User MphLee | chat.stackexchange.com

MphLee

Jun 19, 2021 02:02

Thanks to everyone. Cya.

MphLee

Jun 19, 2021 01:46

OK good to know, so when educated people say "garduate level geometry" they really mean differential g.

MphLee

Jun 19, 2021 01:41

What is graduate level geometry? When I asked you about "geometric" I was only referring to vector spaces, affine spaces, projective spaces...
You and Thor were talking about differential geometry or schemes?

MphLee

Jun 19, 2021 01:39

So, given $N\in\mathbb N$ and $x_0=q$ a first approximation of $\sqrt{N}$ We need to prove that the sequence defined recursively by $x_{n+1}:=(x_n+\frac{N}{x_n})\frac{1}{2}$ converges to $\sqrt{N}$. Is that what your are asking @Omniman ?

MphLee

Jun 19, 2021 01:34

nope, I'm sorry :/

MphLee

Jun 19, 2021 01:29

@TedShifrin are you claiming that for teaching geometry, properly=unformally?

MphLee

Jun 19, 2021 01:27

@Thorgott I'm still surprised by the big divide I felt among first year students. I felt that too since to me convergence of series and asymptotics was like doing random black magic nonsense ... while, to me, algebra was like clean fresh water.... but to many it was like the most difficult thing on earth

MphLee

Jun 19, 2021 01:22

Doesn't seem controversial to me. I heard a professor telling me that ZFC is pretty algebraic in nature, so I'm not gonna be shoked by your claim.

MphLee

Jun 19, 2021 01:10

I ask because I had a weird experience at UNI. I was already well exposed to all sort of set theory, foundational issues and the basic definitions of category made very sense to me.
Back then we had three first year courses: ALG 1(set-theory+ ring theory), GEO 1 (linear algebra+affine spaces+eigentheory) and ANALYSIS.
Everyone was good at analysis, expect for me. Most were amazing doing computations of basis, subspaces and things like that... but, expect few geniuses, most were bad with the Algebra course, and most of them had some problems understanding proofs and the logic behind. I.e. mo…

MphLee

Jun 19, 2021 01:09

Hey I have a question for you. A curiosity I had back when I tried to take my first courses. In your opinion "functor people", I'm gonna stick with the joke, are more geometric ppl or algebraic ones? In your experience.
I used to be at home with abstraction and algebra, even abstraction for it's own sake. But I started to really appreciate and understand CT when I discovered Vector spaces and linear algebra.

MphLee

Jun 19, 2021 01:06

Hi

MphLee

Jun 19, 2021 00:51

Well maybe long ago that was necessary. I wonder if it is still so. With so many online resources.

MphLee

Jun 19, 2021 00:45

To bad I had to steal some more English skill xD, well nvm

MphLee

Jun 19, 2021 00:44

hahah... @TedShifrin that's LITERALLY how I learnt TeX... by stealing here on MSE.

MphLee

Jun 19, 2021 00:42

:thumbs up:
Now I'll steal from you answer the power to draw commutative diagrams on MSE mwhuahahaha >:D

MphLee

Jun 19, 2021 00:38

Reading

MphLee

Jun 19, 2021 00:32

Mhh, ye. In general we can't conclude that $A_{(i)}\cdot A_{(j)}=A^{(i)}\cdot A^{(j)}$

MphLee

Jun 19, 2021 00:25

Probably you are interested in the square matrices case.

MphLee

Jun 19, 2021 00:15

@Thorgott So to reward you for this chat and your time, I hope you can copypaste those two constructions and you useful remark into an answe to q2+opinion on q3. I'll give you the bounty.

MphLee

Jun 19, 2021 00:14

You know what. I'm already satisifed with this combo. With the last generalization.
I know I did half of the work, but you translated that and made some points more clear to me. Q3 can wait. I'll work it myself if it is possible to jist use something more complex than the indiscrete category.
When i'll have more maturity I'll try to study a solution for it my self and see if there is some adjoint. And as long as Q1 goes. when I'll feel ready I'm sure I'll discover the general thing in some lost n-lab page.

MphLee

Jun 19, 2021 00:08

Do you mean $\coprod_{F\colon\mathrm{Ob}(\mathcal{C})\rightarrow\mathrm{Ob}(\mathcal{C})}\prod_{x\in\mathcal{C}}\mathrm{Iso}(F(x),x)$ right?

MphLee

Jun 18, 2021 23:57

Oh @Thorgott If you want you can post this as an answer to Q2. I believe You completely answered that here.
If you can give me some cent on question Q3, that is now basically something like an adjoint situation to Q2, I'll award you the bounty.

MphLee

Jun 18, 2021 23:52

Btw, the deep part I believe is going on I'm referring to is when we stop to consider aoutomorphisms and extend this to isos in general using a and endofunction $f:{\rm Ob}\to{\rm Ob}$. Like in the case of turning cardinality into an endofunctor of Sets, or when we make a non-canonical choice of a basis for every vec. space, making us able to define and endofunctor that is not the identity on the objects.

MphLee

Jun 18, 2021 23:47

Btw, so I'll do this as an exercise: I need to prove that it is an embedding, and to prove that the image is the one you stated, that should be the easy part.

MphLee

Jun 18, 2021 23:47

Well, thank you very much. That already something interesting I believe. I believe there should be something deeper btw.

MphLee

Jun 18, 2021 23:44

ok. Sure, I forgot to write it.

MphLee

Jun 18, 2021 23:41

So, basically, that's literally a function assigning to every choice function and endofunctor. By your previous remark, if I understand it, the image of the functor should be exactly the slice category of Automorphisms of the category over the identity functor.

MphLee

Jun 18, 2021 23:29

@Thorgott I'm sorry If my question was unclear. I'm not good at explaining myself in english. But it is harder when I don't get feed back on my questions.

hahah I remember back in the days when a misplaced comma would result in bad comments, downvotes and votes for closures.

MphLee

Jun 18, 2021 23:27

When you say for each choice there is only...
I'd like to make this precise. Few comments above you stated "
functors naturally isomorphic to the identity functor that act as the identity on objects" I'd hope to phrase this a something regarding objects of the slice category over the identity functor...
I' not sure how: that was question Q2.

MphLee

Jun 18, 2021 23:26

Sure, I agree. I know that I'm defining a functor. And I proved the naturality. It's kinda trivial. But I see some circularity. Probably because... in CT what is mu before I define the functor? It cannot be a natural transformation....because there is no functor yet.

MphLee

Jun 18, 2021 23:23

Is this a theorem, and exercise in some book, a basic result? Something so trivial is not even worth mentioning. Has this a name?
That was the spirit of Q1, in my question.

MphLee

Jun 18, 2021 23:20

What puzzles me is the fact that before I define the functor, \mu can't be a natural transformation, because it has not a functor to be natural with...
And the component on object of the functor needed can't be defined if we haven't made a choice of \mu. When we do that F becomes functorial, and only then \mu become natural.
I see a kind of circularity here.

MphLee

Jun 18, 2021 23:18

Yes. Ok. Because $\mu$'s components are isomorphisms (by definition).

MphLee

Jun 18, 2021 23:16

Sure. I'm sure I don't understading where youre getting at. I'm missing something probably.

MphLee

Jun 18, 2021 23:13

ok, but then if you just chose the identity for each object you're just defining the identity endofunctor... pretty boring isn't it

MphLee

Jun 18, 2021 23:11

@Thorgott The main example, the motivating example which I linked in the question, appears naturally in the context of groupoids, because it is some kind of generalized conjugation. At first sight, I'd say one can extend the argument to category as long as at least one object, or all of them, have at least one non-trivial automorphism.

MphLee

Jun 18, 2021 22:59

Oh, hahah xd ok @TedShifrin thank you anyways. It's pretty late here (EU time)... I was betting on US ppl...
Even functor ppl need abstract nonsense dreams sometimes

MphLee

Jun 18, 2021 22:57

If someone is interested there is a 50rep bounty on it.

MphLee

Jun 18, 2021 22:56

1

Q: Functors making functions natural trasformations and vice-versa.

I apologize in advance if this is naive. In this answer Conjugation in a groupoid it is said that given a groupoid $\mathcal G$, and an arbitrary function $\mu:\mathcal G_0\to \bigcup_{x\in \mathcal G_0}\mathcal G(x,x)$ s.t. $\mu_x\in \mathcal G(x,x)$ we can define an endofunctor $F_\mu\in[\mathc...

MphLee

Jun 18, 2021 22:56

Hi, someone can help me with this question?

MphLee

Jun 7, 2021 11:01

@Thorgott @BalarkaSen Ok thank you very much. I go. See you!

MphLee

Jun 7, 2021 10:52

Well if we consider $C^\infty(I,C)$ a as a category an homotopy is a natural transformation thus components are themselves in $C^\infty(I,C)$

MphLee

Jun 7, 2021 10:50

Oh ... wait. Right because a loop and a point even if homotopic we now know how are homotopic? Is that your point? That homotopy remembers something about the derivative?

MphLee

Jun 7, 2021 10:48

Thx

MphLee

Jun 7, 2021 10:48

Very well. That was fruitful I believe.

MphLee

Jun 7, 2021 10:46

OK, I was aware of n-cats. I was aware of the infinity groupoid of homotopies up to homotopies up to...That is visually very vivid in my mind.
But the simplicial set version is still obscure to me.
Thanks @BalarkaSen I'll look more int this.

Obviously if we want to retain the individual shapes and slopes of the paths we can't mod by homotopy thus, you claim, we will never get a category.

Even if we mod out by reparametrization of paths, as noted, we lose information of higher derivatives. We just retain the set theoretic image of the paths.

MphLee

Jun 7, 2021 10:41

Ok, I'm reading.

MphLee

Jun 7, 2021 10:39

@BalarkaSen Ah... I see xD... Sorry I misunderstood.

MphLee

Jun 7, 2021 10:38

A natural Number Object.

MphLee

Jun 7, 2021 10:38

NNO is a simplicial set?

MphLee

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