Mathematics - 2018-07-28

geocalc33

00:15

Happy friday.

/saturday

Can linear algebra be replaced by algebraic geometry in a systematic way?

Alex Kindel

00:36

I have a numerical approximation problem. Given a rational polynomial $P(x)$, a complex number $n$, and a means of finding rational intervals as small as desired containing the real and imaginary parts of $n$, I need to be able to find rational intervals as small as desired containing the real and imaginary parts of $P(n)$.

Alex Kindel

00:46

I have a way to choose a sufficiently small size for the intervals $\mathrm{Re}(n)\in[re_{lower},re_{upper}]$ that if I evaluate $P$ at the corners of the rectangle demarcated by these bounds and choose the min and max of the real parts from among them, the distance between them is lecss than or equal to the desired distance, and likewise for the imaginary part.

If $P(x)$ were monotonic in that region, then this would suffice as an answer, but I don't know that a priori. I assume I'd have to do something involving using a root-finding algorithm to check whether $P'(x)$ has roots in the rectangle in question. Does that sound right? Or is there some more elegant method that doesn't require the extra complication?

Earlier I meant "the intervals $\mathrm{Re}(n)\in[re_{lower},re_{upper}]$ and $\mathrm{Im}(n)\in[im_{lower},im_{upper}]$."

Secret

07:20

@William I think some kind of lattice will work. We knew that blood types are basically saying whether you have antigen A or B on the surface of the red blood cells, thus putting A and Bs together is bad and putting A or B with O (which has non of the antigens) is also bad. So O being the universal donor, will be some bottom element, while AB being the universal acceptor, will be the top

For example
AB + B = AB
AB + A = AB
AB + O = AB
A + A = A
B + B = B
O + O = O
A + O = A
B + O = B

So you end up with some semigroup where the acceptor is the left entry and the donor is the right entry, and AB will absorb everything on the left, while O will be the right identity element

....

wait a minute, this structure looks familar...

Ok nvm, its just some generic semigroup

Now, one thing that is apparent is that since addition of numbers is commutative, addition does not work.

Factoring might involve O being assigned to 1 since 1 is a factor of every number, and AB being assigned to be a number having A,B,O as its factors

Then the operation a|b should give all the entries and the dead ones will mean "a is not a factor of b"

So preliminary speaking, we could have A=2, B=3, O=1, AB=6

Danu

@TedShifrin So which one is it?!

Secret

07:36

@William So factoring does work, where all combinations that can kill you corresponds to a is not a factor of b

none

08:32

Hey guys, I was thinking about part (b) of this question recently: https://imgur.com/a/6LKHuRo

I can show that if a finite generating set existed, then $1$ must be in that set. All that'd be left to show is that $I \neq R$. However, it seems that any polynomial of $R$ is in the form of polynomials in $I$: $f = x_{t_1}f_{1}+\ldots+x_{t_m}f_{m}$.

namely, if $f$ is in $R$, then simply factor some $x_{t_i}$ from each monomial term the leftover is the $f_i$, then possibly collect some like terms if necessary. Clearly this is wrong, though.

Alessandro Codenotti

what about the polynomial $x_1+1$? is it of that form?

Or the polynomial $1$ actually to keep it simpler

none

Ah yes, good point hahaha, thanks!

Fargle

08:47

I find this is easiest to see via the degree

No polynomial in $I$ can have a term that's degree zero, because it's in the form you gave above; in particular $1 \notin I$

none

Yes yes, of course. Thanks again! I used similar degree arguments before, can't believe I didn't notice this.

Thanks so much!

Alessandro Codenotti

09:08

$I$ however contains every polynomial with zero constant term

user1732

09:49

Behold: The Pythagorean theorem :P

Poline Sandra

10:29

hello

someone help me here

1

Q: Exemple about a set which is not connected but it's interior is connected

As in the title i want to find a set $A$ from a topological space such that $$ A~\text{ is not connected but}~~ \rm int(A)~\text{ is connected} $$ i tryed with this space: $E=\{a,b,c,d,e\}$ with $$\tau=\{\emptyset, E,\{b\},\{c\},\{b,c\},\{c,d\},\{b,c,d\}\}$$ $E$ is connected, $\{b,c,e\}$ is c...

general-topology

Mike Miller

{2} is an open subset of that space.

You need to use the definition of the subspace topology.

The comment gives an example of an open set whose intersection with your subspace is {2}. By definition of the subspace topology {2} is open.

Poline Sandra

to see that (0,1)\cup \{2\} is connected i must find the topology induced by \{2\} ?

Mike Miller

It is not connected. To see something about a topological space you should probably know its topology.

Albas

10:45

@PolineSandra Look at the subset ${2}$ of $(0,1) \cup {2}$ when your space is given the subspace topology as being a subset of the real line which is given the standard topology

Poline Sandra

i think that the example is given in $(\mathbb{R},|.|)$

so i must find the induced topology by $(0,1)\cup\{2\}$

it is to hard

i don't how to find all the topology

$(-\infty,a)\cap (0,1)\cup\{2\}= (0,a), a>0, =\emptyset, a\leq 0,$

$(a,+\infty)\cup (0,1)\cap \{2\}=(a,1), 0\leq a <1,= \emptyset, 1\leq a\leq2, =\{2\}, a>2$

like this

?

someone here?

user1732

11:06

\o @MatheinBoulomenos

MatheinBoulomenos

hi @user1732

Mike Miller

11:48

@MatheinBoulomenos I have been in the Munich airport since 5am :(

Rudi_Birnbaum

@MikeMiller regards to Franz-Josef Strauß!

Alessandro Codenotti

@MikeMiller Are you flying with ryanair?

Mike Miller

12:05

No, there was an incident with someone getting past security without being checked, and they had everyone leave their planes and wait for hours and then go through security again. And now it has been 2.5 hours of waiting in line for a new ticket. Half of the airport (termainal 2) was affected and no flights left

There are barely any employees and it takes forever for a single person to be taken care of at the desk (guessing 20-30 minutes)

Presumably they will refuse the 600 Euro payment for a very late flight as being the fault of the airport

Alessandro Codenotti

I booked my plane ticket to Germany just a couple of days ago and now I kee hearing horror stories of delayed/cancelled flights

Mike Miller

I am sick and need to get to a doctor at home as well

Alessandro Codenotti

12:28

Urgh that sucks

user1732

sounds like there's nothing you can do :(

Mike Miller

Anyway the moral is probably don't go to Bavaria

3

user1732

did you bring some good books with you?

Mike Miller

Nope, only bad books

user1732

don't try to write any formulas down

MatheinBoulomenos

12:41

@MikeMiller wow, that sounds horrible

sad that your first visit to Germany (I assume) has to be like that

Mike Miller

Indeed my first. I got to the desk and the agent took a phone call

They have two people working to deal with a line I would guesstimate around 500

Maybe 300 is better

Alessandro Codenotti

13:44

@MikeMiller Even Germans would agree with that

Mike Miller

On a plane to Vancouver now

Here we go

Looks 2/3 full

Apparently we are leaving a ton of people behind

Silent

14:01

Can $(n,x)$ be principal ideal in $\mathbb Z[x]$ if $n\ne1,-1$ and $n$ is not prime?

Alessandro Codenotti

What do you think?

And what if $n$ is prime?

user2236

Congratulations pal @MikeMiller

Vancouver is sunny and warm :-)

Silent

@AlessandroCodenotti $n$ is prime then $(n,x)$ can't be principle ideal. But since $(n,x)$ is set which consists of integral polynomials with constant term $n$, I think even when $n$ composite, it can't be principal ideal.

Alessandro Codenotti

@Silent How did you prove that it's not principal when $n$ is prime?

@Silent And there are more polynomials in $(n,x)$ than polynomial whose constant term is $n$!

Silent

@AlessandroCodenotti Sorry, I should write constant term is integral multiple of $n$.

Alessandro Codenotti

14:15

agreed now

Silent

@AlessandroCodenotti Its raining here, sorry! I will be back in a while

Alessandro Codenotti

Sure, there's no hurry

Leaky Nun

16:02

(ac)(acde) = (cde)

MatheinBoulomenos

someone wrote in an exam that there are 36 primitive 8th roots of unity in $\Bbb C$

I want to know: why 36?

Alessandro Codenotti

Because there are $36$ numbers between $1$ and $8$ (extremes included), which are coprime with $8$, I'm not sure what kind of explanation you're looking for here

MatheinBoulomenos

idk

loch

duh

Alessandro Codenotti

Jokes apart that's a weird number indeed, I could understand writing that there are $8$ by missing the "primitive" but 36 is a lot

MatheinBoulomenos

16:09

I found out that I won't be able to finish my bachelor next semester as planned (yay!)

Alessandro Codenotti

That sucks, why is that?

MatheinBoulomenos

I have to take one of four courses for my bachelor, none of which is offered next semester

although it should be according to the documents

Alessandro Codenotti

Can't you take the exam without taking the course?

MatheinBoulomenos

no

I have to get enough points on problem sets to take an exam which is hard when there are problem sets

Perturbative

Hey everyone

Leaky Nun

16:17

@MatheinBoulomenos otherwise how many should there be? 4? don't be ridiculous

MatheinBoulomenos

@LeakyNun oh yeah, you're right, 36 is correct

Alessandro Codenotti

Can you still take more courses whose credits will count toward your Master's degree next semester?

MatheinBoulomenos

@AlessandroCodenotti yeah I can do that

Alessandro Codenotti

As annoying as it is at least you're not completely wasting a semester then, but still...

MatheinBoulomenos

I just missed 32 roots of $x^4+1$

@AlessandroCodenotti yeah

Perturbative

16:19

Also that sucks @MatheinBoulomenos, if it holds you back, you should check with your uni if they give concessions (or some equivalent of that) so you can progress anyway without being held back

MatheinBoulomenos

@AlessandroCodenotti well, about "a lot", someone wrote that there are infinitely many primitive 8th roots of unity in $\Bbb C$

Alessandro Codenotti

@MatheinBoulomenos That's not wrong, but could be specified better, assuming CH there's $\aleph_1$ primitive roots, without assuming CH it can be any regular $\kappa$ with $\aleph_0<\kappa\leq 2^{\aleph_0}$

MatheinBoulomenos

lol

Leaky Nun

@MatheinBoulomenos where exactly are your true answers coming from?

are you teaching a class of geniuses?

Alessandro Codenotti

Ok I'll stop writing nonsense now

Just joking, who wants to help me clear some confusion regarding tensor products? (of vector spaces)

MatheinBoulomenos

16:27

ask away!

Alessandro Codenotti

Notation: $V,W$ are the two spaces with basis $e_i$ and $f_j$ respectively

MatheinBoulomenos

though I've had some scotch, so maybe I'll write nonsense

Alessandro Codenotti

So I know I can think about $V\otimes W$ formally as a vector spaces spanned by $e_i\otimes f_j$, there is a nice map $V\times W\to V\otimes W$ and everything, or I can think about $V\otimes W$ in terms of its universal property

Leaky Nun

did you ask this before?

Alessandro Codenotti

Now this thing I'm reading defines $V_1^*\otimes\cdots\otimes V_n^*$ as the space of multilinear maps $V_1\times\cdots\times V_n\to\Bbb R$ (and then defines $V_1\otimes V_2$ as $V_1^{\ast\ast}\otimes V_2^{\ast\ast}$ for finite dimensional vector spaces)

Leaky Nun

16:32

$(cd) (ab) \sigma = 1$, $\sigma = (ab)(cd)$

@AlessandroCodenotti \ast $\ast$

MatheinBoulomenos

@AlessandroCodenotti that's a weird definition

and it doesn't work for infinite-dimensional vector spaces

or modules

Alessandro Codenotti

He wants to talk about differential forms, so he is more interested in the multilinear maps than in the tensor products but it still confuses me a bit

MatheinBoulomenos

are you asking why these are the same? (It's jus that the bidual is naturally isomorphic to the original vector space basically)

or what is your question?

Alessandro Codenotti

@MatheinBoulomenos Oh no that's fine

Let's work with $V_1$ and $V_2$ to save time typing (because I'm lazy)

the space of bilinear maps $V_1\times V_2\to\Bbb R$ and $V_1\otimes V_2$ are obviously isomorphic (same dimension) but is this isomorphism canonical? It seems to me that I have to pick basis of $V_1$ and $V_2$

MatheinBoulomenos

it's not canonical

the dual of $V_1 \otimes V_2$ is the space of bilinear maps $V_1 \times V_2 \to \Bbb R$, by the universal property

the isomorphism of a f.d. vector space and its dual is not canonical

Alessandro Codenotti

16:39

@MatheinBoulomenos So we have $(V_1\otimes V_2)^\ast=V_1^\ast\otimes V_2^\ast$

MatheinBoulomenos

if $V_1$ and $V_2$ are finite-dimensional, then yeah

and you don't need a basis for that

loch

i think this definition is quite common in DG

MatheinBoulomenos

that doesn't make it any better

if $E$ and $F$ are vector bundles over the same smooth manifold $M$, you maybe want to say that $\Gamma (E \otimes F) = \Gamma(E) \otimes_{C^\infty(M)} \Gamma(F)$, but that doesn't make senes if the RHS is not defined because you use a weird definition of tensor products

William

@Secret wow thank you that is awesome

Took a little bit to understand how it works but I believe I do now

Alessandro Codenotti

@MatheinBoulomenos Actually I don't see the universal property argument here

MatheinBoulomenos

16:48

any bilinear map $V_1 \times V_2 \to \Bbb R$ gives you a unique linear map $V_1 \otimes_{\Bbb R} V_2 \to \Bbb R$ that factors over the structure map $V_1 \times V_2 \to V_1 \otimes_{\Bbb R} V_2$. On the other hand, for any linear map $V_1 \otimes_{\Bbb R} V_2 \to \Bbb R$, you get a bilnear map $V_1 \times V_2 \to \Bbb R$ by precomposing with the structure map $V_1 \times V_2 \to V_1 \otimes_{\Bbb R} V_2$

Silent

@AlessandroCodenotti Sorry for taking so long. So, if $n\ne1,-1,0$, $(n,x)$ is proper ideal. Here I assume that $n$ is prime. Suppose $(n,x)$ is a principle ideal, hence $(n,x)=(a(x))$ for $a(x)\in\Bbb Z[x]$. So, $n=p(x)a(x)$ for some $p(x)$ and comparing degrees, and since $n$ prime, $a(x),p(x)\in\{+-1,+-n\}$.

If $a(x)$ was +-1, then $(a(x))$ could not be proper ideal. Hence $a(x)=+-n$, and so $x=nq(x)$, where $q(x)$ has integer coefficients, hence impossible. So, for n prime, $(n,x)$ not principle ideal.

MatheinBoulomenos

these two assignments define a canonical isomorphism

Alessandro Codenotti

@MatheinBoulomenos Oh, I see, but I can't spot anything that goes wrong with infinite dimensional spaces here

MatheinBoulomenos

the part that the dual space of $V_1 \otimes_{\Bbb R} V_2$ is the space of bilinear maps $V_1 \times V_2 \to \Bbb R$ also holds in infinite dimensions

Alessandro Codenotti

Try the same argument with $n$ composite, you get to $a(x),p(x)\in\{\pm 1\pm n_1,\pm n_2,\cdots\pm n\}$ with $n_i$ the factors of $n$ @Silent

Silent

16:53

But, I think that $n$ prime is used only at concluding $a(x),p(x)\in\{+-1,+-n\}=X$, and if $n$ composite, then set $X$ would have more than four integers, too, but, since we could rule out $+1,-1$, then $|n|>1$ and by above argument, $(n,x)$ can't be principle. Am I right? @AlessandroCodenotti

@AlessandroCodenotti oh! just saw

Alessandro Codenotti

Yep

Silent

Thank you very much!

Alessandro Codenotti

So the problem in infinite dimension is that the space of bilinear maps $V_1\times V_2\to\Bbb R$ isn't the same as $V_1^\ast\otimes V_2^\ast$?

MatheinBoulomenos

yes

Alessandro Codenotti

My dislike for this definition has suddenly increased

MatheinBoulomenos

16:58

in finite dimensions you have a trick to get back a space from its dual (just take the dual again) and that is used to define tensor products here, since you can describe the dual via bilinear maps

Alessandro Codenotti

In infinite dimension you can have non isomorphic preduals so there's no hope for anything like that to work

MatheinBoulomenos

given enough category theory, you might say that the space $V_1 \otimes V_2$ is "the" vector space such that for any vector space W, linear maps $V_1 \otimes V_2 \to W$ correspond naturally to bilinear maps $V_1 \times V_2 \to W$, that determines $V_1 \otimes V_2$ up to natural isomorphism, works for infinite dimensions and seems to be kind of in the same spirit as that weird definition

Alessandro Codenotti

Well the author is mostly interested in $\Lambda V^\ast$ at the end of the day, I guess I'll just look up a construction of $\Lambda V^\ast$ as a quotient of the tensor algebra done with standard definitions and come back for the de Rahm stuff after learning about differential forms somewhere else

loch

17:20

I think it's a good idea to understand how this definition of $\wedge V^*$ allows you to identify $k-$forms as alternating $k$ multilinear maps (something I found confusing before)

Alessandro Codenotti

Which definition is "this definition"?

loch

i think lee's smooth manifolds uses this to define the exterior algebra (he defines it as a certain subspace of the tensor product)

the upshot (which he doesn't mention because i don't think he defines the exterior algebra by quotienting) is that this subspace maps isomorphically to the quotient

Perturbative

@AlessandroCodenotti What book are you reading?

Alessandro Codenotti

Some notes written by a professor at my uni for a course @Perturbative

Perturbative

Ahh I see

Abcd

17:36

Hi @LeakyNun

never mind.

Rajesh Dachiraju

18:22

1

Q: What do you call this property involving a function between two complete metric spaces?

I have a notion, for which I am not able to find any reference name, as I am not that familiar with these concepts. Please help me by pointing to a definition for the below scenario. Is there a name for the following property of the setup? There is a an onto function $e : A \to B$, $A$ and $B$ ...

real-analysis analysis functions metric-spaces

Alessandro Codenotti

18:45

Hmm that's a weird property

It's not satisfied by all continuous functions to begin with

Daminark

Hey everybody

Maurizio Moreschi

Hi everybody! @AlessandroCodenotti , yes I am Italian but I live in the Netherlands

MatheinBoulomenos

hi @Daminark

Alessandro Codenotti

I see, did you study in Italy if you don't mind me asking?

Sup @dami

Maurizio Moreschi

I did my bachelor in Padua

Alessandro Codenotti

18:52

Did you end up in the Netherlands through the ALGANT program?

Maurizio Moreschi

Yes, exactly. Are you considering applying for it?

Rajesh Dachiraju

@AlessandroCodenotti I can state it in a bit shrinked form, but with essentially same meaning.

Alessandro Codenotti

@MaurizioMoreschi I considered it but ended up not applying since it's focused in areas I'm not really interested in (I just finished my bachelor in Trento and I'll do my master in Bonn)

Rajesh Dachiraju

Given a onto function between two complete metric spaces $A$ and $B$, for any point $b \in B$, every point in the pre-image of $b$ is a limiting point to every set that is a pre-image of some punctured neighbourhood of $b$.

Maurizio Moreschi

@AlessandroCodenotti It makes sense, it is a very specific program. Is the master in Bonn also thematic? I have to say, I studied in Regensburg for one year and I didn't really like their approach that much, but I have no idea if it also applies to other German universities.

Alessandro Codenotti

19:03

What do you mean with thematic? A classmate of mine did her erasmus in Regensburg and had mixed opinions

Maurizio Moreschi

I mean specifically focused on a certain field

Alessandro Codenotti

Kinda, they divide their courses into 6 main areas and you have to pick a major one to specialize in and a minor one (also a third one to be precise, but you can do just one or two courses from the third area just to fulfill credits requirements). So it is focused on a field but you get to chose which one you want to focus on

Shaun

20:20

Hey everyone :) I'm struggling with my mental health at the moment, so I'm reluctantly considering dropping down from my PhD programme in Mathematics to the MRes programme instead, meaning I'd graduate - if all goes well - in 2019, rather than 2020 or even later, so that I'll have time to concentrate on my health before starting a PhD elsewhere perhaps. I've wanted a PhD for about a decade now but I'm really struggling to do the work. Does anyone have any advice?

A. Hendry

Hi everyone.

I was wondering if I could get some help with a problem. It seems to simple at the outset, yet is driving me mad.

I posted the question here: math.stackexchange.com/questions/2864868/…

Shaun

Hi @A.Hendry. That happens a lot in Mathematics. Don't worry.

A. Hendry

I'm sure!

Even knowing if there is no way to answer this problem would help. At least then I could move on to my next problem!

haha

@Shaun would you happen to have a moment to take a peek?

Shaun

I just did. I specialise in group theory, however, so I won't be able to help, I'm afraid; I'm sorry.

loch

@Shaun I think your (mental/physical) health should be your number one priority

6

A. Hendry

20:30

Oh no worries! Thank you for trying! And sorry to hear about your health, btw. I'm hoping you get some much deserved rest and peace

@loch +1

Abdullah UYU

@Shaun I'm not the person that can give you an advice about the PhD programme process etc., but I wanted to say that the health is way more important than anything else. Secondly, according to Maslow's hierarchy of needs, love/belonging is fairly important. For me, its simply the family.

And I think, usually it's likely to be the source of the problems.

A. Hendry

@loch and @AbdullahUYU any chance you could take a crack at my problem?

Forgive my pertinence. I'm desperate and am reaching out to as many people as I can.

Abdullah UYU

@A.Hendry No problem but I'm not even familiar to those you've written, sorry.

Shaun

That's why I'm considering the change, @loch and @AbdullahUYU. It's still scary though.

Thank you, @A.Hendry.

A. Hendry

20:45

@AbdullahUYU I can send you the link to the question if you like

Abdullah UYU

Well you can, but don't wait a miracle. I don't want to waste your time after all.

A. Hendry

Of course. Any help is appreciated. Certainly not expecting a miracle.

The link is math.stackexchange.com/questions/2864868/…

user193319

Is it true that the topological boundary of a set $E$ equals $\overline{E}- \mbox{int }E$?

Abdullah UYU

@A.Hendry Ah, you meant the question on se. I've took a look of course, I wrote after reading it.

A. Hendry

@user193319 I don't know much topology in regards to set theory. Sorry. Also, just fyi, I'm not able to render your Latex. It may be just an error on my end.

@AbdullahUYU yes, the question on se. thank you!

user193319

20:56

Hmm...Seems to be rendering on my computer.

A. Hendry

@user193319 did you have to set anything up in particular to render the Latex? I'm suspecting the error is on my end

user193319

math.ucla.edu/~robjohn/math/mathjax.html

A. Hendry

@user193319 That was it! Renders perfectly. Thank you!

user193319

No problem!

Nicholas Roberts

Can anyone answer me this? Suppose $F = F_1 - F_2$ is continuous on the real line with $F_1$ also being continuous on the line. Will this imply $F_2$ is continuous as well?

user193319

21:07

@NicholasRoberts Yes.

Nicholas Roberts

Can you give me a quick proof? or quick reasoning

been struggling to prove this for like an hour now

Alessandro Codenotti

@Nicholas Hint:the sum of two continuous functions is continuous

user193319

Are you using the limit definition of continity?

Nicholas Roberts

im using epsilon-delta definition, yes

yes i know the sum of two cts functions is cts, but im asking a partial converse to that

user193319

Since you have that fact available, Alessandro gave you a good hint.

@NicholasRoberts Do you also have that multiples of continuous functions are continuous?

Nicholas Roberts

21:10

yes, i have all arithmetic properties of cts functions

OK, will use the hint. THank you guys

Alessandro Codenotti

@NicholasRoberts Sum $F_1-F_2$ and $F_1$, after changing a sign

Nicholas Roberts

Just got that @AlessandroCodenotti, thank you

Was goin crazy with eps/delta, didnt think to exploit arithmetic properties.

Shaun

21:25

I'd like to reiterate my plea for advice.

I've asked the same question in Ivory Tower, the main chat room of academia.stackexchange.

quallenjäger

How is uniform convergence formulated in $R^n$

?

Do I have to specify a norm on $R^n$

Nicholas Roberts

euclidean metric in R^n

A. Hendry

@Shaun I haven't gone for a doctorate before. I only received my bachelors in mechanical engineering and then went straight to work. That being said, is there any chance you could take a "sebatical" and recharge before pursuing any further?

quallenjäger

So I need to specify a norm on $R^n$?

A. Hendry

Everyone, I believe I found a good stepping stone to solving my problem on se.

I'll post on there.

Nicholas Roberts

21:37

Note that in $\mathbb{R}$ we use absolute values when defining convergence of sequences. This is simply the euclidean metric in dimension 1. For higher dimensions, i believe you just use the euclidean metric

$||p - q|| = \sqrt{(p_i - q_i)^2}$

with a sum over i inside the square root

Alessandro Codenotti

Is $\Bbb R^n$ meant to be the domain or the codomain of the functions whose uniform convergence you want to talk about?

You only need a metric on the codomain to talk about uniform convergence

Rehi @Mathei

MatheinBoulomenos

rehi @Alessandro

Shaun

@A.Hendry Yes, I could take a Leave of Absence. That might be a better option. Thank you.

A. Hendry

@Shaun I'm so glad to hear it! As it were, I resigned from my position at my previous company last Wednesday and am taking time to pursue new options. You are not alone and time is on your side.

Daminark

22:11

Rehi @Mathein and @Alessandro

Alessandro Codenotti

Rehi @Dami

MatheinBoulomenos

rehi @Daminark

Daminark

How's everything going?

MatheinBoulomenos

pretty well

today was the exam for the intro NT class I TA'ed

seems like it went pretty well, although it's not fully graded yet

Daminark

I see, any particularly interesting questions?

MatheinBoulomenos

22:15

nah, it was pretty easy

when I took that class, the exam was harder

there were a few interesting answers though

one guy claimed that there are 36 primitive 8th roots of unity in $\Bbb C$

why 36?

Daminark

Nice, and lol yeah different professors give tests at different levels of quality

Wait 36?

MatheinBoulomenos

also classes are now over, yesterday were the last classes

still have some exams next week though

EnjoysMath

Do you like groups?

MatheinBoulomenos

ah and I found out that I won't be able to finish my bachelor next semester as planned

EnjoysMath

:math.stackexchange.com/questions/2865607/…

Daminark

22:19

Rip, why?

MatheinBoulomenos

I have to take one of four certain classes

none of which is offered next semester

although the syllabus says there should be at least one of those offered

Daminark

Which are the 4 options?

MatheinBoulomenos

all suck

but I have to take one

numerical analysis, statistics, linear optimization, nonlinear optimization

Daminark

22:40

Rip

Well, if none are offered next semester, are you just gonna take the semester off, take fun classes instead and then try to take one of these after that, or what?

MatheinBoulomenos

yeah I'll take fun classes

Daminark

Nice, anything in particular catch your eye?

MatheinBoulomenos

algebraic geometry, Galois representations and Galois cohomology (this will basically be ANT3, although it's not called that), analytic number theory, complex manifolds

and there's a seminar on a special case of Langlands

(for $\mathrm{GL}_2(\Bbb Q_p)$)

Daminark

Oh shit that all sounds like a lot of fun

(Well, as for the analytic number theory... if it's along the lines of modular forms I guess, though what we did in complex analysis was heavy on stuff like integral estimates, which I didn't like quite so much)

(Sad part is, there was barely any reference to complex analysis in that part, except showing that Dirichlet series converge in a half plane and are holomorphic there)

MatheinBoulomenos

having it focus on modular forms would be redundant since we have a separate class for those

it will have a focus on L-series which is fine by me

we already did some Dirichlet series in the modular forms course

we proved an isomorphism between spaces of certain L-functions (so Dirichlet series satisfying some conditions, e.g. a functional equation) and modular forms which was pretty cool

in the analytic number theory course we're going to prove stuff like the prime number theorem, Dirichlet prime number theorem etc.

Daminark

22:53

I see, nice

MatheinBoulomenos

ah and we will also do something with quadratic forms

Daminark

Huh, neat

Adeek

23:18

@MatheinBoulomenos

hi

@MatheinBoulomenos I would like to discuss something in Hartshorne

hi @Daminark

MatheinBoulomenos

@Adeek hi

I haven't read Hartshorne and I've had some scotch, so I'm not sure if I'm so helpful right now

but I can try

Adeek

@MatheinBoulomenos page 73

let me know when your there

Daminark

Hey @Adeek!

Adeek

I don't understand why the maps on the stalks $f^{\sharp}$ is $\phi_p$

@Daminark how are you

Daminark

I'm alright, how about you?

Adeek

23:24

I am good as well. Finished my master thesis. It is pretty nice. It is like 200 pages of hardcore things.

:D

so I am happy about that.

Now I am just preparing for my PhD I am gonna work on algebraic cycles and mirror symmetry.

geocalc33

How do you write $\mapsto $ without the line in the front

loch

@Adeek exciting stuff!

Adeek

yeah @loch

geocalc33

nvm!

Adeek

I am pretty excited. First semester I will just do research. In my second semester I will take number theory and mirror symmetry.

Daminark

23:30

I see, how long is the PhD gonna be?

Also are you staying in the same place, with the same advisor?

loch

sometimes the fact that these things are used in string theory allows me to pretend 'im doing things that's related to physics'

Adeek

I will be same advisor with another one.

@loch I would like to eventually do string theory as well :D :D that would be so cool.

Daminark

Lol I've heard that trying to bring physics into the picture somehow is supposed to be good for getting grants

Adeek

my eventual goal is to work in algebraic geometry and mathematical physics.

@Daminark yeah you will get more money

loch

i also would like to learn a bit about the physics side of the story eventually (although my physics background is... lol)

Adeek

23:33

people like fancy stuff like explaining the universe and sh!t like that.

MatheinBoulomenos

@Adeek ah, that's just commutative algebra so I can probably help. so for basic opens, the induced map on sections is just the map induced by localizations $A_x \to B_{f(x)}$. Now $A_{\mathfrak{p}} = \varinjlim_{x \not \in \mathfrak{p}} A_x$ for a prime $\mathfrak{p}$ and what you're trying to show is that the map that you get out of the universal property of a colimit is the localized map $A_{\mathfrak{p}} \to B_{\mathfrak{q}}$ (where $f^{-1}(\mathfrak{q})=\mathfrak{p}$.
You can just check this by the explicit construction of directed limits of rings. Just think of the directed limit as a …

Adeek

@loch mine too. I know classical mechanics 1,2,3, and 4 but no QM GR or those things.

Daminark

Lol I took a year of physics first year and that was it, and I didn't learn it well because it required a lot more geometric intuition and vector calc than I had

MatheinBoulomenos

it's just a matter of writing down all the constructions and once you've written everything out, you have the result @Adeek

direct limits (or filtered colimits, if you wish) have pretty concrete (and useful!) descriptions in a lot of important categories: sets, rings, groups, modules etc.

they can always be constructed as a quotient of a disjoint union

(it's crucial that the index category is filtered (or the index poset is directed in more classical terminology) for that)

Adeek

@MatheinBoulomenos wait we defined the map f as $f \circ \phi_p$

what I don't understand is that once we take direct limit

that somehow transforms into $\phi_p$?

MatheinBoulomenos

23:38

how do you get a map on stalks (just thinking about universal properties)? notation as above. For $x \not \in \mathfrak{p}$, you have maps $A_x \to B_{f(x)} \to B_{\mathfrak{q}}$, taking all those maps for different $x$ gives you a cocone and that induces a map $A_{\mathfrak{p}} \to B_{\mathfrak{q}}$

I'm just saying that this induced map has a concrete description based on a concrete description of filtered colimits of rings

Adeek

hm 1 sec let me think

wait so the original map that we defined is

$f \circ \phi_p$ ?

MatheinBoulomenos

ah well, this doesn't seem to be what Hartshorne does, he does the maps on stalks first

Adeek

so it is $\phi_p \circ f$ ?

yeah

wait that doesn't make sense

$f^{\sharp} : \mathcal{O}_{Spec(A)}(V) \rightarrow \mathcal{O}_{Spec(B)}(f^{-1}(V))$

yes so it is indeed $\phi_p \circ f$

because elements of the domain are $s : V \rightarrow \bigcup_{x \in V} V_x$

and then we map those to $\phi_p \circ f(s) : f^{-1}(V) \rightarrow \bigcup_{x \in f^{-1}(V)} V_x$ ?

MatheinBoulomenos

oh okay he uses the espace etale construction

Adeek

yeah so what I said above makes sense

MatheinBoulomenos

23:51

I think it's easier to just define it on basic open sets, but this works, too

Adeek

okay now taking direct limit then things collapses to $\phi_p$ agree ?

yeah ok I see what you mean now

okay I agree..

MatheinBoulomenos

yeah

Adeek

actually one way to see this intuitively/geometrically is that taking direct limit means looking at what happens locally which is precisely given by $\phi_p$

the f gives a global picture

thanks @MatheinBoulomenos

MatheinBoulomenos

yes that's the intuition (for morphisms induced on stalks in general), of course you still have to check that the algebra matches that intuition :)

@Adeek np

Adeek

yeah

@MatheinBoulomenos to improve my algebra skills for research in the fall I am planning to solve Matsamura, Allufi, Reid,Michael atiyah, Allufi, and fields and galois theory by Patrick

I need to have very fluid algebra skills

MatheinBoulomenos

23:57

these are a lot of books, but sounds great! being the algebraist I am I always enjoy working through such books

Adeek

nice I will come over to discuss or email you if I have anything :)

MatheinBoulomenos

I'm looking forward to it, I'm always glad to discuss some algebra

Adeek

:D me 2 I like discussing math in general :D. In the fall I am planning to do a topological K-theory introduction in the student seminar.

I am excited for this

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$Mathematics$ Mathematics