On systems of word equations with simple loop sets
On systems of word equations with simple loop sets
On systems of word equations with simple loop sets
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contains one occurrence <strong>of</strong> p. There are nine <strong>word</strong>s α ∈ H ∗ satisfying ϕ(α) =<br />
aa, namely<br />
α 1 = x 1 x 1 , α 2 = x 1 y 1 , α 3 = x 1 z 2 ,<br />
α 4 = y 1 x 1 , α 5 = y 1 y 1 , α 6 = y 1 z 2 ,<br />
α 7 = z 2 x 1 , α 8 = z 2 y 1 , α 9 = z 2 z 2 .<br />
Therefore (13) has the form<br />
9∑<br />
9∑<br />
|ψ(yzxy)| αi = |ψ(xyyz)| αi . (14)<br />
i=1<br />
i=1<br />
The equality holds, since |ψ(yzxy)| αi is equal to one for i = 7, and is zero<br />
otherwise, while |ψ(xyyz)| αi is one just for i = 5.<br />
Informally, we can say that the factor aa comes on the left side <strong>of</strong> the equation<br />
from a different source than on the right side. The formalism <strong>of</strong> the characteristic<br />
equation is designed to express and exploit that fact.<br />
5 <strong>On</strong>e <strong>loop</strong> <strong>systems</strong><br />
We are now ready to prove Theorem 2. The theorem deals <strong>with</strong> the <strong>systems</strong><br />
S and T k when n = 1, hence we define<br />
X = {u 1 ,u 2 , . . .,u m ,v 1 ,x 0 ,x 1 , . . .,x m ,y 0 ,y 1 }.<br />
Fix k ≥ 2, and a morphism ϕ, which solves the system T k . Define H, morphisms<br />
ψ and ϕ as in the previous section. Our task is to show that ϕ<br />
solves S as well. It will be done by showing that the primitive roots <strong>of</strong> all<br />
ϕ(u 1 ), ϕ(u 2 ), . . .,ϕ(u m ) are conjugate. Theorem 1 then applies.<br />
Denote<br />
l i = x 0 u i 1x 1 u i 2x 2 · · ·u i mx m ,<br />
r i = y 0 v i y 1<br />
for i = k, k + 1, k + 2. Recall that, for each i,<br />
(l i , r i ) = (ψ(l i ), ψ(r i ))<br />
is the characteristic equation <strong>of</strong> (l i , r i ) <strong>with</strong> respect to ϕ.<br />
Define the <strong>word</strong> p, whose number <strong>of</strong> occurrences will be counted. Let t be the<br />
shortest among the primitive roots <strong>of</strong> <strong>word</strong>s ϕ(u 1 ), ϕ(u 2 ), . . . , ϕ(u m ). Then p<br />
is defined by the following conditions:<br />
11