rel="nofollow" href="#fb3_img_img_43a97301-9069-5a7d-8e98-006dd503d274.png" alt="figure"/>e1 ∈ E1, e2 ∈ E2, . . . , en ∈ EnThe 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
First, let us take n residue classes of the unit element of G:
Now, let us read the word β. If the first letter is ζi, then all words but ei,ei+1 in the n-systems stay the same, ei becomes equal to t and ei+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 ei+1 stay the same, and ei+1 becomes
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
Actually, if we take the words
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
Actually, let us consider a braid word β, and let the word β1 be defined as
Now let us consider the case
As before, denote f(β) by (. . . Pi . . .), and f(β1) by
Now
