A Course on the Yosida TheoremClassical & Pointfree Versions & Applications

James J. Madden, Louisiana State University

Summer 2020

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Two Postscripts to Lecture 1

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PS. Representations of abelian `-groups

Let A be an abelian `-group. In the last lecture, we proved:

Fact. For every a ∈ A, if a 6= 0, then there is an `-prime `-ideal ofA that does not contain a (e.g., any `-ideal maximal missing a).

Let X (A) denote the set of all prime `-ideals of A.

For each a ∈ A and each p ∈ X (A), let ã(p) := a + p, the residueof a in A/p. This creates a function:

a 7→ ã : A→∏{A/p | p ∈ X (A) }.

a 7→ ã evidently preserves the operations 0,−,+,∨. Because ofthe fact above, it is an injection. Thus . . .

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PS. Representations of abelian `-groups

Theorem. Every abelian `-group is a sub-`-group of a product oftotally-ordered groups.

Remark. ã is a function from X (A) to⋃{A/p | p ∈ X (A) } satisfying

ã(p) ∈ A/p for all p ∈ X (A). The theorem tells us that every abelian `-group isan `-group of functions. This is the spirit of all the representation theorems that

we shall examine. In the Yosida theorem and its variants, the functions take

values in R or R ∪ {±∞}, rather than in a union of totally-ordered `-groups.

Aside. Using ã(p) to denote the coset a + p ∈ A/p is motivated by real algebraic geometry.

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PPS. Positive Cones

Definition. The positive cone of A, denoted A+, is { a ∈ A | 0 ≤ a }.

Note that a+ = a ∨ 0 and a− = (−a) ∨ 0 are both in A+. Sincea = a+ − a−, every element of A can be written as the differencebetween two elements of A+. In particular, if K is sub-`-group of A, then

K = K+ − K+.

Exercise 1.10. Let A and B be abelian `-groups and let S be a subset of A+.Prove the following:

(i) Suppose S ⊆ A+, 0 ∈ S , and S is closed under + and ∨. ThenS − S := { a− b | a, b ∈ S } is a sub-`-group of A.

(ii) Suppose S ⊆ A+, S is a down-set, and S is closed under + and ∨. ThenS − S is an `-ideal of A.

(iii) If φ : A+ → B+ preserves 0, + and ∨, then φ has a unique extension toan `-homomorphism from A to B.

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Lecture 2. The Yosida Space & Archimedean `-Groups

Tuesday, June 16, 2020

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Values and their covers

Suppose A is an abelian `-group and a ∈ A. An `-ideal of A that ismaximal among those not containing a is called a value of a. Theset of all values of a is denoted Val(A, a).

Definition. Let M be a value of a. The cover of M, denoted M∗,is the intersection of all `-ideals of A that properly contain M.

Facts: If M is a value of a:

I a ∈ M∗ \MI For any b ∈ M∗ \M, 〈M, b〉 = M∗.I For any b ∈ M∗ \M, there is n ∈ N and k ∈ M such that|b| ≤ k + n |a|.

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Y (A, a) := Val(A, a) with the spectral topology

Definition. For b ∈ A, coza b := {M ∈ Val(A, a) | b 6∈ M }.

Note that if 0 ≤ b1, b2, then coza b1 ∩ coza b2 = coza(b1 ∧ b2),since:

∀M ∈ Val(A, a) b1 6∈ M & b2 6∈ M ⇔ b1 ∧ b2 6∈ M.

Definition. Y (A, a) denotes Val(A, a) with the weakest topologyin which coza b is open for all b ∈ A.

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Properties of Y (A, a)

Proposition. Y (A, a) is compact and Hausdorff.

Proof. Compact. For any B ⊆ A,⋃{ coza b | b ∈ B } = {M ∈ Y (A, a) | B 6⊆ M }.

Thus, if { coza b | b ∈ B } covers Y (A, a), then B is not contained in any value of a.This implies that 〈B〉 contains a — for otherwise, we could extend 〈B〉 to a value of a.Hence a ∈ 〈B′〉 for some finite subset B′ of B, and { coza b | b ∈ B′ } covers Y (A, a).

Hausdorff. Suppose M1,M2 ∈ Y (A, a) are different. Pick 0 ≤ b1 ∈ M1 \M2 and0 ≤ b2 ∈ M2 \M1, and let u1 = b1 − (b1 ∧ b2) and u2 = b2 − (b1 ∧ b2). Sinceb1 ∧ b2 ∈ M1 ∩M2, u1 ∈ M1 \M2 and u2 ∈ M2 \M1, so M1 ∈ coza u2 andM2 ∈ coza u1. Moreover, u1 ∧ u2 = 0, so coza u1 ∩ coza u2 = ∅.

Lemma. If X ⊆ Y (A, a), then the closure of X is{M |

⋂X ⊆ M }.

Proof. coza b ∩ X = ∅ ⇔[∀M ∈ X , b ∈ M

]⇔ b ∈

⋂X . Therefore,

Y (A, a) \ cl X =⋃{ coza b | b ∈

⋂X } = {M | ∃b ∈

⋂X , b 6∈ M }.

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Archimedean `-groups

Definition. An `-group A is said to be archimedean if: for everya, b ∈ A+ \ {0}, there is n ∈ N, such that n a 6≤ b.

Fact. ([BKW] 11.1.3) An archimedean lattice-ordered group is abelian.

Examples.

I R = 〈R, 0,−,+,∨〉 is archimedean.I If M is a value, M∗/M is totally-ordered and archimedean.

I A sub-`-group of an archimedean `-group is archimedean.

I A product of archimedean `-groups is archimedean.

I A quotient of an archimedean `-groups need not bearchimedean. (E.g., let A be the `-group of all piecewise linearfunctions on [0, 1] ⊆ R and let P be the `-ideal consisting ofall functions that vanish on [0, �) for some � > 0.)

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Hölder’s Theorem

Theorem. Suppose A is a totally-ordered archimedean `-groupand a ∈ A+ \ {0}. Then there is unique `-homomorphism from Ato 〈R,+〉 that sends a to 1, and it is injective.Proof Sketch. For each b ∈ A+, let

La(b) :={ m

n∈ Q≥0 | ma ≤ n b

}ηa(b) := sup La(b) ∈ R.

The following exercises complete the proof.

Exercise 2.1. La(b) is a down-set in Q≥0 containing 0, and its complement is a

non-empty up-set.

Exercise 2.2. The function ηa : A+ → R preserves 0, + and ∨, and therefore, by1.10(iii), it extends to an `-homomorphism from A to R (also called ηa).

Exercise 2.3. ηa is injective.

Exercise 2.4. If ψ : A→ R is an `-homomorphism with ψ(a) = 1, then ψ(b) = ηa(b).

Remark. η is lowercase eta. Uppercase eta is H, reminding one of Hölder.

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Definition of b̂a : Y (A, a)→ R ∪ {±∞} via Hölder

Suppose A is an abelian `-group and 0 6= a ∈ A+. By Hölder’sTheorem, for each M ∈ Y (A, e), there is a unique`-homomorphism

ηa+M : M∗/M → R, with ηa+M(a + M) = 1.

Given b ∈ A, we define b̂a : Y (A, a)→ R ∪ {±∞} as follows:

b̂a(M) :=

ηa+M(b + M), if b ∈ M∗;+∞, if b 6∈ M∗ and 0 ≤ b + M;−∞, if b 6∈ M∗ and b + M ≤ 0.

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How we get b̂a by mowing

Recall that b̃(p) = b + p, when p is an `-prime `-ideal of A.

Let b̃a denote the restriction of b̃ to Y (A, a) ⊆ X (A).

Then, b̂a is what we get when we “mow” b̃, by identifying all theelements of A/M not in M∗/M with either +∞ or −∞.

Media credit: http://philadelphia.cbslocal.com/2018/06/15/rodney-smith-jr-states-mowing-lawns/

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Continuity of b̂a

Lemma. For all b ∈ A, b̂a : Y (A, a)→ R ∪ {±∞} is continuous.

Proof. The sets of the form (p,+∞] or [−∞, q), p, q ∈ Q, form asubbase for the topology of R ∪ {±∞}. Therefore, it suffices toshow that the sets {M ∈ Y (A, a) | m/n < b̂a(M) } and{M ∈ Y (A, a) | b̂a(M) < m/n } are open, whenever m ∈ Z andn ∈ N \ {0}. Now,

m/n < b̂a(M) ⇐⇒ m âa(M) < n b̂a(M)

⇐⇒ 0 < n b̂a(M)−m âa(M)⇐⇒ 0 < (n b −ma) + M⇐⇒ M ∈ coza

((n b −ma) ∨ 0

).

Thus the first of the sets is open. A similar argument shows thatthe second set is equal to coza

((−n b + ma) ∨ 0

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The finite set of b̂a

Definition. For a, b ∈ A, we define

fina b := {M ∈ Y (A, e) | b ∈ M∗ }.

Comment. fina b = (b̂a)−1(R), and thus fina b is open in Y (A, e).

In the next lecture, we will show that if A is archimedean, thenfina b is dense in Y (A, e).

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