1
$\begingroup$

Work in the context of combinatorial games as introduced by Conway. For surreals, the definition of the product is forced by the requirement that surreals should form an ordered field. Say, if $s' < s$ and $t'< t$, then we should have $s-s'>0$, $t-t'>0$, hence $(s-s')(t-t') >0$ and $st > st' + s't - s't'$. In game theoretical terms, if $s'$ is a left option of $s$ and $t'$ is a left option of $t$, then $st' + s't - s't'$ should be a left option of $st$. The argument is similar for the other options defining $st$.

However, the class of general games is not linearly ordered, and it is not necessarily the case that if $g'$ is a left option of a game $g$, then $g'< g$. So we have more possibilities for the definition of the product. For example, we can take as left options of $gh$ only the games of the form $gh' + g'h - g'h'$, with the further assumption that not only $g'$, resp $h'$, are left options of $g$, resp $h$, but also $g' < g$ and $h' < h$. Of course, the other conditions should be treated similarly. It seems that this alternative definition of the product leaves unchanged the product of two surreals and trivializes the product of many games wich are not (surreal) numbers. Thus probably the above definition is not very interesting.

Are there more interesting intermediate possibilities? In particular can we define a product on the class of all games in such a way that (products of surreals remain unchanged and) we get a game which is an imaginarry unit, that is $g^2=-1$? Perhaps the answer is no, but maybe it is worth trying.

$\endgroup$
  • 1
    $\begingroup$ Sums of games are defined and useful. Doesn't Knuth remark that there is no definition of product that will satisfy the desired properties, such as $(a+b)\times c = (a\times c)+(b\times c)$ ??$ $\endgroup$ – Gerald Edgar Aug 7 '18 at 14:05
  • $\begingroup$ Thank you. (I have to say that I am more familiar with surreals than with general games) but as far as I know, Knuth remark concerns games for themselves, but if we consider values of games, instead, the product satisfies all the desired algebraic properties. $\endgroup$ – Paolo Lipparini Aug 7 '18 at 14:31
  • $\begingroup$ @PaoloLipparini What do you mean by "values of games"? $\endgroup$ – Tobias Kildetoft Aug 7 '18 at 14:53
  • $\begingroup$ Two games $g$, $h$ are equal if for every game $x$ the games $g+x$ and $h+x$ give the same outcome (First, Second, Left or Right wins). This is an equivalence class, and the value of a game is its equivalence class (let apart set theoretical complications about classes of classes). Essentially Conway's theory is about values, rather than games (as far as I can see). Yet, the distinction sometimes ingenerates confusion. Think of a Blue and a Red Hackenbush stick, both connected to the ground, it is the same as to have no stick at all. $\endgroup$ – Paolo Lipparini Aug 7 '18 at 15:14
  • $\begingroup$ Ahh, I was not aware of any study of combinatorial games not up to that equivalence (since without it, most properties of sums break down anyway). $\endgroup$ – Tobias Kildetoft Aug 7 '18 at 15:15

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.