Ordinals ordinals ordinals!
Aug. 29th, 2015 01:41 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoySo I have a number of neat things to report about ordinals! In order from least to most exciting:
#1: I can now compute big multichoose, in general
At least, I'm pretty sure I can prove it. More on this in a moment.
(Unfortunately, this means I have no "easy math" I can work on right now -- I need to get back to the hard math, y'know.)
Hm -- I just looked and saw I never defined "big multichoose" here. In short: If α and β are ordinals, we can consider the poset of all weakly decreasing functiosn from β to α; this is a WPO, so we can take the largest ordinal extending it. This is what I mean by "α multichoose β", which I will denote here as ((α|β)) because what else am I going to do.
Point is, I've been puzzling over it for a while, working out special cases, because for obvious reasons this is quite a bit harder to compute than left and right multichoose from earlier.
#2: There really, really, is no natural exponentiation
In appendix B of my ordinal arithmetic paper, I showed that there's no operation satisfying several desiderata that we'd expect of a "natural exponentiation". However, one of the desiderata was a bit less, uh, to be desired than the others. But I've now managed to eliminate the dependence on that hypothesis! Best of all, the new proof is only a slight modification of the existing one. The super short version is, instead of defining f(n) = deg e(n,ω), you define f(n) to be the "leading coefficient" of deg e(n,ω).
I'll update the paper shortly -- I'll also need to notify the journal I submitted it to about the update...
#3: Big multichoose is related to the surreal exponential!
Don't ask me why -- I don't actually understand the surreal exponential at all. But it totally is! The thing is, this doesn't actually depend on any of the new stuff in #1; what's related to the surreal exponential is the rule for computing ((ω|α)), or rather ((ω|α))0. (I'm using the subscript 0 to mean that I'm restricting to functions which are eventually 0.) It turns out for α a limit ordinal,
((ω|α))0 = exp(α)
where exp is the surreal exponential.
Again, I don't have any idea why this is so, I just noticed that the rules for computing them are the same. I didn't notice it before because I hadn't written the rule for computing the former in a way that made it obvious.
So let me go into more detail about #1 now -- the final piece of the puzzle was figuring out how to compute ((ωα|ωβ))0 for α,β>0. What I eventually figured out was that this is equal to ωω^G(α,β) for a certain function G. Specifically, G(α,β) is equal to α⊕β if either α or β is of the form εγ+k for some ordinal γ and some finite k, or if both are infinite. Otherwise -- if one is finite and the other is not of the form εγ+k -- then instead you get (α⊕β)-1, which necessarily makes sense since one of them is finite and nonzero.
Now, Harry Gonshor, in his book "An Introduction to the Theory of Surreal Numbers" -- where he defined the surreal exponential in the first place -- showed that there's a function g such that for a surreal number α>0, exp(ωα)=ωω^g(α), and showed how to compute g. Of course, we only care about the case where α is an ordinal, not just any surreal. And in that case, g(α) is equal to α+1 if α is of the form εγ+k, and is equal to α otherwise.
See the similarity? G(1,β) is exactly equal to g(β). And from this follows the equation above.
Now as I said this only requires knowing how to compute G(1,β), i.e., how to compute ((ω|β))0, which I already knew how to do. But I didn't notice the similarity until I wrote it in this form depending on G. And I wouldn't have written it that way if I weren't trying to determine ((α|β)) more generally. So there is that bit of relation there. I determined the form of G, and then noticed that G(1,β) reminded me of the rule for g(β); and I ran to the library to look it up, and indeed it was the same.
So yay! That's finally solved. Now back to harder math, I guess...
-Harry
#1: I can now compute big multichoose, in general
At least, I'm pretty sure I can prove it. More on this in a moment.
(Unfortunately, this means I have no "easy math" I can work on right now -- I need to get back to the hard math, y'know.)
Hm -- I just looked and saw I never defined "big multichoose" here. In short: If α and β are ordinals, we can consider the poset of all weakly decreasing functiosn from β to α; this is a WPO, so we can take the largest ordinal extending it. This is what I mean by "α multichoose β", which I will denote here as ((α|β)) because what else am I going to do.
Point is, I've been puzzling over it for a while, working out special cases, because for obvious reasons this is quite a bit harder to compute than left and right multichoose from earlier.
#2: There really, really, is no natural exponentiation
In appendix B of my ordinal arithmetic paper, I showed that there's no operation satisfying several desiderata that we'd expect of a "natural exponentiation". However, one of the desiderata was a bit less, uh, to be desired than the others. But I've now managed to eliminate the dependence on that hypothesis! Best of all, the new proof is only a slight modification of the existing one. The super short version is, instead of defining f(n) = deg e(n,ω), you define f(n) to be the "leading coefficient" of deg e(n,ω).
I'll update the paper shortly -- I'll also need to notify the journal I submitted it to about the update...
#3: Big multichoose is related to the surreal exponential!
Don't ask me why -- I don't actually understand the surreal exponential at all. But it totally is! The thing is, this doesn't actually depend on any of the new stuff in #1; what's related to the surreal exponential is the rule for computing ((ω|α)), or rather ((ω|α))0. (I'm using the subscript 0 to mean that I'm restricting to functions which are eventually 0.) It turns out for α a limit ordinal,
((ω|α))0 = exp(α)
where exp is the surreal exponential.
Again, I don't have any idea why this is so, I just noticed that the rules for computing them are the same. I didn't notice it before because I hadn't written the rule for computing the former in a way that made it obvious.
So let me go into more detail about #1 now -- the final piece of the puzzle was figuring out how to compute ((ωα|ωβ))0 for α,β>0. What I eventually figured out was that this is equal to ωω^G(α,β) for a certain function G. Specifically, G(α,β) is equal to α⊕β if either α or β is of the form εγ+k for some ordinal γ and some finite k, or if both are infinite. Otherwise -- if one is finite and the other is not of the form εγ+k -- then instead you get (α⊕β)-1, which necessarily makes sense since one of them is finite and nonzero.
Now, Harry Gonshor, in his book "An Introduction to the Theory of Surreal Numbers" -- where he defined the surreal exponential in the first place -- showed that there's a function g such that for a surreal number α>0, exp(ωα)=ωω^g(α), and showed how to compute g. Of course, we only care about the case where α is an ordinal, not just any surreal. And in that case, g(α) is equal to α+1 if α is of the form εγ+k, and is equal to α otherwise.
See the similarity? G(1,β) is exactly equal to g(β). And from this follows the equation above.
Now as I said this only requires knowing how to compute G(1,β), i.e., how to compute ((ω|β))0, which I already knew how to do. But I didn't notice the similarity until I wrote it in this form depending on G. And I wouldn't have written it that way if I weren't trying to determine ((α|β)) more generally. So there is that bit of relation there. I determined the form of G, and then noticed that G(1,β) reminded me of the rule for g(β); and I ran to the library to look it up, and indeed it was the same.
So yay! That's finally solved. Now back to harder math, I guess...
-Harry
