EDIT 7/27: The statements in this entry may very well be FALSE! It turns out my formula is very wrong in more than two dimensions! Now, admittedly, the case I actually bothered to state here might well still be right; in fact, I'm guessing that more likely than not, it is. (OK, having gone over it again now actually I'm pretty sure it is.) But the more general formula they are a part of is false, and I will have to see about correcting this. More on this in an entry soon to come.
So I wrote this down in an email to John, and then realized I should probably post it here as well.
So let's consider "big multichoose". I want to consider the variant that I denoted ((α multichoose β))0, where we consider all weakly decreasing functions from beta to alpha that are eventually 0.
This can also be considered as the set of lower sets of α×β that are "bounded in each direction", in the sense that neither projection is surjective.
Now of course if α and β are finite -- say n and m -- then this is just counting the number of partitions that fit in an (n-1)x(m-1) box, i.e., it's (n+m-2 choose n-1). (Which is different from multichoose, obviously, but nicer for the purposes I want to consider here.)
So suppose we generalize from two ordinal factors in the product to multiple. What happens now? Can we get a formula similar to my formula for dimension 2?
Answer: We can! It's not hard, either, once you know the dimension 2 case (OK, I haven't sat down and fully proven it, but the proof should be similar).
The thing is that I'd considered this before but always dismissed it because, well -- counting partitions in a given box is easy; counting plane partitions in a given box is less easy; and in dimensions higher than 3, things apparently get pretty wild pretty fast. So I figured, well, if the finite case is intractable, what's the point of attempting the infinite case?
Of course, this is silly, because if you can reduce the infinite case to the finite one, that's still significant. Like really, in the dimension 2 case, if you plug in finite ordinals, my formula doesn't actually say "OK, it's (n+m-2 choose n-1)", rather it says, OK, count the partitions in this box. We just know what that is. So in higher dimensions it will just fail to reduce the problem in the finite case, but in the infinite case, at least now it is a finite description.
So anyway as before the main core of the problem is when all the factors are powers of ω (and not equal to 1). So let's say we're putting in ωα_1, ..., ωα_n. Also, say n≥2; if n=1 (or n=0) the problem A. is trivial and B. is *not* given by the formula below, so let's exclude that. (And yes there is a good reason that dimension≥2 turns out different from dimension 1, but I'm not going to go into that here.)
So what we get out is ωω^g(α_1,...,α_n), for some g. What's g? Well, it's basically the same as before:
If either:
1. one of the αi is of the form epsilon+finite, or
2. at least two of the αi are infinite
then g(α1,...,αn) is just the natural sum of the αi.
But otherwise, it's the natural sum, minus 1.
(Before I did it out, I had assumed that condition (2) would require that all of the αi be infinite, but in fact, only two of them need to be. This is the influence of dimension 2, the starting point for the pattern, being felt all the way up the chain.)
(Note again that in dimensions 1 and 0, the pattern is broken and there is no g; in dimension 0 it's always 1 and in dimension 1 it's the identity. We don't generally get something of the form ωω^α.)
This got me wondering if I could prove some bound on o(Rev(α1×...×αn,X)) for any WPO X by similar means, like how earlier I proved a bound on o(Rev(α,X)), but then I realized I could actually just prove that by taking my earlier bound and iterating it; I don't need to get into anything fancy there (which is good, it's ugly enough in the two-dimensional case I did already need to do by hand).
EDIT 7/25: The above paragraph seems to be false! Yikes, I may have to do that case by hand as well!
So, one more addition to the lore of "big multichoose"...
-Harry
So I wrote this down in an email to John, and then realized I should probably post it here as well.
So let's consider "big multichoose". I want to consider the variant that I denoted ((α multichoose β))0, where we consider all weakly decreasing functions from beta to alpha that are eventually 0.
This can also be considered as the set of lower sets of α×β that are "bounded in each direction", in the sense that neither projection is surjective.
Now of course if α and β are finite -- say n and m -- then this is just counting the number of partitions that fit in an (n-1)x(m-1) box, i.e., it's (n+m-2 choose n-1). (Which is different from multichoose, obviously, but nicer for the purposes I want to consider here.)
So suppose we generalize from two ordinal factors in the product to multiple. What happens now? Can we get a formula similar to my formula for dimension 2?
Answer: We can! It's not hard, either, once you know the dimension 2 case (OK, I haven't sat down and fully proven it, but the proof should be similar).
The thing is that I'd considered this before but always dismissed it because, well -- counting partitions in a given box is easy; counting plane partitions in a given box is less easy; and in dimensions higher than 3, things apparently get pretty wild pretty fast. So I figured, well, if the finite case is intractable, what's the point of attempting the infinite case?
Of course, this is silly, because if you can reduce the infinite case to the finite one, that's still significant. Like really, in the dimension 2 case, if you plug in finite ordinals, my formula doesn't actually say "OK, it's (n+m-2 choose n-1)", rather it says, OK, count the partitions in this box. We just know what that is. So in higher dimensions it will just fail to reduce the problem in the finite case, but in the infinite case, at least now it is a finite description.
So anyway as before the main core of the problem is when all the factors are powers of ω (and not equal to 1). So let's say we're putting in ωα_1, ..., ωα_n. Also, say n≥2; if n=1 (or n=0) the problem A. is trivial and B. is *not* given by the formula below, so let's exclude that. (And yes there is a good reason that dimension≥2 turns out different from dimension 1, but I'm not going to go into that here.)
So what we get out is ωω^g(α_1,...,α_n), for some g. What's g? Well, it's basically the same as before:
If either:
1. one of the αi is of the form epsilon+finite, or
2. at least two of the αi are infinite
then g(α1,...,αn) is just the natural sum of the αi.
But otherwise, it's the natural sum, minus 1.
(Before I did it out, I had assumed that condition (2) would require that all of the αi be infinite, but in fact, only two of them need to be. This is the influence of dimension 2, the starting point for the pattern, being felt all the way up the chain.)
(Note again that in dimensions 1 and 0, the pattern is broken and there is no g; in dimension 0 it's always 1 and in dimension 1 it's the identity. We don't generally get something of the form ωω^α.)
This got me wondering if I could prove some bound on o(Rev(α1×...×αn,X)) for any WPO X by similar means, like how earlier I proved a bound on o(Rev(α,X)), but then I realized I could actually just prove that by taking my earlier bound and iterating it; I don't need to get into anything fancy there (which is good, it's ugly enough in the two-dimensional case I did already need to do by hand).
EDIT 7/25: The above paragraph seems to be false! Yikes, I may have to do that case by hand as well!
So, one more addition to the lore of "big multichoose"...
-Harry
