Just ask; don't ask to ask.

symbolab.com/solver/step-by-step/…

hey@Semiclassical u saw the room?

this is how it looks like

Hi, @AlessandroCodenotti

Is every group the fundamental group of a topological space?

hi pals

@Alessandro Yes.

Just take a bouquet of circles, which give the free groups as fundamental groups

Pick a presentation of the group. Look at the wedge of circles with circles indexed by the set of generators. Attach 2-cells indexed by the relators.

It's more interesting to ask this question for "nice" spaces

Hmm, I don't see how to attach the 2 cells. Let's say I want to get $\Bbb Z_2=\langle a\mid a^2=1\rangle$, so I start with a single $S^1$

Attach $D^2$ to $S^1$ by $\partial D^2 \to S^1$ mapping $z\mapsto z^2$

In general any word (say $a_1^{p_1}a_2^{p_2} \cdots $) in $F_n$ defines a map $S^1 \to \bigvee_n S^1$ by wrapping around $a_1$ ($p_1$ times) then wrapping around $a_2$ ($p_2$ times) then etc etc

The 2-cell provides a homotopy between its boundary and a point; so you have to attach its boundary to "follow the relation"

I can see this with a SvK argument, by taking a disk in the interior of the 2-cell and it's complement slightly enlarged to intersect in an annulus, so that the generator of the intersection goes to the boundary when retracting the second open set

Ya

Visually it's exactly what Steamy said. You just follow the relations so you can push a loop along it off through the disk it bounds

Mhm, ok, I see it now

Also, related: any finitely presented group is the fundamental group of a manifold.

In fact a 4-manifold.

(in particular, you can always do it with a manifold of dimension $\geq 4$

yeah :P

Also, I may be wrong here, but any finite group is the holonomy group of a compact, flat Riemannian manifold.

(in which case the holonomy group is the fundamental group, modded out by $\mathbb{Z}^n$)

Interesting, that I didn't know.

I'm not following you anymore :P

So $F = G/\mathbb{Z}^n$

And for any finite group $F$, there exists such manifold from which you can obtain $F$ in that way

(an open problem is: find the lowest possible dimension $n$ for every finite group)

How do you construct a flat, compact Riemannian manifold out of a given finite holonomy group?

write down $\text{Hom}(G,SO(n))/\sim$, I guess, pick the permutation representation of F, and the Z^n presumably comes from Hom(pi_1,S^1) = Z^{b_1} = Z^n (which comes from some curvature bounds).

Sadly, I don't know. It's somewhere in "Bieberbach Groups and Flat Manifolds" by L.S. Charlap...

hi chat

@BAYMAX nope. but i hadn't checked my messages on the main site yet.

ohi

Is that like a representation variety?

I guess there's an obvious topology which makes it a manifold, coming from SO(n)

What does $\sim$ do?

That is a representation variety, it is almost never a manifold. I'm suggesting pick a representation (where you act on the trivial bundle) and hope it's a flat connection on the tangent bundle.

$\sim$ means mod out by conjugation.

Hmm, I see.

@SteamyRoot Some days I'm shocked we know anything, others I'm shocked we have anything left to find.

Heh

Well, it happens all too often that the difficulty of solving a problem is hard to estimate using the difficulty of the statement of the problem :P

@Fargle You could still ask for what's the class of fundamental groups of 3-manifolds. I suspect it's already done but I don't know an answer. It's probably nontrivial.

Or ask for fundamental groups of smaller subsets of 4-manifolds. Do we know what the fundamental groups of smooth complex algebraic surfaces are? shrug

@BalarkaSen It's probably still an open question whether or not 3D PD groups are exactly the fundamental groups of 3-manifolds.

@BalarkaSen In looking that up, I've found a few much smaller results.

(I suspect we can classify smooth complex algebraic surfaces upto birational equivalence, which then would answer that because the zig-zag theorem of blowup kicks in)

@MikeMiller Ahh.

Like that prime knots are uniquely determined by their knot group even though this isn't true of irreducible 3-manifolds with non-trivial boundary.

@BalarkaSen i.imgur.com/NB3vi07.png

@Fargle That's not true, either.

Left and right trefoil, right?

You need to specify the subgroup the meridian and longitude sit in.

@MikeMiller Damn this paper.

What paper?

What's an example of two primes knots with isomorphic fundamental groups but which are not mirror images of each other?

Huh, seems like you were right

ams.org/journals/proc/1980-080-01/S0002-9939-1980-0574528-5/…

@BalarkaSen I thought that was proven impossible

It's doable for links, though

Duh, this is Gordan-Luecke.

I'm embarassed.

@AkivaWeinberger I suspected, but I dunno

Ah, great

I should be studying for English literature, which I plan to do a little better on

ha ha ok@Semiclassical

@BAYMAX For the first question, it was given a first order differential equation with $f(t,y)$ being continuously differentiable on $\mathbb{R}^2$.

So, Prove or disprove if it is possible that all the solutions are even with domain on $\mathbb{R}$.

that is the question you were struggling with right!

Sort of, the hint was regarding Picard-Lindelof theorem

But in the end it does not require that,

oh

so then

Because the statement is true, an example would be $y = cos(x) + c$.

then the DE would be $y' = -sin(x)$.

what if $y = \sin(x) + c$

Not possible. It says all solutions are even.

or for countering $y = \sin(x)$

That is odd.

but we have to prove it right

we have to prove that the solutions are even is it so ?

Nope, the question actually says Give an example if the statement is true, otherwise give a proof.

ohk

so .. always solutions cannot be even right

Not true for any arbitrary first order DE

But true for my example.

yes true for your example!

hehe 15marks

just 2 lines

wait

the statement is false right?

so you have to give a proof right!

Clearing again

Okay, maybe my phrasing is bad. Prove or disprove if it is possible to have such a DE such that all its solutions are even function with domain $\mathbb{R}$.

Yes THERE EXISTS a differential equation whose all solutions are even.

take $y' = sinx$

Yea thats my answer.

Yeah

and dont forget to mention $y=-cos(x) + c$

or $y' = x^5$

yea

The 2nd question, it was given that a DE is of 1st order linear homogeneous.

Prove or disprove if it is correct that the DE has 2 linearly independent solutions that has the same point of inflection.

Ans: It is not possible. Suppose on the contrary that the 2 linearly independent solutions have the same point of inflection, then the Wronskian is 0. Which implies linearly dependence (contradiction).

Now time to prepare for Numerical Analysis midterms T_T

why wronkian 0

?

point of inflection --> 2nd derivative is 0.

Let $t_{0}$ be the point of inflection.

Ah typo in the question, its 2nd order instead of 1st.

yeah

Point of inflection implies 2nd derivative is 0 is anecessary condition right .. but still right!

Then u can rewrite $y''+p(t_{0})y'+q(t_{0})y=0$ as $p(t_{0})y'+q(t_{0})y=0$

ok

then $-\frac{p(t_{0})}{q(t_{0})}y' = y$

Also in the question, it was given that $p(t)>0 , q(t)>0$

So its fine.

then write out the wronskian

the 1st row is a multiple of 2nd row

So wronskian is 0.

ok..seems nice!

:D i overcame prof's hard question

actually how you proceed after that y' thing

Wronskian involves determinant entries being solutions and their h.o.t right!

or are you using Abels formula

The first row are the solutions at $t_{0}$, the 2nd row is the derivates at $t_{0}$.

yes

yep, i used abel's formula afterwards

wronskian is 0 at a point --> wronskian is 0 for all t

which implies linear dependence

yup!

nice!

thx :D

if i have that $\rm dist(x,A)=0$ then $\inf_{a\in A} \rm dist(x,a)=0$ can i say that $d(x,a)=0$ then $x=a$ then $x\in A\subset \overline{A}$ ?

So numerical analysis

next

Yep

All the best!

Root finding, lagrange interpolation, hermite interpolation, divided difference

Thanks :)

which book are you referring?

Richard L Burden_ J Douglas Faires-Numerical analysis

Hey everybody...

hello @SteamyRoot

The all time favourite

@draks... hello

yeah that's good one!

hi @draks...

I bought the book Friendly introduction to numerical analysis by Brian Bradie also.

For reference

can someone help me ?

me too...

@Vrouvrou What's $a$

a point from A

@AkivaWeinberger

I recently posted a question and it wasn't well received. It is this one:

Consider $A=(0,1)$, the open unit interval. Let $x=0$. What's ${\rm dist}(x,A)$?

nice,ok all the very best@LittleRookie seeya

cya!

Can anyone tell me what's wrong with it?

0

@AkivaWeinberger

@Vrouvrou And yet $x\notin A$

yes

but where is the mistake in what i wrote please

you can't take an $a \in A$

why ? $d(x,A)=\inf_{a\in A} d(x,a)$

@Vrouvrou How did you go from "$\inf_{a\in A}{\rm dist}(x,a)=0$" to "there's an $a$ such that ${\rm dist}(x,a)=0$"?

@Vrouvrou Note the difference between "infimum" and "minimum".

Consider the set $\{\frac{1}{n} \mid n \in \mathbb{N}\}$

by caracterisation of inf

The infimum of a set need not be in the set.

clearly the distance between that set and $0$ is $0$.

But the distance between any element of that set and $0$, will be some fraction $\frac{1}{n}$

$inf_{x\in a} d(x,a)=0$ the caracterisation of inf say that $\forall \varepsilon ?0, \exists a\in A , d(x,a)<\varepsilon$

I improved my post as far as I can. If there is still something unclear please tell me. Reopen votes are also welcome...

Yes, that statement is true.

then going to 0 , d(x,a)=0

so x=a

no?

No.

@Vrouvrou What's $a$ though

For any $\varepsilon$, you find some $a$ to make the distance smaller.

You can never guarantee that this distance will be $0$.

yes

Remember that $A$ need not be closed, so the limit of a sequence of points need not be in $A$

("Need not be" = "might not be")

but this is true for all $\varepsilon>0$

Yes...

then it is true when $\varepsilon\rightarrow0$

What is $\inf \left\{\frac{1}{n} \mid n \in \mathbb{N}\right\}$ ?

0

@Vrouvrou This is false

why?

@AkivaWeinberger

What is $d(0,\frac{1}{n})$ ?

i don't know

Imagine you choose an $a_1$ such that $d(x,a_1)<1$. We can do this, by your definition of infimum (let $\epsilon=1$).

Now choose an $a_2\in A$ such that $d(x,a_2)<\frac12$

Take the standard Euclidean distance, for example.

and an $a_3\in A$ such that $d(x,a_3)<\frac13$

i don't see where is the problem

We can do this forever, right? @Vrouvrou

yes

$d(0,\frac{1}{n}) = \frac{1}{n}$

there is a sequence $(a_{\varepsilon})$

So, you want to say, "let $a=\lim_{n\to\infty}a_n$, then $d(x,a)=0"

yes

So $\inf_{x \in \{\frac{1}{n}\}} d(0,x)= \inf \{\frac{1}{n}\} = 0$

The problem with this is that, even though all of the $a_n$ are in $A$, the limit $a$ might not be in $A$ @Vrouvrou.

none of the $x$ satisfy $d(0,x) = 0$ though.

Only closed sets contain all of their limit points. @Vrouvrou

@AkivaWeinberger ahh ok

i understand

So you can do this if $A$ is closed, but you can't in general.

ok

thank you

But, if $d(x,A) = 0$, then $x \in \bar{A}$ exactly because you can create a sequence in $A$ converging to $x$ (regardless of whether $x \in A$ or not)

ok

yes

thank you

Should I put a more elaborated example? Or is there something technically wrong with it...?

@draks Reading your question, 'm not sure you understand Gödel's incompleteness theorem

It's an often misunderstood theorem, but I think you'll find good explanations somewhere on Math Stack Exchange

by the likes of Asaf Karagila, Noah Schweber, and other set theory / logic people

Hello, i have an other methode about my question of this morning, if $U,V$ are open and $U\cap V=\emptyset$ then $\overset{\circ}{\overline{U}}\cap \overset{\circ}{\overline{V}}=\emptyset$

if i suppose that $\overset{\circ}{\overline{U}}\cap \overset{\circ}{\overline{V}}\neq \emptyset$

there there exits $x\in \overset{\circ}{\overline{U}}$ and $x\in \overset{\circ}{\overline{V}}$ so $\forall W\in \mathcal{V}_{x}, W\cap U\neq \emptyset$ and $\forall W\in \mathcal{V}_{x}, W\cap V\neq \emptyset$

@SteamyRoot really? I thought I got. Whats wrong?

can i say that $W\cap (U\cap V)\neq \emptyset$ then $U\cap V\neq\emptyset $

hey @mike

hi

Hi guys. Question on the multiplicative group of integers modulo $n$; the operation that holds in this group is $\overline a\cdot\overline b=\overline{ab}$. However, does this mean that we can also substract, i.e., is the following allowed: $\overline a-\overline b=\overline{a-b}$? Or $\overline{na}-\overline{ma}=\overline{(n-m)a}$? If so, why is it allowed?

Because, in order to calculate the inverse of some element $\overline a\in(\mathbb Z/n\mathbb Z)^*$, it seems like we do use some for of subtraction, see example:

it is not the group properties, but the ring properties in this case.

Here we're calculating the inverse of $\overline 22\in(\mathbb Z/101\mathbb Z)^*$

I haven't heard of rings yet...

I understand that under the line, we can just use the algebraic properties of the reals, e.g. $\overline{10*8}=\overline{80}$

But what justifies me to do the same if they are 'split', e.g. $\overline8 +\overline {10}=\overline{18}$

$\overline{8}=8+kp$ and $\overline{10}=10+np$ so $\overline{8}+\overline{10}=8+10+(k+n)p=18+(k+n)p=\overline{18}$

same kind of stuff for the product.

Ah great, I have one small question though

The way I've learned it, $\overline 8=\{8+kp:k\in\mathbb Z\}$.

So it's a set

@ShaVuklia If you take $a=b=1$ ($\overline 1$ always being in (Z/nZ)^*) then you'd get $\overline{1}-\overline{1}=\overline{1-1}=\overline{0}$. But $0$ is never in (Z/nZ)^*.

When did we define '+' to be addition in the set?

Shouldn't we work with the multiplication instead? Because the plus simply isn't defined

Well, you're the one who used minus.

yea I used it, asking why it is correct

But how did you defined multiplication then ?

I wouldn't.

Addition/subtraction aren't good operations on a multiplicative group of integers.

It is defined, being the group operation, as follows: $\overline a\cdot\overline b=\overline{a\cdot b}$

When taking inverses in a multiplicative group, you'd just write it as $\overline{a^{-1}}$.

(which looks pretty ugly with the overline on it. oh well.)

it has different thinness though

(I'm taking understanding $a$ to be in Z/nZ here not an integer. Otherwise a^{-1} wouldn't make sense.)

Can't the example I gave (calculating the inverse of $\overline{22}$) be explained without using 'substraction/addition' properties, using only multiplication?

I don't really follow you. Like I said, you can't add/subtract in that multiplicative group.

Hi chat

ok that's great news, but what are they doing here then:

hi @Astyx

What are we doing ?

It seems to me like they're substracting

I frankly don't know wth they're doing.

$\overline{0}$ isn't a sensible expression in (Z/nZ)^*.

well they're trying to write find the $x$ such that $\overline x\cdot \overline{22}=\overline 1$

that's true, actually

Doesn't really matter. 0 isn't in the multiplicative group of units, full stop.

but I guess it would have worked with $\overline 1$ and $\overline 2$ too

So I find the above to be pretty dubious.

Hi all , give me a hell yeah!

hell yeah?

Hahaha

he he

hi Baymax:d

Now, if that's not the multiplicative group of units but rather something else then maybe it works.

Hi @Baymax

Hi @ShaVuklia

But as it stands it doesn't make any sense to me.

but how would you calculate an inverse then, Semiclassical?

Hi @Astyx

for the multiplicative group

sorry for being heavy, but there is supposed to be some logic in the picture posted ? because I do not see it.

^

Well, you've not said something rather important. What's $n$ here?

101

ha ha ha

Okay.

(Z/101Z)^*.

correct

It had to be 101 since 0|101 and 101 is prime

and they say that it's similar to the extended Euclidean algorithm

at least, they referred to it before writing out this calculation

I buy that.

But that makes the first line even goofier, since 101 wouldn't be in (Z/101Z)^* either.

at least, they said they're going to "use the same method"

To me they're just brutally trying values and hoping for the best

I imagine there is a logic to be had, buuut

but how did any of you learn to calculate the inverse then?

because I'm happy to abandon this method and use the correct method

Fermat's little theorem helps there, doesn't it ?

I think the EET route isn't a bad idea, though.

hmm

Maybe try this source instead? www-math.ucdenver.edu/~wcherowi/courses/m5410/exeucalg.html

Not sure wether that's the usual way, I don't often work in $\Bbb Z/101\Bbb Z $

It seems less WTF.

haha:d

ah right, that seems decent :P

I'm going to read it!

But if you use the EET somehow you'll have to use additions, substractions, we are biting our tail.

kk

oh

well, the addition/subtraction won't be taking place in the group itself.

it'll be happening in the integers more generally.

so it's still alright.

btw @pilko you keep writing EET? what does that stand for? Extended Euclidean ... ?

you wrote it first in full words, then semi used the acronym, I copied.

Actually, I used it first. And I was being silly: should've been EEA.

Does this make sense --- $\bar{0}$

Not to my knowledge. (Hence my objection earlier).

It's the class of equivalence of 0 mod n to me

$\bar{0} = 0 + 101 Z$

so

I don't see why it shouldn't make sense

let me type!!!!!

That works if we're doing Z/101Z. But, -again-, we're doing (Z/101Z)^*.

Sorry :(

no no sorry

Maybe just consider Z/101Z as a ring ? I dunno

I ate too much

so

$\bar{0} . \bar{22} = (0 + 101Z).(22 + 101Z)$

Well 0 *22 = 0 =101 mod 101

yo!

and we may meet this

$22.101Z + 101Z$

and that's 101Z

Hey @Semiclassical ?

you saw that paper?

Saw it. Did not try to read it.

ha ha

Because seriously that copy job was awful on the equationso.

I am geting nowhere the article even not in Researchgate!

I will try to manage that as it contains something relevant thogh!

will have to wear a powerful specs

If I did hurt someone now I am sorry...

Ok guys cya...have a nice day!! Unity in diversity!

someone can take a look at my question ? math.stackexchange.com/questions/2187648/…

nice day to you too

@ShaVuklia What such a beautiful girl is doing on a math chat?

@Secret what's your progress so far?

^

...

hi

Hello

whats popping, whats hopping and whats dropping everybody?

tired

test: $\vert$, $\Vert$

Finally done with the final

howd you do

hmmmm

go back to the workshop talks, or head to the math library to grab some books, or work on grading.

decisions.

Back.

It was aight, I def missed some points for stupid mistakes, and missed some other points for things I didn't completely figure out how to do

But overall, it was good

@Semiclassical use $\|$ :)

@Semiclassical hi, i see your answer to my question, did you check if i solved the question correctly? i wasn't sure about it , first question in this subject :P

'sup

I keep forgetting about \|. @s.harp

@Liad You've set it up right.

But unless you know how to actually manipulate the obtained equations to obtain the multipliers and the optimal $u,v$ you haven't solved it.

someone motivate me to do math

What kind of math?

If you math hard enough, you can raise Erdos from the grave, write a paper with him, have it retracted, and have a negative Erdos number!

...