Surreal exponentiation: It gets worse

So in the previous entry I asked the question, what the hell is natural exponentation, and does it agree with surreal exponentiation? This of course assumes that there is such a thing as "surreal exponentiation". After all, the surreals are always described as a place where crazy expressions like (ω+3)^√π make sense, so everyone has to agree what surreal exponentiation means, right?

Well, now I'm not so certain. There's a definition due to Kruskal and a definition due to Gonshor and I have no idea if these are actually the same. They both have the required properties. (Note these are definitions of exp, not of ^, but since both have exp being onto the positive surreals one can therefore easily turn this into a definition of ^.) There may also be one due to Alling? Oy... though that one at least seems to be similar to Gonshor's...

Then Wikipedia -- not citing any source -- lists a recursive definition of 2^x, which, I suppose, could be generalized to a^x by just replacing "2" by "a". Does this agree with the others? Again, no idea.

Let's note here something that doesn't fit in -- the function ω^x, commonly used in studying surreals, is *not* part of what I'm looking for. It agrees with the ordinary ω^x when x is an ordinal, and is not supposed to fit in with any more general notion of exponentiation; the above definitions of exponentiation actually do not agree with it. Wikipedia's definition of 2^x seems rather similar to the recursive definition of ω^x so maybe it agrees with *that* and ordinary ordinal exponentiation instead? :-/ But that would be a surprise -- though there have been some extensions of ordinary ordinal operations to surreals, but these are not actually well-defined on all surreals.

And then on top of that we still have the question of whether any of these agree with "natural exponentiation" -- whatever the hell that is!

Guess I can try reading the original German for an answer to that latter question, anyway. Well, once the people at the library get SpringerLink working here again...

-Harry

Enough on those weird ordinal operations, time for some more natural ones

OK, so I just went and asked Math.SE about the ordinal operations introduced in the previous two posts, since I don't really want to spend my time figuring out more about them.

So! Let's review ordinal arithmetic. Ordinals have their ordinary addition, multiplication, and exponentiation. These can be defined either recursively, or constructively in terms of what order types they represent. The recursive definition allows one to extend this to be extended to a general hyper operation on the ordinals but mostly I won't consider anything further than exponentiation here.

Let's recall the constructive definition of these. α+β is a well-order on α⨿β, with the elements of α coming first and then those of β. αβ is a well-order on α×β, namely, reverse-lexicographic. And α^β is a well-order on finitely-supported functions from β to α, namely, reverse-lexicographic.

(Is there some construction that works for general hyper? Probably not.)

One could also describe these operations in terms of how they work on Cantor normal form, though I don't think you could consider this a "definition". Not sure how exponenation would work on Cantor normal form, actually, maybe that's actually just terrible and you can't sensibly describe it that way. Let's perhaps leave that one out in that case.

Edit: Actually, thinking about it some more, it's not too bad. It's kind of bad, but it's not terrible. The point is, it's doable.

But then we also have the natural operations! These give up the continuity properties of the ordinary operations (in the second argument, anyway) to get better algebraic properties. There are natural sum, natural product, and natural exponentiation. (Natural hyper? I really doubt it. Higher operations don't even have nice properties, except for those that the ordinary hyper would already get you anyway.)

Let's shelve natural exponentiation for now. So we have natural sum and natural product. These can each be defined three ways:

1. Recursively. Note that the recursion for multiplication is usually defined in terms of natural subtraction -- since natural addition is cancellative, one can make a group of formal differences of ordinals under addition. But it isn't necessary to do it this way, because one can just unwind the subtraction and state it purely additively. It's just a little nasty.

(Incidentally, if one changes the (usual) recursive definitions by replacing "smallest ordinal larger than all of..." with "smallest ordinal different from all of..." one gets nimber addition and nimber multiplication. I really doubt there's a nimber exponentiation...)

Edit: Duh, of course there's no such thing as exponentiation in finite characteristic -- or at least not in a domain, anyway. Consider the implications of a^b+c=a^ba^c... well, OK, the results aren't contradictory, it's just that we end up with a^b=1 for all a,b, which is not exactly satisfactory.

2. Constructively. For α⊕β, we take the poset α⨿β, and then take the largest well-order that extends it. It's not obvious that the supremum is indeed a maximum, but it is. For α⊗β, we take the poset α×β, and then take the largest well-order that extends it. Same comment about supremum vs. maximum.

3. "Computationally". To take the natural sum of two ordinals, one just adds their Cantor normal forms in the obvious way. To take the natural product, one just multiplies their Cantor normal forms in the obvious way, using natural sum to add the exponents.

Worth noting is that while the ordinals are often said to embed in the surreals, really it's the ordinals with the natural operations that embed in the surreals. They can't embed with the ordinary operations since those aren't even commutative!

These operations have basically all the properties you would expect of addition and multiplication (they embed in the surreals -- an ordered field -- after all).

Now we come to natural exponentiation. Natural exponentiation is...? I have no idea. According to Wikipedia, Jacobsthal defined it in 1907, and... that's all its says. Well, that and that it's rarely used. I actually dug up Jacobsthal's paper where he defines it, but firstly in German, and secondly even ignoring that it seems like it'll be hard to read. Also, that's just where it's defined, and I'd like to know how it fits into the above framework; I doubt it would even answer all my questions about it. But when I looked for other papers on it, I couldn't find any...

I'll denote the natural exponential as α^β, because that is how Jacobsthal denoted it.

So... can it be defined recursively? Can it be defined "constructively"? There's a nice analogy between how ordinary sum and product are defined, and how natural sum and product are defined. Is this what natural exponentiation is as well? Take finitely-supported functions from β to α, give this the obvious poset structure, and then take the supremum (maximum?) of well-orders extending it? (Note that the category of posets does have an exponential object, but it's the set of all monotonic maps from one to the other, and this is not only totally dissimilar to how ordinary exponential is constructed, but it gives the wrong answer for finite numbers.) Can we give a "computational" definition, too?

Edit: The above "constructive" definition fails very badly. Observe that it only depends on the cardinality of the base and not its order type! Thus it gives answers that are way too large. Still, perhaps it could be salvaged by adding more relations to the poset...

What properties does it have? I would hope it would satisfy:
α⁰=1
α¹=α
0^β=0 for β>0
1^β=1
n^k=n^k for n,k finite
α^β⊕γ=α^β⊗α^γ
α^β⊗γ=(α^β)^γ
(α⊗β)^γ=α^γ⊗β^γ
α^β is strictly increasing in β for α≥2
α^β is strictly increasing in α for β≥2 (This one seems like a stretch, for reasons that should become clear below. But it had damn well better at least be weakly increasing or something's *really* wrong...)

But... does it? No idea. (α⊗β)^γ=α^γ⊗β^γ seems like kind of stretch... except ⊗ is, in fact, commutative, making it not so much of a stretch.

I'd would expect it agrees with the exponential in the surreals, but do the surreals even have an exponential? I had assumed so because they're kind of an "infinity kitchen sink", but Wikipedia only mentions powers of 2 and powers of ω. I should probably look this up. (I had kind of also hoped maybe we might get 2^α>α, but if it agrees with powers of 2 in the surreals, this is false, since that satisfies 2^ω=ω. Oh well -- that one was a real stretch anyway.)

I would also hope for ω^α=ω^α, come to think of it. That would contradict the property in the above paragraph, but like I said, that one was a real stretch in the first place. And certainly, what with it being a natural operation in the first place, it had certainly better satisfy α^β≥α^β. I'd also like αⁿ=αⁿ for finite n, but there's no way that's true; consider α=ω+1. And it's not like we have α⊗n=αn for n finite, either (again, consider α=ω+1).

So... time to either try Math.SE, or try reading Jacobsthal...? Well, quite possibly ask MathOverflow -- even though it's not really a research question, I doubt MathUnderflow will be able to answer it. Maybe I should check out Hausdorff's book on set theory and see if it mentions it by any chance? Well, at least I should be easily able to look up about the exponential in the surreals...

Edit: Having checked out Hausdorff's book, it only discusses natural sum and natural product, not natural exponentiation. But I may finally have found another paper that discusses it... well, it discusses surreal exponentiation, anyway... actually it seems to be easy to find papers discussing that...

Edit again: None of these papers are helpful. Well, at least I now know that the surreals do have exponentiation and logarithms, and these do have the usual properties, and so they therefore have powers in the obvious way, and these have the usual properties. One thing of note: exp and log are both strictly increasing, meaning powers would be strictly increasing in the base as well as the exponent. Hm... a bit odd to have 2^ω=ω but 3^ω>ω... assuming 2^ω=ω in the first place... damn people not bothering to verify compatibility!

-Harry