Can transfinite numbers be extended to more general algebraic structures?

01 Dec 2010

Infinite ordinals and cardinals are beautiful mathematical objects, extending in a natural way $\mathbb{N}$ and sharing most of its properties. I have always wondered whether it would be possible to generalize the constructions of $\mathbb{Z}$ or $\mathbb{Q}$ to get transfinite integers or rationals. In general, putting a group structure on classes extending $\mathrm{Ord}$ or $\mathrm{Card}$ is not possible as I indicated some years ago. Basically, we would have $1+{\aleph }_0={\aleph }_0$ (and also $1+\omega =\omega$) which would imply $1=0$. However, we can consider a slightly modified problem with weaker constraints, as follows. First we assume we have a class of cardinals $C\subseteq \mathrm{Card}$ containing zero and stable by addition.

$$\text{(1)}\{\begin{array}{l}0\in C\\ \forall \kappa ,\lambda \in C,\kappa +\lambda \in C\end{array}$$

We define the class $Z_C=C\cup C^{-}$, where $C^{-}=\{-\kappa \mid \kappa \in C,\kappa \ne 0\}$ is a representation of "negative" cardinals. As we have seen, the class $Z_C$ need not be a group. Nevertheless, we can try to find an equivalence class $R$ over $Z_C$ such that $G_C=\raisebox{1ex}{$Z_C$}\!\left/ \!\raisebox{-1ex}{$R$}\right.$ is a group. Moreover, we want this equivalence class to be compatible with addition and opposite i.e.

$$\text{(2)}\begin{array}{c}\forall \kappa ,\lambda \in C,\overline{\kappa +\lambda }=\overline{\kappa }+\overline{\lambda }\\ \forall \kappa \in C,\overline{-\kappa }=-\overline{\kappa }\end{array}$$

Of course this implies that $\overline{0}$ is the identity element $0$ of the expected group. Similarly, we can settle the same problem for a class $C\subseteq \mathrm{Ord}$ and + is now understood as the ordinal addition. Can we follow this schema to contruct interesting infinite algebraic structures?

For $C\subseteq \mathrm{Card}$, we have for any infinite $\kappa \in C$ the relation $\overline{\kappa }=\overline{\kappa +\kappa }=\overline{\kappa }+\overline{\kappa }$ so $\overline{\kappa }=0$. Hence $G_{C\cap \omega }=G_C$ and the initial problem is reduced to the case where $C\subseteq \omega$ contains only finite cardinals (= finite ordinals) and so the problem is still not really exciting. What about the case $C\subseteq \mathrm{Ord}$? In general it is possible to have ${\beta }_1+\gamma ={\beta }_2+\gamma$ without ${\beta }_1,{\beta }_2$ being equal and so for any $\alpha$ such that ${\beta }_1\alpha ,{\beta }_2\alpha ,\mathrm{\gamma \alpha }\in C$, we get that $\overline{{\beta }_1\alpha }=\overline{{\beta }_2\alpha }$. In particular for any infinite ordinal $\gamma$ and $\alpha \in C$ such that $\mathrm{\gamma \alpha }\in C$ we obtain $\overline{\alpha }=0$ because $1+\gamma =\gamma$ (if this is not obvious to you, a more general statement is proved below). As a consequence, more general assumptions on the class $C$ strongly limit the structure of the group. For example if $C$ is closed under an infinite sum ($C=\mathrm{Ord}$ for example) then ${\text{G}}_{\text{C}}$ is trivial. Similarly, if we require $C$ to be stable by multiplication (in order to define a ring structure for example) then ${\text{G}}_{\text{C}}$ is trivial whenever $C$ contains an infinite ordinal $\gamma$ (hence the remaining case is again $C\subseteq \omega$). If $G_C$ is not trivial, we denote ${\alpha }_0\in C$ the least element such that $\overline{{\alpha }_0}\ne 0$ (in particular ${\alpha }_0>0$).

One additional natural hypothesis should be added. We know that for any ordinal $\alpha <\beta$ there exists a unique ordinal $\gamma$ such that $\alpha =\beta +\gamma$. We would very like $\overline{\gamma }$ to match the difference "$-\overline{\beta }+\overline{\alpha }$". For that purpose, we only require $\gamma$ to belong to $C$. Hence we assume that

$$\text{(3)}\forall \alpha ,\beta \in C,\forall \gamma \in \mathrm{Ord},\alpha =\beta +\gamma \Rightarrow \gamma \in C$$

Let's come back to the case $C\subseteq \omega$ with the new assumption (3) and suppose that $G_C$ is not trivial. Then ${\alpha }_0<\omega$ and any finite $m\in C$ and be written $m={\alpha }_{0}q+r$ with $r<{\alpha }_0$. By (3), $r\in C$ and because $C$ is stable by finite sums, $\overline{m}=\overline{{\alpha }_0}q+\overline{r}=\overline{{\alpha }_0}q$. Thus $G_C$ is the monogenic group generated by $\overline{{\alpha }_0}$.

What about the general case? Unfortunately, it turns out that if $G_C$ is not trivial it is still a monogenic group generated by $\overline{{\alpha }_0}$. To prove this, we need to recall some equalities on ordinals. First, for any $\alpha >0$, we have $1+{\omega }^{\alpha }={\omega }^{\alpha }$. This is clearly true for $\alpha =1$ and we prove the general case by induction on $\alpha$: $1+{\omega }^{\alpha +1}=1+\omega {\omega }^{\alpha }=1+\left(1+\omega \right){\omega }^{\alpha }=\left(1+{\omega }^{\alpha }\right){+\omega }^{\alpha +1}={\omega }^{\alpha }+{\omega }^{\alpha +1}=\left(1+\omega \right){\omega }^{\alpha }=\omega {\omega }^{\alpha }={\omega }^{\alpha +1}$ and $1+{\omega }^{\lambda }=\underset{0<\alpha <\lambda }{sup}\left({1+\omega }^{\alpha }\right)=\underset{0<\alpha <\lambda }{sup}{\omega }^{\alpha }={\omega }^{\lambda }$. Now, if we have two ordinals $\alpha <\beta$ and if $\gamma >0$ is such that $\beta =\alpha +\gamma$ we have ${\omega }^{\alpha }+{\omega }^{\beta }={\omega }^{\alpha }\left(1+{\omega }^{\gamma }\right)={\omega }^{\alpha }{\omega }^{\gamma }={\omega }^{\beta }$. Finally, if we have two ordinals ${\gamma }_1<{\gamma }_2$ we can write their Cantor Normal Forms:

$$\text{(4)}\begin{array}{c}{\gamma }_1={\omega }^{{\beta }_1^1}{k}_1^1+{\omega }^{{\beta }_2^1}{k}_2^1+\dots +{\omega }^{{\beta }_n^1}{k}_n^1\\ {\gamma }_2={\omega }^{{\beta }_1^2}{k}_1^2+{\omega }^{{\beta }_2^2}{k}_2^2+\dots +{\omega }^{{\beta }_n^2}{k}_m^2\end{array}$$

where ${\beta }_1^2\ge {\beta }_1^i>{\beta }_2^i>\dots >{\beta }_2^i$ and $k_j^i$ are positive integers. Using the equality ${\omega }^{\alpha }+{\omega }^{\beta }={\omega }^{\beta }$ for $\alpha <\beta$ it is clear that ${\gamma }_1+{\gamma }_2={\gamma }_2$ if ${\beta }_1^2>{\beta }_1^1$.

Now our element ${\alpha }_0\in C$ can be written ${\alpha }_0={\omega }^{\beta }k+\gamma$ where ${\omega }^{\beta }k$ is the first term of its Cantor Normal Form. For any $\delta \in C$, we can write its euclidean division by ${\alpha }_0$: $\delta ={\alpha }_{0}l+\epsilon$ where $\epsilon <{\alpha }_0$. By the previous discussion, if $l\ge \omega$ then we can write the Cantor Normal Forms of the two elements ${\alpha }_0<\delta$ as in (4) and see that ${\beta }_1^2>{\beta }_1^1$. Hence ${\alpha }_0+\delta =\delta$ and so $\overline{{\alpha }_0}=0$, which is a contradiction. So $l<\omega$ and because $C$ is stable by finite sums, ${\alpha }_{0}l\in C$ and by the property (3) we get $\epsilon \in C$. This means that $\overline{\delta }=\overline{{\alpha }_0}l+\overline{\epsilon }=\overline{{\alpha }_0}l$ and hence $G_C$ is generated by $\overline{{\alpha }_0}$ as claimed above.

As a conclusion, if $C\subseteq \mathrm{Card}$ satisfies the properties (1), (2) above and $G_C=\raisebox{1ex}{$Z_C$}\!\left/ \!\raisebox{-1ex}{$R$}\right.$ is a group then $C\subseteq \omega \subseteq \mathrm{Ord}$. If $C\subseteq \mathrm{Ord}$ satisfies properties (1), (2), (3) above and $G_C=\raisebox{1ex}{$Z_C$}\!\left/ \!\raisebox{-1ex}{$R$}\right.$ is a group then $G_C$ is isomorphic to $\mathbb{Z}$ or to $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$n\mathbb{Z}$}\right.$. Conversely, we can build these groups from $C=\omega$ by defining the relation $R$ as the equality (respectively the equality modulo $n$). But we do not get any new groups...