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October 25

Inverse of a group subset product

Given a group $(G,*)$ and $A,B\subseteq G$, let us define WLOG:

${\begin{aligned}&A^{-1}=\{a^{-1}:a\in A\}\\[3pt]&A\circledast B=\{a*b:a\in A,b\in B\}\end{aligned}}$

Can we prove that $(A\circledast B)^{-1}=B^{-1}\circledast A^{-1}$? יהודה שמחה ולדמן (talk) 17:07, 25 October 2025 (UTC)

First an easy part, an auxiliary result. We establish that $x\in A^{{-}1}\Leftrightarrow x^{{-}1}\in A,$ as follows:

$x\in A^{{-}1}\Leftrightarrow$ $(\exists \,a\in A:x=a^{{-}1})\Leftrightarrow$ $(\exists \,a\in A:x^{{-}1}=a)\Leftrightarrow$ $x^{{-}1}\in A.$

Using this we proceed,

$x\in (A\circledast B)^{{-}1}\Leftrightarrow$ $x^{{-}1}\in A\circledast B\Leftrightarrow$ $(\exists \,a\in A,b\in B:x^{{-}1}=a*b)\Leftrightarrow$$(\exists \,b\in B,a\in A:x=b^{{-}1}*a^{{-}1})\Leftrightarrow$ $x\in B^{{-}1}\circledast A^{{-}1}.$

 ​‑‑Lambiam 18:16, 25 October 2025 (UTC)

November 1

References request: coherence theorem for tetracategories

Crans (2000) wrote, "There is no coherence theorem for tetracategories yet" and the reference list of that paper included "T. Trimble, The definition of tetracategory, handwritten diagrams." Do the handwritten diagrams refer to Trimble's Notes on Tetracategories, published in 2006? Is it correct to think that the statement "There is no coherence theorem for tetracategories yet" refers to the coherence condition explained in Trimble (2006)? I am wondering whether to add that there is currently no coherence theorem for tetracategories to Coherency (homotopy theory). Also, are there any newer references to the coherence theorem for tetracategories?

• Crans, S. (2000). "On braiding, syllapses and symmetries" (PDF). Cahiers de Topologie et Géométrie Différentielle Catégoriques. 41 (1): 2–74.
• Trimble, Todd (2006). "Notes on tetracategories".

--SilverMatsu (talk) 02:11, 1 November 2025 (UTC)

Apparently, Trimble created the handwritten sheets containing the diagrams in 1995, copies of which were in the hands of several mathematicians. They were digitized and made available as pdf files through Baez's website in 2006. Crans's earlier mention refers to these handwritten sheets, one of which, numbered ①, bears the heading "Tetracategories" and starts with the definition, "A tetracategory is given by the following data: ...", followed by six items subject to four conditions expressed in diagrams on further sheets. Taken together, these four sets of in total 51 handwritten sheets constitute the definition.

There have been several papers since then referring to objects by the name of "tetracategory".^[1] Unless they refer to Trimble's definition, it may be non-trivial to verify whether they are talking about the same species.

A coherence condition is something else than a coherence theorem. The former is a requirement that forms part of a definition (as seen, for example, in the formal definition of a monoidal category). The latter is a result of the form that certain coherencies imply further coherencies.  ​‑‑Lambiam 09:09, 1 November 2025 (UTC)

Thank you for your reply. It seems that the handwritten diagrams refer to Trimble (2006). Thank you for the link, I'll check out the paper you linked to. The question was phrased incorrectly, I should have asked: when Crans's said "There is no coherence theorem for tetracategories yet" did he mean that there is no way to prove Trimble's coherence condition?--SilverMatsu (talk) 14:21, 1 November 2025 (UTC)

The use of the adverb "yet" implies that Crans does not deem it impossible that someone will formulate and prove such a theorem in the future, comparable to the existing coherence theorem for tricategories.^[2] He cannot have meant that there is no way to prove Trimble's four coherence conditions, because it is a trivial (but boring) execise to construct objects that have six components as required by the definition, which, however, fail each of the conditions. If a definition contains a condition that can be proved for the general case, the condition is unnecessary baggage.  ​‑‑Lambiam 15:42, 1 November 2025 (UTC)

I see. Thank you so much for explaining everything so thoroughly.--SilverMatsu (talk) 16:02, 1 November 2025 (UTC)

I found one reference from the link you provided. The reference appears to have Trimble's definition typed in LaTeX. I previously merged tetracategory into higher category theory, but since I can find references, it might be a good idea to split the article.

• Hoffnung, Alexander E. (2011). "Spans in 2-Categories: A monoidal tricategory". arXiv:1112.0560 [math.CT].

--SilverMatsu (talk) 00:17, 2 November 2025 (UTC)

The reference is self-published, which may raise some doubts about verifiability. Yet the author was a student of Baez^[3][4] and has published a peer-reviewed article in this area of research ([5]). His contribution is described here as follows:

Despite the complexity of tri-categories, Todd Trimble managed to go beyond and write down the definition of a tetra-category, i.e. a weak 4-category, the result has later been polished [Hof13] by Alexander E. Hoffnung.

([Hof 13] is the arXiv paper). So I guess we may consider this a reliable source.  ​‑‑Lambiam 01:03, 2 November 2025 (UTC)

Thank you for checking if it is a reliable source. We may be able to add the reference to §.External link. I think the previous merge was to merge (strict) 4-category and tetracategory. For example, strict 2-category and weak 2-category are merged into 2-category. There is no coherence theorem for tetracategories yet, so I think (strict) 4-category and tetracategory cannot be merged. I will use Crans (2000) as a reference. --SilverMatsu (talk) 01:41, 2 November 2025 (UTC)

@Lambiam: By the way, Is there a pasting theorem for the Tetracategory? The phrase "pasting diagrams" appears in Notes on Tetracategories (§.History). The pasting theorem for strict n-category version was proved by Power (1991) and Johnson (1989). Regarding the 4-categories, there seems to be a CTRC seminar by Crans, but I could only find the abstract.--SilverMatsu (talk) 07:38, 6 November 2025 (UTC)

I do not know the answer. For 4-categories, the pasting theorem is used to define 4-cells. Would a pasting theorem for tetracategories define "tetracells"? Not unless we can give a sensible definition for the latter. In the Notes on Tetracategories § History we read that "the project got bogged down", Something tells me that this is not one of these cases where everything will just neatly click in place.  ​‑‑Lambiam 09:08, 6 November 2025 (UTC)

I see, thank you for your comment. It seems we will have to wait for further study.--SilverMatsu (talk) 16:32, 6 November 2025 (UTC)

Simultaneous three-way arrival

Consider two people who walk at distinct steady speeds starting at the same time from one end of a straight track. Without any measurement or communication, they can come back to the starting point together if the faster turns round immediately on reaching the far end of the track then, on the pair meeting, both turn round and redo their routes to date in reverse. Can three people with arbitrary distinct steady speeds achieve the same simultaneous arrival back by a similar procedure? It's possible in special cases, e.g. speeds of 1/2/3, but what about in general? If possible, how can it be achieved; if not, how can this be shown? 2A00:23C6:AA0B:3401:DC5D:BE:981F:7BE1 (talk) 14:56, 1 November 2025 (UTC)

Does exactly the same procedure (any time two people meet, they both reverse direction) not work? 173.79.19.248 (talk) 21:09, 1 November 2025 (UTC)

Having done some experimentation, and contrary to my expectations, this question I raised (in response to the original question) seems very difficult to answer! For certain very special combinations of velocities (like 1, 2, 6) it works, but even in very simple cases (like 1, 2, 4) the behavior looks chaotic and it's totally unclear if it is periodic; and this doesn't even touch at all on what happens if the speed ratios are irrational. ~2025-31168-81 (talk) 15:34, 4 November 2025 (UTC)

(OP) Thanks. The suggested procedure certainly works if the two slowest meet some time when the fastest is at the far end, when symmetry says that the reverse of the motion to date will give a simultaneous arrival at the start point. But how/whether this happy state arises for an arbitrary set of speeds seems to be undetermined. My attempts to program the motion got lost in rounding arrors, and to draw a distance/time diagram got lost in angular measurement arrors. ~2025-30900-60 (talk) 20:42, 4 November 2025 (UTC)

Sorry I see that I missed in your original message that "my question" is actually the question you asked. I've reached out to a dynamicist I know to see if this is a known thing/if he has any ideas. When you wrote 1/2/3, you meant the same thing I did by 1, 2, 6, right? (That the fastest person is three times as fast as the middle, and is six times as fast as the slowest.) So far that's the only periodic solution I've found by the rigorous scientific process of "check some random numbers", although one could presumably back-solve to find speeds that do work.

Actually, does that work in general? Like, suppose I choose a sequence of collisions (first the fastest one hits the far end, then the faster two meet, then the fastest one hits the far end, then the slower two meet, then ..., ..., then they all return to the origin simultaneously) -- to what extent can we solve that for the relative speeds of the three walkers? ~2025-31168-81 (talk) 12:59, 5 November 2025 (UTC)

By 1/2/3 I meant that these are the speeds on some scale, so that the fastest is three times as fast as the slowest, while the second fastest is twice as fast as the slowest. When I said that this is possible I meant to arrive back together, not necessarily by the process of all meetings giving a dual change of direction. In this case the fastest and slowest change direction on meeting, but the middle one continues to the far end then returns. ~2025-30900-60 (talk) 16:42, 5 November 2025 (UTC)

Ok I see -- so not precisely the same as my question, you really do want to permit more flexibility.

I haven't heard back from my dynamicist friend, and I haven't worked anything out carefully, but my current suspicion is that if you prescribe the sequence of collisions (first person 3 hits the wall, then person 2 and 3 collide [and either pass through or bounce off], then ..., then they all three arrive simultaneously at where they started), this pattern will actually uniquely define the speed ratios (or perhaps there will be a finite set of possibilities, or perhaps in the worst case there will be some infinite set coming from the solutions of some algebraic equation). But there are only countably many such sequences, so the union over all sequences of the solutions will be nowhere near the full space of speed ratios. ~2025-31168-81 (talk) 01:41, 8 November 2025 (UTC)

Is it possible to design an elliptic curve pairing that is easy to inverse with Miller inversion available?

Recently, Miller inversions algorithms got more generic in 2025 in a way they can be applied to arbitrary pairing friendly curves.

But the hard part that prevents breaking the q strong Diffie Hellman problem is exponentation inversion. The ate pairing doesn t use final exponentiation, but as far I understand it then breaks this criteria $\forall a,b\in \mathbb {F} _{q}^{*},P\in G_{1},Q\in G_{2}:\ e\left(aP,bQ\right)=e\left(P,Q\right)^{ab}$. Several peoples told me I might find how to use Miller inversion for inverting the Weil pairing, but without knowing a point, I fails to see how this can be done.

Now, which type of construction for bn or bls curve is the most likely to find a pairing that respect $\forall a,b\in \mathbb {F} _{q}^{*},P\in G_{1},Q\in G_{2}:\ e\left(aP,bQ\right)=e\left(P,Q\right)^{ab}$ with the step after the Miller algorithm remaining easy to inverse even without knowing 1 of the point involved? 82.67.45.113 (talk) 16:16, 1 November 2025 (UTC)

November 7

Collatz conjecture

I asked a version of this question at the article's talk page but I suspect it won't get much attention.

Is there a function, derived directly from the Collatz conjecture's rule (if even, divide by two; if odd, multiply by 3 and add 1), that will convert an input integer into a sequence length? For example, F(12) = 10, F(19) = 21, F(27) = 111, etc.

I suspect the answer is no, because if such a function did exist, it would be a simple step to then prove or disprove the conjecture. Somewhere in the function must be some instruction about how often the procedure is to be repeated. The simple answer to that is: until you get to 1. But there's the rub - we don't know that it will necessarily get to to 1 in every case; this is the very thing the conjecture seeks to prove. So, we need to repeat until a number is repeated. If that can ever result in a number that's not 1, we've disproved the conjecture. That would mean we've discovered an endless loop OTHER THAN the 4-2-1 loop that every number ever investigated (trillions, literally) gets to.

There can't be any other cases, because if a sequence goes through 1 million steps without any resolution, who's to say the 1,000,001st step won't provide a breakthrough? We just have to keep crunching numbers until we get some result. The idea of a sequence being literally infinite seems very bizarre to me. Is it possible to definitively exclude such a possibility?

Thanks. -- Jack of Oz ^{[pleasantries]} 10:39, 7 November 2025 (UTC)

This is OEIS sequence https://oeis.org/A006577. The programs listed there go through the Collatz procedure and count how many steps have been taken, outputting the result when they reach 1. Iffy★Chat -- 16:54, 7 November 2025 (UTC)

Indeed, given what is known today, it is unknown if such a function exists. However, there is something closely related that Platonic mathematicians can define. Let $C_{i}(n)$ denote the $i$-th iteration of the Collatz transformation with $n$ as starting point. If $f$ denotes the transforming function, $C_{i}(n)=f^{_{\!}i}(n).$ So, for example,

$C_{0}(3)=3,$ $C_{1}(3)=10,$ $C_{2}(3)=5,$ $C_{3}(3)=16,$ $C_{4}(3)=8,$ $C_{5}(3)=4,$ $C_{6}(3)=2,$ $C_{7}(3)=1,$ $C_{8}(3)=4,$ $C_{9}(3)=2,$ $C_{10}(3)=1,~...$

We define another, auxiliary function with two arguments, let's call it $G,$ where we likewise denote one argument as a subscript and one between parentheses, defined by

$G_{i}(n)={\begin{cases}n&{\text{if}}~C_{j}(n)\neq 1~{\text{for all}}~j<i\,;\\G_{i{-}1}(n)&{\text{otherwise}}.\end{cases}}$

For example,

$G_{0}(3)=0,$ $G_{1}(3)=1,$ $G_{2}(3)=2,$ $G_{3}(3)=3,$ $G_{4}(3)=4,$ $G_{5}(3)=5,$ $G_{6}(3)=6,$ $G_{7}(3)=7,$ $G_{8}(3)=7,$ $G_{9}(3)=7,$ $G_{10}(3)=7,~...$

Subsequent values go up each time until the Collatz sequence for $n$ hits a $1$ and then it remains stuck. Now comes some mathematical magic. We apply the limit invocation:

$F(n)=\lim _{i\to \infty }G_{i}(n).$

Since that limit can be infinite, $F$ is a function from the natural numbers $\mathbb {N}$ to the extended natural numbers $\mathbb {N} \cup \{\infty \}.$ When $F(n)$ is finite, it gives the number of transformations needed to hit $1.$ The Collatz conjecture can be formulated as:

$F(n)<\infty ~{\text{for all}}~n.$

The rub is that constructive mathematicians won't have any of this; they do not recognize this as a legitimate definition.  ​‑‑Lambiam 22:07, 7 November 2025 (UTC)