# Has the exponentiation of ordinals a nice geometric model?

It is well known that the sum $$\alpha+\beta$$ of two ordinals $$\alpha,\beta$$ can be defined "geometrically" as the order type of the sum $$(\{0\}\times \alpha)\cup(\{1\}\times\beta)$$ endowed with the lexicographic order.

Also the product $$\alpha\cdot\beta$$ of ordinals $$\alpha,\beta$$ is the order type of the Cartesian product $$\beta\times\alpha$$ endowed with the lexicographic order.

What about the exponentiation of ordinals?

Does $$\alpha^\beta$$ have some nice "geometric'' or combinatorial model?

Maybe as some set of (partial) functions endowed with a suitable well-order?

Ordinal exponentiation is a special case of linear order exponentiation. For any linear order $$L$$, element $$a\in L$$, and ordinal $$\beta$$ we can define the $$\beta$$th power of $$L$$ at $$a$$, which I'll call "$$L_a^\beta$$," as the set of functions $$f:\beta\rightarrow L$$ such that all but finitely many $$\alpha\in\beta$$ have $$f(\alpha)=a$$. The ordering on this set is given by looking at the last point of difference: $$f\trianglelefteq g\iff f=g\mbox{ or } f(\max\{x:f(x)\not=g(x)\})
For ordinals $$\alpha,\beta$$ we have $$\alpha^\beta=\alpha_0^\beta$$. Rosenstein's book treats this in some detail (and is generally an awesome book all-around - it's a huge tragedy that it's so hard to find).
• It's not well-ordered. $(1,0,0,...) > (0,1,0,...) > ... > (0,...,0,1,0,...) > ...$. Maybe replace min with max? Apr 27, 2020 at 9:02
• @MonroeEskew It would make sense to use $\max$ instead if you want to generalize ordinal exponentiation, but it won't be well-ordered if $L$ isn't anyways. Apr 28, 2020 at 9:27
• @nombre That’s ok... I’m just saying that the above definition doesn’t define $\alpha^\beta$ properly. One needs to use max, by my example. (Isn’t it right? The definition given is the same as the lex order, which corresponds to the rationals for $\alpha=2, \beta=\omega$.) But it is easy to show by induction that max works. Apr 28, 2020 at 10:51
• Do you need $\beta$ to be well-ordered? It looks to me like this definition still makes sense when $\beta$ is any linear order, so you've really defined a way of exponentiating with two linear orders . . . right? Apr 29, 2020 at 13:30