Borel Equivalence Relations: Structure and Classification

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BOREL EQUIVALENCE RELATIONS: STRUCTURE AND CLASSIFICATION

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Dual Price 1: Then since Gi is connected and almost Qq -simple, every normal Qq -subgroup is finite, so contained in the center see Margulis [26], 0. Now all the hypotheses of 3. Apply 2. But then, by 2. Then, as in the preceding argument, an application of 2. By Margulis [26], III 5. So, by 3. The argument is similar to that of 3. First since SLn C is connected see, e.

Also SLn R is connected same reference.

Every normal subgroup of SLn K , K any field of characteristic 0, is contained in its center which is finite see Rotman [29], 8. In particular, SLn is semisimple. Then, by repeating the argument in 3. We have to conclude that this cocycle is equivalent to one with values in some compact subgroup of Si k. Applying 3. Let Z[ p1 ] be the ring of rationals whose denominators in reduced form are powers of p.

We will need the following standard result: Lemma 3. Then K is compact. Then by Zimmer [35], Embedding Borel sets under inclusion Our goal in this section is to prove the first main theorem of this paper. Theorem 4. The partial ordering of Borel sets under inclusion can be embed- ded in the partial pre order of countable Borel equivalence relations under Borel reducibility. The proof of 4. We will devote the rest of this section to the proof of Theorem 4. It is clearly invariant under the shift action and it is non-atomic.

The next lemma is quite standard.

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Lemma 4. The following lemma is also standard. For the sake of completeness, we include a proof which does not depend on understanding the mixing property. Then we have the following lemma: Lemma 4. Granting this, we can prove 4. Then i - iii of 4. Since, by 3. By Zimmer [35], B.

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Over the last 20 years, the theory of Borel equivalence relations and related topics have been very active areas of research in set theory and. Borel Equivalence Relations: Structure and Classification Borel reducibility of equivalence relations · Borel ideals in the structure of Borel reducibility.

Using Section 2 paragraph following 2. Then by 2. Now using 2. Hence f maps M1 into a single ET -class. The descriptive complexity of Borel reducibility Let us first fix a parametrization of Borel relations on NN. It is straightforward, using Kechris [21], Theorem 5. By trivially modifying 4. Now assume that [T ] admits a full Borel uniformization, say f. So clearly f is a full uniformization of [T ] and it only remains to check that it is Borel. Clearly the set of trees is Borel.

Following up a conversation with one of the authors, John Steel first found a proof of this using non-standard models. We give below a different proof based on effective descriptive set theory. We assume that the reader is familiar with this theory as exposed for example in Moschovakis [28].

Classification and Orbit Equivalence Relations

Week 5. By the argument in 6. Week 4. Borel Equivalence Relations : Structure and Classification. The argument is similar to that of 3. Embedding Borel sets under inclusion Our goal in this section is to prove the first main theorem of this paper. Cohomology of property T groupoids and applications.

Lemma 5. We will first need some simple facts about trees see Kechris [21], Section 2. Let S, T be trees on sets A, B, resp. Then by inspecting the argument in 5. This is a well-known fact; see, e.

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So we have: Theorem 5. Let S Qn be the space of all subgroups of Qn a closed subset of the space of all subsets of Qn. More precisely, let E be the standard Borel space of ergodic, invariant measures of this action. Miscellanea We mention here some other results that can be proved by the methods used in this paper. Then we show Theorem 7. In fact, in Theorem 7. Theorem 7. We aim for a contradiction. By the argument in 6. If a is totally irrational, then a is in canonical position, with index 0 and with no rational part.

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The conditions that a be totally irrational or in canonical position are both invariant under translation by elements of Qd. In this case, we will say that a is totally irrational resp. Dividing k by the gcd of its entries, we may assume that k is primitive i.