It was only today that it occurred to me -- I say in the paper I'm writing that we're not going to consider "natural" exponentiation, because the one coming from the surreals doesn't work, and so there doesn't seem to be a natural exponentiation (unless you count Jacobsthal or "semi-Jacobsthal" exponentation); but could I sit down and prove that there isn't one, from a list of desiderata, and perhaps add this as an appendix?
(Note that I've already tried the approach of "take surreal exponentiation and then round up to the next ordinal". This has little in the way of nice properties.)
Well, that depends on your desiderata. I wrote down a list of 10 (all of which are satisfied by surreal exponentiation, except for the whole part where it doesn't reuturn an ordinal). Let's use p(a,b) to mean the hypothesized "natural" exponentiation ab.
Then I think we can agree on the following desiderata:
1. p(a,1) = a
2. p(a,b⊕c)=p(a,b)⊗p(a,c)
3. p(a,b⊗c)=p(p(a,b),c)
4. p(a,b) is weakly increasing in a
5. For a>0, p(a,b) is weakly increasing in b
Thing is -- the problem of finding a natural exponentiation is, it seems to me, severely underconstrained. Even with my full list, you could still probably define it in a completely silly way.
But let's add another restriction: A degree law. For an ordinal a>0, I'll define deg(a) to be the largest b such that ωb≤a. I.e. it's
the largest exponent appearing the Cantor normal form.
All the other operations have degree laws, or something like them. In
particular, for ordinary exponentiation and for Jacobsthal exponentiation, we have, assuming a≥ω
deg(ab) = deg(a×b) = deg(a) * b.
And for "semi-Jacobsthal" exponentiation, we have, again assuming a≥ω
deg(a⊗b) = deg(a) × b.
(Let's ignore for now what happens when a<ω; it's easy to describe, but whatever.)
Since this is supposed to be natural exponentiation, let's add the following degree law as a desideratum:
6. For a≥ω, deg(p(a,b)) = deg(a) ⊗ b
And with this, it becomes impossible! Because with this restriction, one can show that if we define f(n) = deg(p(n,ω)), then f(n) is a function from naturals to naturals which A. is weakly increasing, and B. satisfies f(nk)=k*f(n), and these together are sufficient to show that f(n)/f(m) = (log n)/(log m), contradiction.
Whoo. Now I need to decide whether to add this as an appendix.
(Jeff has told me not to worry particularly about whether my paper really is new, and just get it ready and submit it, and if I missed something in the literature and it's not new, I'll find out...)
-Harry
(Note that I've already tried the approach of "take surreal exponentiation and then round up to the next ordinal". This has little in the way of nice properties.)
Well, that depends on your desiderata. I wrote down a list of 10 (all of which are satisfied by surreal exponentiation, except for the whole part where it doesn't reuturn an ordinal). Let's use p(a,b) to mean the hypothesized "natural" exponentiation ab.
Then I think we can agree on the following desiderata:
1. p(a,1) = a
2. p(a,b⊕c)=p(a,b)⊗p(a,c)
3. p(a,b⊗c)=p(p(a,b),c)
4. p(a,b) is weakly increasing in a
5. For a>0, p(a,b) is weakly increasing in b
Thing is -- the problem of finding a natural exponentiation is, it seems to me, severely underconstrained. Even with my full list, you could still probably define it in a completely silly way.
But let's add another restriction: A degree law. For an ordinal a>0, I'll define deg(a) to be the largest b such that ωb≤a. I.e. it's
the largest exponent appearing the Cantor normal form.
All the other operations have degree laws, or something like them. In
particular, for ordinary exponentiation and for Jacobsthal exponentiation, we have, assuming a≥ω
deg(ab) = deg(a×b) = deg(a) * b.
And for "semi-Jacobsthal" exponentiation, we have, again assuming a≥ω
deg(a⊗b) = deg(a) × b.
(Let's ignore for now what happens when a<ω; it's easy to describe, but whatever.)
Since this is supposed to be natural exponentiation, let's add the following degree law as a desideratum:
6. For a≥ω, deg(p(a,b)) = deg(a) ⊗ b
And with this, it becomes impossible! Because with this restriction, one can show that if we define f(n) = deg(p(n,ω)), then f(n) is a function from naturals to naturals which A. is weakly increasing, and B. satisfies f(nk)=k*f(n), and these together are sufficient to show that f(n)/f(m) = (log n)/(log m), contradiction.
Whoo. Now I need to decide whether to add this as an appendix.
(Jeff has told me not to worry particularly about whether my paper really is new, and just get it ready and submit it, and if I missed something in the literature and it's not new, I'll find out...)
-Harry
