7 The Asterix equation
A translation-invariant magma is a magma whose carrier \(G\) is an abelian group \(G = (G,+)\), and whose magma operation takes the form
for some function \(f: G \to G\). Thus the translations on \(G\) become magma isomorphisms.
For translation-invariant magmas, an equational law simplifies to a univariate functional equation. For instance, writing \(x = y+h\), we have
where \(f^2 = f \circ f\), so the Asterix equation (Definition 2.21) for such magmas simplifies to the univariant functional equation
for \(h \in G\).
This equation has some degenerate solutions, for instance we can take \(f(h) = c\) for any constant \(c\) of order \(3\) in \(G\). It is challenging to construct more interesting solutions to this equation; however, we can do this if \(G=\mathbb {Z}\) by a greedy algorithm. We need the following technical definition.
A partial solution \((E_0, E_1, E_2, f)\) to 1 consists of nested finite sets
together with a function \(f: E_1 \to E_2\) with the following properties:
If \(h \in E_0\), then \(f(h) \in E_1\), so that \(f^2(h)\) is well-defined as an element of \(E_2\); furthermore, \(h + f(h) + f^2(h)\) lies in \(E_1\), so that the left-hand side of 1 makes sense; and 1 holds.
The function \(f\) is a bijection from \(E_1 \backslash E_0\) to \(E_2 \backslash E_1\).
We partially order the space of partial solutions to 1 by writing \((E_0, E_1, E_2, f) \leq (E'_0, E'_1, E'_2, f')\) if the following properties hold:
\(E_i \subset E'_i\) for \(i=0,1,2\).
\(f\) agrees with \(f'\) on \(E_0\).
When this occurs we say that the partial solution \((E'_0, E'_1, E'_2, f')\) extends the partial solution \((E_0, E_1, E_2, f)\).
We define the empty partial solution \((E_0,E_1,E_2,f)\) by setting \(E_0=E_1=E_2\) to be the empty set, and \(f\) to be the empty function; it is the minimal element of the above partial order.
We have the following iterative construction, that lets us add arbitrary elements to the core domain \(E_0\):
Because \(f\) maps \(E_1 \backslash E_0\) bijectively to \(E_2 \backslash E_1\), there are three cases:
\(h\) is equal to an element \(h_0\) of \(G \backslash E_2\).
\(h\) is equal to an element \(h_0\) of \(E_1 \backslash E_0\).
\(h\) is equal to \(h_1 = f(h_0)\) for some \(h_0 \in E_1 \backslash E_0\), so that \(h_1 \in E_2 \backslash E_1\).
We deal with these three cases in turn.
First suppose that \(h = h_0\in G \backslash E_2\). We perform the following construction.
Choose an element \(h_1 \in \mathbb {Z}\) that does not lie in \(E_2 \cup \{ h_0\} \); this is possible because \(E_2\) is finite.
Choose an element \(h_2 \in \mathbb {Z}\) such that \(h_2, h_0+h_1+h_2\), and \(-h_1-h_2\) are all distinct from each other and lie outside of \(E_2 \cup \{ h_0, h_1\} \); this is possible because \(E_2\) is finite.
Promote \(h_0\) to \(E_0\), promote \(h_1, h_0+h_1+h_2\) to \(E_1\), and promote \(h_2, -h_1-h_2\) to \(E_2\), creating new sets
\begin{align*} E’_0 & := E_0 \cup \{ h_0\} \\ E’_1 & := E_1 \cup \{ h_0, h_1, h_0+h_1+h_2\} \\ E’_2 & := E_2 \cup \{ h_0, h_1, h_0+h_1+h_2, h_2, -h_1-h_2\} . \end{align*}Clearly we still have nested finite sets \(E'_0 \subset E'_1 \subset E'_2\).
Extend \(f : E_1 \to E_0\) to a function \(f': E'_1 \to E'_0\) by defining
\begin{align*} f’(h_0) & := h_1 \\ f’(h_1) & := h_2 \\ f’(h_0+h_1+h_2) & := -h_1-h_2 \end{align*}while keeping \(f'(h)=f(h)\) for all \(h \in E_1\).
It is then a routine matter to verify that \((E'_0,E'_1,E'_2,f')\) is a partial solution to 1 extending \((E_0,E_1,E_2,f)\) and that \(E'_0\) contains \(h_0\), as required.
Now suppose that \(h = h_0 \in E_1 \backslash E_0\), then the quantity \(h_1 := f(h_0)\) lies in \(E_2 \backslash E_1\). We perform the following variant of the above construction:
Choose an element \(h_2 \in \mathbb {Z}\) such that \(h_2, h_0+h_1+h_2\), and \(-h_1-h_2\) are all distinct and lie outside of \(E_2\). This is possible because \(E_2\) is finite.
Promote \(h_0\) to \(E_0\), promote \(h_1\) and \(h_0+h_1+h_2\) to \(E_1\), and promote \(h_2, -h_1-h_2\) to \(E_2\), thus creating new sets
\begin{align*} E’_0 & := E_0 \cup \{ h_0\} \\ E’_1 & := E_1 \cup \{ h_1, h_0+h_1+h_2\} \\ E’_2 & := E_2 \cup \{ h_0+h_1+h_2, h_2, -h_1-h_2\} . \end{align*}Clearly we still have nested finite sets \(E'_0 \subset E'_1 \subset E'_2\).
Extend \(f : E_1 \to E_0\) to a function \(f': E'_1 \to E'_0\) by defining
\begin{align*} f’(h_1) & := h_2 \\ f’(h_0+h_1+h_2) & := -h_1-h_2 \end{align*}while keeping \(f'(h)=f(h)\) for all \(h \in E_1\).
It is then a routine matter to verify that \((E'_0,E'_1,E'_2,f')\) is a partial solution to 1 extending \((E_0,E_1,E_2,f)\) and that \(E'_0\) contains \(h_0\), as required.
Finally, suppose that \(h = h_1 = f(h_0)\) for some \(h_0 \in E_1 \backslash E_0\), so that \(h_1 \in E_2 \backslash E_1\). Then we perform the following algorithm.
Choose an element \(h_2 \in \mathbb {Z}\) such that \(h_2, h_0+h_1+h_2\), and \(-h_1-h_2\) are all distinct and lie outside of \(E_2\). This is possible because \(E_2\) is finite.
Choose an element \(h_3 \in \mathbb {Z}\) such that \(h_3, h_1+h_2+h_3\), and \(-h_2-h_3\) are all distinct and lie outside of \(E_2 \cup \{ h_2, h_0+h_1+h_2,-h_1-h_2\} \). This is possible because \(E_2\) is finite.
Promote \(h_0, h_1\) to \(E_0\), promote \(h_2, h_0+h_1+h_2, h_1+h_2+h_3\) to \(E_1\), and promote \(h_3, -h_1-h_2,-h_2-h_3\) to \(E_2\), creating new sets
\begin{align*} E’_0 & := E_0 \cup \{ h_0, h_1 \} \\ E’_1 & := E_1 \cup \{ h_1, h_2, h_0+h_1+h_2, h_1+h_2+h_3 \} \\ E’_2 & := E_2 \cup \{ h_2, h_3, h_0+h_1+h_2, h_1+h_2+h_3, -h_1-h_2, -h_2-h_3\} . \end{align*}Clearly we still have nested finite sets \(E'_0 \subset E'_1 \subset E'_2\).
Extend \(f : E_1 \to E_0\) to a function \(f': E'_1 \to E'_0\) by defining
\begin{align*} f’(h_1) & := h_2 \\ f’(h_0+h_1+h_2) & := -h_1-h_2 f’(h_2) & := h_3 \\ f’(h_1+h_2+h_3) & := -h_2-h_3 \end{align*}while keeping \(f'(h)=f(h)\) for all \(h \in E_1\).
It is then a routine matter to verify that \((E'_0,E'_1,E'_2,f')\) is a partial solution to 1 extending \((E_0,E_1,E_2,f)\) and that \(E'_0\) contains \(h_0\), as required.
Every partial solution \((E_0,E_1,E_2,f)\) to 1 can be extended to a full solution \(\tilde f: \mathbb {Z}\to \mathbb {Z}\).
If we arbitrarily well-order the integers, and iterate Lemma 7.2 to add the least element of \(\mathbb {Z}\backslash E_0\) in this well-ordering to \(E_0\), we obtain an increasing sequence \((E^{(n)}_0, E^{(n)}_1, E^{(n)}_2, f^{(n)})\) of partial solutions to 1, where the \(E^{(n)}_0\) exhaust \(\mathbb {Z}\): \(\bigcup _{n=1}^\infty E^{(n)}_0 = \mathbb {Z}\). Taking limits, we obtain a full solution \(\tilde f\).
There exists a solution \(f:\mathbb {Z}\to \mathbb {Z}\) to 1 such that the map \(h \mapsto h + f(h)\) is not injective.
Select integers \(h_0,h_1,h_2,h'_0,h'_1,h'_2\) such that the quantities
are all distinct, but such that
(there are many assignments of variables that accomplish this). Then set
and define \(f: E_1 \to E_2\) by the formulae
One can then check that \((E_0, E_1, E_2, f)\) is a partial solution to 1, and by construction \(h \mapsto h + f(h)\) is not injective on \(E_1\). Using Lemma 7.2 to extend this partial solution to a full solution, we obtain the claim.
There exists a magma obeying the Asterix law (Definition 2.21) with carrier \(\mathbb {Z}\) such that the left-multiplication maps \(L_y: x \mapsto y \diamond x\) are not injective for any \(y \in \mathbb {Z}\). In particular, it does not obey the Obelix law (Definition 2.28)
Note that \(L_y (y+h) = y + h + f(h)\), so the injectivity of the left-multiplication maps is equivalent to the injectivity of the map \(h \mapsto h + f(h)\). The non-injectivity then follows from Corollary 7.4. Note that the Obelix law clearly expresses \(x\) as a function of \(y\) and \(L_y x = y \diamond x\), forcing injectivity of left-multiplication, so the Obelix law fails.