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

Question

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?

Answer 1

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 $f(n)<g(m)$ and $g(n)<f(m)$, then the least fixed points of $f$ and $g$ are equal.

Answer 2

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 $f^n(\omega+1)$.