A Short Proof of the Schröder-Simpson Theorem. Jean Goubault-Larrecq. Summer Topology Conference July 23-26, 2013

Transcription

1 A Short Proof of the Schröder-Simpson Theorem Jean Goubault-Larrecq Summer Topology Conference July 23-26, 2013

2 Continuous Valuations Continuous Valuations Definition Continuous valuation ν = map ν : O(X ) R + such that: ν( ) = 0 (strictness) U V ν(u) ν(v ) (monotonicity) ν(u V ) + ν(u V ) = ν(u) + ν(v ) (modularity) ν( i I U i) = sup i I ν(u i) (continuity) Similar to measures, except give weights to opens. Needed in semantics of programming languages [JonesPlotkin89] Under assumptions on X, cont. valuation=regular measure [KeimelLawson05].

3 Continuous Valuations The Weak Topology Let R + σ = R + with Scott topology (opens = (t, + ]). Definition Let V(X ) = set of continuous valuations of X, with the weak topology, coarsest making ν x h(x)dν continuous, for every continuous h. Subbasic opens: [h > r] = {ν V(X ) h(x)dν > r} x As the usual weak topology, except test functions h are continuous to R + σ (=lsc).

4 A Short Proof of the Schröder-Simpson Theorem A Riesz Representation Theorem Write C(Y ) = space of continuous maps [Y R + ] (=lsc) Y = subspace of linear continuous maps Theorem (Kirch93, Tix95) The following is a homeomorphism: C(X ) G λu O(X ) G(χ U ) V(X ) λh C(X ) x h(x)dν ν No condition on X at all Easy proof

5 A Short Proof of the Schröder-Simpson Theorem The Schröder-Simpson Representation Theorem Write C(Y ) = space of continuous maps [Y R + ] (=lsc) Y = subspace of linear continous maps Theorem (SchröderSimpson05) The following is an isomorphism of cones: h λν V(X ) x X h(x)dν C(X ) V(X ) No condition on X at all Schröder and Simpson s proof [Simpson09] was elaborate

6 Topological Cones Definition A cone (C, +,.) is as a vector space, except scalar product r.c is with non-negative reals r. Topological cone = cone with +: C C C,.: R + σ C C continuous. Example: V(X ), weak topology Non-example: C(X ), Scott topology (unless X core-compact)... product not jointly continuous Linear maps: ψ( i r iν i ) = i r iψ(ν i ) for r i 0

7 A Short Proof of the Schröder-Simpson Theorem The obvious way to a proof C(X ) h λν V(X ) x X h(x)dν V(X )? Main task: given ψ : V(X ) R + linear continuous, find h C(X ) such that ψ(ν) = x X h(x)dν for every ν If h exists, then ψ(δ x ) = x h(x)dδ x = h(x)

8 A Short Proof of the Schröder-Simpson Theorem The obvious way to a proof C(X ) h λν V(X ) x X h(x)dν V(X )? Main task: given ψ : V(X ) R + linear continuous, find h C(X ) such that ψ(ν) = x X h(x)dν for every ν If h exists, then ψ(δ x ) = x h(x)dδ x = h(x) So only one possible choice: h(x) = ψ(δ x ) Need to show for every ν V(X ) x ψ(δ x )dν = ψ(ν) (1)... really hard.

9 The obvious way to a proof (2) Let s try and show (1): x ψ(δ x )dν = ψ(ν) Obvious if ν = m i=1 a iδ xi (simple) Easy for ν quasi-simple (directed sup of simple valuations), by continuity Problem: not all continuous valuations are quasi-simple.

10 Previous proofs [SchröderSimpson 05], [Simpson 09]: elaborate series of deep results and finely bounding inequalities [Keimel 12]: imitates classical proof of similar, measure-theoretic theorem; develops nice theory of quasi-uniform separation in (quasi-uniform) cones.

11 Previous proofs [SchröderSimpson 05], [Simpson 09]: elaborate series of deep results and finely bounding inequalities [Keimel 12]: imitates classical proof of similar, measure-theoretic theorem; develops nice theory of quasi-uniform separation in (quasi-uniform) cones. Our proof: look at the weak open ψ 1 (1, + ]... must be of the form i I n j=1 [h ij > r ij ]... we can take r ij = 1;... use Lemma 1 to eliminate unions... use Lemma 2 to eliminate intersections... so ψ 1 (1, + ] = [h > 1]: this is the right h.

12 The proof Proof, Lemma 1 A surprising observation (mostly due to [Keimel12]) Lemma 1 (Linear cont. ψ cannot tell between sup and sup) Let h i C(X ), i I. TFAE: (1) ψ(ν) sup i x h i(x)dν for every ν (2) ψ(ν) x sup i h i (x)dν for every ν Note: sup i not directed, sup i x h i(x)dν x sup i h i (x)dν. Proof (1/2). Only 1 2 is non-trivial. Trick 1. Every h i directed sup of step functions so can assume each h i step, = k a ikχ Uik Trick 2. sup i I = sup J finite I sup i J by continuity of, can assume I finite.

13 The proof Proof, Lemma 1 A surprising observation (mostly due to [Keimel12]) Lemma 1 (Linear cont. ψ cannot tell between sup and sup) Let h i C(X ), i I. TFAE: (1) ψ(ν) sup i x h i(x)dν for every ν (2) ψ(ν) x sup i h i (x)dν for every ν Proof (2/2). Assume h i = k a ikχ Uik, i I finite. Let C m be atoms of Boolean algebra of sets generated by U ik On each C m, h i constant = a im so x sup i h i (x)dν Cm = max i a im ν(c m ) = sup i x h i(x)dν Cm C1 C3 C2 C4 C5 U12 By 1, ψ(ν Cm ) x sup i h i (x)dν Cm U21 Since ψ additive and ν = m ν C m, 2 follows. U11

14 The proof Proof, Lemma 1 A surprising observation (mostly due to [Keimel12]) Lemma 1 (Linear cont. ψ cannot tell between sup and sup) Let h i C(X ), i I. TFAE: (1) ψ(ν) sup i x h i(x)dν for every ν (2) ψ(ν) x sup i h i (x)dν for every ν Lemma 1, Corollary ψ 1 (1, + ] = i I [h i > 1] ψ 1 (1, + ] = [h > 1] with h = sup i h i. Proof. obvious. For : [h i > 1] ψ 1 (1, + ] implies ψ(ν) x h i(x)dν for all ν, i So 1 holds, hence 2: ψ(ν) x h(x)dν. If ν [h > 1], ψ(ν) x h(x)dν > 1, so ν ψ 1 (1, + ].

15 A Short Proof of the Schro der-simpson Theorem The proof Proof, Lemma 2 Lemma 2 T If nj=1 [hj > 1] ψ 1 (1, + ], Proof, Lemma 2 T there is h C(X ) such that nj=1 [hj > 1] [h > 1] ψ 1 (1,Lemma + ] 2 A Short Proof of the Schro der-simpson Theorem Proof. R R Let F (ν) = ( x h1 (x)dν,, x hn (x)dν) A Short Proof of the Schro der-simpson Theorem T If nj=1 [hj A Short Proof of the Schro der-simpson Theorem A Short Proof of the Schro der-simpson Representation Theorem Theorems U U21 C1 Lemma C2 C3 C42 Proof, U Representation 11 Theorems 12 in R +n 1 (1, +1], > 1] T there is h 2 X such that nj=1 CProof. 5 Let F ( ) = ( Proof, Lemma 2 = 12 1 > 1 Proof, Lemma R x h1 (x)d, 1 A = F (complement of convex non-empty Lemma 2 U = (1, +1]n open T 1 (1, By separation If 2nj=1 [hj > 1] +1], Ptheorem (e.g Lemma 2 Lemma Tc )n = Tnthere is h 2 X1 such that find (~ i ci linear T P [hij > 1] > +1], 1] (1, +1], j=1 If n [h >If1] [h1j(1, P j=1 Separate by Λ(~c ) = j γj cj linear continuous h TnLet h = T is hthat 2 X such that [h1] [h +1] > 1] j > there is h 2 there X such [h > 1] j=1 [h > 1] (1, Proof. U RU U C CR C C (e.g. [Keimel 2006]) Let F ( ) = ( x h1 (x)d,, x hn (x) Proof. Proof. R R R R P Let F ( ) = (LethFA (x)d, x,hu1 (x)d, h (x)d ) in ( ) 1C of, R h1n1(1, (x)d ) U = C +1]) C ==FU ((complement >C1in R xc Let h = j γj hj. Note 1 (1, +1]) A = F (complement of (1, +1]) A = Fconvex (complement of non-empty R P R convex non-empty non-empty n = 1 1 > 1 U = (1, +1] open h(x)dν = j γj x hj (x)dν = Λ(F (ν)) U = (1, +1]convex x U open = (1, +1]n open By separation (e.g. [Keimel 2 Ptheorem By separation (e.g.theorem [Keimel 2006]),[Keimel 2006]), Inequalities follows. Bytheorem separation P find (~ ccontinuous ) = i (e.g. i ci linear continuous P find (~c ) = c linear j=1 j n j=1 x 1 i i i j x12 n U11 U12 U21 =1 2 +n C 2 C C3 C4 C5 U U12 U21 >1 =1 U n U11 U12 U21 C1 C2 A U11 U12 U21= 1C1 C 2 1 C3 =1 Let h = i i i (~ c) = P i ci linear continuous P find LetP h = i i i hi i i hi Let h = i i hi 1 1 >

16 The proof End of proof Let ψ V(X ), write ψ 1 (1, + ] as i I ni j=1 [h ij > 1]. Lemma 2 (recap) There is h i C(X ) s.t. n i j=1 [h ij > 1] [h i > 1] ψ 1 (1, + ] So ψ 1 (1, + ] = i I [h i > 1]

17 The proof End of proof Let ψ V(X ), write ψ 1 (1, + ] as i I ni j=1 [h ij > 1]. Lemma 2 (recap) There is h i C(X ) s.t. n i j=1 [h ij > 1] [h i > 1] ψ 1 (1, + ] So ψ 1 (1, + ] = i I [h i > 1] Lemma 1, Corollary Then ψ 1 (1, + ] = [h > 1] with h = sup i h i. For all ν, t, ψ(ν) > t So ψ(ν) = x h(x)dν. iff ν/t ψ 1 (1, + ] iff ν/t [h > 1] iff x h(x)dν > t.

18 The proof Recap Given ψ V(X ), there is an h C(X ) such that ψ(ν) = x h(x)dν (1) This h derived from the shape of weak opens i I n j=1 [h ij > r ij ]... and taking ν = δ x in (1) implies h(x) = ψ(δ x ).

19 A Short Proof of the Schröder-Simpson Theorem The proof Recap We have proved: Theorem (SchröderSimpson05) The following is an isomorphism of cones: h λν V(X ) x X h(x)dν C(X ) V(X ) Notes: λx ψ(δ x ) ψ V(X ) = C(X ) (here) + V(X ) = C(X ) (Riesz-Kirch-Tix) V(X ) and C(X ) are dual cones (C(X ) with weak topology) No assumption needed on X. Short proof Questions?