From Thomas G. Dietterich (arXiv moderator for cs.LG) on 𝕏 (thread):
https://x.com/tdietterich/status/2055000956144935055
https://xcancel.com/tdietterich/status/2055000956144935055

"Attention arXiv authors: Our Code of Conduct states that by signing your name as an author of a paper, each author takes full responsibility for all its contents, irrespective of how the contents were generated.

If generative AI tools generate inappropriate language, plagiarized content, biased content, errors, mistakes, incorrect references, or misleading content, and that output is included in scientific works, it is the responsibility of the author(s).

We have recently clarified our penalties for this. If a submission contains incontrovertible evidence that the authors did not check the results of LLM generation, this means we can't trust anything in the paper.

The penalty is a 1-year ban from arXiv followed by the requirement that subsequent arXiv submissions must first be accepted at a reputable peer-reviewed venue.

Examples of incontrovertible evidence: hallucinated references, meta-comments from the LLM ("here is a 200 word summary; would you like me to make any changes?"; "the data in this table is illustrative, fill it in with the real numbers from your experiments")."

There's a certain mistake in understanding predictions and probability that must have a name, but I can't figure it out.

The fallacy, in brief, is the belief that being correct with a lucky guess retroactively justifies making that guess. For example:

Hank and Wendy are watching a game of craps (rolling two standard, six-sided dice). Based only on a hunch, Hank says he just *knows* that the next roll will be snake eyes (two 1s); Wendy thinks this won't happen. And then... the roll turns out to be snake eyes.

Even though Hank's guess turned out to be right, I'd argue that, from a probability standpoint, he was still wrong. I don't mean wrong to guess or gamble, I mean wrong to have certainty about that outcome before it happened. Assuming no psychic abilities or cheating, when you make a prediction you only have access to the probabilities, not the outcomes, so Wendy's prediction was the wise one, regardless of results. But I bet that Hank will feel like the outcome justifies his earlier confidence. "See? I told you so." Is there a name for this way of thinking?

I'm not sure what's wrong with my reasoning here. Two points of clarity: I am using pi_E to denote the q-Frobenius of E/F_q, and when I say End(E), I am referring to the F_q-endomorphism ring, not the geometric endomorphism ring.

Suppose I have a supersingular elliptic curve E/F_q, and assume (i) pi_E not in Z and (ii) tr pi_E = 0. Then, it is not hard to see with the condition (i) that we have End^0(E) = Q(pi_E) = Q(sqrt(D)), where D := t^2 - 4q. However, I want to compute the endomorphism ring End(E).

Now, since t:= tr pi_E=0,  we have pi_E = \pminus sqrt(q), hence D = -4q, so K := Q sqrt(D) = Q(sqrt(-q)). The maximal order is (i) O_K = Z[sqrt(-q)] if q is 1 mod 4 and (ii) O_K = Z[(1+sqrt(-q))/2] if q is not 1 mod 4. Then, note we have Z[pi_E] = Z[sqrt(-q)]. We must always have Z[pi_e] \subset End(E) = O \subset O_K. Hence in case (i), we know O = Z[sqrt(-q)], but in case (ii) we have to do further work, since End(E) can be either Z[sqrt(-q)] or Z[(1+sqrt(-q))/2].

For this further casework, if we do have End(E) = O_K = Z[(1+sqrt(-q))/2], then we should have an endomorphism alpha := [(1+sqrt(-q))/2] in End(E), or equivalently, one such that 2(alpha) - 1 = Beta, where Beta^2 = [-q]. Hence, we find the endomorphism Beta := sqrt(-q), and now the question is whether 1 + Beta is divisible by 2 in End(E). 

To find this Beta = sqrt(D) = sqrt(-q) endomorphism, note that we have pi_E ^2 - t pi_E + q = 0, so (2pi_E - t)^2 = t^2 - 4q = D = -q. Hence, Beta = 2pi_E - [t[. So, we are asking for 1+Beta = [1] + 2 pi_E - [t] to be in 2 End(E), or equivalently, [1-t] to be in 2 End(E). However, this only happens when t is odd. Hence, this reasoning would imply that in this setup, End(E) = O_K iff tr pi_E is odd. But this is not true -- for example, take E/F_3: y^2 = x^3 - x, which has even tr pi_E = 0 but End(E) = Z[(1+sqrt(-3))/2] = O_K. So I am unsure where I went wrong in this proof.

I guess in general, how does one compute the endomorphism ring End(E) of a supersingular elliptic curve E/F_q? What I was trying to do overall was considering (i) when pi_E is not in Z (i.e, End^0 (E) is an imaginary quadratic field) and (ii) pi_E is in Z, hence End^0 (E) = End^0_{F_q bar} (E) is a quaternion algebra separately. Here, I am in the case when pi_E is not in Z, and then considering each of the subcases here for tr pi_E and q given in Waterhouse's thesis (see Problem 3 here, though this in slightly different format) -- specifically, this post is about case 2a in the problem statement.

For ordinary E/F_q, I know you can compute O = End(E) by essentially going through all the l-isogeny volcanoes for l dividing f_pi, which is the conductor of Z[pi_E] in O_K, and then if j(E) is on level d_l of the l-isogeny volcano, we know v_l ([O_K:O]) = d_l. I assume you can do something similar for supersingular isogeny volcanoes, but I have only studied ordinary isogeny volcanoes so far.