Frank Olaf Wagner

Plus ultra

Keywords: 2000 Mathematics Subject Classification. 03C45. Key words and phrases. stable, simple, internal, analysable, ultraimaginary, elimination, weak canonical base property

We define a reasonably well-behaved class of ultraimaginaries, i.e. classes modulo ∅-invariant equivalence relations, called tame, and establish some basic simplicity-theoretic facts. We also show feeble elimination of supersimple ultraimaginaries: If e is an ultraimaginary definable over a tuple a with SU (a) < ω α+1 , then e is eliminable up to rank < ω α . Finally, we prove some uniform versions of the weak canonical base property.

Introduction

This paper arose out of an attempt to understand and generalize Chatzidakis' results on the weak canonical base property [6, Proposition 1.14 and Lemma 1.15]. In doing so, we realized that certain stability-theoretic phenomena were best explained using ultraimaginaries. It should be noted that ultraimaginaries occur naturally in simplicity theory and were in fact briefly considered in [START_REF] Ben Yaacov | Constructing an almost hyperdefinable group[END_REF] before specializing to the more restricted class of almost hyperimaginaries. However, they have faded into oblivion since Ben Yaacov [START_REF] Ben | Discouraging Results for Ultraimaginary Independence Theory[END_REF] has shown that no satisfactory independence theory can exist for them, as there are problems with both the finite character and the extension axiom for independence. Nevertheless, at least finite character can be salvaged if one restricts to quasi-finitary ultraimaginaries in a supersimple theory, or more generally to what we call tame ultraimaginaries.

We shall define ultraimaginaries in Section 2 and give various examples. We also give a first example of a natural general result involving them, Proposition 2.12, which for a supersimple theory of finite rank specializes to a theorem of Lascar. In Section 3 we define tame ultraimaginaries and recover certain tools from simplicity theory, even though, due to the lack of extension, canonical bases are not available in our context. One may thus hope to extend the techniques of this section for instance to the superrosy context, where the lack of canonical bases has been one of the main technical problems.

In Section 4 we prove feeble elimination of ultraimaginaries. In particular ultraimaginaries of finite rank are interbounded with hyperimaginaries. This is used in Section 5 to generalize some of Chatzidakis' results [START_REF] Chatzidakis | A note on canonical bases and modular types in supersimple theories[END_REF] on the weak canonical base property from sets of finite SUrank to arbitrary ordinal SU-rank. It is interesting to compare this generalization to the coarser [START_REF] Palacín | Ample thoughts[END_REF]Theorem 5.4] which uses α-closure. We expect that this is a general phenomenon: The use of ultraimaginaries allows for a more direct and more refined proof without explicit use of SU-rank; rank considerations principally intervene via the feeble elimination result of Section 4 and the technical results of Section 3.

All elements, tuples and parameter sets are hyperimaginary, unless stated otherwise. For an introduction to simplicity and hyperimaginaries, the reader is invited to consult [START_REF] Casanovas | Simple Theories and Hyperimaginaries[END_REF], [START_REF] Kim | Simplicity Theory[END_REF] or [START_REF] Frank | Simple Theories[END_REF].

Ultraimaginaries

Definition 2.1. An ultraimaginary is the class a E of a tuple a under an ∅-invariant equivalence relation E.

Note that tuples of ultraimaginaires are again ultraimaginary. Definition 2.2. An ultraimaginary a E is definable over an ultraimaginary b F if any automorphism of the monster model stabilising the F -class of b also stabilises the E-class of a. It is bounded over b F if the orbit of a under the group of automorphisms of the monster model which stabilise the F -class of b is contained in boundedly many Eclasses. A representative of an ultraimaginary e is any hyperimaginary a such that e is definable over a. An ultraimaginary is finitary if it has a finite real representative. Two (tuples of) ultraimaginaries are equivalent over some set A of parameters if they are interdefinable over A.

Note that in contrast to hyperimaginaries, the class of a tuple of size κ modulo an ∅-invariant equivalence relation need not be equivalent to a tuple of ultraimaginaries with representatives of smaller size: Consider the equivalence relation on sequences of length κ of being equal except for a subsequence of smaller length.

Remark 2.3. As usual, if E A (x, y) is an A-invariant equivalence relation, one considers the ∅-invariant relation E(xX, yY ) given by (X = Y ∧ X ≡ A ∧ E X (x, y)) ∨ (X = Y ∧ x = y). This is an equivalence relation, and (aA) E is equivalent to a E A over A.

Remark 2.4. As any ∅-invariant relation, E is given by a union of types over ∅. Definition 2.5. We shall say that two ultraimaginaries have the same (Lascar strong) type over some set A if they have representatives which do. If the ambient theory is simple, we call two ultraimaginaries independent over A if they have representatives which are.

Clearly, two ultraimaginaries are conjugate by a (Lascar strong) automorphism over A if and only if they have the same (Lascar strong) type over A.

Remark 2.6. If e or e ′ is a sequence of ultraimaginaries, for e | ⌣ e ′ to hold we require sequences of representatives which are independent. In particular, it is not clear even for real e ′ that an infinite sequence e of ultraimaginaries is independent of e ′ if every finite subsequence is independent of e ′ .

Ultraimaginaries arise quite naturally in stability and simplicity theory.

Example 2.7. Let p A ∈ S(A) be a regular type in a stable theory. For

A ′ , A ′′ |= tp(A) put E(A ′ , A ′′ ) if p A ′ ⊥ p A ′′ .
Then E is an ∅-invariant equivalence relation, and A E codes the non-orthogonality class of p A .

The work with ultraimaginaries requires caution, as some basic properties become problematic.

Example 2.8. [START_REF] Ben | Discouraging Results for Ultraimaginary Independence Theory[END_REF] Let E be the ∅-invariant equivalence relation on infinite sequences which holds if they differ only on finitely many elements. Consider a sequence I = (a i : i < ω) of elements such that no finite subtuple is bounded over the remaining elements. Then every finite tuple ā ∈ I can be moved to a disjoint conjugate over I E , but I cannot. Similarly, if I is a Morley sequence in a simple theory, then ā | ⌣ I E for any finite ā ∈ I, but

I | ⌣ I E .
Even in the ω-stable context, for classes of finite tuples, the theory is not smooth.

Example 2.9. Let T be the theory of a cycle-free graph (forest) of infinite valency, with predicates P n (x, y) for couples of points of distance n for all < ω. It is easy to see by back-and-forth that T eliminates quantifiers and is ω-stable of rank ω; the formula P n (x, a) has rank n over a. Let E be the ∅-invariant equivalence relation of being in the same connected component. Then existence of non-forking extensions fails over a E , as any two points in the connected component of a have some finite distance n, and hence rank n over one another, but rank ≥ k over a E for all k < ω, since a E is definable over any point of distance at least k.

The same phenomenon can be observed for any type p of rank SU(p) = ω in a simple theory, with the relation E(x, y) on p which holds if SU(x/y) < ω and SU(y/x) < ω (actually, one follows from the other by the Lascar inequalities).

The behaviour of Example 2.9 is inconvenient and signifies that we shall avoid considering types over an ultraimaginary. The behaviour of Example 2.8 is outright vexatious; we shall restrict the class of ultraimaginaries under consideration in order to preserve the finite character of independence. Definition 2.10. An ultraimaginary e is quasi-finitary if there is a finite real tuple a such that e is bounded over a.

For hyperimaginary tuples contained in the bounded closure of a finite set, we shall use quasi-finite rather than quasi-finitary, in order to emphasize the distinction between usual hyperimaginaries and ultraimaginaries. The set of all (quasi-finitary) ultraimaginaries definable over some ultraimaginary set E will be denoted by dcl u (E) (or dcl qf u (E), respectively). Similarly, bdd u (E) and bdd qf u (E) will denote the corresponding bounded closures. If A is a set of representatives for E, then the number of ultraimaginaries with representatives of length κ in bdd u (E) (and a fortiori in the other closures as well) is bounded in terms of the number of Lascar strong types over A of real tuples of length κ, since equality of Lascar strong type over A is the finest bounded A-invariant equivalence relation.

Remark 2.11. If e is a quasi-finitary ultraimaginary then bdd u (e) = bdd qf u (e) and dcl u (e) = dcl qf u (e).

Proposition 2.12. The following are equivalent for two ultraimaginaries a and b:

(1) bdd u (a) ∩ bdd u (b) = bdd u (∅).

(2) For any a ′ ≡ lstp a there is n < ω and a sequence (a i b

i : i ≤ n) such that a 0 = a, b 0 = b, a n = a ′ and for each i < n bdd u (a i )b i+1 ≡ bdd u (a i )b i and a i+1 bdd u (b i+1 ) ≡ a i bdd u (b i+1 ).
If a or b is quasi-finite, this is also eqivalent to bdd qf u (a) ∩ bdd qf u (b) = bdd qf u (∅).

Proof: (1) ⇒ (2) Suppose bdd u (a) ∩ bdd u (b) = bdd u (∅), and define an ∅-invariant equivalence relation on lstp(ab) by E(xy, x ′ y ′ ) if there is a sequence (x i y i : i ≤ n) such that x 0 y 0 = xy, x n y n = x ′ y ′ , and for each i < n we have bdd u (x i )y i+1 ≡ bdd u (x i )y i and x i+1 bdd u (y i+1 ) ≡ x i bdd u (y i+1 ). Now if bdd u (a)b ′ ≡ bdd u (a)b, then |= E(ab, ab ′ ). Hence (ab) E ∈ bdd u (a). Similarly (ab) E ∈ bdd u (b), whence (ab) E ∈ bdd u (∅). But for any a ′ ≡ lstp a there is b ′ with ab ≡ lstp a ′ b ′ . Then |= E(ab, a ′ b ′ ), in particular (2) holds.

(2) ⇒ (1) Suppose not, and consider e ∈ (bdd

u (a) ∩ bdd u (b)) \ bdd u (∅). As e / ∈ bdd u (∅) there is a ′ |= lstp(a) with e / ∈ bdd u (a ′ ). Consider a sequence (a i , b i : i ≤ n) as in (2). Since bdd u (a i )b i+1 ≡ bdd u (a i )b i and a i+1 bdd u (b i+1 ) ≡ a i bdd u (b i+1 ) we have bdd u (a i ) ∩ bdd u (b i ) = bdd u (a i ) ∩ bdd u (b i+1 ) = bdd u (a i+1 ) ∩ bdd u (b i+1 ).
In particular,

e ∈ bdd u (a) ∩ bdd u (b) = bdd u (a 0 ) ∩ bdd u (b 0 ) = bdd u (a n ) ∩ bdd u (b n ) ⊆ bdd u (a ′ ), a contradiction.
The last assertion follows from Remark 2.11. Using weak elimination of ultraimaginaries proven in Section 4, we recover a Lemma of Lascar [START_REF] Lascar | Sous groupes d'automorphismes d'une structure saturée[END_REF] (see also [START_REF] Moosa | On canonical bases and internality criteria[END_REF]Lemma 2.2]), proved originally for stable theories of finite Lascar rank.

Corollary 2.14. Let T be a simple theory of finite SU-rank and a, b finite imaginary tuples. The following are equivalent:

(1) acl eq (a) ∩ acl eq (b) = acl eq (∅).

(2) For any a ′ ≡ lstp a independent of a there are sequences a = a 0 , . . . , a n = a ′ and b = b 0 , .

: i ≤ n) from ab = a 0 b 0 to a n = a ′′ with the sequence (a i b i : n ≤ i ≤ ℓ) from a n b n to a ℓ = a ′ . Hence (2) holds for arbitrary a ′ ≡ lstp
A a, so we can again apply Proposition 2.12.

Ultraimaginaries in simple theories

From now on the ambient theory will be simple. Our notation is standard and follows [START_REF] Frank | Simple Theories[END_REF]. We shall be working in a sufficiently saturated model of the ambient theory. Tuples are again tuples of hyperimaginaries, and closures (definable and bounded closures) will include hyperimaginaries.

Remark 3.1. Since in a simple theory Lascar strong type equals Kim-Pillay strong type, we have bdd u (A) = dcl u (bdd(A)). But of course, as with real and imaginary algebraic closures, bdd(A) ∩ bdd(B) = bdd(∅) does not imply bdd u (A) ∩bdd u (B) = bdd u (∅) unless the theory weakly eliminates ultraimaginaries.

In a simple theory, ultraimaginary independence is clearly symmetric, and satisfies local character and extension (but recall that we only consider hyperimaginary base sets), since this is inherited from suitable representatives. As for transitivity, we have the following.

(Ae) ∩ bdd u (Ae ′ ) = bdd u (A).
Proof: Replacing e and e ′ by A-independent representatives, we may assume that e and e ′ are hyperimaginary. Consider a E ∈ bdd u (Ae) ∩ bdd u (Ae ′ ). We may assume a | ⌣Ae e ′ , whence ae | ⌣A e ′ . Let (a i : i < ω) be a Morley sequence in lstp(a/Ae ′ ). Then E(a i , a j ) for all i, j < ω. But a i | ⌣A a j for i = j, so π(x, a j ) = tp(a i /a j ) does not fork over A, and neither does π(x, a). Note that π(x, y) implies E(x, y).

Now suppose a E /

∈ bdd u (A). We can then find a long sequence (a

′ i : i < α) of A-conjugates of a such that ¬E(a ′ i , a ′ j ) for i = j.
By the Erdős-Rado theorem and compactness (see e.g. [5, Proposition 1.6]) there is an infinite A-indiscernible sequence (a ′′ i : i < ω) whose 2-type over A is among the 2-types of (a ′ i : i < α). In particular ¬E(a ′′ i , a ′′ j ) for i = j, and (π(x, a ′′ i ) : i < ω) is 2-inconsistent. Since a ′′ 0 |= tp(a/A), we see that π(x, a) divides over A, a contradiction.

As we have seen in Remark 2.6, finite character may fail for ultraimaginaries. The next definition singles out the subclass of ultraimaginaries where this does not happen, at least for hyperimaginary sets. Hence A ⊆ bdd(b) and e is bounded over b, so e is quasi-finitary. In a supersimple theory the converse is obvious.

We are really interested in the set of tame ultraimaginaries. However, we do not have a good criterion when an ultraimaginary is tame; moreover, an ultraimaginary definable over a tame ultraimaginary need not be tame itself. For instance, the sequence I in Example 2.8 is tame (since it is real), but I E is not. Clearly, an ultraimaginary definable (or even bounded) over a quasi-finitary/supersimple ultraimaginary is itself quasi-finitary/supersimple. In a supersimple theory quasi-finitary ultraimaginaries are the correct ones to consider: Due to elimination of hyperimaginaries all parameters consist of imaginaries of ordinal SU-rank; as canonical bases of such imaginaries are finite, we can always reduce to a quasi-finitary situation.

Another kind of tame ultraimaginaries arose in the generalization of the group configuration theorem to simple theories [START_REF] Ben Yaacov | The group configuration in simple theories and its applications[END_REF][START_REF] Ben Yaacov | Constructing an almost hyperdefinable group[END_REF]. We shall now consider how to obtain invariant equivalence relations, and hence ultraimaginaries.

∃xy [Xx ≡ Y y ≡ Ab ∧ x | ⌣ X Y ∧ y | ⌣ Y X ∧ E(x, y)].
Then R is an ∅-invariant equivalence relation on tp(A).

Proof: Clearly, R is ∅-invariant, reflexive and symmetric. So suppose that R(A, A ′ ) and R(A ′ , A ′′ ) both hold, and let this be witnessed by b, b ′ and b

* , b ′′ . Let b 1 |= tp(b ′ /A ′ ) = tp(b * /A ′ ) with b 1 | ⌣A ′ AA ′′ . Since A ′ is algebraically closed, b ′ | ⌣A ′ A and b * | ⌣A ′ A ′′ we have b 1 ≡ AA ′ b ′ and b 1 ≡ A ′
| ⌣AA ′ b 1 A ′′ and b 2 | ⌣A ′ A ′′ b 1 A ′ . Now b 1 | ⌣A ′ AA ′′ implies b 0 | ⌣AA ′ A ′′ and b 2 | ⌣A ′ A ′′ A. Then b 0 | ⌣A A ′ and b 2 | ⌣A ′′ A ′ imply b 0 | ⌣A A ′′ and b 2 | ⌣A ′′ A, whence R(A, A ′′ ) holds. So R is transitive.
Remark 3.11. Recall that a reflexive and symmetric binary relation R(x, y) on a partial type π(x) is generically transitive if whenever x, y, z |= π and x | ⌣y z, then R(x, y) and R(y, z) together imply R(x, z). If T is merely simple, the relation R in Proposition 3.10 is still generically transitive. However, contrary to the type-definable case [13, Lemma 3.3.1], the two-step iterate of an ∅-invariant, reflexive, symmetric and generically transitive relation on a Lascar strong type need not be transitive.

Example 3.12. Consider on the forest of Example 2.9 the relation R(a, b) which holds if 3 divides the distance between a and b. This is generically transitive, as for a ′ | ⌣a a ′′ the distance between a ′ and a ′′ is the sum of the distances between a ′ and a and between a and a ′′ . However, two points of distance 2 are easily seen to be R 2 -related, so the transitive closure E of R is just the relation of being in the same connected component. But no two points of distance 1 are R 2 -related.

Clearly, in the above example the three-step iterate R 3 is transitive, as it is just the relation of being connected. Is there an example of a generically transitive symmetric and reflexive ∅-invariant relation R such that R n is not transitive for any n < ω ?

The next proposition shows that in a simple theory and under some conditions, if R is a generically transitive reflexive and symmetric relation, then at least its transitive closure is not bounded, unless R holds for two independent elements. We first recall the definitions of SU p -rank and p-closure. Definition 3.13. [13, Remark 5.1.19] Let P be an ∅-invariant family of regular types closed under nonforking extensions. The SU P -rank is the smallest function from the collection of all types to the ordinals together with infinity, such that SU(a/A) ≥ α + 1 if there is some B ⊇ A and some c | ⌣B a with tp(c/B) ∈ P . The P -closure of a set A is given by cl P (A) = {a : SU P (a/A) = 0}.

Then SU P -rank satisfies the Lascar inequalities [13, Exercise 5.1.20]. Note that unless P contains all non-orthogonality classes of types of SU-rank 1, the P -closure has the size of the monster model. (1) tp(a/A) is foreign to all types q with SU P (q

) = 0. (2) a | ⌣A cl P (A). (3) a |
⌣A dcl(aA) ∩ cl P (A). ( 4) dcl(aA) ∩ cl P (A) ⊆ bdd(A).

Note that P -closure is well-behaved with respect to independence: Remark 3.16. If p ∈ S(∅) is regular and P is the family of all nonforking extensions of p, we shall write SU p and cl p . Another choice for P is the family of all regular types of SU-rank ω α used in the proof of Theorem 4.6. One can also take P to be the family of types foreign to some ∅-invariant collection Σ of partial types; in this case P -closure cl P is equal to the Σ-closure cl Σ defined in Definition 5.2 (see [START_REF] Frank | Some remarks on one-basedness[END_REF]). Lemma 3.17. Let P be an ∅-invariant family of regular types closed under nonforking extensions. Suppose SU P (a/bc) = SU P (a/b) = n is finite. Then cl P (a) | ⌣cl P (b) cl P (c).

Proof: Let (a i : i ≤ ω) be a Morley sequence in lstp(a/bc) with a = a ω , and put c k = (a i : i < n) and d k = Cb(a/bc k ). We shall show first that

d k ∈ cl P (b).
Let (a ′ i : i < ω) be a Morley sequence in lstp(a/bc k ). Since

SU P (d k /b) ≤ SU P (c k /b) ≤ nk is finite, there is ℓ < ω and r ≥ 0 with SU P (d k /b, a ′ i : i < ℓ ′ ) = r for all ℓ ′ ≥ ℓ. Suppose r > 0. Then there is B ⊇ {b, a ′ i : i < ℓ} and b ′ with tp(b ′ /B) ∈ P such that b ′ | ⌣B d k ; we may assume Bb ′ | ⌣ b,d k ,(a ′ i :i<ℓ) (a ′ i : i < ω).
As

d k ∈ dcl(a ′ i : i < ω), there is minimal ℓ ′ ≥ ℓ such that b ′ | ⌣B (a ′ i : i ≤ ℓ ′ ). So SU P (a ′ i : i ≤ ℓ ′ /b, d k , a ′ i : i < ℓ) = SU P (a ′ i : i ≤ ℓ ′ /Bb ′ ) < SU P (a i : i ≤ ℓ ′ /B) ≤ SU P (a ′ i : i ≤ ℓ ′ /b, a ′ i : i < ℓ) ≤ ℓn.
By Lascar symmetry (the second Lascar inequality), Then a E = a ′ E , so there is n < ω and a chain a = a 0 , a 1 , . . . , a n = a ′ such that R(a i , a i+1 ) holds for all i < n. Put a ′ 0 = a 0 , and for 0

r = SU P (d k /b, a ′ i : i ≤ ℓ ′ ) < SU P (d k /b, a ′ i : i < ℓ) = r, a contradiction. Thus r = 0. By the Lascar inequalites, ℓn + SU P (d k /b) = SU P (a i : i < ℓ/b, d k ) + SU P (d k /b) ≤ SU P (d k , a i : i < ℓ/b) ≤ SU P (d k /b, a i : i < ℓ) ⊕ SU P (a i : i < ℓ/b) = 0 ⊕ ℓn, whence SU P (d k /b) = 0 and d k ∈ cl P (b). Since a | ⌣bd k (a i : i < k) Fact 3.15 yields a | ⌣cl P (b) (a i : i < k) for all k, whence a | ⌣cl P (b) (a i : i < ω). If d = Cb(a/
< i < n let a ′ i ≡ lstp a i a ′ i-1 with a ′ i | ⌣ a i a i+1 . Claim. bdd u (a ′ i ) ∩ bdd u (a i+1 ) ⊆ bdd u (a 0 ).

Proof of Claim:

For i = 0 this is trivial. For i > 0, as

a ′ i ≡ lstp a i a ′ i-1 and bdd u (a i ) = dcl u (bdd(a i )), we get bdd u (a ′ i ) ∩ bdd u (a i ) = bdd u (a ′ i-1 ) ∩ bdd u (a i ). Next, a ′ i | ⌣a i a i+1 implies bdd u (a ′ i a i ) ∩ bdd u (a i a i+1 ) = bdd u (a i ) by Lemma 3.4. Hence inductively bdd u (a ′ i ) ∩ bdd u (a i+1 ) ⊆ bdd u (a ′ i ) ∩ bdd u (a i ) = bdd u (a ′ i-1 ) ∩ bdd u (a i ) ⊆ bdd u (a 0 ).

Now by generic transitivity and induction, R(a ′

i , a i+1 ) holds for all i < n. In particular R(a ′ n-1 , a n ) holds, and by Lemma 3.

4 bdd u (a ′ n-1 ) ∩ bdd u (a n ) ⊆ bdd u (a 0 ) ∩ bdd u (a n ) = bdd u (c). Choose a ′′ with R(a ′′ , a ′ n-1 ) such that SU P (a ′′ /a ′ n-1 ) is maximal possi- ble. We may choose it such that a ′′ | ⌣a ′ n-1 a n . Then bdd u (a ′′ ) ∩ bdd u (a n ) ⊆ bdd u (a n ) ∩ bdd u (a ′ n-1 ) ⊆ bdd u (c) and SU P (a ′′ /a n ) ≥ SU P (a ′′ /a ′ n-1 a n ) = SU P (a ′′ /a ′ n-1
). Rename a ′′ a n as a 1 a 2 , and note that bdd u (a 1 ) ∩ bdd(a 2 ) ⊆ bdd u (c), c ⊆ bdd(a 2 ), and SU P (a 1 /a 2 ) is maximal possible among tuples (x, y) with R(x, y). Moreover, SU P (a 2 /a 1 ) = SU P (a 1 a 2 ) -SU P (a 1 ) = SU P (a 1 a 2 ) -SU P (a 2 ) = SU P (a 1 /a 2 ), so this is also maximal.

Choose a 3 | ⌣a 2 a 1 with a 3 ≡ lstp a 2 a 1 . By generic transitivity R(a 1 , a 3 ) holds. Moreover, SU P (a 3 /a 1 ) ≥ SU P (a 3 /a 1 a 2 ) = SU P (a 3 /a 2 ), so equality holds. Similarly, SU P (a 1 /a 3 ) = SU P (a 1 /a 2 a 3 ) = SU P (a 1 /a 2 ). Now SU P (a i /a j ) = SU P (a i /a j a k ) for {i, j, k} = {1, 2, 3} implies by Lemma 3.17 that cl

P (a i ) | ⌣ cl P (a j ) cl P (a k ).
In particular, cl P (a i ) ∩ cl P (a k ) = cl P (a 1 ) ∩ cl P (a 2 ) ∩ cl P (a 3 ).

Let b = cl P (a 1 )∩cl P (a 2 )∩bdd(a 1 a 2 ). Then cl P (a

But now

Cb(a 3 /cl P (a 1 )cl P (a 2 )) ⊆ cl P (a 1 ) ∩ cl P (a 2 ) = cl P (b) = cl P (∅), so a 3 | ⌣cl P (∅) a 2 , as required. We shall illustrate the use of the proposition in Propositions 3.26 and 3.28, whose proof in the hyperimaginary case uses canonical bases. From now on, let Σ be an ∅-invariant family of partial types. Recall first the definitions of internality, analysability and orthogonality for hyperimaginaries. • Σ-analysable if for any realization a of π there is a sequence (a i : i < α) such that tp(a i /A, a j : j < i) is Σ-internal for all i < α, and a ∈ bdd(A, a i : i < α). Note that in the definition of analysability, we may in addition require a i ∈ dcl(Aa) for all i < α. We now generalize these notions to ultraimaginaries. Definition 3.20. We shall say that an ultraimaginary e is (almost) Σinternal, or is Σ-analysable, if it has a representative which is. Similarly, e is orthogonal over A to some type p if for all B | ⌣A e such that p is over B and for any realization b |= p|B we have e | ⌣A Bb. Remark 3.21. This definition does not imply that we define the notion of an analysis of an ultraimaginary. Moreover, e orthogonal to p over A need not imply that e has a representative which is orthogonal to p.

And unless e is tame, orthogonality of e over A to p need not imply orthogonality to p (ω) . Definition 3.22. For an ordinal α the α-th Σ-level of a over A is defined inductively by ℓ Σ 0 (a/A) = bdd(A), and for α > 0

ℓ Σ α (a/A) = {b ∈ bdd(aA) : tp(b/ β<α ℓ β (a/A)) is almost Σ-internal}.
We shall write ℓ Σ ∞ (a/A) for α ℓ Σ α (a/A), i.e. the set of all hyperimaginaries b ∈ bdd(aA) such that tp(b/A) is Σ-analysable.

α = ∞, noting that ℓ Σ ∞ (b/c) = ℓ Σ ∞ (b) = c.
The next proposition is well-known for hyperimaginaries in simple theories, even without the restriction on SU p -rank: If tp(a/A) is nonorthogonal to a regular type p ∈ S(∅), then there is a p-internal a 0 ∈ bdd(aA) \ cl p (A): If B | ⌣A a and b |= p|AB with b | ⌣AB a, just take a 0 = Cb(bB/aA). In fact, if we just require a 0 ∈ bdd(aA) \ bdd(A), we do not even need p to be regular [START_REF] Frank | Simple Theories[END_REF]Propositon 3.4.14]. However, as for ultraimaginary a the canonical base does not make sense, we have to work harder. Proof:

Let c = ℓ p 1 (b). Define an ∅-invariant relation R on tp(c) by R(c ′ , c ′′ ) ⇔ ∃b ′ b ′′ [b ′ c ′ ≡ b ′′ c ′′ ≡ bc ∧ E(b ′ , b ′′ )].
qf u (A 1 D) ∩ bdd qf u (A 2 D) = bdd qf u (BD) ∩ bdd qf u (A 1 D) = bdd qf u (BD) ∩ bdd qf u (AD), we have e ′ ∈ dcl qf u (B ′ D) ∩ bdd qf u (AD). Moreover, B ′ | ⌣BD A, whence B ′ | ⌣B A and bdd qf u (A) ∩ bdd qf u (B ′ ) ⊆ bdd qf u (A) ∩ bdd qf u (B) = bdd qf u (∅). Choose A ′ ≡ lstp AD B ′ with A ′ | ⌣AD B ′ . Then e ′ ∈ dcl qf u (A ′ D) ∩ dcl qf u (B ′ D). Now, D | ⌣B A implies D | ⌣B AB ′ ; as D | ⌣ B we get D | ⌣ ABB ′ . Hence D | ⌣A B ′ , whence A ′ | ⌣A B ′ and bdd qf u (A ′ ) ∩ bdd qf u (B ′ ) ⊆ bdd qf u (A) ∩ bdd qf u (B ′ ) = bdd qf u (∅).
We may assume e ′ = (A ′ D) E for some ∅-invariant equivalence relation E. Define a ∅-invariant reflexive and symmetric relation R on lstp(A ′ ) by

R(X, Y ) ⇔ ∃ Z [XZ ≡ Y Z ≡ A ′ D ∧ Z | ⌣ XY ∧ E(XZ, Y Z)].

By the independence theorem, if

A 1 | ⌣A 2 A 3 such that R(A 1 , A 2 ) and R(A 2 , A 3 ) hold, we have R(A 1 , A 3 ). Hence R is generically transitive; let E ′ be the transitive closure of R. Clearly A ′ E ′ is quasi-finitary. Next, consider A ′′ ≡ B ′ A ′ with A ′′ | ⌣B ′ A ′ . By the independence the- orem there is D ′ with A ′ D ≡ B ′ A ′ D ′ ≡ B ′ A ′′ D ′ and D ′ | ⌣B ′ A ′ A ′′ . Then D ′ | ⌣ B ′ , whence D ′ | ⌣ A ′ A ′′ and (A ′ D ′ ) E = (A ′′ D ′ ) E ∈ dcl qf u (B ′ D ′ ). Therefore E ′ (A ′ , A ′′ ) holds and A ′ E ′ ∈ dcl qf u (B ′ ). Thus A ′ E ′ ∈ dcl qf u (A ′ ) ∩ dcl qf u (B ′ ) ⊆ bdd qf u (∅). By Proposition 3.18 there is A ′′ | ⌣cl p (∅) A ′ with R(A ′ , A ′′ ). Let D ′ wit- ness R(A ′ , A ′′ ). Then D ′ ≡ A ′ D, so we may assume D ′ = D. Since cl p (D) | ⌣cl p(∅) cl p (A ′ A ′′ ) and cl p (A ′ ) | ⌣cl p(∅) cl p (A ′′ ) we obtain cl p (A ′ ) | ⌣ clp(∅) cl p (A ′′ )cl p (D)
and hence

A ′ | ⌣cl p (D) A ′′ . But now e ′ = (A ′ D) E = (A ′′ D) E ∈ dcl qf u (A ′ D) ∩ dcl qf u (A ′′ D) ⊆ bdd qf u (cl p (D))
by Lemma 3.4. Since e ∈ bdd qf u (e ′ ), this contradicts non-orthogonality of e to p over D.

Remark 3.29. Again, the proof of the hyperimaginary analogue of Proposition 3.28 for simple theories uses canonical bases and does not generalize.

Elimination of ultraimaginaries

One cannot avoid the non-tame ultraimaginaries of Example 2.8 which do not satisfy finite character and hence cannot be eliminated. Similarly, on a type of rank ω we cannot eliminate the relation of having mutually finite rank over each other (Example 2.9), since the rank over a class modulo such a relation is not defined. We thus content ourselves with elimination of supersimple ultraimaginaries in a simple theory (and in particular of quasi-finitary ultraimaginaries in a supersimple theory) up to rank of lower order of magnitude. This seems to be optimal, given the examples cited. An ultraimaginary e can be α-eliminated if there is a representative a with SU(a/e) < ω α . A supersimple theory has feeble elimination of ultraimaginaries if for all ordinals α, all quasi-finitary ultraimaginaries of rank < ω α+1 can be α-eliminated. In particular bdd(c) is a representative for e of lower rank, a contradiction.

Remark 4.7. Let p be a regular type (or type of weight 1). Then two realizations a and b of p are independent if and only if bdd qf u (a) ∩ bdd qf u (b) = bdd qf u (∅): One direction is Lemma 3.4, the other follows from the observation that dependence is an invariant equivalence relation on realizations of p. However, this does not hold for all types: By elimination of quasifinite ultraimaginaries, it is in particular false in non one-based theories of finite rank.

Decomposition

In this section we shall give ultraimaginary proofs of some of Chatzidakis' results from [START_REF] Chatzidakis | A note on canonical bases and modular types in supersimple theories[END_REF] around the weak canonical base property, and suitable generalisations to the supersimple case. As before, Σ will be an ∅-invariant family of partial types in a simple theory.

Recall that a and b are domination-equivalent over A, denoted a A b, if for any c we have c | ⌣A a ⇔ c | ⌣A b. The following lemma is folklore, but we give a proof for completeness.

Since tp(Aa

/A ′ ) is foreign to Σ\Σ ′ , we have Aa | ⌣A ′ B ′ , whence a A ′ B ′ A by Lemma 5.1(1). Similarly b A ′ B ′ B. But A | ⌣A ′ B ′ B, and thus a | ⌣A ′ B ′ b. Let a 0 = bdd(A ′ a) ∩ bdd(A ′ B ′ b) and b 0 = bdd(B ′ b) ∩ bdd(A ′ B ′ a). By one-basedness of tp(a/A ′ ) and tp(b/B ′ ), a | ⌣ A ′ a 0 B ′ b and b | ⌣ B ′ b 0 A ′ a. Hence A ′ B ′ a | ⌣ A ′ B ′ a 0 b 0 and A ′ B ′ b | ⌣ A ′ B ′ b 0 a 0 . Suppose A ⊆ bdd(A i : i ∈ I). Then B = Cb(A/B) ⊆ bdd(B i : i ∈ I); moreover bdd(A) = bdd(Cb(B/A)) = bdd(Cb(B i /A) : i ∈ I) = bdd(Cb(B i /A i ) : i ∈ I) ⊆ bdd(A i : i ∈ I) = bdd(A)
again by perpendicularily. Hence bdd(Cb(B i /A i )) = A i , and similarly bdd(Cb

(A i /B i )) = B i . But if Σ i is one-based, then B i = bdd(Cb(A i /B i )) ⊆ bdd(A i ) ∩ bdd(B i ) = bdd(∅) ; similarly A i = bdd(∅).
= Cb(B/c ā). Since A is not contained in Ā, neither is C. Hence there is ā ∈ A such that c = c ā / ∈ Ā.
As the maximal Σ i -analysable subset of bdd(c) is equal to bdd(c) ∩ A i we may replace A by c and thus assume that A is quasi-finite. Similarly, we may assume that B is quasi-finite.

Since A = Cb(B/A) ⊆ Ā, we have A | ⌣ Ā B; as A | ⌣ Ā B we obtain A | ⌣ Ā B B.
Let (b j : j < α) be an analysis of B over B such that for every j < α the type tp(b j / B, b ℓ : ℓ < j) is Σ i j -analysable for some i j ∈ I. Let k be minimal with

A | ⌣ Ā B (b j : j ≤ k). Then A | ⌣ Ā B, (b j : j < k) and Cb( B, (b j : j ≤ k)/A) is almost Σ i k -internal over Ā. Put A ′ = ℓ Σ i k 1 (A/ Ā) and B ′ = ℓ Σ i k 1 (B/ B). Then A ′ ⊆ Ā, and Cb(A ′ /B) ⊆ B ′ since Ā | ⌣ B B. Similarly Cb(B ′ /A) ⊆ A ′ . Moreover A ′ | ⌣ Ā B B, whence A ′ | ⌣ Ā B B ′ .
Replacing A by Cb(B ′ /A) = Cb(B ′ /A ′ ) and B by Cb(A ′ /B) = Cb(A ′ /B ′ ) we may assume that tp(A/ Ā) and tp(B/ B) are both almost Σ k -internal (where we write k instead of i k for ease of notation).

Claim. bdd qf u (AB k ) ∩ bdd qf u (B) = bdd qf u (B k ).
Proof of Claim: Suppose not. As B is analysable in i∈I Σ i , Corollary 3.27 yields some i ∈ I and Remark 5.15. In the finite rank context, it is easy to achieve the hypothesis of Theorem 5.13, as it suffices work over bdd(A) ∩ bdd(B). In general, however, if bdd qf u (A) ∩ bdd qf u (B) bdd qf u (bdd(A) ∩ bdd(B)), there is no hyperimaginary set C with bdd qf u (A) ∩ bdd qf u (B) = bdd qf u (C), as this equality implies bdd(C) = bdd(A) ∩ bdd(B). Thus, we cannot work over bdd qf u (A) ∩ bdd qf u (B), which is not eliminable. If SU(A/bdd(A)∩bdd(B)) < ω α+1 , feeble elimination nevertheless yields bdd qf u (A) ∩ bdd qf u (B) ⊂ bdd qf u (cl α (A) ∩ cl α (B)), so we can work over α-closed sets, as is done in [START_REF] Palacín | Ample thoughts[END_REF]Theorem 5.4].

Remark 2 . 13 .

 213 For hyperimaginary a and ultraimaginary b and c the condition bdd u (a)b ≡ bdd u (a)c is equivalent to b ≡ lstp a c.

Fact

Definition 3 . 5 .

 35 Let T be simple. An ultraimaginary e is tame if for all sets A, B of hyperimaginaries we have e | ⌣A B if and only if e | ⌣A B 0 for all finite subsets B 0 ⊆ B. It is supersimple if it has a representative of ordinal SU-rank. Remark 3.6. A supersimple ultraimaginary in a simple theory is quasi-finitary; in a supersimple theory the converse holds as well. Proof: Suppose A is a representative for an ultraimaginary e with SU(A) < ∞, and let B be a real tuple with A ∈ bdd(B). Let b ∈ B be a finite subtuple with SU(A/b) minimal; it follows that A | ⌣b B.

Lemma 3 . 7 .

 37 A supersimple ultraimaginary is tame. In particular, quasi-finitary ultraimaginaries in a supersimple theory are tame. Proof: Let e be a supersimple ultraimaginary, and a a representative with SU(a) < ∞. Consider sets A and B. There is a finite b ∈ B with a | ⌣Ab B. So e | ⌣A B if and only if e | ⌣A b by Fact 3.2. Thus e is tame.

Definition 3 . 8 .

 38 An invariant equivalence relation E is almost typedefinable if there is a type-definable symmetric and reflexive relation R finer than E such that any E-class can be covered by boundedly many R-balls (i.e. sets of the form {x : xRa} for varying a). A class modulo an almost type-definable equivalence relation is called an almost hyperimaginary. Fact 3.9. [3, page 188] Almost hyperimaginaries are tame. In fact, they satisfy finite character.

Proposition 3 . 10 .

 310 Let T be stable. For algebraically closed A and an ∅-invariant equivalence relation E on tp(b), consider the relation R(X, Y ) given by

  A ′′ b * by stationarity. Hence there are b 0 , b 2 with bb ′ ≡ AA ′ b 0 b 1 and b * b ′′ ≡ A ′ A ′′ b 1 b 2 . In particular E(b 0 , b 1 ) and E(b 1 , b 2 ) hold, and so does E(b 0 , b 2 ). Moreover, we may assume b 0

Fact 3 .

 3 [START_REF] Frank | Some remarks on one-basedness[END_REF].[START_REF] Frank | Simple Theories[END_REF] Lemma 3.5.3] The following are equivalent:

Fact 3 .

 3 15. [13, Lemma 3.5.5] and [14, Lemma 3] Suppose A | ⌣B C. Then cl P (A) | ⌣cl P (B) cl P (C). More precisely, for any A 0 ⊆ cl P (A) we have A 0 | ⌣B 0 cl P (C), where B 0 = dcl(A 0 B) ∩ cl P (B). In particular, cl P (AB) ∩ cl P (BC) = cl P (B).

  bc), then d ∈ dcl(a i : i < ω) and a | ⌣bd c. So cl P (a) | ⌣cl P (b) cl P (bd) and cl P (a) | ⌣cl P (bd) cl P (c) by Fact 3.15; the result follows by transitivity. Proposition 3.18. Let T be simple. Suppose R is an ∅-invariant, reflexive, symmetric and generically transitive relation on lstp(a), and P is an ∅-invariant family of regular types closed under non-forking extensions such that SU P (a) is finite. Let E be the transitive closure of R, and suppose a E ∈ bdd u (cl P (∅)). Then there is a ′ | ⌣cl P (∅) a with R(a, a ′ ). Proof: Put c = bdd(a)∩cl P (∅). Then a | ⌣c cl P (∅) by Fact 3.14, whence a E ∈ bdd u (c) by Lemma 3.4. Let a ′ ≡ lstp c a with a ′ | ⌣c a.

| ⌣a 1 a 2 with a ′ 3 ≡ lstp a 1 a 2

 32 1 )∩cl P (a 2 ) = cl P (b) by [10, Lemma 3.18]. Let F (x, y) be the ∅-invariant equivalence relation on lstp(b) given by cl P (x) = cl P (y). As b F is fixed by the bdd(a 2 )automorphism moving a 1 to a 3 and a 1 | ⌣a 2 a 3 , we get b F ∈ bdd u (a 2 ) by Lemma 3.4. Similarly, considering an a ′ 3 we obtain b F ∈ bdd u (a 1 ), whence b F ∈ bdd u (a 1 ) ∩ bdd u (a 2 ) ⊆ bdd u (c). So if b ′ |= lstp(b/c) with b ′ | ⌣c b, then b ′ F = b F and cl P (b ′ ) = cl P (b) = cl P (c) = cl P (∅).

Definition 3 .

 3 19. Let π be a partial type over A. Then π is • (almost) Σ-internal if for every realization a of π there is B | ⌣A a and a tuple b of realizations of types in Σ based on B, such that a ∈ dcl(B b) (or a ∈ bdd(B b), respectively).

Finally, p

 p ∈ S(A) is orthogonal to q ∈ S(B) if for all C ⊇ AB, a |= p, and b |= q with a | ⌣A C and b | ⌣B C we have a | ⌣C b. The type p is orthogonal to B if it is orthogonal to all types over B.

Remark 3 .

 3 23. So a ∈ ℓ Σ α (a/A) is and only if tp(a/A) is Σ-analysable in α steps. Lemma 3.24. If tp(a/A) is Σ-analysable in α steps for some ordinal α or α = ∞ and A | ⌣ b, put c = ℓ Σ α (b). Then Aa | ⌣c b. Proof: Cb(Aa/b) is definable over a Morley sequence (A i a i : i < ω) in lstp(Aa/b). Then (A i : i < ω) | ⌣ b and tp(a i /A i ) is Σ-analysable in α steps for all i < ω. The union of these analyses level-by-level gives us a Σ-analysis of Cb(Aa/b) over (A i : i < ω) in α steps. As (A i : i < ω) | ⌣ Cb(Aa/b), we obtain that Cb(Aa/b) is analysable over ∅ in α steps, and must be contained in c. Thus Aa | ⌣c b. Corollary 3.25. If c = ℓ Σ ∞ (b), then tp(b/c) is foreign to all Σ-analysable types. Proof: We apply Lemma 3.24 over c for

Proposition 3 .

 3 26. Let T be simple. Suppose b E is an ultraimaginary non-orthogonal to some regular type p ∈ S(∅), and SU p (ℓ p 1 (b)) < ω. Then there is an almost p-internal ultraimaginary e ∈ bdd u (b E ) \ bdd u (cl p (∅)). Moreover, e ∈ bdd u (ℓ p 1 (b)).

  This is reflexive and symmetric; moreover for c ′ | ⌣c ′′ c ′′′ with R(c ′ , c ′′ ) and R(c ′′ , c ′′′ ) we can find b ′ , b ′′ , b * , b ′′′ with b ′ c ′ ≡ b ′′ c ′′ ≡ b * c ′′ ≡ b ′′′ c ′′′ ≡ bc, such that E(b ′ , b ′′ ) and E(b * , b ′′′ ) hold. Since c ′′ is boundedly closed, b ′′ ≡ lstp c ′′ b * ; moreover b ′′ | ⌣c ′′ c ′ and b * | ⌣c ′′ c ′′′ by Lemma 3.24. By the Independence Theorem we can assume b ′′ = b * , so E(b ′ , b ′′′ ) and R(c ′ , c ′′′ ) hold. Hence R is generically transitive; let F be its transitive closure. The class c F is clearly almost p-internal. Moreover, if E(b ′ , b) holds there is c ′ with b ′ c ′ ≡ bc. Thus F (c ′ , c) holds, so c F is bounded over b E . Finally, suppose c F ∈ bdd u (cl p (∅)). By Proposition 3.18 there is c ′ | ⌣cl p(∅) c with R(c ′ , c). Hence there are b ′ , b * with b ′ c ′ ≡ b * c ≡ bc and |= E(b ′ , b * ). Applying a c-automorphism (and moving c ′ ), we may assume b = b * . Let A | ⌣ b be some parameters and a some realization of p over A with a | ⌣A b E ; we may assume Aa | ⌣b b ′ , whence A | ⌣ bb ′ . Moreover b | ⌣c Aa by Lemma 3.24, whence b ′ | ⌣c Aa. Thus b ′ | ⌣cl p(c) Aa by Fact 3.15. Now c ′ | ⌣cl p(∅) c yields c ′ | ⌣cl p(∅) cl p (c), and hence c ′ | ⌣cl p(∅) Aa. Then a | ⌣A cl p (∅) implies a | ⌣A c ′ . Now b ′ | ⌣c ′ Aa by Lemma 3.24, whence b ′ | ⌣A a. As b E = b ′ E we obtain a | ⌣A b E , a contradiction. Corollary 3.27. Let e be a supersimple ultraimaginary. Suppose e is non-orthogonal to some regular type p over some set B. Then there is an almost p-internal supersimple e ′ ∈ bdd qf u (Be) \ bdd qf u (cl p (B)). Proof: Let a be a representative of e with SU(a) < ∞ and put b = Cb(a/B). Then SU(b) < ∞, as b is bounded over a finite initial segment of a Morley sequence in lstp(a/B). Now e | ⌣b B, so tp(e/b) is non-orthogonal to p. Note that SU p (ℓ p 1 (a/b)/b) is finite by supersimplicity. By Proposition 3.26 applied over b there is an almost p-internal ultraimaginary e ′ ∈ bdd u (be) \ bdd u (cl p (b)); moreover e ′ ∈ bdd u (ℓ p 1 (a/b)) ⊆ bdd(ab). Thus e ′ is supersimple, almost p-internal over b and thus over B; it is quasi-finitary by Remark 3.6. Proposition 3.28. Let T be supersimple. If AB | ⌣ D and bdd qf u (A)∩ bdd qf u (B) = bdd qf u (∅), then bdd qf u (AD) ∩bdd qf u (BD) = bdd qf u (D). Proof: We may assume that A, B and D are boundedly closed. Consider e ∈ (bdd qf u (AD) ∩ bdd qf u (BD)) \ bdd qf u (D). Let p be a regular type of least SU-rank non-orthogonal to e over D. This exists by transitivity since e is tame. By Corollary 3.27 we may assume that e is almost p-internal of finite SU p -rank over D; let a ′ be a representative which is almost p-internal over D. Put a = Cb(a ′ D/A). As a | ⌣ D we obtain that tp(a) is almost p-internal; note that SU(a) < ∞. Since e | ⌣aD A, Lemma 3.4 implies that e ∈ bdd qf u (aD). So we may assume that A = bdd(a) and SU p (A) < ω. Moreover, we may assume that D = bdd(Cb(aa ′ /D)) is the bounded closure of a finite set. Let (A i : i < ω) be a Morley sequence in lstp(A/BD) with A 0 = A, and put B ′ = bdd(A 1 A 2 ). Then B ′ is almost p-internal of finite SU prank. Since e ∈ bdd qf u (AD) ∩ bdd qf u (BD) we have e ∈ bdd qf u (A i D) for all i < ω. Let e ′ be the set of B ′ D-conjugates of e, again a quasifinitary ultraimaginary. Since any B ′ D-conjugate of e is again in bdd

Definition 4 . 1 .Remark 4 . 2 .

 4142 Let e be ultraimaginary. We shall say that SU(a/e) < ω α if for all representatives b of e we have SU(a/b) < ω α . Conversely, SU(e/a) < ω α if there is a representative b with SU(b/a) < ω α . This does not mean that we define the value of SU(a/e) or of SU(e/a). In fact, one might define SU(e/a) = min{SU(b/a) : b a representative of e}, but this suggests a precision I am not sure exists.On the other hand, Example 2.9 shows that SU(a/e) = sup{SU(a/b) : b a representative of e} is not a good definition, as the rank of a point a over its connected component e = a E would be ω, i.e. the same as SU(a).

Lemma 4 . 3 .

 43 Let e be ultraimaginary. SU(e/a) < ω 0 if and only if e ∈ bdd u (a), and SU(a/e) < ω 0 if and only if a ∈ bdd(e). Proof: If b is a representative of e with SU(b/a) < ω 0 , then b ∈ bdd(a), so e ∈ bdd u (a). If e ∈ bdd u (a), then e ∈ dcl u (bdd(a)), so b = bdd(a) is a representative of e with SU(b/a) < ω 0 . If a / ∈ bdd(e), then there are arbitrarily many e-conjugates of a. Then for any representative b of e there is some e-conjugate a ′ of a which is not in bdd(b). Let b ′ be the image of b under an e-automorphism mapping a ′ to a. Then b ′ is a representative of e, and SU(a/b ′ ) ≥ ω 0 . On the other hand, if a ∈ bdd(e), then a ∈ bdd(b) for any representative b of e, whence SU(a/b) < ω 0 . Definition 4.4.

Remark 4 . 5 .Theorem 4 . 6 .

 4546 0-elimination is usually called weak elimination; in the presence of imaginaries this equals full elimination. I do not know what the definition of feeble elimination of ultraimaginaries should be in general for simple theories -but then their whole theory is much more problematic. If e is ultraimaginary with SU(e) < ω α+1 , then e can be α-eliminated. A supersimple theory has feeble elimination of ultraimaginaries; a supersimple theory of finite rank has elimination of quasi-finitary ultraimaginaries.Proof: Let a be a representative of e of minimal rank. Since SU(e) < ω α+1 we have SU(a) < ω α+1 . Suppose SU(a/e) ≥ ω α , so there is some representative b of e with SU(a/b) ≥ ω α . Let P be the family of regular types of SU-rank ω α . Then SU P (a) < ω and SU P (a/b) = n > 0; we choose b such that n is maximal. Consider a ′ ≡ lstp b a with a ′ | ⌣b a. Since e ∈ dcl u (b) we have e ∈ dcl u (a ′ ). By maximality of n, SU P (a/a ′ ) ≤ n = SU P (a/b) = SU(a/a ′ b) ≤ SU P (a/a ′ ), so equality holds. By lemma 3.17 we have a | ⌣ cl P (a ′ ) cl P (b). On the other hand, a | ⌣b a ′ implies by the anlogue of Fact 3.15 that a | ⌣ cl P (b) cl P (a ′ ), so c = Cb(a/cl P (b)cl P (a ′ )) ⊆ cl P (b) ∩ cl P (a ′ ). Then a | ⌣c b, so e ∈ bdd u (c) by Lemma 3.4. On the other hand, c ∈ cl P (a ′ ) ∩ cl P (b) implies SU(c/a ′ ) < ω α , and SU(c/b) < ω α . Then SU(a ′ /c) ≥ SU(a ′ /cb) ≥ ω α since SU(a ′ /b) ≥ ω α . It follows that SU(a) = SU(a ′ ) ≥ SU(c) + ω α .

Lemma 5. 1 . ( 1 ) 2 ) 3 )( 1 )( 2 )Definition 5 . 2 .

 11231252 Suppose a b. If c | ⌣ a and c | ⌣ b, then a c b. (Suppose a c b. If c | ⌣ ab then a b. (Suppose a c b. If tp(a) and tp(b) are foreign to tp(c), then a b. Proof: Consider any d with d | ⌣c a. Then cd | ⌣ a, whence cd | ⌣ b. Now b | ⌣ c implies b | ⌣c d. The converse follows by symmetry. Consider any d with d | ⌣ a. Clearly we may assume d | ⌣ab c, whence abd | ⌣ c. Since a | ⌣ c we get d | ⌣c a, whence d | ⌣c b and cd | ⌣ b. But c | ⌣d b, so d | ⌣ b; the converse follows by symmetry. (3) Consider any d with d | ⌣ a. Since a | ⌣ c we get d | ⌣c a, whence d | ⌣c b and cd | ⌣ b. If b | ⌣ d, then b | ⌣d c by foreigness, whence b | ⌣ cd, a contradiction. So b | ⌣ d; the converse follows by symmetry.For the following definitions, we require the notion of Σ-closure alluded to in Remark 3.16. For an ordinal α we putcl α Σ (A) = {a : tp(a/A) is Σ-analysable in α steps}. The Σ-closure of A is cl Σ (A) = cl ∞ Σ (A) = α cl α Σ (A). Remark 5.3. Note that ℓ Σ α (a/A) = cl α Σ (A) ∩ bdd(aA). In particular, a | ⌣ℓ Σ α (a/A) cl α Σ(A) by Lemma 3.24. Proposition 5.4. Let a and b be domination-equivalent over cl α Σ (∅), where a is quasi-finite and bdd qf u (a) ∩ bdd qf u (b) = bdd qf u (∅). Then ab ∈ cl α Σ (∅).

Proof: 4 . 5 . 6 .Fact 5 . 7 .

 45657 Note that by Lemma 3.24 the domination-equivalence of a and b over cl α Σ (∅) means that for any d and D = ℓ Σ α (abd) we have a | ⌣ D d ⇔ b | ⌣ D d. Clearly, domination-equivalence over cl α Σ (∅) is an ∅-invariant equivalence relation E on lstp(a). Let a ′ ≡ lstp b a with a ′ | ⌣b a. Then E(a ′ , a) holds. But bdd qf u (a) ∩ bdd qf u (a ′ ) ⊆ bdd qf u (a) ∩ bdd qf u (b) = bdd qf u (∅).Hence (a) E = (a ′ ) E ∈ bdd qf u (∅), and there is a ′′ | ⌣ a with E(a, a ′′ ). By Lemma 3.24 we have a| ⌣ℓ Σ α (a) a ′′ ℓ Σ α (aa ′′ ), whence a | ⌣ℓ Σ α (aa ′′ ) a ′′ . By domination-equivalence, a | ⌣ℓ Σ α (aa ′′ ) a, that is a ∈ ℓ Σ α (aa ′′ ), whence a ∈ cl α Σ (∅). Similarly, b ∈ cl α Σ (∅). Corollary 5.5. Let A, B, a, b be (hyperimaginary) sets, such that a is quasi-finite, bdd qf u (Aa) ∩ bdd qf u (Bb) = bdd qf u (∅),and a and b are interbounded over AB. Suppose AB is Σ-analysable in α steps for some ordinal α or α = ∞. Then a and b are Σ-analysable in α steps. Proof: Since a and b are interbounded over AB, they are dominationequivalent over cl α Σ (∅). Now apply Proposition 5.Remark By Theorem 4.6, if SU(Aa) or SU(Bb) is finite, then bdd(Aa) ∩ bdd(Bb) = bdd(∅) implies bdd qf u (Aa) ∩ bdd qf u (Bb) = bdd qf u (∅), and we recover [6, Lemma 1.15 and Lemma 1.22] for α = ∞ and α = 1. [10, Theorem 3.4(3)] Let Σ ′ be an ∅-invariant subfamily of Σ. Suppose tp(a) is Σ-analysable, but foreign to Σ \ Σ ′ . Then a and ℓ Σ ′ 1 (a) are domination-equivalent. Corollary 5.8. Let A ⊆ bdd(Cb(B/A)) consist of quasi-finite hyperimaginaries, with bdd qf u (A) ∩ bdd qf u (B) = bdd qf u (∅). If A is Σanalysable and Σ ′ is the subset of one-based partial types in Σ, then A is analysable in Σ \ Σ ′ . Proof: Suppose A is not analysable in Σ \ Σ ′ . For every finite tuple ā ∈ A put c ā = Cb(B/ā), and let C = {c ā : ā ∈ A}. Then A | ⌣C B, as for any ā ∈ A and C-indiscernible sequence (B i : i < ω) in tp(B/C) the set {π(x, B i ) : i < ω} is consistent, where π(x, B) = tp(ā/B), since π(x, B) does not fork over c ā ⊆ C. So A ⊆ bdd(C); as A is not analysable in Σ \ Σ ′ , neither is C, and there is ā ∈ A such that c = c ā is not analysable in Σ \ Σ ′ . Clearly c ⊆ bdd(ā) is quasi-finite and c = Cb(B/c). Replacing A by c we may thus assume that A is quasi-finite. Let A ′ ⊆ bdd(A) and B ′ ⊆ bdd(B) be maximally analysable in Σ \ Σ ′ . So tp(A/A ′ ) and tp(B/B ′ ) are foreign to Σ \ Σ ′ , and A ⊆ A ′ . Since A = Cb(B/A) we get A | ⌣A ′ B; as A | ⌣A ′ B ′ by foreignness to Σ \ Σ ′ , we obtain A | ⌣A ′ B ′ B. In particular B ⊆ B ′ . By Fact 5.7 the first Σ ′ -levels a = ℓ Σ ′ 1 (A/A ′ ) and b = ℓ Σ ′ 1 (B/B ′ ) are non-trivial, one-based, and a A ′ A and b B ′ B.

  Put Ā = bdd(A i : i ∈ I) and B = bdd(B i : i ∈ I). It remains to show that A ⊆ Ā. So suppose not. As in the proof of Corollary 5.8 put c ā = Cb(B/ā) for every finite tuple ā ∈ A, and let C = {c ā : ā ∈ A}. Then again A | ⌣C B and A ⊆ bdd(C); moreover c ā

d

  ∈ (bdd qf u (AB k ) ∩ bdd qf u (B)) \ bdd qf u (B k ) such that d is almost Σ i -internal over B k ; since tp(B/B k ) is foreign to Σ k we have i = k. Hence A | ⌣ ĀB k d, whence d ∈ bdd qf u ( ĀB k ) by Lemma 3.4. But Ā = bdd(A i : i ∈ I) and d |⌣A i B k Ā by almost Σ iinternality of d over B k , whence d ∈ bdd qf u (A i B k ). If B k d | ⌣ A i , then d | ⌣B k A i and d ∈ bdd qf u B k by Lemma 3.4, contradicting the choice of d. Therefore B k d | ⌣ A i ; by Corollary 3.27 there is almost Σ i -internal d ′ ∈ bdd qf u (B k d) \ bdd qf u (∅). Note that d ′ ∈ bdd qf u (A i B k )∩bdd qf u (B). But then d ′ A i | ⌣ B k , whence d ′ | ⌣A i B k and d ′ ∈ bdd qf u (A i ) ∩ bdd qf u (B) = bdd qf u (∅), a contradiction.Claim. We may assume B k = bdd(∅).Proof of Claim:Put A ′ = Cb(B/AB k ). Then B k ⊂ A ′ = Cb(B/A ′ ), and bdd(A ′ ) qf u ∩ bdd(B) qf u = bdd qf u (B k ). If B ′ = Cb(A ′ /B) = Cb(A ′ /B ′ ), then A ′ | ⌣B ′ B and A | ⌣A ′ B yield B | ⌣B ′ A by transitivity, since B ′ ⊆ bdd(B). Thus B ⊂ bdd(B ′ ). We add B k to the language; note that B k = bdd(∅) implies B | ⌣ B k , whence SU(B ′ /B k ) < SU(B). By induction it is thus sufficient to show that A ′ , B ′ is still a counterexample over B k .So suppose not, and let bdd(A′ ) = bdd(A ′ i : i ∈ I) and bdd(B ′ ) = bdd(B ′ i : i ∈ I) be decompositions, where A ′ i and B ′ i are maximally Σ ianalysable over B k in bdd(A ′ ) and bdd(B ′ ), respectively. So B ′ k is Σ kanalysable, whence B ′ k = B k ⊆ B by maximality. Since B ⊆ bdd(B ′ ) is almost Σ k -internal over B and (B ′ i : i = k) is foreign to Σ k , we get B | ⌣ B (B ′ i : i = k), whence B ⊂ B, a contradiction. By symmetry, we may also assume A k = bdd(∅).Put B ′ = Cb(B/A B). Then B ⊆ bdd(B ′ ), and since B is almostΣ k -internal over B, so is B ′ . If A ′ = Cb(B ′ /A), then B ′ | ⌣A ′ A and A | ⌣B ′ B yield A | ⌣A ′ B, since A ′ ⊆ bdd(A). Thus A ⊆ bdd(A ′ ). Put B ′′ = Cb(A/B ′ ) = Cb(A/B ′′ ). Then B ′′ ⊆ bdd(B ′ ) ⊆ bdd(A B), and B ′′ is almost Σ k -internal over B. Moreover, Ā | ⌣ B B ′ , whence B = Cb( Ā/ B) = Cb( Ā/B ′ ) ⊆ Cb(A/B ′ ) = B ′′ . Finally, A | ⌣B ′′ B ′ implies A ⊆ bdd(Cb(B ′ /A)) ⊆ bdd(Cb(B ′′ /A)).Claim. bdd qf u (A) ∩ bdd qf u (B ′′ ) = bdd qf u (∅).Proof of Claim: Suppose not. By Corollary 3.27 there is i ∈ I andd ∈ (bdd qf u (A) ∩ bdd qf u (B ′′ )) \ bdd qf u (∅) which is almost Σ i -internal; since A is foreign to Σ k we have i = k. As B ′′ is almost Σ k -internal over B we have d | ⌣ B B ′′ , whence d ∈ bdd qf u (A) ∩ bdd qf u ( B) ⊆ bdd qf u (A) ∩ bdd qf u (B) = bdd qf u (∅), a contradiction.Thus A, B ′′ is another counterexample; by induction on SU(AB/ Ā B) we may assumeSU(A/ Ā B) = SU(AB ′′ / Ā B) = SU(AB/ Ā B).Similarly, there is another counterexample A ′′ , B ′′ with Ā ⊂ A ′′ ⊆ bdd( ĀB ′′ ), whenceSU(B ′′ / Ā B) = SU(A ′′ B ′′ / Ā B) = SU(AB ′′ / Ā B) = SU(A/ Ā B). But if SU(X/Y ) = SU(Z/Y ) with X ∈ bdd(Y Z), then X Y Z. Thus B ′′ Ā B A, so B ′′ and A are domination-equivalent over cl i =k Σ i (∅) by perpendicularity and Lemma 5.1[START_REF] Ben | Discouraging Results for Ultraimaginary Independence Theory[END_REF]. So AB ′′ is analysable in i =k Σ i by Proposition 5.4. But tp(A/ Ā) is almost Σ k -internal, whence foreign to i =k Σ i , yielding the final contradiction. Remark 5.14. If C | ⌣ AB, then bdd qf u (A) ∩ bdd qf u (B) = bdd qf u (∅) implies bdd qf u (AC)∩bdd qf u (BC) = bdd qf u (C) by Lemma 3.28. Hence Theorem 5.13 applies over C; this can serve to refine the decomposition.

  3.2. [3, Lemma 1.10]Let A, a be hyperimaginary, and e, e ′ , e ′′ ultraimaginary.′′ , then e ∈ bdd u (e ′′ ). Lemma 3.4. If A is hyperimaginary and e, e ′ ultraimaginary with e | ⌣A e ′ , then bdd u

	Fact 3.3. [3, page 189] Let A be hyperimaginary and e, e ′ ultraimagi-
	nary with e | ⌣A e ′ .
	• If f, f ′ are ultraimaginary with f | ⌣A e, f ′ | ⌣A e ′ and f ≡ Lstp A then there is f ′′ | ⌣A ee ′ with ef ′′ ≡ Lstp A ef and e ′ f ′′ ≡ Lstp e ′ f ′ . f ′ , A • If e ′′ ∈ bdd u (Ae) then e ′′ | ⌣A e ′ . Moreover, if e | ⌣a e for every representative a of an ultraimaginary e Next, ultraimaginary bounded closures of independent sets intersect
	trivially.
	• If e | ⌣A e ′ e ′′ and e ′ | ⌣A e ′′ , then ee ′ | ⌣A e ′′ and e | ⌣A e ′ . • e | ⌣A ae ′ if and only if e | ⌣A a and e | ⌣Aa e ′ .
	The Independence Theorem and Boundedness axiom also hold.

Partially supported by ANR-09-BLAN-0047 Modig and ANR-13-BS01-0006.

It follows that a 0 and b 0 are interbounded over A ′ B ′ . We can now apply Corollary 5.5 to see that a 0 is analysable in Σ \ Σ ′ , whence a 0 ∈ A ′ . But then a | ⌣A ′ B ′ b, a contradiction. Remark 5.9. In a theory of finite SU-rank, due to weak elimination of quasi-finitary ultraimaginaries, we obtain that for any A, B

is analysable in the collection of non one-based types of SU-rank 1.

Remark 5.10. Without the quasi-finite hypothesis in Proposition 5.4, Corollary 5.5 and Corollary 5.8, the conclusions still hold if we assume that the full ultraimaginary bounded closures intersect trivially.

The following Theorem generalizes [START_REF] Chatzidakis | A note on canonical bases and modular types in supersimple theories[END_REF]Proposition 1.16] to supersimple theories of infinite rank, at the price of demanding that the quasifinite ultraimaginary bounded closures intersect trivially, rather than just the bounded closures. The proof is essentially the same, but we have to work with ultraimaginaries at key steps. Of course, in finite rank this is equivalent, due to elimination of quasifinite hyperimaginaries; moreover, the families Σ i in the Theorem are just different orthogonality classes of regular types of rank 1.

Definition 5.11. Two ∅-invariant families Σ and Σ ′ are perpendicular if no realization of a type in Σ can fork with a realisation of a type in Σ ′ .

Example 5.12. If p and p ′ are two orthogonal types of SU-rank 1 non-orthogonal to ∅ (or whose ∅-conjugates remain orthogonal), then the families of ∅-conjugates of p and of p ′ are perpendicular. Theorem 5.13. Let T be supersimple. Suppose A ⊆ bdd(Cb(B/A)) and B ⊆ bdd(Cb(A/B)), with bdd qf u (A)∩bdd qf u (B) = bdd qf u (∅). Let (Σ i : i ∈ I) be a family of pairwise perpendicular ∅-invariant families of partial types such that A is analysable in i∈I Σ i . For i ∈ I let A i and B i be the maximal Σ i -analysable subset of bdd(A) and bdd(B), respectively. Then A ⊆ bdd(A i : i < α) and B ⊆ bdd(B i : i < α);

Proof: Since Cb(A i /B) is tp(A i )-analysable and hence Σ i -analysable, we have Cb(A i /B) ⊆ B i ; similarly Cb(B i /A) ⊆ A i . As the families in (Σ i : i ∈ I) are perpendicular, we obtain

A.

Corollary 5.16. Let T be supersimple, and Σ 1 and Σ 2 two perpendicular ∅-invariant families of partial types. Suppose a is quasi-finite, tp(a) is analysable in Σ 1 ∪ Σ 2 and tp(a/A) is Σ 1 -analysable, with bdd qf u (a) ∩ bdd qf u (A) = bdd qf u (∅). Then tp(a) is Σ 1 -analysable.

Proof: Clearly we may assume that A = Cb(a/A). If a ′ = Cb(A/a), then A is interbounded with Cb(a ′ /A). Moreover, as tp(a/a ′ ) is Σ 1analysable, tp(a ′ ) is Σ 1 -analysable if and only if tp(a) is. So we may assume in addition that a = Cb(A/a).

By Theorem 5.13 we have bdd(a) = bdd(ℓ Σ 1 ∞ (a), ℓ Σ 2 ∞ (a)). Hence tp(ℓ Σ 2 ∞ (a)/A) is Σ 1 -analysable. By perpendicularity, ℓ Σ 2 ∞ (a) ∈ bdd(A) ∩ bdd(a) = bdd(∅). Hence a ∈ ℓ Σ 1 ∞ (a) is Σ 1 -analysable. For SU(a) finite, this specialises to [START_REF] Chatzidakis | A note on canonical bases and modular types in supersimple theories[END_REF]Proposition 1.20]


Jonathan Grizou
email: jonathan.grizou@inria.fr

Manuel Lopes
email: manuel.lopes@inria.fr

Pierre-Yves Oudeyer
email: pierre-yves.oudeyer@inria.fr

Robot Learning from Unlabelled Teaching Signals

INTRODUCTION

Human Robot Interaction (HRI) is the multidisciplinary study of interactions between humans and robots. Researchers in the field investigate how robots properties -such as embodiment, speech, gesture, facial expression, personalitycan change the behaviour, feelings or attitudes of humans; how robots can produce and recognise communicative acts and how robots and humans can learn from each other [START_REF] Dautenhahn | Socially intelligent robots: dimensions of human-robot interaction[END_REF][START_REF] Fong | A survey of socially interactive robots[END_REF]; among others. Among them (a) perception and (b) task learning are vastly investigated areas which rely mainly on the machine learning and artificial intelligence community. While impressive results have been obtained in each domain [START_REF] Argall | A survey of robot learning from demonstration[END_REF][START_REF] Zeng | A survey of affect recognition methods: Audio, visual, and spontaneous expressions[END_REF], most work considered them separately making important assumptions:

(a) In perceptual learning research applied to HRI, human's communicative signals are learnt to be associated to their respective meanings. Such learning often rely on supervised classification or regression methods, which implies collecting signals samples associated with their respective meanings. This latter requirement implies the robot is aware of the communicative goal of the human.

(b) In robot learning from human interaction most work has focused on how to extract statistical task models from human teaching signals [START_REF] Kaplan | Robotic clicker training[END_REF]. Therefore a usual assumption is that the robot understands the meanings of human's communicative signals.

Their respective assumptions makes those two lines of research incompatible. On the one hand, working on perceptual learning, i.e. learning the signal-to-meaning mapping, requires the robot to know the task. On the other hand, teaching a new task to a robot requires the robot to already know the signal-to-meaning mapping. Consequently it is impossible for a user to interact -from scratch -with a robot using his own preferred teaching signals, the user must comply to the use of pre-defined ones. This paper describes a preliminary approach allowing a robot to be instructed a new unknown task by a human using communicative signals initially totally unknown to the robot. In other words, we address the problem of removing the need for calibration.

The question of how a robot can learn to interpret personalised and potentially noisy teaching signals has not been much explored. In a preliminary work [START_REF] Lopes | Simultaneous acquisition of task and feedback models[END_REF], we presented a computational approach addressing this problem by considering a finite space of teaching signals in simulation but requiring to bootstrap the system with known symbols. Later [START_REF] Grizou | Robot Learning Simultaneously a Task and How to Interpret Human Instructions[END_REF], we released the need for bootstrapping and allow the teacher to use any kind of signal that can be represented as a fixed length feature vector, which is better suited for HRI scenarios.

We provide an intuition on how the algorithm works and report preliminary results.

INTUITION

For the present discussion, we restrain our analysis to scenarios where the user does not actively deliver commands to the robot, but only delivers feedback about actions performed by the robot. Such feedback is given by an unlabelled communicative signals than can be either symbolic, e.g. button presses, or represented as a fixed length feature vector, e.g. spoken words. Signals are not converted to a symbolic meaning. The robot needs to actively execute actions to solve the tasks which implies learning the signal-to-meaning association.

This control can be exemplified for a reaching task, where the user wants the robot to reach a target position unknown at start. The robot performs several actions (e.g. moving left or right), and receives unlabelled feedback signals from the user. The feedback signals are generated as a response to the execution of an action in a state according to the true unknown task the user wants to solve. For instance, binary feedback signals simply encode whether the action executed is correct or incorrect according to the user intended task.

The key point is that these signals are generated from an underlying model that for binary signals has two different classes.

To solve this problem, we must exploit an other source of information, namely task constraints. Task constraints are properties of the environment that limit the space of possible task, e.g. of possible target positions. This set of hypothetic tasks enable us to create a set of hypothetic signal-tomeaning models. Since the right task will assign the right labels to the signals, while the other tasks will gradually mix them, estimating models' likelihood is a good measure to identify the user's intended task.

PRELIMINARY RESULTS

We construct a small size pick-and-place task with a real robot that has 4 actions available: rotate left, rotate right, grasp cube and release cube. This robot is going to be programmed using natural speech with words a priori unknown to the robot. The teacher is facing the robot and chooses a specific arrangement of cube, i.e. a specific task, it wants the robot to build. It then decides one word to use as positive feedback and one as negative feedback and starts teaching the robot. Once the robot has understood the first task, using the method presented above, it has also understood the signal-to-meaning model. We can freeze the corresponding classifier and start learning a new task. Fig. 1 shows results from this setting. In the first run it takes about 100 iterations for the robot to learn both the task and the signal-to-meaning mapping. Whereas in the second run, when reusing knowledge from the first one, the robot is able to learn a new task faster, in about 30 iterations, meaning that it has well found the two clusters in our feature space as well as the mapping to their corresponding meanings.

LIMITATIONS AND PERSPECTIVES

We presented a new approach for HRI which enables a user to use his own preferred teaching signals when interacting with a robot, e.g. using his own language or defining his own finger movements. The claim that it would be more intuitive or natural to interact with a robot this way is far from being obvious and should be tested with care. When interacting with machine, including robots, we are often, if not always, told how to use them. Do people want to have an open-ended choice about how to interact with machines? Would they be more efficient? Investigating such question is part of our future work.

Our method remove the necessity of a calibration phase. Among the many application requiring calibration, Brain Computer Interaction (BCI) is one of the most challenging. We are currently investigating in this direction [START_REF] Grizou | Zero-calibration bmis for sequential tasks using error-related potentials[END_REF]. BCI has the advantage to make obvious the fact that we cannot ask the user to comply to the use of pre-defined signals.

An important assumption of our method is that it is possible to define a finite -and reasonable -set of task hypothesis.

This assumption is limiting for many theoretical problems but many useful real word application can still benefit from it. An eligible scenario is the problem of grasping, on a table, one object among a finite set of objects. In this scenario the set of hypothesis consist of all the objects on the table.

While this is not the main target, the work presented in this paper is also relevant with regards to infant social development and learning, as well as in adult mutual adaptation of social cues. This has been the subject of experiments in experimental semiotics [START_REF] Galantucci | Experimental semiotics: a new approach for studying communication as a form of joint action[END_REF], such as in the work of Griffiths et al. [START_REF] Griffiths | Bottom-up learning of feedback in a categorization task[END_REF] who conducted an experiment with human learners learning the meaning of unknown symbolic teaching signals.

An innovative direction would be to embed the algorithmic principles introduced in this paper for experimental linguistics studies.

Figure 1 :

 1 Figure 1: Evolution of the probability of the taught task. 1) The robot learns a task from unlabelled speech feedback. 2) By freezing the classifier corresponding to the first task, the user teaches the robot a new task faster.

Acknowledgement Work (partially) supported by INRIA, Conseil Régional d'Aquitaine and the ERC grant EXPLORERS 24007.