# Are Conway's omnific integers the Grothendieck group of the ordinals under commutative addition?

This is a question in two parts.

Say that $\mathbf{On}$ is the proper class of all ordinal numbers in ZFC. We can define a binary operator over $\mathbf{On}$ which corresponds to the commutative version of ordinal addition; this has been called "Hessenberg addition" and "natural addition" before. It's also the operation you get by restriction of the $+$ operation from Conway's surreals to the subchain of ordinals (e.g. surreals with empty right set). I'll use the $+$ symbol for this operation over the ordinals.

$\langle\mathbf{On},+\rangle$ is a commutative monoid, which hence admits the notion of constructing a Grothendieck group $\mathrm{K}(\mathbf{On})$. The group $\langle\mathrm{K}(\mathbf{On}),+\rangle$ hence adds expressions such as $\omega$, $\omega-1$, $\omega^\omega - \omega^2 + 5$, etc. to the ordinals.

• Question 1: is $\mathrm{K}(\mathbf{On})$ equivalent to Conway's "omnific integers" $\mathbf{Oz}$? In Conway's "On Numbers and Games," he defines an omnific integer $x$ as one which can be represented as a surreal number $\left \{ x-1 \mid x+1 \right \}$. Are these two classes isomorphic to one another?

It's also noteworthy that the field of fractions $Quot(\mathbf{Oz})$ is the full field $\mathbf{No}$ of surreal numbers. We can further turn $\mathrm{K}(\mathbf{On})$ into a ring $\langle\mathrm{R}(\mathbf{On}),+,\times\rangle$ by defining a new commutative operation called $\times$, called the "Hessenberg product", "Hausdorff product" or "natural product" of ordinals, which is commutative, associative, has an identity of 1, and distributes over the Conway normal form of the ordinal. A good definition for the Hessenberg product can be found on pages 24-25 of Ehrlich 2006.

• Question 2: even if $\mathrm{K}(\mathbf{On})$ isn't isomorphic to $\mathbf{Oz}$, is $Quot(\mathrm{R}(\mathbf{On}))$ isomorphic to $\mathbf{No}$?

I'm tempted to answer in the negative for #1, as $\sqrt{\omega}$ is in $\mathbf{Oz}$, but is it in $\mathrm{R}(\mathbf{On})$? That is, given $\mathrm{K}(\mathbf{On})$ and ordinary commutative multiplication, is it the case that $\omega$ becomes a perfect square?

(Also, a last note - I'm aware that $\\mathbf{On}$ is a proper class. I'm not sure what foundational issues arise specifically in the above question, but I don't care how you want to handle them - NBG set theory, Grothendieck universes, whatever.)

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This is just the standard construction given here: en.wikipedia.org/wiki/Grothendieck_group Is there some subtlety I'm not noticing? – Will Sawin Jul 25 '12 at 2:41
Added for clarity. – Mike Battaglia Jul 25 '12 at 3:52
Vaguely related, although I'm not sure if directly. mathoverflow.net/questions/72691/… – Asaf Karagila Jul 27 '12 at 20:38
Thanks for the link, Asaf. I've indeed seen that question before; it actually is what sparked this one. – Mike Battaglia Jul 27 '12 at 23:15

There is an obvious extension of Cantor normal form to the Grothendieck group of the ordinals. Then the standard argument that $\sqrt{x}$ does not lie in the ring $\mathbb Z[x]$ applies to $\sqrt{\omega}$. Specifically, $\sqrt{\omega}$ must have a Cantor normal form $a + b \omega +$ higher-order terms, which squares to $a^2 + 2ab \omega +$ higher-order terms. For this to equal $\omega$ we need $a^2=0$ but $2ab=1$, which is of course impossible.

So your first question has a negative answer.

For your second, the equation $a^2=b^2\omega$ is equally problematic. Again apply the standard argument:

Write $a=k \omega^x +$ higher-order terms, and $b=l \omega^y +$ higher-order terms. Then the lowest term of $a^2$ and $b^2\omega$ must be equal, so $\omega^{x+_H x}=\omega^{y+_H y+_H 1}$, so $x+_H x = y+_H y+_H 1$, which cannot be because the $1$s coefficient of the first expression must be even while the $1s$ coefficient of the second expression must be odd.

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Hi Will, thanks for the answer. I'd considered that, but it still seemed to me that there might be some way that some extremely strange number which is some expression of ordinals might end up somehow squaring to $\omega$. Perhaps it really is as simple as you wrote, even despite the weird proper class stuff going on. As for the second part, natural multiplication is commutative, associative, has 1 as identity, and distributes over the normal form. Ehrlich has a good definition on pages 24-25 of ohio.edu/people/ehrlich/AHES.pdf, which I've added to my question for clarity's sake. – Mike Battaglia Jul 25 '12 at 4:22
But there isn't "some expression of ordinals". The Grothendieck group contains only differences of ordinals. Consider the ideal in the ring of ordinals generated by $\omega^2$. In this ideal, all possible Cantor normal form terms but $1$ and $\omega$ vanish. Thus, the quotient is isomorphic to $Z[x]/x^2$, which does not contain a square root of $x$. – Will Sawin Jul 25 '12 at 14:00
Will, are you talking about the semiring of actual ordinals, or the ring constructed from the Grothendieck group I mentioned above? In either case, I'm not entirely sure how to interpret your reasoning here. How do I interpret the quotient of either one of these structures by $\omega^2$? It's clear that $\omega^3$ would map to $\omega^1$ under the quotient, and that $\omega^n$ for any natural number would map to n mod 2 under the quotient. What about $\omega^\omega$, what does that map to? – Mike Battaglia Jul 26 '12 at 0:33
The reason I find your analogy with $Z[x]/x^2$ to be confusing is that the exponent in $Z[x]$ must be a natural number, whereas the exponents involved with ordinal numbers are themselves ordinals. Another way to formulate my above question is to ask what $\omega$ mod 2 is. – Mike Battaglia Jul 26 '12 at 1:55
Actually they all map to $0$ because they are all multiples of $\omega^2$. Every ordinal is a number plus $\omega$ times a number plus an element of the ideal $(\omega^2)$. Same for the grothendieck group. – Will Sawin Jul 26 '12 at 2:16