Hessenberg power of ordinals (redux)

exponentiationorder-theoryordinalsset-theorywell-orders

This is a follow-up to this question, in which the given definition failed.

Let $f : \varepsilon_0 \rightarrow \mathbb{N} \rightarrow \mathbb{N}$ recursively defined by
\begin{align}
f\left(\sum_{i=1}^n \omega^{\beta_i} c_i\right)
&= \sum_{i=1}^n \mathrm{id}_\mathbb{N}^{f(\beta_i)} c_i
\end{align}

where the argument to $f$ is in Cantor normal form and all arithmetic operations between functions are defined pointwise. Notice that

\begin{align}
\alpha +_H \beta &= f^{-1}(f(\alpha) + f(\beta)) \\
\alpha \cdot_H \beta &= f^{-1}(f(\alpha) \cdot f(\beta))
\end{align}

where $+_H$ and $\cdot_H$ are the Hessenberg sum and product of ordinals. Let
\begin{align}
\alpha \uparrow_H \beta = f^{-1}(f(\alpha)^{f(\beta)})
\end{align}

This seems to be the natural candidate for "Hessenberg exponentiation" (at least for ordinals below $\varepsilon_0$), satisfying properties one would expect, such as

\begin{align}
\alpha \uparrow_H (\beta +_H \gamma)
&= f^{-1}(f(\alpha)^{f(\beta +_H \gamma)}) \\
&= f^{-1}(f(\alpha)^{f(\beta) + f(\gamma)}) \\
&= f^{-1}(f(\alpha)^{f(\beta)} \cdot f(\alpha)^{f(\gamma)}) \\
&= f^{-1}(f(\alpha \uparrow_H \beta) \cdot f(\alpha \uparrow_H \gamma)) \\
&= (\alpha \uparrow_H \beta) \cdot_H (\alpha \uparrow_H \gamma) \\
\end{align}

Does it have a purely order-theoretic definition, like the Hessenberg sum and product? Can it be extended to all ordinals?

Best Answer

Unfortunately this exponentiation formula is not defined for all ordinals.

For example, assume $$\alpha=2\uparrow_H\omega$$ According to the formula, this would imply that $$f(\alpha)(n) = 2^n$$ Now since ordinals are totally ordered, either $\alpha<\omega^\omega$ or $\alpha\ge\omega^\omega$.

If $\alpha<\omega^\omega$, it is a polynomial in $\omega$, therefore $f(\omega)(n)$ is a polynomial in $n$. Clearly $2^n$ is not a polynomial in $n$.

On the other hand, if $\alpha\ge\omega^\omega$, then for sufficiently large $n$, $f(\alpha)(n)\ge n^n$. However $2^n<n^n$ for $n>2$.

Thus $\alpha$ can neither be less than $\omega^\omega$ nor greater or equal, which is a contradiction.

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