# Hessenberg power of ordinals

According to these notes on ordinal arithmetic:

• The Hessenberg sum $$\alpha + \beta$$ is the supremum of ordinals that are isomorphic to some well-order on $$\alpha \sqcup \beta = (\{0\} \times \alpha) \cup (\{1\} \times \beta)$$ extending the union partial-order: $$x \leq y \Longleftrightarrow (x_1 = y_1) \land (x_2 \leq y_2)$$

• The Hessenberg product $$\alpha \times \beta$$ is the supremum of ordinals that are isomorphic to some well-order on $$\alpha \times \beta$$ extending the product partial-order: $$x \leq y \Longleftrightarrow (x_1 \leq y_1) \land (x_2 \leq y_2)$$

Does it make sense to define the Hessenberg power $$\alpha^\beta$$ as the supremum of ordinals that are isomorphic to some well-order on $$\beta \rightarrow \alpha$$ extending the following partial-order? $$x \leq y \Longleftrightarrow \forall z \in \beta: x(z) \leq y(z)$$

If so, does it satisfy the following properties?

1. $$\alpha^{\beta + \gamma} = \alpha^\beta \times \alpha^\gamma$$
2. $$\alpha^{\beta \times \gamma} = (\alpha^\beta)^\gamma$$

And what would be, for example, $$2^\omega$$, $$3^\omega$$, and $$\omega^\omega$$?

• There may be some way to define a commutative exponentiation but I doubt it would behave well. The problem with this one is that any well-founded order on a set extends to a well-order. Any such well-order on say $2^{\omega}$ must be larger than $2^{\aleph_0}$, which I suppose is not what you are looking for. Feb 14, 2020 at 19:37
• @nombre Sorry, what do you mean by a commutative exponentiation? Feb 14, 2020 at 20:40
• Poor choice of words on my part: I mean with $\alpha^{\beta \gamma}=\alpha^{\gamma \beta}$ (I tend to think of Hessenberg operations as "commutative versions of ordinal operations", although this is misleading). Feb 14, 2020 at 20:50
• @nombre Ah, but isn't $\beta \gamma = \gamma \beta$ always true (in Hessenberg arithmetic)? Feb 14, 2020 at 21:01
• Hehe, indeed it is, I guess I don't even know what I meant! Feb 14, 2020 at 21:09

Consider the case $$2^\omega$$. The functions in that case would be the characteristic functions of subsets of $$\omega$$, and the partial order you give then simply refers to the subset order of the corresponding sets.
Now consider the set $$M=\{\omega\setminus n\mid n\in\omega\}$$. The elements of this set have the form $$S_n=\omega\setminus n=\{n,n+1,n+2,n+3,\ldots\}$$. Now it is easily seen that the inclusion partial order already gives a total order on $$S_n$$, namely $$S_m\subseteq S_n\iff m\ge n$$.
Note the reversal of the order direction here, which means that because $$\omega$$ has no maximal element, $$V$$ has no minimal element. And since $$V$$ is already totally ordered, no extension of the partial order on $$2^\omega$$ will cause its order to change.
Therefore no extension of that partial order on $$2^\omega$$ to a well-order can exist, which means your exponentiation function is not defined for $$2^\omega$$.