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Finite Dimensional Riesz Spaces and Their Automorphisms R.M. Jongen Finite dimensional Riesz spaces and their automorphisms Bachelor thesis, August 2012 Supervisor: Dr. M.F.E. de Jeu Mathematisch Instituut Universiteit Leiden Contents Introduction 3 1 Lattices and Riesz spaces 4 2 Lattice homomorphisms 12 3 Structure of finite dimensional Riesz spaces 18 4 Automorphism groups of finite dimensional Riesz spaces 21 References 25 2 Introduction In this thesis finite dimensional Riesz spaces will be studied. Riesz spaces are real vector spaces equipped with a partial order. Under this partial order the Riesz space must satisfy some axioms, including the axiom that it is a lattice. In the first chapter the concept of a lattice and a Riesz space will be introduced and some examples will be given. Also some general definitions and properties will be given. Furthermore two partial orders on Rn, the pointwise and the lexicographical order, will be introduced. In the second chapter homomorphisms between Riesz spaces will be discussed. In chapter three we will use these homomorphisms to understand the structure of finite dimensional Riesz spaces. Here we will prove that each finite dimensional Ln Riesz space E can be built up as the direct sum j=1 Ij of a finite number of Riesz spaces, where every Ij can be constructed out of a lower dimensional Riesz space, using a variant of the lexicographical order. In the last chapter we will use these results to find the automorphism groups of finite dimensional Riesz spaces. 3 1 Lattices and Riesz spaces Before studying Riesz spaces, we need to know the definition of a lattice. Definition 1.1. A partially ordered set (X; ≤) is called a lattice if for every x; y 2 X, the set fx; yg has an infimum and a supremum in X. To make the notation simpler, we will write x _ y for supfx; yg and x ^ y for inffx; yg. Example 1.2. The space (R2; ≤) equipped with the coordinatewise order- ing (i.e. (x1; x2) ≤ (y1; y2) if and only if x1 ≤ y1 and x2 ≤ y2) is a lat- 2 tice. For each (x1; x2); (y1; y2) 2 R the infimum and supremum are given by (x1; x2) ^ (y1; y2) = (minfx1; y1g; minfx2; y2g) and (x1; x2) _ (y1; y2) = (maxfx1; y1g; maxfx2; y2g). 2 2 2 However the closed unit disc U = f(x1; x2) 2 R : x1 + x2 ≤ 1g equipped with the same coordinatewise ordering is not a lattice. To see this, consider (1; 0); (0; 1) 2 U. We see that every upper bound of f(1; 0); (0; 1)g must be greater than or equal to 1 in both its coordinates. Therefore U does not contain an upper bound of f(1; 0); (0; 1)g and thus it does not contain its supremum. Definition 1.3. A lattice (X; ≤) is called distributive if x ^ (y _ z) = (x ^ y) _ (x ^ z) for all x; y; z 2 X. Lemma 1.4. A lattice (X; ≤) is distributive if and only if x _ (y ^ z) = (x _ y) ^ (x _ z) for all x; y; z 2 X. Proof. Suppose that (X; ≤) is distributive. Let a := x_(y^z), b := (x_y)^(x_z) and w := x _ y. We have to prove that a = b. Now we find: b = w^(x_z) = (w^x)_(w^z) = ((x_y)^x)_((x_y)^z) = x_(x^y)_(x^z)_(y^z): Notice that x ^ y ≤ x and x ^ z ≤ x, so x _ (x ^ y) _ (x ^ z) = x. So we have indeed a = x _ (y ^ z) = b: Now suppose that x _ (y ^ z) = (x _ y) ^ (x _ z) for all x; y; z 2 X. Let a := x ^ (y _ z), b := (x ^ y) _ (x ^ z) and w := x ^ z. We have to prove that a = b. We find: 4 b = (x^y)_w = (x_w)^(y_w) = (x_(x^z))^(y_(x^z)) = x^(x_z)^(x_y)^(y_z): Notice that x _ z ≥ x and x _ y ≥ x, so x ^ (x _ z) ^ (x _ y) = x. So we can conclude a = x ^ (y _ z) = b: The idea of lattices can be extended to partially ordered real vector spaces. Definition 1.5. Let (E; ≤) be a partially ordered real vector space. E is called an ordered vector space if for all f; g; h 2 E and for all λ 2 R≥0 the following properties hold: 1. f ≤ g ) f + h ≤ g + h; 2. f ≥ 0 ) λf ≥ 0; If E is also a lattice with respect to its partial order, E is called a Riesz space. We will write E+ for the positive cone of a Riesz space E, which is defined to be the collection ff 2 E : f ≥ 0g. Now let us look at some examples. Example 1.6. It is clear that the properties of an ordered vector space hold n n coordinatewise in R , for n 2 Z>0. Therefore R with coordinatewise order is an ordered vectorspace. Moreover it is a Riesz space, since the infimum of two elements is equal to the pointwise minimum and the supremum to the pointwise maximum. Besides this coordinatewise ordering, we can also look at the lexicographical 2 order. If we take R , this order is defined in the following way: (x1; x2) ≤ (y1; y2) if and only if x1 < y1, or x1 = y1 and x2 ≤ y2. In the same way we can order n n R for any n 2 Z>0. Together with this order R becomes a totally ordered Riesz space. Example 1.7. The vector space (C[0; 1]; ≤) of all real continuous functions on the unit interval together with the pointwise order (i.e. f ≤ g if and only if 8x 2 [0; 1] : f(x) ≤ g(x)) is a Riesz space. For f; g 2 C[0; 1] the supremum f _ g is given by the function h : [0; 1] ! R, x 7! maxff(x); g(x)g. The infimum f ^g is given in the same way. Lemma 1.8. Let (E; ≤) be a Riesz space and let f; g; h 2 E and λ 2 R≥0. Then the following equalities are valid: 1. (f _ g) + h = (f + h) _ (g + h); 2. (f ^ g) + h = (f + h) ^ (g + h); 5 3. (λf _ λg) = λ(f _ g); 4. (λf ^ λg) = λ(f ^ g); 5. (f _ g) = −((−f) ^ (−g)); 6. (f ^ g) = −((−f) _ (−g)). Proof. 1. From f ≤ f _ g and g ≤ f _ g it follows that f + h ≤ (f _ g) + h and g + h ≤ (f _ g) + h. Therefore (f _ g) + h is an upper bound of f(f + h); (g + h)g. Now suppose that u is also an upper bound. Then it follows that f ≤ u−h and g ≤ u−h, thus (f _g) ≤ u−h and (f _g)+h ≤ u. 2. This proof is analogous to the proof of 1. 3. From f ≤ f_g and g ≤ f_g it follows that λf ≤ λ(f_g) and λg ≤ λ(f_g). Therefore λ(f _ g) is an upper bound of fλf; λgg. Now suppose that u is also an upper bound. Then it follows that f ≤ λ−1u and g ≤ λ−1u, thus f _ g ≤ λ−1u and λ(f _ g) ≤ u. 4. This proof is analogous to the proof of 3. 5. From f ≤ f _ g and g ≤ f _ g it follows that −f ≥ −(f _ g) and −g ≥ −(f _ g). Therefore −(f _ g) is a lower bound of {−f; −gg. Now suppose that u is also a lower bound. Then it follows that f ≤ −u and g ≤ −u, thus (f _ g) ≤ −u and −(f _ g) ≥ u. 6. This proof is analogous to the proof of 5. Definition 1.9. Let (E; ≤) be a Riesz space. For any f 2 E we will use the following notation: 1. f + := f _ 0; 2. f − := (−f) _ 0; 3. jfj := f _ (−f). Definition 1.10. Let (E; ≤) be a Riesz space and let x; y 2 E. The order interval [x; y] is the set fz 2 E : x ≤ z ≤ yg: Lemma 1.11. Let (E; ≤) be a Riesz space and f; g 2 E. Then the following equalities are valid: 1. f = f + − f −; 2. jfj = f + + f −; 3. (f _ g) + (f ^ g) = f + g; 6 4. (f _ g) − (f ^ g) = jf − gj; 1 1 5. f _ g = 2 (f + g) + 2 jf − gj; 1 1 6. f ^ g = 2 (f + g) − 2 jf − gj. Proof. 1. f + − f = (f _ 0) − f = 0 _ (−f) = f −, so f = f + − f −; 2. jfj = f _ (−f) = ((2f) _ 0) − f = 2f + − (f + − f −) = f + + f −; 3. f _ g = ((f − g) _ 0) + g = (f − g)+ + g and f ^ g = f + (0 ^ (g − f)) = f − (0 _ (f − g)) = f − (f − g)+, so (f _ g) + (f ^ g) = f + g; 4. Using f ^ g = f − (f − g)+ and f _ g = ((g − f) _ 0) + f = (g − f)+ + f, we find: (f _g)−(f ^g) = (g −f)+ +(f −g)+ = (f −g)+ +(f −g)− = jf −gj; 5. This follows from adding 3 and 4; 6. This follows from subtracting 4 from 3. Lemma 1.12 (Riesz decomposition property). Let (E; ≤) be a Riesz space and + + let g; f1; f2 2 E such that g ≤ f1 + f2. Then there exist g1; g2 2 E such that g1 ≤ f1, g2 ≤ f2 and g = g1 + g2. + Proof. Let g1 = g ^ f1 and g2 = g − g1.
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