How to calculate multiplication of transfinite nimbers with a Cantor normal form

Question

I failed to calculate nimber multplication in the form of $[\omega^\alpha]*[\omega^\beta]$, according to the "mex" definition.

The cases when $\alpha<3,\beta<3$ are easy, while $[\omega^3]*[\omega^3]$ and others seem to be too difficult, not to mention the ordinals that are way too big to be represented by $\omega$, so I hope to have a result for ordinals with Cantor normal forms.

Answer 1

The discussion in ONAG (On Numbers and Games), second edition, Theorem 49 and its proof should be helpful. In particular, the following comment at the bottom of p.61 enables us to do computations with nimbers below $[\omega^{\omega^\omega}]$. I have changed the notation in the text from $\alpha_i$ to $\beta_i$ to avoid an unfortunate clash of notation with the previously defined $\alpha_p$ (for odd primes $p$) in Theorem 49.

Observe that the theorem enables us to compute with numbers below $[\omega^{\omega^\omega}]$, using the expansion $$\Omega^{p-1} \beta_{p-1} + \ldots + \Omega \beta_1 + \beta_0 = [\Omega^{p-1} \beta_{p-1} + \ldots + \Omega \beta_1 + \beta_0]$$ for $\Omega = [\omega^{\omega^k \cdot p^n}]$, $\beta_0, \ldots, \beta_{p-1} < \Omega$, which follows from Theorem 46.

Here $p$ is the $(k+1)$st odd prime. Note that the convention in ONAG is that the use of square brackets $[]$ (as on the right side of the equation above) indicates the usual (Cantor normal form) ordinal addition and multiplication, while without square brackets (as on the left side of the equation above), we are doing nimber addition and multiplication.

So polynomials of degree < $p$ in $\Omega$ are computed just by the usual ordinal multiplication (Cantor normal form). At degree $p$, one must use the values $\alpha_p$ described in Theorem 49. The value $\alpha_p$ is defined as the least number in $[\omega^{\omega^k}]$ with no $p$th root in $[\omega^{\omega^k}]$. By an earlier theorem in ONAG, this implies that $$[\omega^{\omega^k}]^p = \alpha_p.$$

This means that if you know the value of $\alpha_p$ for odd primes $p$ up to the $(k+1)$st odd prime, then you can compute with all nimbers less than $[\omega^{\omega^{k+1}}]$. Theorem 49 tabulates the first few values $$\alpha_3 = 2, \alpha_5 = 4, \alpha_7 = \omega+1.$$

In particular, for $k=0$ and $p=3$, we have $\omega^3 = \alpha_3 = 2$.

Let's look at your example $\beta = [\omega^3]$. We want to compute $\beta^2$. Here the exponent of $\omega$ in $\beta$ is 3, a natural number, so it is the case $k=0$ and hence $p=3$. Hence $\beta$ is just the $\Omega$ in the theorem with $n=1$. Since $\beta^2$ has degree 2 which is less than $p=3$, we just do ordinary multiplication so $\beta^2 = [\omega^6]$.

For $\beta^3$, note that the first sequence in Theorem 49 says that $\beta^3 = \omega$. This enables us to express higher powers of $\beta$ entirely in terms of 1, $\beta$, $\beta^2$.

For higher natural number [powers] of $\omega$, we still have $k=0$ and $p=3$, but we use the base 3 expansion of the exponent. For example, $[\omega^4] = [\omega^3 \omega] = [\omega^3][\omega]$ so that $[\omega^4]^3 = [\omega^3]^3[\omega]^3 = \omega \cdot 2 = [\omega \cdot 2]$.

I hope this is helpful to illustrate how nimber calculations with higher ordinals is done.