Mathematics - 2017-10-10 (page 1 of 5)

Lucas Henrique

00:08

@TedShifrin what do you mean by $T$?

Like, a... bijective map?

Semiclassical

A mapping from R^2 to R^2

Lucas Henrique

00:21

can I say that it's bijective?

Semiclassical

well, is $x\mapsto ax$ a bijection?

Lucas Henrique

I'm not acquainted with that notation

Semiclassical

$f(x)=ax$, then.

Lucas Henrique

But it looks like a line, so I'd say yes

Semiclassical

right. and same for $y\mapsto by$.

Lucas Henrique

00:23

@Semiclassical exactly what I was thinking. yes.

Semiclassical

so T is a bijection in each argument, and that makes it a bijection

ALannister

00:40

Back with another question. And I've had a downvote.

-1

Q: Explain how the partition $A_{i}$ is chosen to make this chain of equalities true

If there is only positive measure at a discrete point $\omega_{i} \in \Omega$, $i = 1, 2, \cdots$, then for any function defined on $(\Omega,\mu)$, $$\int f d\mu = \int_{\Omega}\sum_{i=0}^{\infty} f(\omega) I_{A_{i}}(\omega) d\mu = \sum_{i=0}^{\infty} f(\omega_{i})\mu(\{ \omega_{i} \}) = \sum_{i=...

I mean noboy is even looking at it!! What's so wrong with this question??

Could somebody please at least upvote it so people don't keep ignoring it???

mdave16

00:58

i think your question is fine

although you could put it in the constructive criticism room if you wanted to improve it (i don't know how effective it is)

ALannister

I think someone was on a downvoting rampage, because a bunch of other questions asked around the same time also had downvotes.

I just hope that I will get some answers soon!

mdave16

anyone up for a brainteaser?

Simply Beautiful Art

Depends if I can brain

mdave16

if i have a 3 by 3 by 3 cube, what's the minimum number of cuts to turn it into 27 unit cubes? (i can rearrange after i cut)

and what is the simplest argument for showing it (I know 2 answers, i want to see if someone has a different answer)

hows your calc course going btw?

Simply Beautiful Art

@mdave16 Slowly, but still not dead xD

mdave16

01:13

where is the room for it? i couldn't find it

Simply Beautiful Art

$Mathematics$ Calculus and analysis

For questions about calculus, real analysis, functional analys...

@mdave16

mdave16

I have just learned that you can star a room

technology, everything is possible now

2

ALannister

01:33

I wear out my welcome really easily when people answer my questions.

Because I ask TONS of follow-ups, and most people on MSE can't stand someone as disgustingly needy as I am.

I'm like dirt on the bottom of your shoe.

If people were just more forthcoming in the beginning, maybe I wouldn't have to bother them so much.

Because get used to it, the unwashed like me are not leaving mathematics any time soon.

Jorge Fernández

04:28

Some people don't like brats that want everything spoonfed

Daminark

05:24

Hey everyone!

@Jorge look there's no need for aggression

Jorge Fernández

that isn't aggression

Daminark

Calling people brats is kinda rude

Jorge Fernández

I think being kinda rude is way more acceptable than being agressive

Daminark

I mean I'd rather have 2 years to live than 6 months, that isn't saying anything. You can tell that to people without being insulting

Jorge Fernández

?

can you check what you wrote? I didn't understand

Jasper

05:35

Hello @JorgeFernández and @Daminark.

Jorge Fernández

I understood the first sentence, but not the second one

@Jasper Hello, how are you doing?

are you better?

Jasper

@JorgeFernández Well, recently, I went back to meds again, and I will be seeing a therapist soon as well. I don't know what will happen to me. I'll just keep trying.

@mdave16 Starring a room is nothing. You can do way cooler things, lol.

Jorge Fernández

I don't know what will happen to me either. Of course I understand your struggle is very hard, even though I have never suffered something like that. Stay strong.

Daminark

You can tell people that they shouldn't be incessantly posting without calling them brats

Jorge Fernández

@Daminark Oh, I understand now

Daminark

05:40

Like if only to avoid a fight, and also to not make people unnecessarily upset, that's much preferred

Jasper

@Daminark I hope the first sentence is not true, just an example.

Jorge Fernández

is avoiding a fight desirable in general?

Daminark

Oh yeah it's just an example, I think/hope I'm not terminally ill

In general maybe not but the vast majority of the time, yes. It only serves to hurt and rarely actually gets you want you want

Jasper

@JorgeFernández I miss your photo, lol.

@Daminark I thought you were going to show your face, lol.

Hello @LeakyNun how is school so far?

Leaky Nun

@Jasper not bad

Jasper

05:45

@LeakyNun Do you have a favourite English dictionary?

Leaky Nun

@Jasper I don't

Jorge Fernández

@Daminark I try to say my opinion as truthfully as possible. Of course, I wont do it if I think there will be a large enough negative repercussion.

Daminark

You can definitely be truthful, like expressing disapproval about about a practice is absolutely okay, and in fact encouraged

But being mean just doesn't do any good

Jorge Fernández

I think you are being condescending right now for example

"expressing disapproval about about a practice is absolutely okay, and in fact encouraged"

Daminark

That isn't meant to be condescending at all

Like I think we should be open if we think there's something not good. I also am less than totally happy about, say, homework questions with no effort involved

I don't think the case above is of the sort but that's another discussion I won't open

My point is that it's very much a good thing to mention the stuff, just don't do it rudely

Jasper

05:51

There used to be plenty of unicorn avatars on SE, because someone created a unicorn pic generator.

Jorge Fernández

wordreference.com/es/translation.asp?tranword=rude

Daminark

Anyway I think I've made my case now, pushing this further probably won't help anyone

Leaky Nun

non-real roots of a rational polynomial must be related via conjugates?

Jorge Fernández

@LeakyNun what is a rational polynomial?

Leaky Nun

@JorgeFernández a polynomial with rational coefficients

Jasper

05:54

@LeakyNun The complex conjugate of a non-real root of a real polynomial is a root.

Jorge Fernández

the statement is true for any real polynomial

Leaky Nun

@Jasper and they must be different from itself

alright

$X^5-X-1$ factors into 3 irreducible polynomials in $\Bbb R[X]$?

Jorge Fernández

if $\alpha$ is a root of $P$ with multiplicity $m$ then $\overeline \alpha$ is also a root with multiplicity $m$ of $P$.

Leaky Nun

@JorgeFernández thanks

Jorge Fernández

Yes, it does, the number of factors is $roots+(deg-roots)/2$. In your case $x^5-x-1$ has one root. To see this notice the derivative is $x^4-1$. Use this to look at what happens in each of the ranges $(-infty,1),[-1,1](1,infty)$

Leaky Nun

05:59

@JorgeFernández right, I got that part already

Jorge Fernández

well, I got the derivative wrong

but just check in the three ranges in which the derivative only has one sign

in fact its easy to see that any root must be at least $1$. And $x^5-x-1$ is clearly monotone in that range.

so the number of roots is $1$.

Leaky Nun

So the roots are $a,b,\overline b,c,\overline c$ where $a$ is real?

Jorge Fernández

yes, that is certainly true for suitable $a,b,c$.

You can generalize this to other field extensions by the way

if $F$ is a field, $E$ is an extension of $F$ and $P$ is a polynomial on $F$ with root $\alpha\in E$ then the minimal polynomial of $\alpha$ divides $P$. It follows that every root $\beta\in E$ of $P$ is also a root of $P$.

Silent

@LeakyNun, how to pick a point so that the ball of radius $ϵ$ centered at $x$ has some points in $A$ here?

Jorge Fernández

the minimal polynomial of $c$ is $(x-c)(x-\overline c)$

@Silent ??

Silent

06:12

@JorgeFernández I can't find a way to show that every point on boundary of a closed ball in Euclidean metric space is a limit point to open ball with same radius and center.

Leaky Nun

well we're on a Euclidean metric space

the nicest space possible

Jorge Fernández

is it fine by you if we assume the ball has radius $1$ and center at origin?

and the point $a$ we want to show is in the boundary is $(1,0,0,\dots,0 )$?

or does that sound like cheating?

Leaky Nun

well, we want to prove that $B(x,\varepsilon) \cap B(z,d(x,z)) \ne \varnothing$

i.e. $\exists a[|a-x| < \varepsilon \land |a-z| < |x-z|]$

drawing a graph will certainly help your intuition

Silent

@LeakyNun Why $B(z,d(x,z)$?

Leaky Nun

@Silent that's the original ball

Silent

06:18

oh

Leaky Nun

I claim that $a=z+\left(1-\dfrac\varepsilon{2\|z-x\|}\right)(x-z)$ is the witness

Silent

Thank you very much! I will check it.

I tried using some $\lambda z+(1-\lambda)x$ argument.@LeakyNun

Leaky Nun

@JorgeFernández por que $x^5-x-1 \equiv (x^2+x+1)(x^3+x+1) \pmod 2$ significa que el Galois group contiene un cyclo de typo 2,3?

Jorge Fernández

what does the author mean by a cycle of type 2,3?

I don't know, what is the source?

Leaky Nun

(12)(345)

under suitable renaming

See Example 4.17 here

Jorge Fernández

06:49

hmm, Sorry I don't remember that stuff very well

Leaky Nun

@JorgeFernández never mind

I'm getting strange results in $\Bbb F_3$, namely $1+2\sqrt[4]2+2\sqrt[4]8=0$ @JorgeFernández

Leaky Nun

07:45

0

Q: Factoring $x^3-3x-1\in \Bbb Q[x]$ in terms of a unknown root

So $f=x^3-3x-1\in \Bbb Q[x]$ is irreducible by rational roots test. The notes say that let $\alpha$ be a root, and then we get the roots $\alpha^2-\alpha-2$ and $2-\alpha^2$. I feel like an idiot being unable to solve for these two, no idea why I can't get them. So letting $\alpha$ be a root ...

abstract-algebra polynomials ring-theory irreducible-polynomials

This is dank

What is $\operatorname{Gal}(\Bbb Q(\ln 2)/\Bbb Q)$?

@TobiasKildetoft

Silent

@LeakyNun This was awesome! I could see that $a=x-\frac{\epsilon(x-z)}{2||z-x||}$, where by $-\frac{(x-z)}{||z-x||}$, we are travelling unit distance in the direction from $x$ to $z$, and by $\epsilon/2$ we are remaining in the ball centered at $x$, right?

Leaky Nun

@Silent that's the correct intuition

Silent

So quick! @LeakyNun thanks.

Alessandro Codenotti

@LeakyNun that's the same as $\text{Aut}(\Bbb Q((X)))$ right?

Leaky Nun

@AlessandroCodenotti then what is it?

@AlessandroCodenotti My guess is $\Bbb Z_2$...

Leaky Nun

08:31

Must $\operatorname{Gal}(\Bbb Q(\sqrt t)/\Bbb Q(t)) = \Bbb Z_2$ be true for any $t$?

anon

true as long as t is not a perfect square

@AlessandroCodenotti $\Bbb Q(\ln 2)$ would be isomorphic to $\Bbb Q(X)$ not $\Bbb Q((X))$ no?

in which case the automorphism group should be ${\rm PGL}(2,\Bbb Q)$ acting by mobius transformations on $X$

usually the term "galois" is reserved for algebraic extensions

Leaky Nun

@anon wat

anon

all automorphisms of $\Bbb Q(X)$ are of the form $f(X)\mapsto f((aX+b)/(cX+d))$

Leaky Nun

I understand what you mean

Alessandro Codenotti

08:51

@anon indeed, too many brackets there

Jasper

09:02

You can perform a long division first @Abcd.

Abcd

@Jasper I had to find limit...

I used L Hospital Rule and did it.

coz limit of a constant is the constant itself.

I found it's double derivative, then used the above property.

One of my messages isn't sent.

How do I decompose $\dfrac{n}{(1+n^2+n^4)}$ into partial fractions?

It's not possible acc to me.

s.harp

09:41

The following appears to be a general case of the hairy ball theorem:

Let $n$ be uneven, if $U\subset \Bbb R^n$ is open and bounded with $0\in U$, $f\in C(\partial U, \Bbb R^n-\{0\})$ then there must be an $x\in \partial U$, $\lambda\in\Bbb R, \lambda\neq0$ with $f(x)=\lambda x$

The hairy ball theorem follows since it tells you that you cannot have a non-zero vectorfield on $S^{n-1}$ that always lies in the tangent bundle (at one point it must lie in the normal bundle)

Does this theorem have a name? I wnat to find a proof, does it follow from Brouwer?

Balarka Sen

@TedShifrin Yeah, this would not be a geodesic below.

Alessandro Codenotti

Hi @Balarka

Balarka Sen

Hi @Alessandro

@LeakyNun lmao learn some actual galois theory instead of shitposting galois theory all over the chat

4

Alessandro Codenotti

We did some interesting stuff about group characters in Galois Theory today

And the Hilbert basis theorem in commutative algebra

Balarka Sen

09:58

@AlessandroCodenotti Ah, tell me about it

Hilbert basis theorem is very cool. I think about it as a theorem about algebraic varieties

Leaky Nun

@BalarkaSen I was practising it in the chat

Balarka Sen

ur practicing something while learning it from wikipedia? lol

Leaky Nun

13 hours ago, by MatheiBoulomenos

I know you didn't ask me, but you could try to compute the Galois group of $X^6-2tX^3+1$ over $\mathbb{Q}(t)$

@BalarkaSen come on

I'm not learning it from wikipedia

Balarka Sen

then why are you spamming people requesting galois theory problems lolkek

Leaky Nun

@BalarkaSen I read this pdf

@BalarkaSen because I need to practise?

Balarka Sen

10:05

get a textbook

Leaky Nun

I will

Balarka Sen

lots of problems to practice on in textbooks

Alessandro Codenotti

@BalarkaSen We didn't do that much, we showed that distinct characters are linearly independent and then used that fact to show things about $|E:E^G|$ where $E$ is a field, $G$ is a set of automorphism of $E$ and $E^G$ is the subfield fixed by all elements of $G$

Balarka Sen

@Alessandro mmm. I barely remember these

Oh, I see

You're trying to get one direction of the Galois correspondence to work

Given a group of automorphisms, produce a field extension with that Galois group

Alessandro Codenotti

We just defined what a Galois extension is, but I think that's where we're headed

I'll find out in the next lecture I guess

Balarka Sen

10:14

I have forgotten how you characterize Galois extensions. I think the extension K/F is Galois iff |Gal(K/F)| = [K : F], but there are actual descriptions of K/F

I think normal + separable or something

Leaky Nun

@BalarkaSen give me a group

@BalarkaSen yes

Balarka Sen

@LeakyNun no, how about i put you on ignore instead

Leaky Nun

@BalarkaSen :c

FreeMind

@BalarkaSen Hey man!

Balarka Sen

@FreeMind Hi

How's it going

MatheiBoulomenos

10:16

Hi there

FreeMind

@BalarkaSen I am fine and you?

Balarka Sen

Hi @MatheiBoulomenos... you're the complex analysis guy, right?

@FreeMind More or less quite good.

MatheiBoulomenos

Well, I did ask questions about complex analysis here once, but I wouldn't want to be reduced to a "complex analysis guy" :P

FreeMind

@BalarkaSen Good. Are you still obsessed with Algebra stuff? Or you have changed your study subject?

Balarka Sen

@MatheiBoulomenos Oh, that is exactly what I meant. I see you also know a lot of Galois theory, from reading the transcript.

Alessandro Codenotti

10:19

@BalarkaSen that's how we defined it

Leaky Nun

@MatheiBoulomenos oh hey

Balarka Sen

It's good to have such a regular, because I and @Alessandro are learning Galois theory

Leaky Nun

thanks for your problem yesterday

MatheiBoulomenos

Yeah, actually algebra is more my thing than complex analysis

Balarka Sen

hahah algebraists are taking over the chat

I'm from the topology fortress

so I don't know shit about algebra

Alessandro Codenotti

10:20

Complex analysis is algebraic topology with a (not even that good) disguise

Balarka Sen

@AlessandroCodenotti Indeed, that's how I think about it

Alessandro Codenotti

That's enough to call it algebra :P

MatheiBoulomenos

Well, covering space theory is really similiar to Galois theory

Balarka Sen

Normal + separable never was much intuitive

@MatheiBoulomenos Yep!

Tobias Kildetoft

@BalarkaSen Also because the whole separable thing is automatic in so many cases

Leaky Nun

10:21

@MatheiBoulomenos do you have any problem for me?

MatheiBoulomenos

You could try to compute the Galois group of $x^5-2$ as a semi-direct product

Alessandro Codenotti

@BalarkaSen I have 3 algebra courses and no geometry/topology ones this semester :/

MatheiBoulomenos

@AlessandroCodenotti that sounds like a dream

Balarka Sen

frowns

Alessandro Codenotti

Next semester it'll be even with differential geometry and algebraic number theory

Leaky Nun

10:25

@MatheiBoulomenos $\Bbb Q(\sqrt[5]2,\zeta_5)$, hmm

MatheiBoulomenos

@BalarkaSen the way I think about normal+separable is that both conditions ensure in some sense, that we have "enough" automorphisms: separable ensures that there are as much homomorphisms into an algebraic closure as possible and normality ensures that all these homomorphisms into the algebraic closure restrict to automorphisms of the field

@Alessandro Ah, I'll be taking ANT next semester, too! I'm really looking forward to it

Balarka Sen

Oh wow, that sounds like an interesting description

Tobias Kildetoft

@MatheiBoulomenos Interesting. I would have thought it was the other way around with those conditions

Balarka Sen

K/F is normal if every polynomial over F splits completely into linear factors in K if it has one root in K, right?

MatheiBoulomenos

Yes

But there is an equivalent description in terms of homomorphisms

$K/F$ is normal if for every $F$-algebra homomorphism $\varphi:K \to \bar{F}$, we have $\varphi(K) = K$

Leaky Nun

10:31

@MatheiBoulomenos it's $\Bbb Z_5 \rtimes \Bbb Z_5^*$ innit

Balarka Sen

Hmm.

MatheiBoulomenos

@LeakyNun that's correct

Leaky Nun

yay

generated by (12345) and (2453)? @MatheiBoulomenos

Tobias Kildetoft

@MatheiBoulomenos Ahh, I see how you mean the descripton now. Makes sense

Balarka Sen

I think one way to compute $\text{Gal}(\Bbb Q(\sqrt[5]{2}, \zeta_5)/\Bbb Q)$ by considering the tower of Galois extensions $\Bbb Q(\sqrt[5]{2}, \zeta_5)/\Bbb Q(\zeta_5)/\Bbb Q$, which gives a short exact sequence of Galois groups $1 \to \Bbb Z/5 \to G \to \Bbb Z/4 \to 1$.

Now all I need is a section of that

Leaky Nun

10:36

@BalarkaSen I like to think of the image of the generators under the automorphism

$\sqrt[5]2$ must map to another root of $x^5-2$ and $\zeta_5$ to $\zeta_5^k$ where $k \ne 0$

Balarka Sen

sure, I'm doing it non-explicitly by invoking the Galois correspondence

Leaky Nun

sure, both work

MatheiBoulomenos

@LeakyNun Yup, $(12345),(2453)$ seems right

Leaky Nun

@MatheiBoulomenos thanks

MatheiBoulomenos

@BalarkaSen that seems like a nice approach

Leaky Nun

10:38

@MatheiBoulomenos I just used $i \mapsto \sqrt[5]2 \zeta_5^{i-1}$

Balarka Sen

$\text{Gal}(\Bbb Q(\zeta_5)/\Bbb Q)$ is already a subgroup of $G$, so I feel like the inclusion map is exactly the section

MatheiBoulomenos

@BalarkaSen I think it's not difficult to show that a section exists mostly group-theoretically. Since $G \to \mathbb Z/4$ is surjective, we can lift a generator of $\mathbb Z/4$. The order of this lift must by a multiple of $4$, but as the whole group has order $20$ and is evidently non-cyclic (as there is a non-normal subextension), the order of the lift must by exactly $4$, so we just need to map the generator of $\mathbb Z /4$ to the lift

Balarka Sen

Ah, you're right.

Clever

user84215

11:54

I want to ask a silly question (Although I believe that no question is silly).

user84215

The standard curvature tensor is a curvature tensor?

Leaky Nun

@MatheiBoulomenos anything more?

user84215

You want some exercises in Galois theory?

user84215

It seems that nobody here is interested in my silly question.

MatheiBoulomenos

12:11

@LeakyNun Let $K$ be a field and $f$ be a separable polynomial over $K$. Let $G = \operatorname{Gal}(f)$ show that $f$ is irreducible if and only if $G$ acts transitively on the roots of $f$

0ßelö7

12:54

@MathematicsAminPhysics I have no idea what you're asking

user84215

13:06

@0ßelö7 I want to know whether the standard curvature is a curvature tensor or not.

0ßelö7

@MathematicsAminPhysics I don't even know what to say

It's an algebraic curvature tensor in the sense of Besse/Berger, sure

user84215

@0ßelö7 The standard tensor curvature is the tensor curvature of the unit sphere. Right?

0ßelö7

Maybe? I have no idea. I've never seen anyone say "standard curvature"

Most likely whoever said that means the Riemann tensor

user84215

@0ßelö7 I mean the curvature we use to define the sectional curvature.

0ßelö7

then that's the Riemann tenso

what is the question then?

user84215

13:16

@0ßelö7 I have read somewhere that the standard curvature tensor is multiplied by $k$ when the metric is scaled by $k$, but the curvature tensor remains invariant under the scaling of the metric. I can not understand it.

Eric Silva

@MathematicsAminPhysics you're probably talking about the $(3,1)$ Riemann tensor vs the one you get by lowering an index cause they have the scaling properties you mentioned

0ßelö7

^this

when you raise/lower an index you change the conformal transformation properties

user84215

@EricSilva @0ßelö7 I can not understand what you mean. I know that the standard curvature tensor and the curvature tensor are the same kind of tensor, so they must behave the same under scaling (multiplying the metric by a positive scalar) the metric.

Eric Silva

can you define the words you're using. Standard curvature tensor isn't common language so if you want to expect people here to know what you're talking about you should do more work to make sure you can be understood.

0ßelö7

$R^a{}_{bcd}$ and $R_{abcd}$ are not the same "kind" when it comes to scaling

to go between then you use the object you're scaling!

user84215

13:27

17 mins ago, by MathematicsAminPhysics

@0ßelö7 The standard tensor curvature is the tensor curvature of the unit sphere. Right?

0ßelö7

I give up

Eric Silva

lol "tensor curvature" is also not standard language

user84215

I mean "curvature tensor"

Mary Star

Hello!! Suppose we have the equations $u=xy$ and $v=x^2-y^2$. How could we solve for $x, y$ (as functions of $u,v$) ? Could you give me a hint?

user84215

Hello

Eric Silva

13:30

i mean the fact that you asked "right?" indicates to me that you have no idea what the definitions are yourself, so you should probably be looking at another source to clarify your confusion before asking people here

user84215

I have to go.

Martin Sleziak

13:44

@AlessandroCodenotti Functional analysis chat room does not have too much activity. So it will be nice if you occasionally stop by. (But it seems that there are more people who are asking questions in that room. than people who are able to answer them.

If you look at the starboard or bookmarked conversations, you can see that most of the discussions were about rather elementary stuff.

@MatsGranvik Wikipedia says that there are numbers which are called strong prime numbers. I am not sure whether there are also stable primes numbers. (If yes, it is natural to ask whether they can be strong a steble.

Since you mention British prime minister - are you from UK?

Mats Granvik

14:01

@MartinSleziak No I am not from the UK. I spent a year at Imperial College London back in the year 2000-2001 and for less than a year I attended a British elementary school as a child in Riyadh Saudi-Arabia, when I was 7 years old. That joke about the prime minister was my variation of something I saw in the comment section of a youtube video by numberphile.

Martin Sleziak

14:14

I see. Now with Brexit and election not too long ago, the politics seems to be rather hot topic in UK. (It seems that it is discussed enough for the slogan "strong and stable" getting also to me, despite the fact that I have nothing to do with Great Britain.)

Faust

14:51

they changed it

it bothers me

Abcd

15:13

Stack exchange changed .... I saw it rn.

Balarka Sen

just the top bar

Abcd

Yeah, do you like it?

Balarka Sen

meh its fine

MatheiBoulomenos

it will take some time to get used to

Faust

It changed...

Balarka Sen

15:22

I have to write my ODE notes... gah

I hate LaTeXing

MatheiBoulomenos

Are these notes about solving ODEs or theory of ODEs?

Faust

Morning @BalarkaSen

Balarka Sen

@MatheiBoulomenos well, it's a reading course I'm doing on ODEs and dynamics. It's definitely more about theory of ODEs than solving ODEs, for sure

MatheiBoulomenos

Oh, that's nice. Solving ODEs is so boring

Balarka Sen

Morning @Faust

Yeah I don't like that

Faust

15:25

Odes arent that bad PDEs are a pain

Balarka Sen

I can definitely believe that

MatheiBoulomenos

theory of ODEs actually came quite up in my differential geometry class in a quite important way

Balarka Sen

the geodesic equation is an ODE :)

MatheiBoulomenos

exactly! We used theorems about ODEs like Picard-Lindelöf or the fact that the solution to an ODE smoothly depends on the initial conditions to reason about properties of geodesics

Balarka Sen

Right

MatheiBoulomenos

15:31

that kind of changed my view on ODEs

Eric Silva

sup @Balarka

Balarka Sen

Hi @EricSilva

Eric Silva

ODEs come up when ur doing comparison geometry too

But so do PDEs, which are the actual interesting things

Balarka Sen

I need to learn comparison geometry before I die

Eric Silva

@Balarka there's some really dope theorems like Schoen-Yau, 10/10 highly recommend

Balarka Sen

15:38

any textbook recommendations that I can note down for the future?

Eric Silva

Uhh I recall Peter Li's book on geometric analysis had some stuff in it and there's of course cheeger ebin which is the classic

Balarka Sen

Oh sick I have heard of Cheeger Ebin

Eric Silva

There's a little for metric spaces in bridson haefliger too

Balarka Sen

@MatheiBoulomenos Actually another important (also second order) ODE that pops up in geometry is the Jacobi equation. The solutions to that determine variations of a specific geodesic ray starting at a point "orthogonally" from itself

MatheiBoulomenos

Oh, I vaguely remember that one

Balarka Sen

15:45

Which is important because, roughly, variations of a geodesic should tell you about curvature of the manifold you are in

If you are in negatively curved manifolds if you variate a given geodesic ray a tiny bit the variated ray diverges exponentially out of the original ray

Indeed, a consequence of that picture is Hadamard's theorem. The exponential map from the tangent space of a complete Riemannian manifold with everywhere nonpositive sectional curvature is a universal cover of the manifold

BAYMAX

Hi!

MatheiBoulomenos

Wow, I knew about Hadamard's theorem, but I never viewed it like that

BAYMAX

Is spatial direction means time direction?

Balarka Sen

@MatheiBoulomenos Ya that's actually the heart of the proof of Hadamard. The point is, critical points of the exponential map are precisely points where two geodesic rays emanating from the same point (image of origin by the exponential map) bump into each other

Which only happens in positive curvature; think S^2

But doesn't in negative curvature; think H^2, which is why the exponential map is a local diffeomorphism there

It's a small walk from local diffeomorphism to covering space.

MatheiBoulomenos

Really cool. I know very little about geometry

Alex K Chen

15:53

hi all

MatheiBoulomenos

hi @AlexKChen

Balarka Sen

@Mathei Me too, tbh. There's way too much to learn

so you're into the algebraic world of math?

MatheiBoulomenos

Yes, algebra is what I can do best

Balarka Sen

cool, I suck at that

MatheiBoulomenos

I'm also interested in number theory, mostly when I can use algebra to reason about number theory

I suck at differential geometry

and analysis

Tobias Kildetoft

15:58

finally, we are starting to outnumber those geometers

Balarka Sen

oh nice, @Daminark (another chat user who is into algebra/number theory) would be excited to know that then

hahah I can't do geometry/analysis either, just topology

$Mathematics$ Calculus and analysis

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$Mathematics$ Mathematics