Mathematics - 2019-07-09

Ultradark

00:02

Ye

Isiah Meadows

00:25

Sorry. DFA = deterministic finite automaton

It's more on-topic to CS.SE anyways. Sorry for the noise and confusion.

Ted Shifrin

Yup, I personally am clueless.

Albas

05:01

@TedShifrin Thanks a lot for that. Quite a few books say that you should "row reduce it". Whereas they should actually talk about diagonalization of the matrix. And the job is easier for symmetric matrices.

ÍgjøgnumMeg

07:49

Mornin' all

Alessandro Codenotti

07:59

Morning @ÍgjøgnumMeg

ÍgjøgnumMeg

What's up :)

Alessandro Codenotti

Just studying for exams

ÍgjøgnumMeg

Fun

What courses do you have exams for this semester?

Alessandro Codenotti

Models of set theory I, noncommutative geometry and type theory

On Monday, Tuesday and Wednesday next week, in this order

ÍgjøgnumMeg

Cool, have fun

hahaha

you feeling prepared?

Alessandro Codenotti

08:04

Ehhh... Kinda hahaha

ÍgjøgnumMeg

lol, I'm sure you'll be fine

Alessandro Codenotti

I don't feel too good about noncommutative geometry, we did a bunch of Riemannian geometry stuff I'm very unfamiliar with

Leaky Nun

@AlessandroCodenotti they don't exist

Alessandro Codenotti

what

Leaky Nun

models of set theory

Alessandro Codenotti

08:09

Yeah, well, you know, that's just, like, your opinion, man.

cit.

ÍgjøgnumMeg

08:21

lol

I'm just trying to remember how to do mathematics

Alessandro Codenotti

With tears and pain

ÍgjøgnumMeg

indeed

Just haven't done any calculus in so long

lol

Kenkar

10:04

Are complex numbers ordered pairs with aditional properties or something we can represent with ordered pairs ?

ÍgjøgnumMeg

10:23

@Kenkar sure, you have ordered pairs of real numbers with a specific addition and multiplication attached to them

Secret

11:27

Claim: The zero ordinal is a limit ordinal

Proof: Let $[]$ be the empty sequence

Then its subsequence is itself, which is also a subset of itself

We can pick elements out of $[]$ by iterating the operation on its subsequences

all these elements are inside empty sets which has no elements

Therefore, we can take the supremum of this constant sequence, and get $0$

Alternately, take a bipartition on [], which gives [],[]. Rinse and repeat on every subset that is generated. Then use the axiom of countable choice to produce a sequence {[],[],[],[],...}. The supremum of this is clearly [] which is 0

Sebastian Nielsen

12:03

These three expressions are equivalent, right? By the way, the context is asymptotic notation.

f(n) is Θ(g(n))

f(n)=Θ(g(n))

f(n) ∈ Θ(g(n))

Leaky Nun

13:01

yes

the first two are abuse of notation

Sebastian Nielsen

Why do you think the two first are abuse of notation?

Leaky Nun

ok the first one can count as ambiguous

the second one is clear: the left hand side is a function and the right hand side is a set of functions

so clearly they mean the third one

Abuse of notation

In mathematics, abuse of notation occurs when an author uses a mathematical notation in a way that is not formally correct but that seems likely to simplify the exposition or suggest the correct intuition (while being unlikely to introduce errors or cause confusion). However, the concept of formal correctness depends on time and on the context. Therefore, many notations in mathematics are qualified as abuse of notation in some context and are formally correct in other contexts; as many notations were introduced a long time before any formalization of the theory in which they are used, the q...

Sebastian Nielsen

15:01

In the context of asymptotic notation, are there infinitely many functions $f(n)$ that is a superset of $\Theta( g(n) )$.

In other words, are there infinitely many functions $f(n)$ that satisfy the following:
$$f(n)=\Theta( g(n) )$$

.

In other other words, is it possible to come up with infinitely many functions $f(n)$ that, from some $x_0$ value and beyond on the x-axis, stays in the white area (between $c_1*g(x)$ and $c_2*g(x)$:

* I just realized I set "superset", I meant "member of"

Thorgott

15:50

Note that you are asking two different questions. The constants $c_1,c_2$ in the definition of $\Theta(g(n))$ can be chosen dependent on $f$, so the second question is more restrictive. However, the answer to both is "yes", granted $c_1\neq c_2$ and $g\neq0$ (hopefully I'm not missing another degenerate case).

Balarka Sen

16:49

@MatheinBoulomenos I summon thee

Alessandro Codenotti

Hi @Balarka

Balarka Sen

Hey

Julian Gonzalez

Hello

Alessandro Codenotti

17:55

The chat is not very active lately

J. Doe

18:41

Hello

Alessandro Codenotti

18:54

Hi

J. Doe

Hey

Astyx

Hey

J. Doe

Hi

Sebastian Nielsen

19:26

Howdy

GFauxPas

20:00

This exercise asks to prove that $f(x) = \sum_{n=1}^\infty 2^{-n} \cos(10^n \pi x)$ is nowhere differentiable. The hint is to prove that

$\displaystyle \left \vert {f\left({\frac{j+1}{10^k}}\right)- f\left({\frac{j}{10^k}}\right)}\right\vert \ge 2^{-k}$ for all integers $j, k$ with $k \ge 1$

how does that prove the non-differentiability at values that can't be written in the form $\dfrac{j}{10^k}$?

Astyx

This is smaller than $|f({j+1\over 10^k})-f(x)|+|f(x)-f({j\over 10^k})|$

When ${j\over 10^k}\le x\le{j+1\over 10^k}$ this tells you something

GFauxPas

20:16

hm

are we holding $k$ fixed here and looking at it as a sequence in $j$?

or the other way around?

Balarka Sen

20:38

@Alessandro Are you around?

Alessandro Codenotti

I am

Balarka Sen

I have a question. If $E$ is a vector bundle over $M$, then I can take the sheaf $\Gamma_E$ of sections of $E$ over $M$. This is a (projective) $C^\infty(M)$-module valued sheaf, and Serre-Swan theorem states that the global section functor is an equivalence of categories between $\text{Vect}_E$ and $C^\infty(M)-\text{PMod}$, right?

Finitely generated whatever

So how do I reconstruct the vector bundle given a finitely generated coherent projective $C^\infty(M)$-module valued sheaf on $M$?

Alessandro Codenotti

Uh I think you want a compact $M$

Balarka Sen

OK

I want to explicitly understand what the functor in the other direction is, I guess. Given a f.g. projective $C^\infty$-module I can take the stalks at every point on $M$; projective local rings are free. This should be the "total space" of the vector bundle. But how do you get the topology back?

It's not the espace etale; that gives the stalks discrete topology! (Which is useful, but for a different correspondence, between locally constant sheaves and representations of $\pi_1 M$)

Alessandro Codenotti

Also I'm not sure about $C^\infty(M)$. The statement of Serre-Swan I'm familiar with just asks for a compact Hausdorff $X$ and gets an equivalence between finitely generated projective $C(X)$ modules and vector bundles over $X$. This should be a minor difference

Balarka Sen

20:48

I am thinking smooth category. Tell me how to construct a vector bundle on $X$ from a f.g. projective $C(X)$-module

Alessandro Codenotti

Ok so we are given a fpg $C(X)$-module $A$ and we want a bundle $E$ such that $\Gamma(E)\simeq A$

We can find a projection in $M_n(C(X))$ with $A\simeq p(C(X)^{\oplus m})$

$p\colon C(X)^{\oplus m}\to C(X)^{\oplus m}$ induces a bundle morphism $\tau\colon X\times\Bbb C^m\to X\times\Bbb C^m$

Balarka Sen

Hold on. $M_n(C(X))$ is the matrix ring over $C(X)$? What is $m$?

I don't follow

Alessandro Codenotti

$m$ is some integer

Balarka Sen

How is $C(X)^{\oplus m}$ embedded in $M_n(C(X))$?

Alessandro Codenotti

Oh ok, wait, I was skipping a lot of steps

Astyx

20:54

@GFauxPas we're looking at $j\over 10^k$ approximating $x$

Alessandro Codenotti

A module $A$ over $R$ is finitely generated projective iff there is some $n$ and some $p=p^2\in M_n(R)$ with $A\simeq p(M_n(R))$

GFauxPas

hm

Alessandro Codenotti

The general statement is that $A$ is projective over $R$ iff there is a free module $F$ and $p=p^2\in\mathrm{End}_R(F)$ with $A\simeq p(F)$

Balarka Sen

@Alessandro Yes, because $A \oplus B = R^n$ for some $B$ and $n$. There is a map $p : R^n \to R^n$ which projects to $A$. Got it.

Alessandro Codenotti

Why is the second summand free?

Balarka Sen

20:58

Whoops

GFauxPas

ah, I see. thanks Astyx

Alessandro Codenotti

$0\to B\to R^n\to A\to 0$ is split exact, and $p$ is the map $R^n\to A\to R^n$, where the second map is the splitting

Balarka Sen

Just $R^n =A \oplus B \to A \oplus 0 \subset R^n$, no? :P

GFauxPas

now I just have to figure out how to analyze $\cos(10^{n-k} \pi n)$ for when $k > n$

Alessandro Codenotti

Fair enough :P

Balarka Sen

21:00

I don't understand your notation. $A \simeq p(M_n(R))$??

Alessandro Codenotti

Ok so we have the map $p\colon C(X)^m\to C(X)^m$ now

GFauxPas

when $k \le n$ it's $1$ or $-1$

Alessandro Codenotti

@BalarkaSen A is isomorphic to the image of the matrix ring through $p$

Of course I wanted $p(R^n)$ instead

woops

sorry

When the free module is finitely generated we can identify $\mathrm{End}_R(R^n)$ and $M_n(R)$

Is this clear now?

Balarka Sen

Yep, got it.

Now you'll realize $p$ as a bundle map $X \times \Bbb R^n \to X \times \Bbb R^n$ and $E$ will be image of this bundle map, I suppose.

Alessandro Codenotti

Ok so we have a map $p\colon C(X)^m\to C(X)^m$, by taking $\Gamma$ of everything we get a bundle morphism $\tau\colon X\times \Bbb C^m\to X\times \Bbb C^m$

@BalarkaSen Exactly

Balarka Sen

21:06

Nice, that's exactly what I wanted. What a beautiful way to recover the topology.

GFauxPas

I know $\vert \cos x - \cos y \vert \le \vert x - y \vert$ but I'm not sure that will help

since I want a lower bound not an upper bound

Alessandro Codenotti

(Do you see a simple reason why the image of $\tau$ is a subbundle of $X\times \Bbb C^m$? Because I don't and I don't really follow the argument we did in class either)

Balarka Sen

I'd try to prove that the kernel of $\tau$ corresponds to $B$, the factor the projection $p$ kills.

I'll have to think carefully about it

Once we prove that $E$ is a quotient bundle of $X \times \Bbb R^n$, basically

It's 2:40 AM so I will sleep now, but I will ponder on this and tell you what my thoughts are tomorrow @Alessandro. Thanks a ton!

Alessandro Codenotti

No problem

The professor argued that $x\mapsto\mathrm{rank}(\tau_x)$ is continuous (hence locally constant), which is a condition strong enough to guarantee that $\ker\tau$ is a subbundle of the first one

Balarka Sen

Right, so you need to argue $\tau$ is a bundle morphism in the first place.

Alessandro Codenotti

21:14

Well $\tau=\Gamma(p)$, so if we know that $\Gamma$ is a functor we're fine

Balarka Sen

I'd say $p = \Gamma(\tau)$ rather, the global section functor applied to the bundle morphism. You're trying to prove the inverse guy is a functor

That's not clear apriori to me

Alessandro Codenotti

Oh yeah sorry I got it backward

I should go to sleep too lol

Balarka Sen

I might need to see your professor's argument, I like the rank idea a lot. In my case it won't be continuous, because my bundles are stratified :3

It will be continuous on each strata

But yeah let's hit the sack for today lmao

Alessandro Codenotti

Ok but if know that $\Gamma$ is full (and it is) we can find some $\tau$ with $\Gamma(\tau)=p$

Balarka Sen

I see, OK

I'll try to understand it better tomorrow. Thanks again!

Alessandro Codenotti

21:18

Ah that's actually easy to see after proving that $\Gamma$ commutes with tensor products and duals

By writing $\mathrm{Hom}(E,F)$ for two bundles $E,F$ as the bundle $E^\ast\otimes F$

Sure, I'll be happy to talk more about this tomorrow

(It's among the stuff I need to know for an exam next Tuesday lol)

Balarka Sen

Hah great we can talk it out then

G'night!

Alessandro Codenotti

Good night!

(one last remark, the other direction of the implication $A$ projective iff there is $p=p^2\in\mathrm{End}_R(F)$ for a free $F$. Suppose you're given such a $p$, then $F\simeq A\oplus (I-p)(F)$, so $A$ is a direct summand of a free module)

Kalua

21:42

I don't know much about K-theory and stumbled upon the expression "the K-theory class of p in K_0(B) does not come from K_0(A)" for C*-algebras A,B, what does that mean? There is a canonical inclusion homomorphism phi:A->B. Could this mean that p is not in the image of the induced morphism K_0(phi):K_0(A)->K_0(B)?

Alessandro Codenotti

Yes that's how I would read it

Kalua

Great thanks!

Alessandro Codenotti

You have a pushforward map $\varphi_\ast$ induced by $\varphi:A\to B$ that sends finitely generated projective modules over $A$ to fgp modules over $B$. The induced map on K-theory is then $[E]\mapsto[\varphi_\ast E]$ for an fgp $A$-module $E$. Saying that $p$ does not come from the K-theory of $A$ is just saying that its class not in the image of this map

Kalua

Followup: The paper says that $K_0(C_0(N x G) \rtimes_r G) = \oplus_{n\in N} Z$ for a countable discrete group G. I know that $K_0(C_0(G)\rtimes_r G)=K_0(CompactOperators(\ell^2(G)))=Z$. Is there an easy way to see this first equation?

Ryan Unger

@AlessandroCodenotti since when are you a K theorist

Alessandro Codenotti

21:54

@Kalua Now that's way beyond what I can help with, sorry

Kalua

Np, you still helped me there

Alessandro Codenotti

@RyanUnger I only know a little tiny bit about K-theory for C*-algebras, it was part of the noncommutative geometry course

By the way @Ryan do you know how analysts construct tensor products of vector spaces?

Kalua

@AlessandroCodenotti Why would analysts define them any differently, is that a trick question? :D

Alessandro Codenotti

It's easy, you take $V\times W$, equip it with the discrete topology and then look at $C_c(V\times W)$, the vector space of compactly supported functions, so that $\{\chi_{(v,w)}\mid v\in V, w\in W\}$ is a generating set, and then you quotient by the subspace generated by $\chi_{(v_1+v_2,w)}-\chi_{(v_1,w)}-\chi_{(v_2,w)}$ and the other tensor relations

Kalua

22:16

Hmmm, that started really ugly with the $C_c(V\times W)$ and I was determined to dislike this approach, but there may be some beaty in it the more I think about it... But I don't see the appeal, universal property or span{v cross w | v,w in chosen bases of V,W} seem more elegant

Alessandro Codenotti

I know, it was a joke

(but I actually found that definition in a book)

Kalua

I hope I will never have to practically use this xD

Daminark

22:31

Oh boy

Ryan Unger

@AlessandroCodenotti your “analyst” is some twisted mutant algebraist

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$Mathematics$ Mathematics