It's well-known that if you take the definition of surreal multiplication and attempts to generalize it to all games, the result is not well-defined, in that it does not respect equivalence of games.

What I'm wondering is this: What if we consider games that are equivalent to numbers, such as $\{*|*\}$, which is equivalent to $0$? Or to put it differently, what if we consider surreal multiplication, but we allow representing our numbers in non-numeric ways, such as representing a $0$ by $\{*|*\}$?

Is multiplication well-defined (equivalence-respecting) in this intermediate setting?

(I ask because I noticed that, although multiplication is well-defined for impartial games, it's not well-defined for games equivalent to impartial games. I'm wondering if something similar happens with numbers, or whether it remains well-defined; I haven't been able to find a counterexample.)

Thank you!


(the counterexample has been worked out in collaboration with Harry Altman. All errors are mine)

The counterexample provides a game $K$ such that $K=0$ but $K^2\neq 0$. Here $\cong$ is identity of games, $=$ is Conway equivalence and juxtaposition (or sometimes $\cdot$) denote Conway product, and $K^2$ is $K \cdot K$.

Consider $G \cong^{def} \{-1,0|0,1\}$ and $H \cong^{def} * \cong^{def} \{0|0\}$.

$G$ and $H$ are Conway equivalent, and moreover $*+*=0$, hence $K\cong^{def} G + H =0$.

Let's compute

$* \cdot * \cong *$

$GH \cong G \cdot * \cong \{*|*\} =0 $

$K^2 \cong (G+H)(G+H) = G^2 + GH + HG + H^2 = G^2 + \{0|0\}$.

Considering the left options $-1$, $0$ in the definition of the Conway product $G \cdot G$ we get that

$ G \cdot (-1) + 0 \cdot G - 0 \cdot (-1) = -G = G = * $

is a left option of $G^2$; similarly

$G =*$ is a right option of $G^2$.

There is no need to fully compute $G^2$ to see that $G^2 + \{0|0\}$ is a first-winner game (hence not $=0$), since the first player can always move in $G^2$ choosing the option equivalent to $*$. The second player is left to play with a game equivalent to $*+*$, so he loses.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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