Mathematics

JeSuis

5:00 PM

@BalarkaSen i do not know $\pi_n$ for $n>1$ lol. I will learn it this summer I hope

Dodsy

That is, X must be.

Eric Silva

GMT is like waaaaaaay more techinical than the stuff you did in the first few quarters @Daminark, I think it might be good to gloss over things in that case, idt it's the right time to be very technical, my point is, idt that approach is conducive to doing things.

Alessandro Codenotti

@BalarkaSen Is $\pi_n(S^n)=\Bbb Z$ always true?

Balarka Sen

@Dodsy The only compact connected 1-dimensional manifold is the circle, correct. I don't know why this is relevant to what I wrote though.

@Alessandro Yep.

Dodsy

X must also be this in order to be homeomorphic, is what I'm getting at.

Eric Silva

5:01 PM

@Alessandro the homotopy groups of $n$-spheres bigger than $n$ are fucking crazy

JeSuis

@BalarkaSen do you know homology ?

Semiclassical

Is $\pi_n(S^n)=\mathbb{Z}$? I may be entirely wrong

Dodsy

However, I have limited knowledge.

Alessandro Codenotti

@EricSilva I've heard there are a few open problems about them as well?

JeSuis

limited why ?

Balarka Sen

5:02 PM

@Dodsy $X$ must be a circle in order for $X \times X$ to be homeomorphic to $S^2$? Why?

Zee

@Daminark fair. I still think you should not glaze too much. Epsilon delta is the workhorse of analysis

Balarka Sen

(Indeed, $S^1 \times S^1$ is the torus, not the sphere)

2

Abdullah UYU

$a$ and $b$ have opposite signs. About the function $y=ax^3+bx^2+cx+d$:
I)It have local minimum.
II)It has local maximum.
III)It has inflection point.

Select the ones that always true. @Semiclassical

Eric Silva

@Alessandro definitely, I've been told that basically entire branches of algebraic topology exist historically to study the problem of computing them

Semiclassical

Ew

Actually, though, it's not too bad

Dodsy

5:05 PM

@BalarkaSen No, $X$ must be a compact, connected one dimensional manifold in order to be homeomorphic to $S^2$ (so I thought, with my limited knowledge, I'll add again) Which means that $X x X$ may change these properties of $X$?

Semiclassical

What's the condition for having an inflection point, for instance?

Daminark

I mean for what it's worth I may not do too many more things along those lines, so I liked that Marianna's wrap up was very enjoyable

Also like, if you do grad analysis they teach you measure theory again anyway

Dodsy

Anyways as I said, I've only had a passing encounter with any of this and I am just going off a vague memory.

Abdullah UYU

$y^{''}$ must have roots

Eric Silva

written: again; read:*for the first time*

Semiclassical

5:06 PM

Right. What's y''(x) in this case?

Abdullah UYU

$6ax+2b$ have roots. So III always true.

Daminark

I mean, we learned a reasonable amount, and for what it's worth we had a number of important theorems that were actually careful. Fubini, for example, and Marianna was significantly more cautious about stuff like Radon-Nikodym

Abdullah UYU

for I and II $y^{'}$ 's discriminant is important

Semiclassical

More precisely, it's always got a positive root at x=-b/(3a)

Balarka Sen

@Dodsy Oh, hm, I see what you're saying. $X \times X$ is a 2-manifold, does that necessarily mean $X$ is a 1-manifold?

If it is you'd be done by what you said.

Abdullah UYU

5:08 PM

does positivity required?

Eric Silva

I mean you say that @Daminark but I really do not believe people tend to come out of HA having learned much of anything, even if it's covered in class

Semiclassical

You're told that a,b have opposite signs

Abdullah UYU

sorry, again i wrote it wrong, $a$ and $c$ have opposite signs

Semiclassical

Oh.

Balarka Sen

I guess that boils down to saying if $X \times X$ is homeomorphic to $\Bbb R^2$, is $X$ homeomorphic to $\Bbb R$?

SteamyRoot

5:09 PM

Actually, is $\left(\pi_{n+k}(S^n)\right)_{n \in \mathbb{N}}$ eventually constant for all $k \in \mathbb{N}$ ? Or am I misremembering something?

Semiclassical

Then no, strike the positive remark

Abdullah UYU

so, III is a true statement

Daminark

I mean, define much. By the standards of graduate school of course it's pathetic, as far as an undergraduate level is concerned it's pretty decent

Dodsy

@BalarkaSen I don't really know what $X x X$ means. But if you were to look at what that does to a space, perhaps you'd find that there can't exist a space $X$ that when $X x X$ has the property of being a one dimensional manifold.

Alessandro Codenotti

@dodsy JeSuis is asking for $X$ a space, he never specified that $X$ needs to be a manifold so there might be some dogbone style weirdery?

Studentmath

5:11 PM

@EricSilva That's neat, right. Have to show $k(x,y)$ is the same as $\bar{k(y,x)}$

Thanks!

Dodsy

@AlessandroCodenotti Hey, you could be right, for sure.

Eric Silva

Honestly @Daminark my year was pretty good but I think a minority of people came out of the class with workable knowledge. They just knew a bunch of words.

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$Mathematics$ Mathematics