Important unpublished works in mathematics

I know Harrington is quite famous for unpublished works in computability theory and a friend of mine specializing in set theory says Woodin is also quite similar which led me to the following question:

What are some of the most important/interesting works that were never published by the authors?

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Inequality on the $n$th composite number

Conjecture. Assume that $A_n$ denotes the $n$th composite number. For all integers $n \geq m \geq 2$, $$A_{\lfloor n/m\rfloor}-\frac{A_n}{m}-\frac{n}{m}\left(\frac{\log m}{\log^2 n}+\frac{\log^2 m+4\log m}{\log^3n}\right)+11\geq 0.$$

This conjecture is motivated by the weaker bound that $A_{\lfloor n/m\rfloor} \geq \frac{A_n}{m}$ for all $n \geq m \geq 1$.

Now to get my stronger conjecture, you use the fact that the number of composites up to $n$ is $n-\pi(n)-1$ and so $n = A_n-\pi(A_n)-1 \implies A_n = n+\pi(A_n)+1$. Now since $$\pi(x) = \dfrac{x}{\log(x)}+\dfrac{x}{\log^2(x)}+O\left(\dfrac{x}{\log^3(x)}\right)$$ and $A_n = n+o(n)$ we try to find an asymptotic expansion of the form $$A_n = n+\dfrac{\alpha n}{\log(n)}+\dfrac{\beta n}{\log^2(n)}+O\left(\dfrac{n}{\log^3(n)}\right)$$ to get asymptotic bound $A_n \approx n+\frac{n}{\log(n)}+\frac{2n}{\log(n)^2}$.

Note that if you use the asymptotic expansion $A_n \approx n+\frac{n}{\log(n)}+\frac{2n}{\log(n)^2}+\frac{3n}{\log(n)^3}$ and substitute it into $A_{n/m}-\frac{A_n}{m}$ it is unbounded from below so it is not an ideal lower bound.

Now we can use this to approximate $A_{n/m}-\frac{A_n}{m}$ and get the bound in my conjecture.

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Has there been progress on Hamiltonicity in 4-connected claw-free graphs with a constant maximum degree?

In 1984, Matthews and Sumner [1] conjectured that every 4-connected claw-free graph is Hamiltonian, and this conjecture is still wide open.

I would like to know if there has been any progress on this conjecture when the maximum degree of these graphs is restricted to a small range. For example, can it be asserted that every 4-connected claw-free graph with the maximum degree of 10 is Hamiltonian?

[1] M. M. Matthews and D. P. Sumner, Hamiltonian results in K1,3 -free graphs,J Graph Theory 8(1) (1984), 139–146.

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What is longest domino-respecting path on a grid?

Assume that someone placed a bunch of dominos on an $n \times n$ grid.
We say that a path is domino-respecting if whenever it contains one half of a domino, that half is connected to the other half of the domino in the path.
(This is kind of like a pairing strategy in combinatorial game theory.)

What is the longest domino-respecting path we can surely find?

Is it $O(n)$ or $\Omega(n^2)$ or something else?

My motivation is to solve this problem.

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On construction of certain class of analytic functions with given properties

(I don't know how naïve this question is / could be.)

Is there a class of function with following properties:

1. $f: \mathbb{R} \to \mathbb{R}$ is analytic.

2. $f$ is such that each positive integer input produces reciprocal of positive integer once, not necessarily the same integer,

This is equivalent to fact that

$f: \mathbb{N} \to \mathbb{N}^{-1}$ is bijective not necessarily in particular order (as "pseudo random"as possible).

3. Is closed form possible for such function?

I'm not very much concerned about the 3rd question.

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When Instant-on weights for A2 sub-algebra embedded in E8 via E8 ⊃ E6 × A2 — what selects the vacuum angle?

Consider the maximal subgroup decomposition of $E_8$:

$$\mathbf{248} = (\mathbf{78},\mathbf{1}) + (\mathbf{1},\mathbf{8}) + (\mathbf{27},\mathbf{3}) + (\overline{\mathbf{27}},\bar{\mathbf{3}})$$

under $E_6 \times A_2$, where $A_2 = SU(3)$.

The $A_2$ instanton potential.

For an $A_2$ gauge theory, the instanton-induced effective potential for the vacuum angle $\varphi$ takes the form:

$$V(\varphi) = -K_1\cos\varphi - K_2\cos\!\left(\varphi - \frac{2\pi}{3}\right) - K_3\cos\!\left(\varphi - \frac{4\pi}{3}\right)$$

Under exact $Z_3$ symmetry ($K_1 = K_2 = K_3$), this potential is identically zero — the vacuum angle $\varphi$ is completely undetermined. Any $Z_3$ breaking produces a non-trivial minimum.

The naive calculation and why it fails.

A natural attempt is to weight each $A_2$ representation $R$ in the $E_8$ decomposition by its Dynkin index $T(R)$ and its $Z_3$ triality phase $e^{2\pi i q(R)/3}$:

$$S = \sum_R n_R \cdot T(R) \cdot e^{2\pi i\, q(R)/3}$$

With the decomposition above:

• $(78,\mathbf{1})$ singlets: $n_R = 78$, $T = 0$, $q = 0$ — contribution $= 0$
• $(\mathbf{1},\mathbf{8})$ adjoint: $n_R = 1$, $T = 3$, $q = 0$ — contribution $= +3$
• $(\mathbf{27},\mathbf{3})$ fundamentals: $n_R = 27$, $T = 1/2$, $q = 1$ — contribution $= 13.5\,e^{2\pi i/3}$
• $(\overline{\mathbf{27}},\bar{\mathbf{3}})$ anti-fundamentals: $n_R = 27$, $T = 1/2$, $q = 2$ — contribution $= 13.5\,e^{4\pi i/3}$

The total:

$$S = 3 + 13.5\left(e^{2\pi i/3} + e^{4\pi i/3}\right) = 3 - 13.5 = -10.5$$

giving $\arg(S) = 180°$. The potential minimum sits at $\varphi = 180°$.

The observed value and the question.

The Koide lepton mass formula $$Q = \frac{m_e + m_\mu + m_\tau}{(\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau})^2} = \frac{2}{3}$$ holds experimentally to $6 \times 10^{-6}$ precision. It follows exactly from a $Z_3$ cosine ansatz for the square-root masses: $$\sqrt{m_k} = M\left(1 + \sqrt{2}\cos\!\left(\varphi + \frac{2\pi k}{3}\right)\right), \quad k = 0,1,2$$ with phase $\varphi = 132.7328°$. This splits as: $$\varphi = \underbrace{120°}_{A_2\ \text{Weyl chamber wall}} + \underbrace{12.73°}_{\delta}$$ where $\delta \approx \theta_C = 13.00°$ (Cabibbo angle) to 2.1%.

Three questions:

(1) The naive Dynkin index sum gives $180°$, not $132.73°$. What is the correct physical weighting of $A_2$ representations in an $E_8 \supset E_6 \times A_2$ instanton calculation? Should the $(\mathbf{27},\mathbf{3})$ matter sector and the $(\mathbf{1},\mathbf{8})$ generation sector receive different weights, and if so, what determines the ratio?

(2) Is there a known result for the $A_2$ vacuum angle in the context of $E_8$ gauge theory broken to $E_6 \times A_2$? Any reference to the instanton-induced superpotential in this breaking would be very helpful.

(3) More generally: for a gauge group $G$ broken to $H \times A_2$, what determines the $A_2$ vacuum angle from the branching rules of the $G$ adjoint? Is there a standard formula in the literature?

References checked:

• Slansky (1981), Physics Reports 79 — branching rules
• Witten, arXiv:2601.01587 — axion wormhole and vacuum angle formalism

Any pointers to the relevant literature on vacuum angle selection in partially broken exceptional gauge theories would be greatly appreciated.

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A series related to $\log192-\sqrt{3}\pi$

Let $$H_n:=\sum_{0<k\le n}\frac1k\ \ \ \ (n=0,1,2,\ldots).$$ Inspired by Question 507603 and Question 507592, I discovered the following identity $$\sum_{k=0}^\infty\frac{\binom{2k}k^2\binom{3k}k}{(-192)^k}\left(3H_{3k}+2H_{2k}-5H_{k}\right)=\frac{\Gamma(\frac{1}{3})^6}{4\pi^4}(\log192-\sqrt{3}\pi),\tag{1}$$ i.e.,

$$\sum_{k=0}^\infty\frac{\binom{2k}k^2\binom{3k}k}{(-192)^k}\left(3H_{3k}+2H_{2k}-5H_{k}\right)=(\log192-\sqrt{3}\pi)\sum_{k=0}^\infty\frac{\binom{2k}k^2\binom{3k}k}{(-192)^k}.\tag{2}$$

My questions are:

1. Is the series (1) known?

2. How can it be proven?

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Project Hail Mary, question? (Spoiler)

I just watched the movie Project Hail Mary yesterday, and I have a question about the ending of the movie.

(Spoilers)

At the end, they show the alien nicknamed “Rocky” in a protective clear polyhedral suit that is pretty close-fitting and seemingly quite flexible, allowing him to walk around freely. This is maybe not the most unbelievable aspect of the movie, but it made me wonder if such a suit was mathematically possible, seemingly as it was portrayed?

Flexible non-convex polyhedra were shown to exist by Robert Connelly with one degree of freedom, although very little range of motion.

Let $R$ be a closed piecewise-smooth embedded surface in Euclidean 3-space homeomorphic to a 2-sphere, and assume we are given $n\in \mathbb{N}, \epsilon >0$. Define a Rockyhedron to be an embedded polyhedral surface $S$, also homeomorphic to $S^2$, which is $\epsilon$ close to $R$ in the Hausdorff topology, and such that the interior of $S$ contains $R$, and which has $n$ degrees of freedom.

Do Rockyhedra exist, question?

Addendum: My daughter reminded me that Rocky makes a polyhedral spacesuit for Ryland Grace at one point earlier in the movie so he can visit Rocky’s ship, so presumably that was another Rockyhedron. It’s possible that the CGI effects people modeled the front of the suits without having to worry about the deformation of the rear out of view, but the impression is given of a flexible polyhedron suit.

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Is the limit of this hybrid sum-integral sequence always transcendental?

Let $f: \mathbb{R}^+ \to \mathbb{R}$ be a $C^\infty$ function where all moments $M_n = \int_0^\infty x^n f(x) dx$ are rational.

Consider the limit: $$L = \lim_{N \to \infty} \left( \sum_{k=1}^N \frac{f(k)}{k} - \int_1^N \frac{f(x)}{x} dx + \sum_{k=1}^N \frac{(-1)^k f(k)}{k^2} \right)$$

Is this limit $L$ necessarily transcendental? Does there exist a non-vanishing $f$ where $L$ is algebraic (e.g., $L = \sqrt{2}$)?

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Convergence of Kaplan-Meier estimator for pooled sample

I am new to survival analysis. Recently I have been thinking about the Kaplan-Meier estimator for pooled sample. Suppose we have two group of samples, group 1 has $n_1$ samples from the survival distribution $S_1(t)$, while group has $n_2$ samples from the survival distribution $S_2(t)$. Let $n=n_1+n_2$. We also have the limit like $$ \lim_{n\to\infty} \frac{n_1}{n}=a_1,\quad \lim_{n\to\infty} \frac{n_2}{n}=a_2. $$ Let $\hat{S}(t),\hat{S}_1(t),\hat{S}_2(t)$ respectively denote the Kaplan-meier estimators for the pooled samples, group 1 samples and group 2 samples with the standard definition like $$ \hat{S}(t)=\prod_{s\leq t}\left(1-\frac{d(s)}{Y(s)}\right), $$ where $d(s)$ is the number of events at time $s$ and $Y(s)$ is number of samples at risk at time $s$. From Theorem 3.4.2 of 'Counting processes and survival analysis' by T.R.Fleming and D.P.Harrington, we know that even if the samples are censored, under simple condition we still have the consistency of the single group Kaplan-meier estimator $\hat{S}_i(t)$ like $$ \sup_{0\leq s\leq t}|\hat{S}_i(s)-S_i(s)|\to 0 \quad \text{in probability as} \;n_i\to\infty. $$ My question is do we have the similar results under the same condition for the pooled sample Kaplan-meier estimator $\hat{S}(t)$ like $$ \sup_{0\leq s\leq t} |\hat{S}(t)-S(t)|\to 0 \quad \text{in probability as} \;n\to\infty, $$ in which $S(t)=a_1S_1(t)+a_2S_2(t)$? If so, how to prove it then? Or could we prove something like $$ \sup_{0\leq s\leq t} \left|\hat{S}(t)-\left(\frac{n_1}{n}\hat{S}_1(t)+\frac{n_2}{n}\hat{S}_2(t)\right)\right|\to 0 \quad \text{in probability as} \;n\to\infty? $$

Thank you for your attention and help.

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Discriminant of a cubic and CM elliptic curves/Heegner points

A cubic polynomial

$f = ax^3 + bx^2 + cx + d \in \mathbb{Z}[x]$

has discriminant

$\Delta = 18abcd−4b^3d+b^2c^2−4ac^3−27a^2d^2$.

Suppose we are trying to find cubic polynomials with a given discriminant $N \in \mathbb{Z}$. Using the idea that finding Heegner points on an elliptic curve with complex multiplication and rank 1 is relatively efficient, how likely are we to get a rank 1 elliptic curve with CM by choosing say $a,b \in \mathbb{Z}$ at random, then homogenizing the equation

$18abxy−4b^3y+b^2x^2−4ax^3−27a^2y^2 = N$

and transforming the resulting general cubic

$18abXYZ−4b^3YZ^2+b^2X^2Z−4aX^3−27a^2Y^2Z = NZ^3$

into Weierstrass form (if possible)?

Since CM elliptic curves are relatively rare, we would be restricted in the choice of $a,b$ but maybe there is a way to choose them wisely?

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Derivative bounds for self convolution of the spherical measure in $R^d$

While reading this article on near $L^1$ estimates for the spherical lacunary maximal function, I came across the estimate $$ |\partial^{\gamma} (\widetilde{\sigma} \ast \sigma)(x)| \lesssim |x|^{-(1 + |\gamma|)} \chi_{|x| \leq 2}(x) \qquad \text{for $\gamma \in \mathbb{N}_{0}^{d}$ and $x \in \mathbb{R}^d$}.$$ Here, $\sigma$ is a measure on the unit sphere $\mathbb{S}^{d-1}$, given by $$\langle{\sigma, \phi}\rangle := \int \phi(y) \chi(y) d \theta(y),$$ with $\chi$ being a smooth function with a compact support contained inside a ball of radius $\leq 1/2$. Moreover, its refection $\widetilde{\sigma}$ is defined by the action $$\langle{\widetilde{\sigma}, \phi}\rangle := \langle{\sigma, \tilde{\phi}}\rangle, \qquad \text{where} \qquad \tilde{\phi}(x) = \phi(-x).$$ The authors mention that the formula is a standard one, but I have been unable to find a reference or a proof. Any help would be appreciated. Thanks!

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Local class field theory and inflation--restriction for CM elliptic curves at a totally ramified prime

Let $K$ be an imaginary quadratic field, and let $E/\mathbb{Q}$ be an elliptic curve with complex multiplication by $\mathcal{O}_K$.

Fix a rational prime $p$, and let $\mathfrak{p} \mid p$ be a prime of $K$. Assume that $E$ has good reduction at $p$.

Set $$ F := K(E[\mathfrak{p}]), \qquad \Delta := \mathrm{Gal}(F/K). $$

Let $\mathfrak{P}$ be the unique prime of $F$ above $\mathfrak{p}$, and assume that $\mathfrak{p}$ is totally ramified in $F/K$. Let $ \Phi := F_{\mathfrak{P}}. $

Then we may identify $$ \Delta \cong \mathrm{Gal}(\Phi/K_{\mathfrak{p}}). $$

Let $$ U = \mathcal{O}_{\Phi}^{\times}. $$

Question 1

By local class field theory, we have the reciprocity isomorphism $$ \mathrm{rec}_\Phi : \Phi^\times \xrightarrow{\sim} G_\Phi^{\mathrm{ab}}. $$

Since $E[\mathfrak{p}]$ is an abelian Galois module, any continuous homomorphism $$ G_\Phi \to E[\mathfrak{p}] $$ factors through $G_\Phi^{\mathrm{ab}}$, hence $$ \mathrm{Hom}(G_\Phi, E[\mathfrak{p}]) \cong \mathrm{Hom}(\Phi^\times, E[\mathfrak{p}]). $$

Now using the decomposition $$ \Phi^\times \cong \pi^{\mathbb{Z}} \times U, $$ one expects that the contribution from $\pi^{\mathbb{Z}}$ is trivial (or controlled via Frobenius), so that the main contribution comes from $U$.

Thus one should obtain $$ \mathrm{Hom}(G_\Phi, E[\mathfrak{p}]) \cong \mathrm{Hom}(U, E[\mathfrak{p}]). $$

Taking $\Delta$-invariants gives the desired isomorphism.

Question: What is the cleanest way to rigorously justify the passage from $\Phi^\times$ to $U$, and the compatibility with the $\Delta$-action?

Question 2

I would like to justify the isomorphism $$ H^1(G_{\Phi}, E(\overline{\Phi}))[\mathfrak{p}]^{\Delta} \;\cong\; H^1(G_{K_{\mathfrak{p}}}, E(\overline{K_{\mathfrak{p}}}))[\mathfrak{p}], $$ using inflation--restriction.

We have the exact sequence $$ 1 \to G_{\Phi} \to G_{K_{\mathfrak{p}}} \to \Delta \to 1, $$ which gives rise to the inflation--restriction exact sequence $$ 0 \to H^1(\Delta, E(\overline{\Phi})) \to H^1(G_{K_{\mathfrak{p}}}, E(\overline{K_{\mathfrak{p}}})) \to H^1(G_{\Phi}, E(\overline{\Phi}))^{\Delta}. $$

How should one pass to the $\mathfrak{p}$-primary parts in this sequence to obtain the desired isomorphism?

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Why does $M(|\nabla u_m|^2)\nabla u_m \to M(|\nabla u|^2)\nabla u$ a.e.?

In this paper, the author proves $M(|\nabla u_m|^2)\nabla u_m \to M(|\nabla u|^2)\nabla u$ a.e. in $Q_T$. Could someone explain the exact idea behind this step?

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Free coproduct cocompletions respecting countable coproducts

Where can I read about universal properties of free coproduct cocompletions of (additive) categories and on their modifications?

For a subcategory $A$ that is closed with respect to countable coproducts inside an additive category $A'$ that is closed with respect to small coproducts I would like to have criterion to check that $A\to A'$ is a universal functor that respects countable coproducts whose target category is closed with respect to all small coproducts. I suspect that it is as follows: any object of $A'$ is a coproduct of objects of $A'$ and any object of $A$ is countably small in $A'$. Any related references would be very welcome!

I dont' think that this functor can be described as a cocompletion with respect to a given class of diagrams (cf. Cocompletion with respect to a given class of diagrams); can it?

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Proof attempt: the Collatz map has no non-trivial cycles [closed]

We prove that the Collatz map $T(n) = (3n+1)/2^{v_2(3n+1)}$ on odd positive integers admits no non-trivial cycles. We ask whether the argument is correct.

Notation. Write $P(n) = v_2(3n+1)$. A non-trivial cycle is an odd integer $e > 1$ satisfying $T^C(e) = e$ for some $C \geq 1$.

Lemma 1 (Arithmetic of steps). For odd $n$:

$$n \equiv 1 \pmod{4} \implies P(n) \geq 2 \implies T(n) < n$$

$$n \equiv 3 \pmod{4} \implies P(n) = 1 \implies T(n) > n$$

Proof. If $n = 4k+1$ then $3n+1 = 12k+4 = 4(3k+1)$, so $v_2(3n+1) \geq 2$ and $T(n) \leq (3n+1)/4 < n$. If $n = 4k+3$ then $3n+1 = 12k+10 = 2(6k+5)$ with $6k+5$ odd, so $P(n) = 1$ and $T(n) = (3n+1)/2 > n$. $\square$

Lemma 2 (Block structure). Any cycle decomposes into consecutive blocks, where each block consists of one or more steps with $P \geq 2$ followed by $j \geq 1$ steps with $P = 1$. The last $P \geq 2$ step in each block lands on $3 \pmod{4}$, and all preceding $P \geq 2$ steps in the block land on $1 \pmod{4}$.

Proof. A cycle of only $P \geq 2$ steps is impossible: each such step strictly decreases the value (Lemma 1), giving $\ell(0) > \ell(1) > \cdots > \ell(C) = \ell(0)$, a contradiction. Therefore every cycle contains at least one $P = 1$ step. By Lemma 1, a $P = 1$ step occurs exactly when $n \equiv 3 \pmod 4$, which can only be reached via a $P \geq 2$ step landing on $3 \pmod{4}$. Grouping each such landing step together with all preceding consecutive $P \geq 2$ steps and all following $P = 1$ steps gives the block decomposition. $\square$

Theorem. The Collatz map has no non-trivial cycles.

Proof. By Lemma 2, any cycle consists of $m$ blocks. Each block is a composition of affine maps of the form $\ell \mapsto (3\ell+1)/2^P$, which is affine in $\ell$. The entire block therefore acts as $g_i(\ell) = \alpha_i \ell + \beta_i$ with $\beta_i > 0$, where $\alpha_i = 3^{s_i}/2^{N_i}$ with $s_i$ the number of steps in block $i$ and $N_i$ the total number of divisions by 2 in block $i$.

The composition $g = g_m \circ \cdots \circ g_1$ is affine with slope

$$A = \prod_{i=1}^{m} \alpha_i = \frac{3^K}{2^N}, \qquad K = \sum_{i=1}^m s_i = C, \quad N = \sum_{i=1}^m N_i.$$

Since $3^K$ is odd and $2^N$ is even, $A \neq 1$, so $g$ has a unique fixed point satisfying

$$\ell \cdot (2^N - 3^K) = B \cdot 2^N$$

for some $B > 0$. Therefore $2^N \mid \ell \cdot 3^K$. Since $\gcd(2^N, 3^K) = 1$, we get $2^N \mid \ell$. But $\ell$ is odd, so $2 \nmid \ell$. Contradiction. $\square$

Question. Is this argument correct? We are particularly uncertain about two points: (1) Is Lemma 2 complete — are there cyclic patterns not captured by the block decomposition? (2) Is the fixed point equation correctly derived from the composition of affine maps?

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Geometric intuition of Gorenstein rings

As part of a course in commutative algebra or algebraic geometry, one will generally learn that, for a Noetherian local ring:

regular $\Rightarrow$ complete intersection $\Rightarrow$ Gorenstein $\Rightarrow$ Cohen-Macaulay

with none of these arrows being an equivalence.

The very basic geometric intuition one is typically given is that each of these properties corresponds to the exclusion of certain bad forms of singularities.

Now “regular” is clear enough intuitively (there is a tangent space, there is no singularity). “Complete intersection” I think I have something of a feel for (though it might certainly be improved) by counting dimensions and equations. For “Cohen-Macaulay”, this question and that one have some very useful answers.

But I am left utterly without intuition as to what “Gorenstein” means. (I have seen counterexamples to the converse of the two implications around it but, failing to visualize what they look like, I don't find them so enlightening.) I realized while reading this recent answer that I didn't know if $k[[t^2,t^3]]$ was Gorenstein (it is, whereas my immediate intuition would have been no). So:

Question: What are some intuitions that are helpful to keep in mind as to the geometrical meaning of the “Gorenstein” property for local rings? What are some examples or theorems that can be useful to develop an intuition? How might one hand-wave an explanation or draw a picture of what a Gorenstein ring “looks like”?

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Conservativity of stalks of ind-constructible sheaves

I have a simple question about the conservativity of stalks of ind-constructible sheaves. Let $X$ be a topologically noetherian scheme, $S$ a set of geometric points of $X$ corresponding bijectively to the scheme-theoretic points of $X$, $\Lambda$ an algebraic extension of $\mathbf{Q}_\ell$ or the ring of integers of such, $I$ a filtered category (which I will assume is a directed poset), $\operatorname{Cons}_X(\Lambda)$ the abelian category of constructible $\Lambda$-sheaves on $X_{\text{pro-étale}}$ in the sense of Bhatt-Scholze §6.8, and let $(\mathcal F_i)_{i\in I}$ correspond to a functor $I\to \operatorname{Cons}_X(\Lambda)$ with colimit $\mathcal F$ computed in $\operatorname{Mod}(X_{\text{pro-étale}},\Lambda)$, the category of sheaves of $\Lambda$-modules, where $\Lambda$ here and often below also denotes the sheaf $\Lambda_X$ associated to the topological ring $\Lambda$.

My question is, if $\mathcal F_x=0$ for every $x\in S$, is $\mathcal F=0$? (Here $\mathcal F_x:=\Gamma(x_{\text{pro-étale}},x^*\mathcal F$.)

I believe I can prove this is true (argument below), but what is giving me some doubts is Warning 2.3.4.10 of the paper Weil’s conjecture for function fields [PDF], which says that there is a non-zero ind-constructible sheaf which vanishes at all closed points. This is relevant since Corollary 3.51 of the recent paper Constructible sheaves on schemes and a categorical Künneth formula connects $\operatorname{Ind}(D_{\mathrm{cons}}(X,\Lambda))$ to filtered colimits of pro-étale sheaves. Admittedly, this warning only deals with closed points, but so far I haven’t been able to construct a non-zero ind-constructible sheaf that vanishes at all closed points. Here is my purported (simple) ‘proof’ that the answer to my question is ‘yes’:

Proof It would suffice to show that for every $i\in I$, there is a $j\geq i$ so that $\varphi_{ij}:\mathcal F_i\to\mathcal F_j$ is zero. Suppose we could find a nonempty open $U\subset X$ over which this is true. Then we could replace $i$ by $j$ and $X$ by $\operatorname{supp}\operatorname{coker}(\ker\varphi_{ij}\to\mathcal F_i)$, and we'd be done by noetherian induction.

To find $U$: let $\eta$ be a geometric generic point of $X$. We can find a $j\geq i$ so that $\varphi_{ij}:\mathcal F_i\to\mathcal F_j$ induces zero on stalks at $\eta.^\dagger$ There is some connected neighborhood $U$ of $\eta$ so that both $\mathcal F_i$ and $\mathcal F_j$ are locally constant of finite presentation over $U$. Then the same is true of $\ker\varphi_{ij}$, and as $(\ker\varphi_{ij})_\eta=(\mathcal F_i)_\eta$, $\ker\varphi_{ij}|_U=\mathcal F_i|_U.^\ddagger$ $\square$

$\dagger$: Both $\eta^*$ and $\Gamma(\eta_\text{pro-étale},-)$ commute with colimits, the latter since $\eta$ is connected and w-contractible. For each $i$, there is some neighborhood $U_i$ of $\eta$ restricted to which $\mathcal F_i$ is (pro-étale) locally of the form $M_i\otimes_\Lambda\Lambda_X$, where $M$ is a $\Lambda$-module of finite presentation and I've dropped the underlines for constant sheaves ($\Lambda_X$ is not literally a constant sheaf). (This is what I mean by ‘locally constant of finite presentation.’) Note $(M_i\otimes_\Lambda\Lambda_X)_\eta=M_i$ as $\Lambda_X(\eta)=\Lambda$ and $\eta$ is w-contractible.

$\ddagger$: When $\Lambda$ is a field, this follows from Corollary 6.8.5 of [BS] and that $(-)_x$ is an exact functor (of abelian categories). Otherwise, modulo possibly shrinking $U$, replace Corollary 6.8.5 with Proposition 6.8.11.

It remains unclear to me how this doesn’t contradict Warning 2.3.4.10. Thanks for any clarification of the matter.

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Injectivity of the map from prefixes of Coxeter elements to reflection subgroups

Let $(W,S)$ be a Coxeter group and denote $T$ its set of reflections. Let $c$ a Coxeter element of $W$. The poset $$[1,c]_T = \{x\in W \ \vert \ 1 \leq_T x \leq_T c\}$$ of prefixes of $c$ for the absolute order $\leq_T$ on $W$ appears in several contexts; it is a lattice when $W$ is finite, isomorphic to the noncrossing partition lattice when $W$ is the symmetric group. It is an important tool also in the study of (dual) Artin groups; it allows to build (quasi-)Garside structures on such groups in some cases, and was an important ingredient in Paolini and Salvetti's recent solution to the $K(\pi, 1)$-conjecture for Artin groups of affine type.

The posets $[1,c]_T$ for general Coxeter groups remain quite mysterious. The lattice property fails in general but there seems to be no reasonable conjecture on for which $W$ and $c$ it should be a lattice.

A property that appears to be useful when $W$ is finite is the following: let $x\in [1,c]_T$ and let $P(x) = \langle t\in T \ \vert \ t \leq_T x \rangle$. Then $P(x)$ is a parabolic subgroup of $W$, of rank $\ell_T(x)$, in which $x$ is a Coxeter element. Moreover, the map from $[1,c]_T$ to reflection subgroups of $W$ (partially ordered by inclusion) sending $x$ to $P(x)$ is injective (and even an isomorphism of posets on its image). This is a reformulation of a result of Brady and Watt.

For arbitrary $W$, in general $P(x)$ fails to be parabolic (but is probably "parabolic in some weak sense"), but $x$ is still conjectured to be a Coxeter element in $P(x)$ (known for affine, rank three, universal Coxeter systems).

Question : is the map from $[1,c]_T$ to reflection subgroups of $W$, sending $x$ to $P(x)$, still injective in general ?

It is not even clear that $P(x)$ has rank $\ell_T(x)$ (it would hold if the Hurwitz action on $T$-reduced decompositions of prefixes of Coxeter elements was transitive, which is still conjectural).

(Here as well, for some families of infinite Coxeter groups this is folkloric or easy to handle, but I am wondering about the general case.)

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Principal ideals under integral closure

Let $(R, \mathfrak{m}, \kappa)$ be a one dimensional Noetherian Gorenstein local domain such that $\kappa$ is algebraically closed and let $a,b \in R$. Let $\overline{R}$ be the integral closure of $R$, and assume that $a$ and $b$ generate the same principal ideal over $\overline{R}$ (that is $a\overline{R} = b\overline{R}$ ). Is it true that $a$ and $b$ generate the same principal ideals over $R$ (i.e. $aR=bR$)?

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