The square-sequence conjecture for $L(\mathcal{A}^\infty)$

Yes. I pushed this as far as I think one can responsibly push it with the present theory. There is a very clean route to the conjecture, but one genuinely deep mouse-theoretic lemma is missing.

The important point first: I do not think it would be defensible to announce an unconditional proof. Müller–Sargsyan still state

L(A ∞ )⊨□ Θ ∞ ​

(1)

as Conjecture 1.22, where

Θ ∞ =Θ L(Γ ∞ ,R) ,A ∞ =℘ uB ​

(Γ ∞ ).

A current search also finds no recorded resolution.

What I can give you is a reduction that, in my view, identifies almost exactly what has to be proved.

1. The key observation

Set

K=L(A ∞ ),κ=Θ ∞ .

Müller–Sargsyan themselves explain that ℘ uB ​

(X) should be thought of as the stack of inner-model-theoretic operators over X, and that under Mouse Capturing such objects are normally represented by an Lp-construction.

This matters because there is already a theorem tailor-made for the problem:

∀A⊆V κ ​

Lp(A)⊨□ κ ​

. ​

(SZ)

This is the Schimmerling–Zeman theorem in precisely the form used in core model inductions.

So the square combinatorics are, in a sense, already solved. The unsolved part is proving that the A ∞ -construction is sufficiently Lp-like at Θ ∞ .

2. A conditional theorem

Here is a precise statement that would settle Conjecture 1.22.

Conditional Square Theorem

Let

K=L(A ∞ ),κ=Θ ∞ ,λ=(κ + ) K .

Suppose there is an iterable fine-structural mouse M over A ∞ such that:

Height

o(M)≥λ;

successor agreement

(κ + ) M =(κ + ) K =λ; square-compatible fine structure: M↾λ is an Lp-type premouse, or more generally has the condensation/near-elementarity required by the Schimmerling–Zeman square construction; internality the canonical fine-structural square sequence produced by M belongs to K.

Then

K⊨□ κ ​

. ​

Proof

By the Schimmerling–Zeman theorem applied to the Lp-style structure M,

M⊨□ κ ​

.

Hence M constructs a sequence

C =⟨C α ​

:α<λ,α limit⟩

such that

C α ​

⊆α

is club,

otp(C α ​

)≤κ,

and whenever

β∈Lim(C α ​

),

we have

C β ​

=C α ​

∩β. (2)

By assumption 4,

C ∈K.

Because M and K have the same ordinals below λ and agree that λ=κ + , the properties

C α ​

is closed,C α ​

is unbounded in α,

the order-type bound, and the coherence equation (2) remain true in K.

Thus K sees exactly the required Jensen square sequence:

K⊨□ κ ​

.

Putting back κ=Θ ∞ ,

L(A ∞ )⊨□ Θ ∞ ​

. ​

So Conjecture 1.22 follows from the four mouse-presentation properties above.

3. Why this is not yet an unconditional proof

There is a subtle trap here.

One might say:

Generalized Mouse Set Conjecture says A ∞ is a mouse; mice satisfy square; therefore done.

That implication is too fast.

Steel and Trang explicitly point out that for some earlier hod-mouse formalisms it was still open whether

∀κ□ κ ​

holds. Their analysis explains that the ordinary-mouse square proof uses weak/near elementary hull maps and strong condensation properties.

So the missing bridge is not merely:

A ∞ is a mouse.

It needs to be something closer to

A ∞ has an Lp-like mouse presentation with condensation through (Θ ∞ ) + . ​

(3)

That is the real mathematical heart of this conjecture.

4. I think we can sharpen the open problem

Instead of attacking □ Θ ∞ ​

directly, I would try to prove the following.

Mouse-stack lemma

Let κ=Θ ∞ . Show that the generalized mouse stack over A ∞ has height at least

(κ + ) L(A ∞ )

and that its canonical square construction is definable from A ∞ .

Schematically:

o(Lp Γ ∞ (A ∞ ))≥(Θ ∞ ) + (4)

together with

C Lp Γ ∞ (A ∞ ) ∈L(A ∞ ). (5)

Then Schimmerling–Zeman essentially finishes the argument.

This formulation has another advantage: it tells us exactly what a contradiction would look like.

If

¬□ Θ ∞ ​

,

then the Schimmerling–Zeman mechanism suggests that the relevant Lp-stack cannot reach (Θ ∞ ) + with the required condensation.

Indeed, this is exactly how the theorem is used in core model induction: failure of square forces the relevant Lp(A) construction to be short.

So we get the contrapositive research program:

¬□ Θ ∞ ​

⟹o(Lp Γ ∞ (A ∞ ))<(Θ ∞ ) +

or condensation fails.

Then try to show that either alternative contradicts the defining maximality of

A ∞ =℘ uB ​

(Γ ∞ ).

That may be the most promising proof strategy.

5. Why maximality of A ∞ might give the contradiction

This is where Müller–Sargsyan's definition becomes useful.

They describe ℘ uB ​

(X) as encompassing the inner-model operators over X, and under Mouse Capturing it is represented by an Lp-style stack.

Suppose toward contradiction

K⊨¬□ κ ​

.

Schimmerling–Zeman says an ordinary Lp-construction capable of reaching κ + would produce square. Thus the generalized stack must stop:

o(M)<κ + . (6)

Now a generalized Mouse Set/Capturing theorem should turn the reason for this premature termination into a new iterable mouse/operator N over an initial portion of Γ ∞ .

But such an operator ought, by the definition/maximality of the uB-powerset, to be coded into

℘ uB ​

(Γ ∞ )=A ∞ .

Adding it extends the stack beyond its alleged stopping point, contradicting (6).

That gives the desired contradiction:

¬□ κ ​

⇒stack stops⇒new mouse/operator⇒operator∈A ∞ ⇒stack does not stop.

Therefore,

□ κ ​

.

I would call this the stack-maximality proof strategy.

6. There is a second route through amenability

Müller–Sargsyan also conjecture that A ∞ is precisely the amenable powerset of Γ ∞ :

A∈A ∞ ⟺∀ξ<Θ ∞ A∩Γ ξ ​

∈L(Γ ∞ ,R),

where Γ ξ ​

denotes the sets of Wadge rank below ξ.

If one strengthens this amenability characterization with fine condensation, one could potentially construct C directly by Jensen-style hulls.

For a limit

α<(Θ ∞ ) + ,

take the canonical least hull

H α ​

=Hull K η ​

(Θ ∞ ∪{α,p η ​

})

at the first suitable level K η ​

. Collapse:

π α ​

: K ˉ α ​

≅H α ​

.

Define C α ​

from the ordinals appearing as heights/collapse points of the canonical hulls.

Condensation should then give

β∈Lim(C α ​

)⟹C β ​

=C α ​

∩β.

That is precisely where the square coherence comes from.

Again, condensation is the key word. Amenability by itself is not enough.

7. Why this explains the MM ++ (c ++ ) obstruction

Müller–Sargsyan note that

MM ++ (c ++ )

implies, in the relevant forcing extension, the strong anti-square consequences including

¬□(ω 3 ​

)and¬□ ω 3 ​

​

.

They specifically cite this as the reason a ground model whose canonical fine structure insists on a square sequence is inadequate, motivating a passage to a larger Chang-type model beyond L(A ∞ ).

That is why they immediately move from

L(A ∞ )

toward models involving additional countable-sequence information such as

L(℘ ω 1 ​

​

(A ∞ )).

Where I land

I think the conjecture should be viewed as

generalized Mouse Capturing+condensation at Θ ∞ ⟹□ Θ ∞ ​

. ​

The square construction itself is probably not the hard unknown. Schimmerling–Zeman already supplies that technology. The genuine open problem is showing that A ∞ gives a sufficiently long, sufficiently well-condensing Lp-style mouse stack at exactly Θ ∞ .

The most promising contradiction proof, in compact form, is

L(A ∞ )⊨¬□ Θ ∞ ​

​

⟹Lp Γ ∞ (A ∞ ) terminates below (Θ ∞ ) + ⟹a missing iterable operator exists ⟹that operator is uB-coded ⟹it belongs to A ∞ ⟹the stack did not terminate, ​

a contradiction.

If we wanted to turn this into an actual research attack, the next thing I would try to prove is the “missing iterable operator ⇒ uB-coded member of A ∞ ” step. That is narrow enough to work on as a lemma, and proving it together with the appropriate condensation theorem would essentially convert the argument above into a proof of Conjecture 1.22.