# Fixed points of ordinal exponentiation for bases besides $\omega$

The first fixed point of the map $$x \to \omega^x$$ is the first epsilon number $$\epsilon_0$$, which is the supremum of $$\omega, \omega^\omega, \omega^{\omega^\omega}, ... = \omega^{\omega^{\omega^{.^{.^{.}}}}}$$.

How does this change if we use a different base for ordinal exponentiation, for instance $$x \to (\omega+1)^x$$? Then the first fixed point would be the supremum of $$\omega+1, (\omega+1)^{(\omega+1)}, (\omega+1)^{(\omega+1)^{(\omega+1)}}, ... = (\omega+1)^{(\omega+1)^{(\omega+1)^{.^{.^{.}}}}}$$.

In general, my questions are:

1. Does changing the base to $$\omega+1$$ yield a different class of fixed points than the epsilon numbers?
2. What about changing the base to $$\omega\cdot2$$, or $$\omega^2$$, or in general, any countable ordinal less than $$\epsilon_0$$? (Or more than $$\epsilon_0$$?)
3. Do you get a different class of fixed points for every base? If not, when do you get a different class?

A nice way to put this is to ask about the fixed points of the general function $$f(\alpha, \beta) \to \alpha^\beta$$, which should be continuous and strictly increasing (a normal function) in the argument $$\beta$$. Are these characterized and well-known?

If $$\omega\le\alpha<\epsilon_0$$, then we get a "dovetailing" phenomenon (and so identical fixed points) - namely, any finite exponential tower of $$\alpha$$s is bounded by some finite exponential tower of $$\omega$$s (since $$\alpha<\epsilon_0$$) and vice versa (since $$\omega\le\alpha$$). For example, taking $$\alpha=\omega^2+1$$ we have $$\alpha<\omega^\omega$$, so $$\alpha^{\alpha^\alpha}\le\omega^{\omega^{\omega^{\omega^{\omega^{\omega}}}}}.$$ In general, it's easy to show that if $$f, g$$ are normal and for every finite $$n$$ there is some finite $$m$$ such that $$f(n) and $$g(n), then the least fixed points of $$f$$ and $$g$$ are equal.

• Thanks - I can see how this leads to the same least fixed point of $\epsilon_0$. But does it lead to the same class of fixed points? Also, what about for bases greater than $\epsilon_0$? – Mike Battaglia Feb 13 at 3:55
• @MikeBattaglia The same argument shows that the least fixed point of one above a mutual fixed point will be the same as the least fixed point of the other above that mutual fixed point. Meanwhile, a limit of mutual fixed points is clearly a mutual fixed point. So they have the same classes of fixed points. – Noah Schweber Feb 13 at 4:06
• Now what about $\alpha\ge\epsilon_0$? Well, given $\alpha\ge\epsilon_0$ let $\beta=\max\{\gamma: \epsilon_\gamma\le\alpha\}$; the tower-comparison picture above gives that $\alpha$ and $\epsilon_\beta$ yield the same classes of fixed points (note: $(\epsilon_\beta)^{\epsilon_\beta}$ is an $\epsilon$-number bigger than $\epsilon_\beta$). So everything is characterized by "where in the $\epsilon$-hierarchy" our $\alpha$ is. – Noah Schweber Feb 13 at 4:12
• Ah, ok. One more thing that I just thought of; what if $\alpha$ is finite? Then we have $2^\omega = \omega$, but after that I can't find a fixed point until $2^{\epsilon_0} = \epsilon_0$. Is this just the same as the epsilon numbers, but with an $\omega$ prepended as an initial point? – Mike Battaglia Feb 13 at 5:16

This is a bit of clarification regarding the finite case of $$x \mapsto 2^x$$ mentioned in comments. I think you are right regarding the point that the next fixed point of this function after $$\omega$$ should be $$\epsilon_0$$. To see this let $$f(x)=2^x$$.

We already know that $$f(\omega)=\omega$$. So, due to this, we know that the next fixed point of $$f$$ is $$sup\{f^n(\omega+1):n \in \omega\}$$. We have:

$$f(\omega+1)=2^{(\omega+1)}=w \cdot 2$$ $$f^2(\omega+1)=f(\omega \cdot 2)=2^{(\omega\cdot 2)}=\omega^2$$ $$f^3(\omega+1)=f(\omega^2)=2^{(\omega^2)}=\omega^\omega$$

The last equation does require a bit of convincing. To proceed further from this, let $$g(x)=\omega^x$$. We can observe that $$g(\omega)=f(\omega^2)$$.

With a bit of further thought, we can convince ourself that as we move $$1$$ step further for function $$g$$ (in the input), we have to proceed $$\omega$$ steps further (in the input) for function $$f$$. For example, $$g(\omega+1)=f(\omega^2+\omega)$$, $$g(\omega+2)=f(\omega^2+\omega \cdot 2)$$ etc. With this we are lead to $$g(\omega^2)=f(\omega^3)$$, $$g(\omega^3)=f(\omega^4)$$, $$g(\omega^4)=f(\omega^5)$$ etc. right up till: $$\omega^{\omega^\omega}=g(\omega^\omega)=f(\omega^\omega)=f^4(\omega+1)$$

So now we have $$f^5(\omega+1)=f(\omega^{\omega^\omega})$$.

But now a simple observation is sufficient to conclude that $$f^n(\omega+1)$$ (for all $$n \geq 5$$) is just $$n-1$$ towers of $$\omega$$. The main thing to note is that when we concluded $$g(\omega^\omega)=f(\omega^\omega)$$ we might as well have concluded a more general equality such as $$g(\omega^\omega \cdot \alpha)=f(\omega^\omega \cdot \alpha)$$ for all $$\alpha \geq 1$$. Let's see for example how we can calculate $$f(\omega^{\omega^\omega})$$ with the previous equality. We have: $$f(\omega^{\omega^\omega})=2^{\omega^{\omega^\omega}}=2^{\omega^\omega \cdot \omega^{\omega^\omega}}=w^{\omega^\omega \cdot \omega^{\omega^\omega}}=\omega^{(\omega^{\omega^\omega})}=\omega^{\omega^{\omega^\omega}}$$

Though in going through these equalities, we have used the fact that $$\omega^{\omega^\omega}$$ is a fixed point $$x \mapsto \omega^\omega \cdot x$$. I think the same equality can be carried through for all subsequent values of $$f^n(\omega+1)$$.