"Does a bowtie monodromy construction produce a Hodge class with no classical algebraic representative?"

I'm an informal explorer investigating a potential angle on the Hodge conjecture. I have no formal background in algebraic geometry so please bear with the informal language.

Let X be a smooth complex projective fourfold fibered over a base curve B. Suppose the monodromy around a special point p ∈ B performs a "bowtie twist" — two complex loops γ₁, γ₂ sharing exactly one crossing point, pushed forward into X via a figure-8 map.

This produces a class: α=[γ1​]∪[γ2​]∈H^2,2 (X) ∩ H^4(X,Q)

The Question is Does the self-cancellation of orientations at the crossing point force a violation of Hodge Index Theorem constraints, such that no classical algebraic cycle Z can satisfy the intersection behavior of α on this fourfold?

Specifically — is α's only possible representative a derived object in D^b(Coh(X)) rather than a classical subvariety? And if so, does Deligne's original formulation exclude this?

I'm curious whether this construction has been attempted before and where it breaks down.

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On the Arithmetic Structure of Harmonic Cores: A $3/2$-Power Algorithm Approach to Collatz Trajectories

I am presenting a non-stereotypical analysis of the Collatz phenomenon, diverging from traditional heuristic or probabilistic approaches. My research identifies a brutal-structural line within $\mathbb{N}_{odd}$ based on the discovery of what I term Harmonic Forms.By introducing the 3/2-Power Algorithm, I have established a deterministic engine for the "unwrapping" of these harmonic cores. The methodology utilizes three specific numeric classifications—Type 1, Type 2, and Type 3—to navigate the trajectory space. These tools demonstrate that every odd integer undergoes a core traversal that converges toward a terminal wave at states $\{\mathcal{T}_1, \mathcal{T}_2\}$. This work suggests that the conjecture is not merely a problem of iteration, but a structural algebraic discipline inherent to $3x+1$ mappings within $\mathbb{N}$.I invite the community to scrutinize the complete call graph and the algebraic criteria established in my formal publication: Full Proof (Archived at Zenodo): https://doi.org/10.5281/zenodo.19957753

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On $H^2(\mathbb{M},\mathbb{T}^{196883})$

What, if anything, is known about $H^2(\mathbb{M},\mathbb{T}^{196883})$?

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Mathematical Model alternative

The current derivative of f(x)=x has problem in that no matter how small x changes it produces an immediate effect on f(x) and since x is usually placed in physics for time variable and the most fundamental phenomenon is SHM at which it takes some time for the effect. Would mathematicians be interested in a framework which reproduces current continuous framework at high frequencies of sines and cosines at high frequencies (one frequency at a time not fourier transform) . The Mathematics there is rigorous in the article I am genuinely looking for a proper feedback and will remove if the content don't hold up to the standards. https://zenodo.org/records/19959723

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What are some of the earliest examples of analytic continuation?

I'm wondering how Riemann knew that $\zeta(z)$ could be extended to a larger domain. In particular, who was the first person to explicitly extend the domain of a complex valued function and what was the function?

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The “voluble witness” argument and $L$

John Steel has argued that inner model theory gives evidence for the consistency of large cardinals, saying, “Such a thorough and detailed description of what a universe satisfying [a given theory] H might look like provides evidence that H is indeed consis- tent, for a voluble witness with an inconsistent story is more likely to contradict himself than a reticent one.”

Does the construction and analysis of $L$ similarly provide evidence for the consistency of ZF? It is surely a voluble witness to some picture of ZF.

On the other hand, this sounds somewhat suspicious. ZF is used in the construction and analysis of $L$. So why should the properties of $L$ give us more confidence in Con(ZF)? Isn’t it circular? Perhaps we just find more and more consequences of ZF (or Con(ZF)), which can also be done in other ways. Not that $L$ is unimportant but that it is not of a fundamentally different epistemic status. Similarly, large cardinal hypotheses are used in the construction of inner models of these hypotheses, so perhaps the models just show some more consequences of the assumptions rather than providing some justification of the assumptions.

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GAP versus SageMath for branching to Lie subgroups

Which computer package is better, GAP or SageMath, for decomposing an irreducible representation of a (simple) Lie group $G$ into representations of a Lie subgroup. I am most interested when branching to Levi, or parabolic, subgroups.

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Are all complex zeros of $Li_s(z)\, \pm \, Li_{1-s}(z)$ on the critical line or outside the critical strip for $z \le -1$?

This question loosely builds on this one, however is a bit simpler and I found the results to be more robust.

It seems that all zeros in the critical strip $0 \lt \Re(s) < 1$ of:

$$Li_s(z)\, \pm \, Li_{1-s}(z)$$

reside on the critical line $\Re(s)=\frac12$ for all $z \le -1$.

Below is a graph that shows where this function (using $\pm = -$) vanishes for $s=\frac12 \pm t\,i$ at different values of $z$. The lines marked in red all lead to a non-trivial zero of $\zeta(s)$, since $Li_s(-1)=\eta(s)$, with $\eta(s)$ being the Dirichlet Eta-function.

I have extended the graph towards $z \rightarrow 0^-$ to also show how some lines continue in that domain, however there clearly are zeros off the critical line when $z > -1$ (note: the lines coming from the left don't have a 'hard stop' at $z=-1$ and actually continue a short bit further to the right).

For the root finding process I used the following expression of the PolyLog function that I found to be evaluating much faster for higher values than the standard PolyLog function (in Maple):

$$Li_s(z) = \frac{\Gamma(1-s)}{(2\,\pi)^{1-s}} \left(i^{1-s}\,\zeta_H\left(1-s,\frac12+\frac{\ln(-z)}{2 \,\pi \, i}\right)+i^{s-1}\,\zeta_H\left(1-s,\frac12-\frac{\ln(-z)}{2 \,\pi \, i}\right)\right)$$

where $\zeta_H(s,q)$ is the Hurwitz zeta function.

Addition:

Numerical evidence suggests that the claim could be extended further into:

All complex zeros in the critical strip $0 \lt \Re(s) < 1$ of:

$$Li_s(z)\, \pm \, Li_{2a-s}(z)$$

reside on the critical line $\Re(s)=a$ for all $z \le -1$ and all $a \le \frac12$ (this might be related to this paper).

The patterns of lines of zeros at $s=a + t i$ are very similar to the graph above and values of $t$ only shift slightly at different $a$. However, the zeros at $z \rightarrow 0^-$ always remain $s=a \pm \frac{k \, \pi \, i}{\ln(2)},k \in \mathbb{N}$ and their imaginary part $t$ seems independent of the choice of $a$. This might be a clue for answering my second question (note that $2^s-2^{2a-s}$ has the same roots, however I did not find any connection).

Questions:

1) Is there a counterexample with a zero lying off the critical line, but in the critical strip for $z \le -1$?

2) When $z \rightarrow 0^-$ the function $Li_s(z)\, - \, Li_{1-s}(z)$ vanishes at $s=\frac12 \pm \frac{k \, \pi \, i}{\ln(2)}$ with $k \in \mathbb{N}$, however I failed to derive this apparently trivial result from the known formulae. Could this be proven?

Thanks.

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Continuous $(m-1)$-psh + $m$-qpsh + vanishing $m$-Hessian measure $\implies m$-psh?

This is a question about the definition and characterization of $m$-subharmonic ($m$-psh) functions in pluripotential theory (complex Hessian equations). Cross posted at https://math.stackexchange.com/questions/5135220/continuous-m-1-psh-m-qpsh-vanishing-m-hessian-measure-implies-m-p

Let $\Omega\subset\mathbb{C}^n$ be a domain, and let $\omega=dd^c|z|^2$ be the standard Euclidean Kähler form. Let $w\in C^0(\Omega)$ be a continuous function on $\Omega$ such that:

1. $w$ is $(m-1)$-psh;
2. $w$ is $m$-qpsh (i.e., $w+C|z|^2$ is $m$-psh for some constant $C>0$);
3. The $m$-th complex Hessian measure vanishes: $$ (dd^c w)^m \wedge \omega^{n-m} = 0 $$ as positive Borel measures on $\Omega$.

Question: Is $w$ necessarily $m$-psh?

Remarks

• For $C^2$ smooth functions, this is trivially true.
• For merely continuous functions, the usual tools break down if we only assume $(m-1)$-psh. I suspect an extra condition like $\Delta w\leq C$ (in the distribution sense) might be needed, but I am not certain.

Background definitions (standard in the field)

The positivity cone for the $\sigma_m$-equation is $$ \Gamma_m = \bigl\{\lambda\in\mathbb{R}^n : \sigma_1(\lambda)>0,\,\dots,\sigma_m(\lambda)>0\bigr\}, $$ where $\sigma_j(\lambda)=\sum_{1\leq i_1<\cdots<i_j\leq n}\lambda_{i_1}\cdots\lambda_{i_j}$ is the $j$-th elementary symmetric polynomial.

A smooth function $u$ is $m$-psh if and only if $$ (dd^c u)^k \wedge \omega^{n-k} \geq 0,\quad k=1,\dots,m. $$

Following Bedford–Taylor and Błocki, the weak definition for non-smooth functions is:

Definition 1.1. A subharmonic function $u$ on $\Omega$ is $m$-subharmonic ($m$-psh) if for any $C^2$ smooth $m$-psh functions $v_1,\dots,v_{m-1}$, the current inequality $$ dd^c u \wedge dd^c v_1 \wedge\cdots\wedge dd^c v_{m-1} \wedge \omega^{n-m} \geq 0 $$ holds in the weak sense of positive currents.

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Does the PNT-scale asymptotic justify this derivation for prime gaps? [closed]

Let $p_n$ be the $n$th prime, and define

$$ g_n:=p_{n+1}-p_n. $$

The Prime Number Theorem gives

$$ p_n\sim n\log n. $$

My question is whether this asymptotic justifies the derivation below.

We start from the exact identity

$$ g_n=p_{n+1}-p_n. $$

Dividing by $p_n$ gives

$$ \frac{g_n}{p_n} = \frac{p_{n+1}}{p_n}-1. $$

Using the PNT scale, one writes

$$ p_n\sim n\log n $$

and

$$ p_{n+1}\sim (n+1)\log(n+1). $$

Thus the proposed asymptotic lifting is

$$ \frac{p_{n+1}}{p_n} \sim \frac{(n+1)\log(n+1)}{n\log n}. $$

Now compute

$$ \frac{(n+1)\log(n+1)}{n\log n} = \left(1+\frac1n\right) \frac{\log(n+1)}{\log n}. $$

Since

$$ \log(n+1) = \log n+\log\left(1+\frac1n\right), $$

we have

$$ \frac{\log(n+1)}{\log n} = 1+ \frac{\log(1+1/n)}{\log n}. $$

Using

$$ \log\left(1+\frac1n\right) = \frac1n+O\left(\frac1{n^2}\right), $$

we get

$$ \frac{\log(n+1)}{\log n} = 1+ O\left(\frac1{n\log n}\right). $$

Therefore

$$ \frac{(n+1)\log(n+1)}{n\log n} = \left(1+\frac1n\right) \left( 1+O\left(\frac1{n\log n}\right) \right). $$

Expanding gives

$$ \frac{(n+1)\log(n+1)}{n\log n} = 1+\frac1n+O\left(\frac1{n\log n}\right). $$

Hence

$$ \frac{(n+1)\log(n+1)}{n\log n} = 1+\frac1n+o\left(\frac1n\right). $$

If the asymptotic lifting is valid, then

$$ \frac{p_{n+1}}{p_n} = 1+\frac1n+o\left(\frac1n\right), $$

so

$$ \frac{p_{n+1}}{p_n}-1 \sim \frac1n. $$

Substituting into the exact identity

$$ \frac{g_n}{p_n} = \frac{p_{n+1}}{p_n}-1 $$

gives

$$ \frac{g_n}{p_n} \sim \frac1n. $$

Therefore

$$ g_n \sim \frac{p_n}{n}. $$

Since

$$ p_n\sim n\log n, $$

this gives

$$ g_n\sim \log n. $$

Equivalently, since $\log p_n\sim \log n$,

$$ g_n\sim \log p_n. $$

Now square the exact identity

$$ \frac{g_n}{p_n} = \frac{p_{n+1}}{p_n}-1. $$

This gives

$$ \left(\frac{g_n}{p_n}\right)^2 = \left(\frac{p_{n+1}}{p_n}-1\right)^2. $$

Multiplying by $p_n$ gives

$$ \frac{g_n^2}{p_n} = p_n \left( \frac{p_{n+1}}{p_n}-1 \right)^2. $$

Using

$$ \frac{p_{n+1}}{p_n}-1\sim \frac1n $$

and

$$ p_n\sim n\log n, $$

we get

$$ \frac{g_n^2}{p_n} \sim p_n\frac1{n^2} \sim \frac{n\log n}{n^2} = \frac{\log n}{n}. $$

Hence

$$ \frac{g_n^2}{p_n}\to 0. $$

Therefore

$$ \left(\frac{g_n}{\sqrt{p_n}}\right)^2\to 0, $$

and so

$$ \frac{g_n}{\sqrt{p_n}}\to 0. $$

Thus

$$ g_n=o(\sqrt{p_n}). $$

In particular,

$$ g_n=o(\sqrt{p_n}\log p_n). $$

Question 1.

Is the asymptotic lifting

$$ p_n\sim n\log n \quad\Longrightarrow\quad \frac{p_{n+1}}{p_n} \sim \frac{(n+1)\log(n+1)}{n\log n} $$

valid in this derivation?

Equivalently, does the PNT asymptotic justify the step

$$ \frac{p_{n+1}}{p_n}-1\sim \frac1n, $$

and hence the conclusions

$$ g_n\sim \log p_n $$

and

$$ g_n=o(\sqrt{p_n}\log p_n)? $$

Addendum: logarithmic normalization

There is also a related logarithmic normalization.

Define $\alpha_n\in(0,1)$ by

$$ \log(p_n) \int_{\alpha_n}^{1}\frac{dx}{x} = g_n. $$

Since

$$ \int_{\alpha_n}^{1}\frac{dx}{x} = -\log(\alpha_n), $$

we get

$$ g_n = -\log(p_n)\log(\alpha_n). $$

Therefore

$$ -\log(\alpha_n) = \frac{g_n}{\log(p_n)}. $$

Equivalently,

$$ \alpha_n = \exp\left( -\frac{g_n}{\log(p_n)} \right). $$

Now introduce a real parameter $\beta_n$ by requiring

$$ \int_{\alpha_n}^{1}\frac{dx}{x} = \int_{1}^{p_n}\frac{dx}{x} + \int_{p_n}^{p_{n+1}-\beta_n}\frac{dx}{x}. $$

Evaluating the integrals gives

$$ -\log(\alpha_n) = \log(p_n) + \log\left( \frac{p_{n+1}-\beta_n}{p_n} \right). $$

The right-hand side simplifies to

$$ -\log(\alpha_n) = \log(p_{n+1}-\beta_n). $$

Hence

$$ \alpha_n = \frac1{p_{n+1}-\beta_n}, $$

or equivalently,

$$ \beta_n = p_{n+1}-\frac1{\alpha_n}. $$

Multiplying the defining identity by $\log(p_n)$ gives

$$ g_n = \log(p_n) \left[ \log(p_n) + \log\left( \frac{p_{n+1}-\beta_n}{p_n} \right) \right]. $$

Thus

$$ g_n = \log^2(p_n) + \log(p_n) \log\left( \frac{p_{n+1}-\beta_n}{p_n} \right). $$

Dividing by $\log^2(p_n)$ gives

$$ \frac{g_n}{\log^2(p_n)} = 1+ \frac{ \log\left( \frac{p_{n+1}-\beta_n}{p_n} \right) }{ \log(p_n) }. $$

Using the earlier derived scale

$$ g_n\sim \log(p_n), $$

we get

$$ \frac{g_n}{\log(p_n)}\to 1. $$

Since

$$ -\log(\alpha_n) = \frac{g_n}{\log(p_n)}, $$

this gives

$$ -\log(\alpha_n)\to 1. $$

Therefore

$$ \alpha_n\to e^{-1}. $$

Using

$$ -\log(\alpha_n) = \log(p_{n+1}-\beta_n), $$

we obtain

$$ \log(p_{n+1}-\beta_n)\to 1. $$

Hence

$$ p_{n+1}-\beta_n\to e. $$

Therefore

$$ \log\left( \frac{p_{n+1}-\beta_n}{p_n} \right) = \log(p_{n+1}-\beta_n)-\log(p_n). $$

Since

$$ \log(p_{n+1}-\beta_n)\to 1, $$

we get

$$ \log\left( \frac{p_{n+1}-\beta_n}{p_n} \right) = 1-\log(p_n)+o(1). $$

Dividing by $\log(p_n)$ gives

$$ \frac{ \log\left( \frac{p_{n+1}-\beta_n}{p_n} \right) }{ \log(p_n) } \to -1. $$

Substituting this into

$$ \frac{g_n}{\log^2(p_n)} = 1+ \frac{ \log\left( \frac{p_{n+1}-\beta_n}{p_n} \right) }{ \log(p_n) } $$

gives

$$ \frac{g_n}{\log^2(p_n)} \to 1-1 = 0. $$

Hence

$$ g_n=o(\log^2(p_n)). $$

Question 2.

Does this logarithmic normalization depend on the same asymptotic lifting step as Question 1?

That is, is the derivation of

$$ g_n=o(\log^2(p_n)) $$

valid once the preceding PNT-scale derivation of

$$ g_n\sim \log p_n $$

is valid?

(Comment on this)

When does the full restricted semidirect product of $A$ by $B$ have a split Billhardt congruence given by the projection on $B$?

Theorem 12 in Chapter 5 of Lawson's Inverse Semigroups: The Theory of Partial Symmetries states: If $\rho$ is a split Billhardt congruence on an inverse semigroup $S$ then $S \cong \operatorname{Ker} \rho \bowtie S/\rho$.

I am interested in the converse situation. What conditions need to be imposed on $A$ and $B$ along with the action such that there is a split Billhardt congruence on $A \bowtie B$ given by $\pi_B$. So I can write $A \bowtie B \cong \operatorname{Ker} \pi_B \bowtie ((A \bowtie B)/\pi_B)$. Along with having $A \cong \operatorname{Ker}\pi_B$ and $B \cong (A \bowtie B)/\pi_B$.

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Is there any benefit from writing homomorphisms on the same side as the module/group action?

Let $R$ be a ring, and let $M_R$ and $N_R$ be right $R$-modules. I was taught to always write module homomorphisms on the opposite side from scalars. Thus, a map $\varphi\in {\rm Hom}_R(M,N)$ would be written on the left side of elements of $M$. The homomorphism rules then become $$ \varphi(m_1+m_2)=\varphi(m_1)+\varphi(m_2) $$ and $$ \varphi(m\cdot r)=\varphi(m)\cdot r. $$

The benefits of this notational choice are numerous. Let me list just a few.

1. Terms are easier to parse.

Writing both module actions using concatenation for simplicity, then the second homomorphism rule tells us that the term $\varphi m r$ makes sense no matter how we parenthesize it. It is simply well-defined.

If we have written homomorphisms on the same side as the action, and similarly remove parentheses, the second rule become $\varphi rm=r\varphi m$. This looks like a sort of commutativity rule. Moving symbols just adds extra unnecessary complication. It is better to have a single well-defined unparenthesized term than to identify two different terms (which we might want to use to express other things anyway, or to use as sanity checks--see below).

2. Bimodule naturality is more easily expressed.

Suppose moreover that $_SM_R$ is an $S$-$R$-bimodule, and that $_TN_R$ is a $T$-$R$-bimodule. Then ${\rm Hom}_R(M,N)$ is naturally a $T$-$S$-bimodule! This happen by defining the actions of $t\varphi s$ on $M$ via: $$ (t\varphi s)(m)=t(\varphi(sm)). $$ After verifying that this makes the hom set a bimodule, we again get a simple term parsing fact as in #1.

It is quite convenient, and gives a sort of sanity check, to think of the hom set as "using up" the $R$-module structures, but leaving the $S$ and $T$ structures unused.

Another benefit is that if symbol order is switched, we know an error was made (unless more information, like $R$ being a commutative ring, is being used). Such an easy visual cue is not available when homs are on the same side as the action.

Another benefit is that it leads to more natural sounding statements. For example, the hom-tensor adjunction is often phrased in a way where the sides of the actions are all over the place. But it is actually quite simple when phrased as: There is a natural isomorphism of $U$-$R$-bimodules $$ {\rm Hom}_T(_RA_S\otimes_S\,{}_SB_T,{}_UC_T)\cong {\rm Hom}_S(_RA_S,{\rm Hom}_T(_SB_T,{}_UC_T)) $$ When one doesn't pay attention to which side homomorphisms are written on, it is much harder to parse the statement. The proof becomes an absolute mess. (Try it both ways!) It is quite convenient to think of tensors as "using up" the part of the bimodule structure in the middle (in our case, the two $S$-module structures).

3. It makes unnaturality and opposite ring constructions more apparent.

Recall that if $D$ is a division ring and $V_D$ is a (right) vector space over $D$, then the dual space is $V^{\ast}:={\rm Hom}_D(V,D)$. Noting that $D$ is naturally an $D$-$D$-bimodule, then from what we said in #2, the set $V^{\ast}$ is naturally a left $D$-module (i.e., a left vector space). In other words, the hom set "used up" the right $D$-module structures on $V$ and $D$, but did not use up the left $D$-module structure on $D$.

When $D$ is a (commutative) field, one might think that caring about which side is which doesn't matter too much. And yet, if we do pay attention, this mostly explains (even in the commutative case) why there is no natural map $V\to V^{\ast}$; the categories are just wrong! It also gives intuition about why $V\to V^{\ast\ast}$ will make sense; they are both right $D$-spaces.

When $M_R$ is a right $R$-module, then it is automatically a left $R^{op}$-module, via the action $$ r^{op}m:=mr. $$ But it is not automatically an $R^{op}$-$R$-bimodule unless $R$ is commutative. A term $r_1^{op}mr_2$, without parentheses, is not necessarily well-defined.

We can transfer (unnaturally) a commutative action. But noncommutative actions don't transfer from one side to the other (even unnaturally) unless we pass to the opposite ring and we forget about the original action. This is most apparent when working with group actions (for a non-abelian group). In representation theory, you will see all sorts of extra inverses appearing. These come from writing homs on the same side as the action, and further identifying $R[G]^{op}$ with $R[G]$ after inverting elements of $G$. Such oddities often disappear when writing homs on the opposite side.

My question is now:

Is there any good reason to write homs on the same side as the scalars?

Of course, some undergraduate students are taught linear algebra using scaling on the left, and they are also used to functional notation on the left. But I'm talking about for graduate-level or higher mathematics. I'm talking about yellow Springer books. Is there any good reason when teaching representation theory to complicate the subject by not following the nice notational scheme of writing homs on the opposite side from actions?

Extra note: Even for linear algebra taught to undergraduates it makes pedagogical sense to follow this convention, and teach this notational choice early. If $V$ is an $n\times 1$ column space, scalars are $1\times 1$ matrices, so should go on the right when multiplying. If $W$ is an $m\times 1$ column space, then ${\rm Hom}(V,W)$ is the set of $m\times n$ matrices, and those go on the left of $V$ when multiplying--the opposite side as the scalars. When instead working with (the dual) row spaces, then scalars go on the left and homs on the right.

(Comment on this)

Can we paraphrase constructible sets as object predicates?

This question is linked to a prior one.

Language: Mono-sorted first order logic with equality and additional primitives of a total unary function $\xi$ signifying is the order of, and the binary relation $ [ \ ] $ signifying predication that is $p[x]$ to be read as $p$ predicates $x$, or equivalently as $x$ is predicated by $p$. The domain of this theory only range over predicates, which would be called as object predicates. So, this theory is a first order theory speaking about a sort of high order logic. An order of an object predicate in this theory can be as big as any ordinal provable to exist in ZF. Care to be exercised not to confuse $p[x]$ for $P(x)$, the first $p$ only range over objects of this theory, while the second $P$ range only over predicates that are not objects of this theory. We'll use upper cases for the latter, while the former would be written in lower casse.

Define: $q \subseteq p \iff \forall x (q[x] \to p[x])$

Define: $q \subsetneq p \iff q \subseteq p \land \neg p \subseteq q$

Define: $\operatorname {preord} (\alpha) \iff \forall \beta \, (\alpha [\beta] \to \beta \subsetneq \alpha \land \forall \gamma \, (\beta[\gamma] \to \gamma \subsetneq \beta))$

In English: $\alpha$ is a preordinal if and only if its a properly transitive (whatever predicated by it is a proper subpredicate of it) predicate predicating properly transitive predicates.

Define: $\operatorname {ext}(\alpha) \iff \forall \gamma \, (\gamma \equiv \alpha \to \gamma = \alpha)$

Where $\equiv$ is the co-predication relation, that is predicating the same objects.

Define: $\operatorname {ord}(\alpha) \iff \operatorname {preord}(\alpha) \land \operatorname {ext}(\alpha)$

Define: $\alpha < \beta \iff \operatorname {ord} (\alpha) \land \operatorname {ord}(\beta) \land \beta [ \alpha]$

Axioms:

Ordinals: The order of every object predicate is an ordinal.

$$\forall p \forall \alpha: \xi(p)=\alpha \to \operatorname {ord} (\alpha) $$

Ordering: the order of an object predicate is the minimal strict upper bound on the orders of all predicates it predicates.

$$\xi (p) = \min \alpha: \forall q \,( p [q] \to \xi (q) < \alpha)$$

Comprehension: For every order of predication $\alpha$, and any formula $\psi$, there exists an object predicate predicating all object predicates satisfying $\psi$ having their predication orders lower or equal to $\alpha$.

$$\forall \alpha: \operatorname {ord} (\alpha) \to \exists p \, \forall q \, (p [q ] \leftrightarrow \xi (q) \leq \ \alpha \land \psi(q))$$

Infinity: There is a strict upper limit order of predication.

$$\exists \operatorname {ord} \lambda > \varnothing: \forall \alpha < \lambda \, \exists \beta \, ( \alpha < \beta < \lambda)$$

Replacement: Any definable one-to-one replacement of predicates predicated by an object predicate, by object predicates; defines an object predicate.

$$\forall p \exists q \forall y (q[y] \leftrightarrow \exists x: p[x] ∧ y=F(x)) $$

Extensionality: All object predicates are extensional:

$$\forall p: \operatorname {ext}(p)$$

Constructability: written by the same constructability formula "V=L" but one that replaces each membership relation $y \in x $ by $x[y]$.

The idea here is that sets are nothing but object predicates, i.e. predicates that are objects but of a higher type (order) than what it predicates. A concrete example is like saying that Red is itself an object but of a higher type than the red objects it predicates. So, ZFC+(V=L) would be paraphrased as a first order theory about an infinite kind of higher order logic. The reason why this can be seen as a kind of logic is because its primitives are nothing but ordered predicates and their satisfaction, and the axioms seems to be natural for the altitudes of predication it aims to capture. This paraphrases set theory as a kind of logicism. I need however to make sure of two matters:

Is this theory synonymous with ZFC+(V=L).

A similar question but for the case V=HOD?

I need this synonymy in order to establish the logicism argument.

(Comment on this)

Rational solutions to $x^3+y^3=z^3+1$ with $xyz=\square$, and the equation $X^6+h^3Y^6=Z^6+h^3W^6$

In my investigation on higher power Diophantine triples (arxiv, related MO question), I considered the following equation in rationals $$\tag{1} \label{cubic_sum_square_product} x^3+y^3=z^3+1, \\ xyz=\square, $$ where by $\square$ I mean a rational square. (This may be homogenized to integers for symmetry.) There are trivial solutions by taking $x=z$ or $x=1$, but we disregard those. A complete parametric solution is known to the first equation (due to Euler, cf. Elkies) but substituing that in the second condition - that the product be a square - didn't reveal much.

If we could ask for x,y,z to all be squares, then the second equation would be satisfied automatically, but this would lead to an integer solution of $X^6+Y^6=Z^6+W^6$. No such is currently known.

Brute force search through rational $x,y,z$ of height <30000 though demonstrated two solutions of (\ref{cubic_sum_square_product}) (in homogenized form): $$ 1600^3+243^3 = 1587^3+484^3, \\ 2809^3+78^3 = 2808^3+289^3.$$ The corresponding products are indeed squares. Dehomogenizing, we get $x=1600/484$, in the first equation and $x=2809/289$, both squares! Thus the only two examples I could find, satisfy the added restriction that $x=\square$ as well. Moreover, we may write the above as $$ 40^6 + 3^3\cdot 9^6 = 22^6+3^3\cdot 23^6, \\ 53^6 + 78^3\cdot 1^6 = 17^6 + 78^3\cdot 6^6. $$

If we add to (\ref{cubic_sum_square_product}) the condition that $x=\square$, one may derive with a little algebra that such a (non-trivial) solution of (\ref{cubic_sum_square_product}) is equivalent to finding a non-trivial solution of $$ \tag{2} \label{two_representations} X^6+h^3Y^6=Z^6+h^3W^6, $$ just like the two examples above.

My questions are:

1. Are there infinitely many square-free $h$ that admit non-trivial solutions to (\ref{two_representations})?
2. For a fixed $h$ (like $h=3$ or $78$), is there a reason to expect finitely or infinitely many solutions $(X, Y, Z, W)$? (Except for computational evidence or lack thereof.)
3. Do there exist any solutions to the original system (\ref{cubic_sum_square_product}) where none of $x, y, z$ are squares?

Extra question: for a fixed integer $k$, are there infinitely many numbers (non-trivially) representable in two ways as $X^6+kY^6$?

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Sequence of MMP with scaling cannot be isomorphism

Let $(X,B)$ be a projective klt pair, H is an $\mathbb{R}$-divisor s.t. $K_X+B+H$ is nef. Suppose that we can run $(K_X+B)$-MMP with scaling $H$(that is, the flip exists) and denote $\alpha_i:X \dashrightarrow X_i$ to be the map to the i-step of the MMP. Then my question is how can I show that $\alpha_i\neq\alpha_j$ for any $i\neq j$ ?(That is, the induced birational map $\alpha_i\cdot\alpha_j^{-1}:X_j\dashrightarrow X_i$ is not an isomorphism.

Of course we can assume each step is the flip. Then from the construction of MMP with scaling we obtain a sequence of real numbers $1\geq t_0\geq t_1\geq t_2\geq...\geq 0$ . Can we prove the strict inequality : $t_i> t_{i+1}$?

Thanks for your comments!

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Sentences with unbounded universal quantifier

It is well known that for the following statement containig a bounded universal quantifier: $$∀y<b∃x_{1},…,x_{m}[F(a,y,x_{1},…,x_{m})=0]....................(1)$$ where $a$ is the parameter(s) of the polynomial $F$, the Chinese remainder theorem method results in the following system of Diophantine conditions solvable in the unknowns $q,w,z_{0},…,z_{m}$ provided that (1) holds: $$F(a,z_0,z_1,…,z_m)\equiv 0 \pmod {\binom {q}b}$$

$$z_0=q$$

$b!×( b+w+B(a,b,w)) !$ divide $q+1$

$\binom {q}b $ divide $\binom {z_1}w$

$\binom {q}b $ divide $\binom {z_2}w$

....

$\binom {q}b $ divide $\binom {z_m}w$

where the polynomial $B(a,b,w)$ is obtained from $F(a,y,x₁,…,x_{m})$ by changing the signs of all its negative coefficients and systematically replacing $y$ by $b$ and $x₁,…,x_{m}$ by $w$.

The statement (1) is hold for all $y<b$, i.e., $0≤y<b$. Now, my question is:

How one can transform the statement:

$$∀y>b∃x_{1},…,x_{m}[F(a,y,x_{1},…,x_{m})=0]....................(2)$$

to a system of Diophantine conditions where $b>0$. Statement (2) is modeled by an unbounded universal quantifier. An unbounded universal quantifier $∀$ denotes that a property holds for all elements in a domain without restriction, often ranging over infinite sets.

Several searching return no results even in classical books of logic.

For example this sentence: For all $y>b$, there exist $x,z$: $$y=(x+2)(z+2)$$

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For compact manifolds, is the space of homeomorphisms homotopy equivalent to a CW complex?

It is known since 70s for a compact PL manifold the space of PL homeomorphisms are ANR, hence homotopy equivalent to a CW complex, due to Haver (1973). The same can be said to diffeomorphisms, with respect to Whitney topology.

However, it is open whether or not the space of homeomorphisms is ANR. In dimension 1 and 2 this is known to hold. Is it also unknown, for a compact manifold $X$, whether $\mathrm{Homeo}(X)$ has CW homotopy type? (If $X$ has boundary, we require homeomorphisms restrict to identity on boundary.)

At least, maybe we can expect this statement to be true, for standard spaces like $\mathbb{S}^n$, $D^n$?

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Is there a "slow" version of $0^\sharp$?

Given a theory $A$, let $S_T$ be the set of reals $r$ such that, for some $s\le_T r$, $s$ codes (in some fixed appropriate way) an end extension of $L_\alpha$ where $\alpha$ is the smallest ordinal $>\omega$ such that $L_\alpha[r]\models A$. Let $(*)_A$ be the statement that $S_A$, viewed as a game in the usual way, is determined.

Question: Is there a (ideally natural and entailed by $\mathsf{KP}$) theory $A$ such that (over $\mathsf{ZFC}$) $(*)_A$ has consistency strength strictly weaker than $0^\sharp$ but strictly greater than $\mathsf{ZFC}$ itself?

This question is a special case of this earlier question of mine. $0^\sharp$ is equivalent to $(*)_{\mathsf{KP}}$, and it seems natural to wonder whether using some weaker closure condition might yield such an intermediate game.

If $A$ says (only) that the closure of the domain under primitive recursive set functions, I believe that the resulting principle is outright provable in $\mathsf{ZFC}$. On the other hand, if $A$ says that the universe is $E$-closed then this gives $0^\sharp$ again: if $r$ wins for player $1$ then the relevant $\alpha$ is countable in $L_{\alpha+1}[r]$ and so $E$-closedness is just admissibility in this case.

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Regular polygon shadows of convex polyhedra

Fix a finite subset $S$ of the natural numbers $\mathbb{N}$, each element $\ge 3$.

Is there a convex polyhedron $P$ that has among its shadows regular $n$-gons for each $n \in S$? Does such a $P$ exist for every $S$?

By shadow I mean the orthogonal projection of $P$ onto a 2D plane, for example, the shadow on the $xy$-plane, with $P$ above ($z>0$) that plane and the light at $L=(0,0,+\infty)$. By "among its shadows" I mean that you can rotate $P$ as you wish, and the goal is to find orientations of $P$ that cast the requisite shadows. What shadows are cast at other orientations is not relevant.

I believe the answer to my question is No. But the answer is Yes for a (seemingly minor) variation of the question, exactly the same except that the light is not infinitely far away, but rather say it is at $L=(0,0,h)$, so that the projection is central projection from $L$ rather than orthogonal projection, and one can translate and rotate $P$ (or equivalently, move $L$ and the projection plane around). Then one could build a $P$ for a given $S$ as follows. Let me use $S=\{5,7,11\}$.

Form $P$ as the convex hull of a regular 5-gon, a regular 7-gon, and a regular 11-gon, stacked in parallel planes, each scaled so that the hull includes all the vertices and edges of each regular $n$-gon:

Then placing $L$ just above the center of the pentagon produces a regular pentagon shadow, moving $L$ straight up enough yields a regular heptagon shadow, and raising $L$ high enough vertically, a regular hendecagon shadow. It seems not difficult to extend the construction to any $S$.

But I cannot see how to accomplish this with the light at $L=(0,0,+\infty)$. My first idea was to arrange the regular $n$-gons roughly as inscribed in longitudinal great circles of a sphere, so that rotating $P$ on a North-South "spit" will project each in turn.
                   
But I think this approach succeeds only with $S$ of small cardinality.

[This question is a small bite out of an earlier MO question, "What is determined by the combinatorics of the shadows of a convex polyhedron?", and it can be viewed as a lower-dimensional version of another MO question, "Are the Platonic solids shadows of 4-polytopes?".]

Either a construction of a $P$ for any $S$, with $L=(0,0,+\infty)$, or an argument why this is possible only for some $S$, would be much appreciated. In fact, a single counterexample $S$ would enlighten. Thanks for ideas!

Addendum. Here is an illustration of the elegant idea of Andreas Blass, for regular 4-, 8-, and 12-gons (blue, green, and red respectively):

His construction raises interesting questions. Can any three regular polygons be achieved as shadows (without the (mod 4) restriction)? Is there another construction that achieves more than three regular shadows, for perhaps similarly restricted $n$-gons?

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Rank unimodality of "generalized crown posets"

We shall define the $k$-crown poset $C_{n,k}$ on $2n$ elements by a set of minimal elements {$v_1,v_2,...,v_n$} and maximal elements {$u_1,u_2,...,u_n$}, where each $u_i$ covers the following set of elements: {$v_i,v_{i+1}\bmod{n},...,v_{i+k}\bmod{n}$}.

$C_{n,1}$ is the well-studied crown poset, as described in this paper by Emanuele Munarini, Norma Zagaglia Salvi, and $C_{n,n-1}$ is a "complete bipartite" poset $P_{n,n}$, which can be defined as a poset where the Hasse diagram is a complete bipartite graph. Rank unimodality was shown for the lattice of order ideals of crown posets in 2, and I recently obtained a short result that shows rank unimodality fails for the lattice of order ideals of complete bipartite posets $P_{n,m}$ with $n,m\ge2$. The goal of my generalization to $k$-crowns was to attack "intermediary" rank-two posets, as the converse of my result does not hold.

First, I'm curious if $k$-crowns have been previously studied under a different name? The generalization feels natural for my purposes here, but I hope to get a sense of how these objects contribute to other ideas as well.

Primarily, I am curious if anyone has previously come across a surprising claim I came up with recently, regarding these $k$-crowns and rank unimodality.

Claim: Let $C_{n,k}$ be a $k$-crown poset. Then the lattice of order ideals $\mathcal{J}(C_{n,k})$ is rank unimodal for $n>k^2$.

I have computational evidence through $k=5$, and am currently writing some code to offer insight for larger cases. Of course, if such a result is already known, I would be thrilled to read about the work. Thank you for your time, and certainly let me know if anything I have written is unclear.

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