My paper with Tatiana Gateva-Ivanova is to be published. I’ll describe it here to demonstrate that one can find permutation groups almost anywhere.
The Yang–Baxter equation (YBE) is a kind of braiding equation for a linear map R on V⊗V. We define two linear maps on V⊗V⊗V, namely R^{12}, acting on the first two components of a 3-tensor and fixing the third, and R^{23}, acting on the second and third component and fixing the first. The Yang–Baxter equation is R^{12}R^{23}R^{12} = R^{23}R^{12}R^{23}. It has connections with mathematical physics which I don’t feel competent to describe here.
The combinatorial (or set-theoretic) YBE is defined for a map r on X×X, where X is a set; it asserts that r^{12}r^{23}r^{12} = r^{23}r^{12}r^{23}, where r^{12} and r^{23} are similarly defined. Of course, a solution of this equation gives a solution of the “linear” equation over any field F, on setting V = FX.
We impose three further conditions on the function r, namely
- r is an involution;
- r fixes pointwise the diagonal of X×X;
- r is non-degenerate (see below).
We can describe the function r in a different way: if r(x,y) = (u,v), we set u = f_{x}(y) and v = g_{y}(x); for each x, the function f_{x} is a map from X to itself, and similarly for g_{y}. Non-degeneracy is the requirement that each of the functions f_{x} and g_{y} is a bijection on x.
I will use the short term solution for a function r that satisfies the combinatorial Yang–Baxter equation and our extra three conditions.
Here are two examples.
- The function r(x,y) = (y,x) satisfies the Yang–Baxter equation (essentially by the usual proof that the transpositions (1,2) and (2,3) in the symmetric group S_{3} satisfy the braid relation), and also our three additional conditions. In this case, the functions f_{x} and g_{y} are the identity, for all choices of x and y. This is referred to as the trivial solution.
- The function which swaps (1,2) with (3,1), (1,3) with (2,1), and (2,3) with (3,2) (and fixes all diagonal pairs) is a solution on X = {1,2,3}.
If r is non-degenerate then the functions f and g can be regarded as maps from X into the symmetric group on X. It can be shown that a pair of maps f,g: X → Sym(X) arise in this way from a solution if and only if they satisfy the following conditions:
- f_{x}(x) = x;
- f_{fx(y)}g_{y}(x) = x;
- f_{fx(y)}f_{gy(x)} = f_{x}f_{y}.
(Unlike my usual convention, maps act on the left and compose right-to-left.)
Note that the second equation shows that g is determined by f. So everything can be expressed in terms of f. In our second example above, f_{1} and g_{1} are equal to the transposition (2,3), while the other functions are the identity.
One of the most important unsolved problems about this set-up is the following.
Problem. If f and g satisfy the above conditions and |X| > 1, is it true that there exist two points x and y with f_{x} = f_{y}?
I will say something about the significance of this problem later.
The commonest way to turn this situation into algebra is to define the Yang–Baxter group corresponding to r to be the group
G(r) = ⟨X | xy=uv whenever r(x,y)=(u,v)⟩
For the trivial solution, this is a free abelian group of rank |X|.
One can also consider the semigroup, or the F-algebra (for a field F) generated by X with the same relations.
However, to me a more natural construction is to let G* be the group of permutations of X generated by the functions f_{x}. This is the Yang–Baxter permutation group of the solution r. It turns out that the map x → f_{x} extends to a homomorphism from G to G*.
There are some slightly surprising things that can be said. For example,
- The YB permutation group cannot be transitive if |X|>1.
- The YB group (and hence the YB permutation group) of every finite solution is a solvable group.
In fact, one of the things we prove is that the derived length of G is
always one greater than that of the finite group G*.
Define an equivalence relation on X by x ≡ y if f_{x} = f_{y}. An affirmative answer to the problem stated earlier would say that this equivalence relation is not the relation of equality whenever |X|>1. It is easy to see that the given solution induces a solution (called a retract) on the set of equivalence classes. We can repeatedly take retracts; if the corresponding congruences are always non-trivial, eventually we get a solution on a 1-element set. Such a solution is called a multipermutation solution; its level is the number of retractions required to reach the 1-element set.
For a multipermutation solution, the YB group is solvable with derived length bounded by the multipermutation length; so the derived length of the YB permutation group is bounded by one less than the multipermutation length.
Our constructions of solutions with finite multipermutation level indicate that the cardinality of X grows exponentially with the multipermutation level; our smallest example of level m has cardinality 2^{m-1}+1.
We also have a structure theorem for finite solutions with abelian YB permutation group; they are necessarily multipermutation, with level at most the number of orbits of G*, and they can be constructed from the solutions corresponding to the orbits by a construction known as strong twisted union. Every finite abelian group is isomorphic to the YB permutation group of such a solution.
There is surely much more to say about this situation!
I believe the answer to the problem is “no”. Here there is an example:
Let X={1,2,…,8} and
f_1 = g_1 = (5,7),
f_2 = g_2 = (6,8),
f_3 = g_3 = (2,6)(4,8)(5,7),
f_4 = g_4 = (1,5)(3,7)(6,8),
f_5 = g_5 = (1,3),
f_6 = g_6 = (2,4),
f_7 = g_7 = (1,3)(2,6)(4,8),
f_8 = g_8 = (1,5)(2,4)(3,7).
Then all the conditions you mentioned are satisfied. However, f_x=f_y if and only if x=y.