∕=푗(휆푖 − 휆푗). An element in 픤픩푛 is therefore regular∏ if and only if all its eigenvalues are distinct. Finally we compute the discriminant 푐 for 픤픩 (픽): (휆 − 휆 ) = (휆 − 휆 )(휆 − 휆 ) = −(휆 + 휆 )2 + 4휆 휆 . If we write a matrix (2 ) 2 푖∕=푗 푖 푗 1 2 2 1 1 2 1 2 푎 푏 , then the discriminant is −(푎 + 푑)2 + 4(푎푑 − 푏푐). This a clearly a homogeneous polynomial 푐 푑 of degree 푛2 − 푛 = 22 − 2 = 2.