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One of the way a person shows their wealth is by having many diamonds. The same can be said about models of $\sf ZFC$. We can add generic diamond sequences, while preserving the old ones, so in some sense, some models have more diamonds than others. And some have none at all.

Recall that a diamond sequence is a sequence $\langle A_\alpha\subseteq\alpha\mid\alpha<\omega_1\rangle$ such that $A_\alpha\subseteq\alpha$, and for any $A\subseteq\omega_1$, $\{\alpha\mid A\cap\alpha=A_\alpha\}$ is stationary.

Very very clearly, we can replace any one particular $A_\alpha$, in fact any countably many of them, and in fact any non-stationary set of them, and the sequence is still a diamond sequence. By that trivial observation alone, if there is one diamond sequence, there are $2^{\aleph_1}$ of them.

So the "plain counting argument" is the wrong one. But we can talk about equivalence up to a non-stationary set. That is, two sequence are equivalent, if they differ on a non-stationary set.

Okay, but maybe you can find a permutation of $\omega_1$ which maps stationary sets to stationary sets. And maybe that permutation is not the identity (mod. non-stationary, that is), so maybe that gives you a new sequence, where in fact you just moved that diamond from an earring to a necklace.

So it is unclear to me that agreeing on a club is the right notion here. But maybe it is.

Questions:

  1. Is there a better notion of making diamond sequences distinct?
  2. How wealthy are canonical inner models? In particular does $L$ have only one diamond sequence (up to a reasonable notion of equivalence)?

The reason to ask is that normally we argue that a canonical inner model has a diamond sequence by constructing said sequence from some fine structural properties of the model. But that gives us a very concrete sequence, which may hint at the relative poverty in which canonical inner models live.

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    $\begingroup$ I just want to clarify that diamonds in canonical inner models are ethical diamonds, i.e. diamonds that were not obtained by forcing. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 29, 2020 at 18:19
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    $\begingroup$ I think that equivalence up to a club is still too strong (you still have $2^{\aleph_1}$ many such diamond sequences from one). Take $X \subseteq \omega_1$ and ask the diamond to guess some coding of the pair (A, X), where $A \subseteq \omega_1$. By taking the guess for the A-part, in ordinals in which we guessed the $X$ part correctly, we get a diamond sequence. For any $X$ you get a different diamond sequence and the non-trivial parts of those sequences can agree only on a bounded piece. $\endgroup$
    – Yair Hayut
    Commented Apr 29, 2020 at 19:17
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    $\begingroup$ @Asaf I fear that we need to clarify the notion first before much can be said. I was hoping that wealth could be expected because of combinatorial consequences: in this model we can build this and that diamond sequences, the first gives us A, the second B, and "there does not seem to be a reasonable way" of getting B from the first or A from the second. $\endgroup$
    – Andrés E. Caicedo
    Commented Apr 29, 2020 at 19:30
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    $\begingroup$ Since you asked about the situation in $L$, I'll conjecture that, for any "reasonable" notion of equivalence, $L$ will have $\aleph_2$ inequivalent diamond-sequences and that this should be provable by tweaking standard constructions of one diamond sequence in $L$. (Here "reasonable" is intended to mean something like "combinatorial" and to exclude relative constructibility.) $\endgroup$
    – Andreas Blass
    Commented Apr 29, 2020 at 19:54
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    $\begingroup$ Another random thought. Replace your diamond sequences by allowing countably guesses at each $\delta$ instead of one. Then you can distinguish diamond sequences by whether or not they are "somewhere" a $\diamond^*$-sequence, that is, whether there is a stationary set $S$ where any set is guessed "almost always". Once you've got a $\diamondsuit^*(S)$ sequence, any run-of-the-mill single-guessing $\diamondsuit$ sequence on $S$ is built from it in some sense. This would push towards $L$ having only a single type of $\diamondsuit$ sequence, built from the hierarchy in standard way. $\endgroup$
    – Todd Eisworth
    Commented May 2, 2020 at 17:59

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