Читать книгу Invariants And Pictures: Low-dimensional Topology And Combinatorial Group Theory - Vassily Olegovich Manturov - Страница 27

In this section, we are going to present an invariant of virtual braids proposed by the first-named author in [Manturov, 2008] and show that the classical braid group is a subgroup of the virtual one. For an elementary proof of this fact see [Manturov, 2016b]. More precisely, we give a generalisation of the complete braid invariant described before for the case of virtual braids. The new “virtual invariant” is quite strong. The question of whether the invariant is complete was answered negatively by O. Chterental [Chterental, 2015]. The completeness of the multi-variable extension of the invariant (see [Manturov, 2003]) is unknown.

Thus the main question is the word problem for the virtual braid group: how to recognise whether two different (regular) virtual braid diagrams β₁ and β₂ represent the same braid B.

Remark 2.2. The recognition problem for virtual braids was solved by O. Chterental [Chterental, 2017].

Given two braid diagrams one can apply the virtual braid group relations to one of those diagrams without getting the other diagram and one does not know whether he has to stop and say that they are not isomorphic or he has to continue.

A partial answer to this question is the construction of a virtual braid group invariant; i.e., a function on virtual braid diagrams (or braid words) that is invariant under all virtual braid group relations. In this case, if for an invariant f we have f(β₁) ≠ f(β₂), then β₁ and β₂ represent two different braids.

Here we give a generalisation of the complete classical braid group invariant for the case of virtual braids.

Let G be the free group in generators a₁ . . . , an, t. Let Ei be the quotient set of right residue classes {ai}\G for i = 1, . . . , n.

Definition 2.17. A virtual n-system is a set of elements e₁ ∈ E₁, e₂ ∈ E₂, . . . , en ∈ En.

The aim of this subsection is to construct an invariant map f (non-homomorphic) from the set of all virtual n-strand braids to the set of virtual n-systems.

Let β be a braid word. Let us construct the corresponding virtual n-system f(β) step-by-step. Namely, we shall reconstruct the function f(βψ) from the function f(β), where ψ is σi or or ζi.

First, let us take n residue classes of the unit element of G: e, e, . . . , e. This means that we have defined

Now, let us read the word β. If the first letter is ζi, then all words but ei,e_i+1 in the n-systems stay the same, ei becomes equal to t and e_i+1 becomes t⁻¹ (here and in the sequel, we mean, of course, residue classes, e.g. [t] and [t⁻¹]. But we write just t and t⁻¹ for the sake of simplicity).

Now, if the first letter of our braid word is σi, then all classes but e_i+1 stay the same, and e_i+1 becomes . Finally, if the first letter is , then the only changing element is ei: it becomes a_i+1.

The procedure for each next letter (generator) is the following. Denote the index of this letter (the generator or its inverse) by i. Assume that the left strand of this crossing originates from the point (p, 1), and the right one originates from the point (q, 1). Let ep = P, eq = Q, where P, Q are some words representing the corresponding residue classes. After the crossing all residue classes but ep, eq should stay the same.

Then if the letter is ζi then ep becomes P · t, and eq becomes Q · t⁻¹. If the letter is σi, then ep stays the same, and eq becomes . Finally, if the letter is , then eq stays the same, ep becomes PQ⁻¹aqQ. Note that this operation is well defined.

Actually, if we take the words instead of the words P, Q, then we get: in the first case

and in the second case we obtain

In the third case we obtain

Thus, we have defined the map f from the set of all virtual braid diagrams to the set of virtual n-systems.

Theorem 2.5. The function f, defined above, is a braid invariant. Namely, if β₁ and β₂ represent the same braid β, then f(β₁) = f(β₂).

Proof. We have to demonstrate that the function f defined on virtual braid diagrams is invariant under all virtual braid group relations. It suffices to prove that, for the words β₁ = βγ₁ and β₂ = βγ₂ where γ₁ = γ₂ is a relation we have proved, we can also prove f(β₁) = f(β₂). During the proof of the theorem, we shall call it the A-statement.

Indeed, having proved this claim, we also have f(β₁δ) = f(β₂δ) for arbitrary δ because the invariant f(β₁δ) (as well as f(β₂δ)) is constructed step-by-step; i.e., knowing the value f(β₁) and the braid word δ, we easily obtain the value of f(β₁δ). Hence, for braid words β, δ and for each braid group relation γ₁ = γ₂ we prove that f(βγ₁δ) = f(βγ₂δ). This completes the proof of the theorem.

Let us return to the A-statement.

To prove the A-statement, we must consider all virtual braid group relations. The commutation relation σiσj = σjσi for “far” i, j is obvious: all four strands involved in this relation are different, so the order of applying the operation does not affect on the final result. The same can be said about the other commutation relations, involving one σ and one ζ or two ζ’s.

Now let us consider the relation which is pretty simple too.

Actually, let us consider a braid word β, and let the word β₁ be defined as for some i. Let f(β) = (P₁, . . . , Pn), . Let p and q be the numbers of strands coming to the crossing from the left side and from the right side. Obviously, for j ≠ p, q we have . Besides, .

Now let us consider the case (obviously, the case is quite analogous to this one).

As before, denote f(β) by (. . . Pi . . .), and f(β₁) by , and the corresponding strand numbers by p and q. Again, we have: for j ≠ p, q : . Moreover, by definition of f(since the p-th strand makes an overcrossing twice), and .

Now let us check the invariance under the third Reidemeister move. Let β be a braid word, β₁ = βζiζ_i+1ζi and β₂ = βζ_i+1,ζiζ_i+1. Let p, q, r be the global numbers of strands occupying positions n, n + 1, n + 2 at the bottom of β.

Denote f(β) by (P₁, . . . , Pn), f(β₁) by , and f(β₂) by . Obviously, ∀i ≠ p, q, r we have . Direct calculations show that and .

Now, let us consider the mixed move by using the same notation: β₁ = βζiζ_i+1σi, β₂ = σ_i+1ζiζ_i+1. As before, for all j ≠ p, q, r. Now, direct calculation shows that

and

Finally, consider the “classical” case ; the notation is the same. Again . Besides this, since the p-th strand forms two overcrossings in both cases then . Then,

and

As we see, the final results coincide and this completes the proof of the theorem.

Thus, we have proved that f is a virtual braid invariant; i.e., for a given braid B the value of f does not depend on the diagram representing B. So, we can write simply f(B).

Remark 2.3. In fact, we can think of f as a function valued not in (E₁, . . . , En), but in n copies of G: all these invariants were proved for the general case of (G, . . . , G). The present construction of (E₁, . . . , En) is considered for the sake of simplicity.

Classical braids (i.e., braids without virtual crossings) can be considered up to two equivalences: classical (modulo only classical moves) and virtual (modulo all moves). Now, we prove that they are the same. This fact is not new. It follows from [Fenn, Rimanyi and Rourke, 1997]. An elementary proof was given in [Manturov, 2016b].

Theorem 2.6. Two virtually equal classical braids B₁ and B₂ are classically equal.

Proof. Since B₁ is virtually equal to B₂, we have f(B₁) = f(B₂). Now, taking into account that f is a complete invariant on the set of classical braids, we have B₁ = B₂ (in the classical sense).

As in the case of virtual knots, in the case of virtual braids there exists a forbidden move, namely, . Now, we are going to show that it cannot be represented by a finite sequence of the virtual braid group relations.

Theorem 2.7. A forbidden move (relation) cannot be represented by a finite sequence of legal moves (relations).

Proof. Actually, let us calculate the values f(σ₁σ₂ζ₁) and f(ζ₂σ₁σ₂). In the first case we have:

In the second case we have:

As we see, the final results are not the same (i.e., they represent different virtual n-systems); thus, the forbidden move changes the virtual braid.

Remark 2.4. If we put t = 1, the results f(X) and f(Y) become the same. Thus that is the variable t that “feels” the forbidden move.