I have proported twin prime conjecture proof. Attempt number 99

Recall an algebra $\mathcal A$ on a set $E$ is a set of functions on $E$ which is closed under addition, multiplication and scalar multiplication. Moreover, if for each $x_1\neq x_2$ there is a function $f\in\mathcal A$ such that $f(x_1)\neq f(x_2)$, then it is said to separate points. If for each $x\in E$, there's a function $g\in\mathcal A$ such that $g(x)\neq 0$, then $\mathcal A$ vanishes at no point of $E$.

Now, is it obvious that the trigonometric polynomials $$f(x)=\sum_{-N}^N c_ne^{inx}$$form an algebra on the unit circle? I'm a bit concerned about scalar multiplication. Do we require that $cf$ is a function on the unit circle? I'm confused.

I'd be really grateful if someone could help me confirm this is an algebra indeed (which separates points and vanishes at no points on the unit circle).

@psie if $f$ is a complex-valued function on the unit circle, then $cf$ is a complex-valued function on the unit circle for any complex scalar $c$

the question you then have to ask yourself whether $cf$ is then also a trigonometric polynomial if $f$ is assumed to be one

I believe you're capable of answering that on your own

alright, thanks 👍

@Ben probably I haven't thought this through enough, but what is the purpose of working with partially defined maps in Lurie's HA formalism?

as in, e.g. if $\mathcal{C}$ is symmetric monoidal, there are a bunch of morphisms in $\mathcal{C}^{\otimes}$ that do not "come from" $\mathcal{C}$, why do we need them?

I'm not sure I understand what you mean by "partially defined maps"

to borrow Lurie's notation, a pointed map $\langle n\rangle\rightarrow\langle m\rangle$ is the same as a "partially defined" map $\langle n\rangle^{\circ}\rightarrow\langle m\rangle^{\circ}$

you're basically forgetting a bunch of induces and then mapping

ah yes

don't think of it that way, is the answer :^)

so when defines $\mathcal{C}^{\otimes}$, there are maps $[C_1,\dotsc,C_n]\rightarrow[C_1^{\prime},\dotsc,C_m^{\prime}]$ that do not correspond to maps $C_1\otimes\dots\otimes C_n\rightarrow C_1^{\prime}\otimes\dotsc\otimes C_m^{\prime}$

cause we're first simply forgetting some of the elements and then taking a nice map

but I'm not appreciating what structural purpose these additional morphisms serve

give me a second to think about this

I suppose one point is that Lurie's $\rho^i$ only makes sense as a partially defined map

Yeah

I was about to say this

you really need the $\rho^i$ to encode the Segal condition

these correspond to sort of "phantom projections" $C_1\otimes\dots\otimes C_n\rightarrow C_i$ that obviously don't make sense in $\mathcal{C}$

and once you have those, don't you get all the other non-inert maps automatically?

@BenSteffan right

@BenSteffan should by taking coproducts, yeah

it still feels a bit clumsy

which is probably to say I don't fully grasp this Segal condition

the Segal condition ensures that the "space of $n$-ary operations" $\mathcal{C}_{\langle n\rangle}^\otimes$ really consists of something you can call $n$-ary operations

did you read section 2.1.1?

Lurie does a pretty good job at motivating where this definition comes from and why it looks like it does

yeah, it's what I'm reading right now

but these "phantom maps" are a bit trippy

by the way I hate the "colored" terminology

they should call it an operadoid

yes, and then they should call generalized colored operads operadoidoids :)

@Ben another thing I'm confused and that's probably even more silly is the following: what's the difference between E_0 and Triv?

they're obviously different categories but they seem to encode the same n-ary operations?

Have you seen Lurie's computation of $\mathrm{Alg}_{\mathbb{E}_0}(\mathcal{C})$?

...to spoil it for you: Triv does not encode any information

$\mathbb{E}_0$ encodes unitality

an $\mathbb{E}_0$-algebra is an object + a map from the unit object, that's it

oh, I guess I was looking at n>=1

I suppose E_0 has a 0-ary operation and Triv does not

which should correspond to what you just said

yes

Triv is the most obvious example of a non-unital $\infty$-operad

in the sense of section 2.3.1

ah, that makes sense, thanks

hi

@leslietownes Oh I can see it now

But then I'd say okay, what if we would define it for any path gamma instead of for any sequence

it turns out not to be equivalent

X4J: well, depending on your definition of "path" i guess. if "path" includes anything like differentiability there's still intuition there, which is that locally a differentiable path might as well be a straight line path

X4J: if "path" is just "continuous function" from R into whatever R^n, you do get the usual notion of limit. see e.g. math.stackexchange.com/questions/2416520/… but for the same intuitive reasons, any proof of that will envision a world of "paths" that are not required to be (and generally are not) differentiable

i always choose a different path.

Sorry for this dumb question, but in Rudin's definition 8.17, he says that the integral that defines the Gamma function converges for $0<x<\infty$. Then he says in parenthesis that "When $x<1$, both $0$ and $\infty$ have to be looked at." What does he mean by this?

By $0$ and $\infty$, I think he means the integration limits, though I still don't quite understand his phrase.

Maybe he means that when $x<1$, then the integral is an improper integral.

Though I'd expect this to be the case for any $x$.

I think I'm on the right track. He means that when $x<1$, both $0$ and $\infty$ are points where the integral is improper.

Follow-up question maybe; when $x<1$, what does the integral turn into then? A double limit of some sort?

@psie correct

@psie it's just an integral

well, $\int_0^\infty=\lim_{n\to\infty}\int_0^n$, but now it is improper also at $0$, so $$\int_0^\infty=\lim_{b\to\infty}\int_0^b=\lim_{b\to\infty}\lim_{a\to0}\int_a^b\,?$$

just think of it as a Lebesgue integral

ah ok, yes that simplifies it

that said, the integrand is non-negative, so it will be absolutely integrable and the double limit description will be accurate and independent of the order in which you take the limits

Why $[AB]_{16}=171_{10}$?

ok 👍

@Binky that's literally just a computation

just do it

and by "computation" I mean "two multiplications and one addition"

You could write it PLS

No.

I doubt that would be helping you.

Rectific

I do 10*16+11*1

=171

that's correct, so why do you come here and ask?

Why is the base 10?

because you chose it as your output base?

how would the same computation look if your output base was 2? what would you have to multiply?

@BenSteffan Where ?

there's an algorithm to convert from base X to base Y. You applied the algorithm with X = 16, Y = 10

do as per my second comment and you will understand this better

$X = \sum_{i=0}^{n-1} a_i \cdot b^i$

AB=1010 1011

This should be base 2

Hi

@BenSteffan Is it possible to say more here (I am just seeing this)

I would hazard a guess but I seem to have proved to myself that my claim is true, so something is wrong

It's very silly

the claim is correct if you assume $X \neq \emptyset$

:)

For $X = \emptyset$ it's clearly false, since $Y^\emptyset$ is a singleton and any operation on a singleton is trivially associative and commutative

I'm also wondering about the following: an exercise asks me "Show that composition ◦ is not, in general, a commutative operation on Funct(X, X)." I did this by taking $X$ to be a two-element set and defining the constant functions thereon, which don't commute. But what can we say more generally? Can we give conditions on $X$ such that this claim is true in general? I feel like "most" Xs should not admit a commutative composition operation

@BenSteffan Going to think about this right now

Thank you

@EE18 Convince yourself that your argument applies to all $X$ with $|X| \geq 2$, and then treat the remaining two special cases :)

Ah I see. So you're saying we could have some noncommutative/nonassociative operation on Y but if the domain is the empty set then we have trivial commutativity/associativity on the induced operation

yes

there is only one operation on a singleton set

@BenSteffan The empty set and the singleton set both have one function defined thereon, and these are trivially (in different ways, admittedly) commutative/associative

Thanks so much Ben!

You're welcome :)

a more interesting question, I suppose, is when the monoid End(X) is a group for X an object in a category C

not sure it's really more interesting in that generality

it's the case iff $\mathrm{End}(X) = \mathrm{Aut}(X)$, and what more can you say...

why did I say group

I meant commutative

ah, so you're asking whether the $A_\infty$-algebra structure on $\mathrm{End}(X)$ can be extended to an $\mathbb{E}_\infty$-algebra structure? :)

that's indeed interesting, if very difficult in general

of course, just as with any other problem, we should first lift it to a homotopy-coherent setting and then think about it

hmm

Yup.

Could someone explain the conclusion of this proof? Rudin wants to show a function satisfying (a), (b) and (c) is uniquely determined. Then at the end, he just says $\varphi(x)$ is determined. What does he mean?

$\varphi(x) - \log(\dots)$ is zero. So $\varphi(x)$ is equal to that expression.

@XanderHenderson that shows existence, right? Not uniqueness.

The limit is unique, no?

Is that not a property of limits?

Ah ok, so he is showing that the function $f$ is the limit of a sequence of functions, and by the uniqueness of limits, there can only be one limiting function $f$.

@psie He is showing that $\varphi$ is the pointwise limit of a sequence of functions, but $f$ is related to $\varphi$ in a nice one-to-one way.

Yes, you're right :)

This proof strategy is new to me. Showing something is a limit of some sequence to show uniqueness...hmm, maybe I'll have that for breakfast tomorrow. Let's see.

Usually when you show uniqueness of something, you assume two things exist that satisfy in this case (a), (b) and (c), and then you show they equal.

@psie I mean, he is showing that any such thing must be equal (pointwise) to the limit given.