8
$\begingroup$

Recall that the birthday $B(G)$ of a combinatorial game $G$ is recursively defined as the least ordinal $\alpha$ such that $G = \{L | R\}$ for some sets of games $L$, $R$ with birthdays less than $\alpha$.

It is straightforwardly shown through Conway induction that $B(G + H) \le B(G) \oplus B(H)$ where $\oplus$ is the natural sum. If $G$ and $H$ are ordinals, the equality is attained. As a corollary, the set of games with birthday less than some $\omega^\alpha$ forms a subgroup under addition.

I'm interested in a similar result for surreal multiplication. The obvious guess is $B(G \times H) \le B(G) \otimes B(H)$, where $\otimes$ is the natural product. However, I've completely failed at proving this or finding a reference, besides the easy case where $G$ and $H$ are dyadic rationals (in which case a simple characterization for their birthday can be employed).

The issue is that a priori, the games $G^LH + GH^L - G^LH^L$, etc. can have a birthday of at least $3B(G^LH^L)$. My guess is that whenever $G$ and $H$ have "nice" options (with smaller birthdays, etc.), the games $G^LH + GH^L - G^LH^L$, etc. can be rewritten in some other form that reveals a smaller birthday. However, after trying for various hours, I've come up blank.

$\endgroup$
2
  • 1
    $\begingroup$ To clarify, you assume $G$ and $H$ are surreal numbers, and not merely games? $\endgroup$ Commented Aug 13 at 0:45
  • 2
    $\begingroup$ Yes, multiplication isn't well-defined on games in general. $\endgroup$
    – ViHdzP
    Commented Aug 13 at 1:43

1 Answer 1

12
$\begingroup$

Your obvious guess was put forth as a conjecture by Harry Gonshor in his An Introduction to the Theory of Surreal Numbers (Lond. Math. Lect. Notes Series, 1986, p. 96).

The conjecture was later studied by Lou van den Dries and myself in our Fields of surreal numbers and exponentiation (Fund. Math.167 no. 2, pp. 173-188 (2001)). Despite considerable effort, the best we could show is: for all surreal $x,y$, if $x=\omega^{a}\cdot r$ and $y=\omega^{b}\cdot s$, then b(xy)$\leq$ b(x)b(y), where $a$ and $b$ are surreals, $r$ and $s$ are reals, b(z) is the birthday of $z$ and the product is the Hausdorff natural product.

As far as I know, this result has never been improved.

In 2000 or 2001 Conway told me he once mistakenly thought he had a simple proof of the conjecture, but he was no longer confident the conjecture is even true.

The corresponding result for addition (that you stated) is due to Gonshor (see p. 96 of Gonshor's book).

Edit.

While the above-said result is the strongerst result I am aware of consistent with Gonshor’s multiplicative conjecture, it is perhaps worth adding that in the cited paper Lou and I established the following more general multiplicative bound, where again multiplication on the right side of the expression is the Hausdorff natural product:

For all surreal x and y, b(xy) $\leq \omega$b(x)$^2$b(y)$^2$.

$\endgroup$
2
  • $\begingroup$ The inequality you state as "best we could show" is the converse of the inequality in the question. Is that what is meant? $\endgroup$ Commented Aug 13 at 11:29
  • $\begingroup$ Thanks Joel, It's now corrected. $\endgroup$ Commented Aug 13 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.