The following definitions are fairly standard, but reworded in a way that will be more appropriate for my question (so what follows is fairly long, but should be easy to read for the experts and might even serve as an introduction to the subject):

A

**partizan combinatorial game**is a set $G$ (of**positions**or**states**), together with an element $x_0 \in G$ (the**starting position**or**initial state**) and two relations, $L, R \subseteq G^2$ which we call the**blue**and**red**edges of the graph $(G, L\cup R)$ (an edge which is both blue and red can be called**green**), such that the graph in question is well-founded (=progressively finite, =there is no infinite sequence $z_0,z_1,z_2,\ldots$ of elements of $G$ with $(z_i,z_{i+1}) \in (L\cup R)$ for each $i$). The game is played as follows: starting with $x = x_0$, each player (called**Left**and**Right**) selects an edge $(x,y)$ in his respective set ($L$ for Left and $R$ for Right), replacing the position $x$ by its out-neighbor $y$; the player who cannot play loses (something which always happens in finite time by the well-foundedness assumption).Define the

**extensional collapse**of a game $G$ as follows: first, remove all vertices unreachable from the starting position; then, repeatedly (transfinitely) identify two vertices $x,x'$ which have the same sets of blue-neighbors and the same sets of red-neighbors (i.e., identify $x,x'$ if $\{y : (x,y) \in L\} = \{y : (x',y) \in L\}$ and the same is true for $R$). Call two games $G,G'$**extensionally equivalent**when they have isomorphic extensional collapses and write $G \cong G'$. For example, any two games in which no player has any legal play ($L=R=\varnothing$) are extensionally equivalent to the trivial game $0$ with only one position. Essentially everything will be studied up to extensional equivalence.The

**sum**$G'' := G \oplus G'$ of two games $G,G'$ is the Cartesian product $G\times G'$ with starting position $x''_0 := (x_0,x'_0)$ and edge relations $L'' := (L\times\Delta_{G'}) \cup (\Delta_G\times L')$ and similarly for $R''$, where $\Delta$ stands for the diagonal: i.e., playing a move in $G\oplus G'$ consists of playing a move in $G$ xor playing one in $G'$. Clearly this is well-founded, and clearly this operation is compatible with extensional equivalence, and it is commutative and associative (up to extensional equivalence, in fact, even isomorphism). The**opposite**$-G$ of a game $G$ is the same set $G$ but with $L$ and $R$ exchanged; we write $G \ominus G'$ for $G \oplus (-G')$.A game $G$ is

**Conway-positive**when Left has a winning strategy no matter who plays first,**Conway-negative**when Right has (i.e., $-G$ is positive), and**Conway-zero**when the*second*player has a winning strategy. Two games $G,G'$ are**Conway-equal**when $G-G'$ is Conway-zero. Let me write $G \doteq G'$ for this. It is not hard to see that $\doteq$ is an equivalence relation, which is weaker (=larger) than extensional equivalence.The

**product**$G'' := G \otimes G'$ of two games $G,G'$ is defined informally as follows. Its positions will be finite combinations of "plus" and "minus" tokens on the vertices of $G \times G'$. The starting position has a single "plus" token on $(x_0,x'_0)$ (the starting positions of $G$ and $G'$). At each turn, each player chooses a single token ("plus" or "minus"), at $(x,x')$ say, and chooses edges $(x,y)$ from $G$ and $(x',y')$ from $G'$: for a "plus" token, Left must choose two blue edges or two red edges (i.e., $(x,y) \in L$ and $(x',y') \in L'$, or else $(x,y) \in R$ and $(x',y') \in R'$) and Right must choose a blue edge and a red edge; for a "minus" token, it's the reverse (Left chooses a blue edge and a red edge, and Right chooses two blue edges or two red edges); in either case, the token that was at $(x,x')$ is removed and*three*new tokens are created: two of the same sign as the original token at $(x,y')$ and $(y,x')$, and one of opposite sign at $(y,y')$. In brief, we move from $x\otimes x'$ to $(x\otimes y') \oplus (y\otimes x') \ominus (y\otimes y')$.

The product has some nice properties: it is commutative and associative (up to extensional equivalence), it is distributive over the sum (ditto), has $1$ (defined below) as unit element (again, up to extensional equivalence) and $-G \cong (-1) \otimes G$.

For **impartial** games (those with $L=R=:E$), Conway-equality is exactly the equality of the Grundy ordinals (the Grundy ordinal of an impartial game $(G,x)$ being recursively defined as the least ordinal not equal to the Grundy ordinal of $(G,y)$ for any out-neighbor $y$ of $x$, i.e., $(x,y) \in E$), and the sum and product operations are compatible with this equivalence, and define over the ordinals two operations called "nim sum" and "nim product" which have plenty of nice properties (e.g., each infinite cardinal is an algebraically closed field of characteristic $2$; Conway calls these "nimbers", a portmanteau of "nim" and "number").

Sadly, whereas the sum is always compatible with Conway equality (if $G_1 \doteq G_2$ then $G_1 \oplus G' \doteq G_2 \oplus G'$), product is not in general. For a counterexample, leg $1$ be the game with two positions $0,1$ and a blue edge between $1$ and $0$, let $2$ be the game with three positions $0,1,2$ and blue edges $L = \{(1,0),(2,0),(2,1)\}$, and let $1\oplus 1$ be the same game as $2$ but without the $(2,0)$ edge (this is, indeed, extensionally equivalent to the sum $1\oplus 1$ as defined above): then $1\oplus 1 \doteq 2$ yet $(1\oplus 1)\otimes * \cong * \oplus * \doteq 0$ is not Conway-equal to $2 \otimes * =: * 2$, where here $*$ and $* 2$ are the same games as $1$ and $2$ but with green (=both blue and red) edges instead of just blue (multiplying by $*$ is extensionally the same as making the game impartial by making all edges green): the Grundy ordinals of $*$ and $*2$ are $1$ and $2$ respectively, so they are not Conway-equal.

So there are two obvious ways to try to solve this problem:

Restrict the definition of the product $G \otimes N$ for certain games $N$ for which $G \doteq G'$ implies $(G\otimes N) \doteq (G'\otimes N)$. This is Conway's approach in

*On Numbers and Games*, the games $N$ in question being called "numbers" (I'm not sure whether all the games for which $G \doteq G'$ implies $(G\otimes N) \doteq (G'\otimes N)$ are numbers, but numbers satisfy this property). This gives a nice theory where the numbers are a field and the games a partially ordered group, and we can multiply games and numbers, but we can't multiply two games.But the other would be to change the equivalence relation $\doteq$, and define a stronger (=smaller, =finer) equivalence relation, say $\triangleq$, by $G\triangleq G'$ iff $(G\otimes H) \doteq (G'\otimes H)$ for all games $H$ (i.e. not only does the second player win $G' \ominus G$, but in fact $(G'\ominus G) \otimes H$ for all games $H$; in particular, this is required for $H = *$, which demands that the impartial games $G \otimes *$ and $G' \otimes *$ have the same Grundy ordinals). It seems that this should still give the games (up to a certain size…) a ring structure, of which the nimbers are a factor or something like that.

My question (at last) is whether this second approach has been studied, and whether it leads to an interesting theory or whether on the contrary there is something obviously wrong with it.

finerequivalence relation would probably have been a bad idea, since equality is always supposed to be "the finest equivalence relation that will ever need to be considered in any given domain". $\endgroup$ – Gro-Tsen Mar 27 '16 at 0:09