Are all contractible spaces semilocally simple connected? I think so. All contractible spaces are simply connected. Given $x \in X$, $X$ is an open neighborhood of $x$ such that the inclusion-induced map $\pi_1(X,x) \to \pi_1(X,x)$ is trivial, because $\pi_1(X,x)$ is the trivial group. Is this right?

Hi everyone

I have a very simple equation that I'm having trouble with. Consider $x,y,p \in [0,1]$. If $px+(1-p)y = 0$ must hold for every $p$, what are the possible solutions to $x$ and $y$?

@jonem They must both be $0$

Hi @loch

@santimirandarp what do you define as an "amateur"?

Like what is the mathematical background you are looking for?

Do you like shawarma, @TedShifrin ?

hi @AlessandroCodenotti

I'm thinking about an exercise, among other things it asked me to decide whether $\mathrm{Spec}\Bbb Z[i]$ is a normal and regular scheme. Normal is easy since the stalks are all PIDs (being localizations of a PID) hence integrally closed, but I'm unsure about the regularity, I guess there is some CA trick to avoid actually checking it?

It probably shouldn't be too hard to check regularity by hand, but maybe think about what's nice about $\mathbb{Z}[i]$ (think about some of the adjectives that one might use to describe $\mathbb{Z}[i]$)

Oh, it is a Dedekind domain and Dedekind domains are regular rings

yeah

Nice, thanks

Sanity check: If $X=\mathrm{Spec}A$ is a regular affine scheme and $A$ is a Noetherian ring of finite Krull dimension $n$ then the Zariski (co)tangent space has dimension $n$ at all closed points of $X$ because if $\mathfrak p$ is a closed point, hence a maximal ideal, $\dim A=\dim A_\mathfrak p$ and by regularity this is also the dimension of the (co)tangent space

"Let f be a ring homomorphism from a ring R to a ring S. If R has a unity 1, S$\ne${0}, and f is onto, then f(1) is the unity of S." Why do we need onto ?

Is it because $f(x)=0$ is a possible ring homomorphism? Or any other reason?

If you don't require rings to be unital then the zero map is a ring homomorphism

But there are more interesting examples, if your rings are not unital then ideals are subrings and the inclusions are further counterexamples

But here R unital is assumed

So why do we need onto assumption?

Consider the map $\Bbb Z\to M_2(\Bbb Z)$ sending $n\mapsto\begin{pmatrix}n & 0 \\ 0 & 0\end{pmatrix}$

The unit of $\mathbb Z$ is not being mapped to the unit of $M_2(\Bbb Z)$

Wow!!! Thank you so much!

Another example is to take any nontrivial product $A\times B$ and consider the inclusion of $A$ as $a\mapsto(a,0)$

Basically the idea is that if your rings are not required to be unital you can have subrings with different units

@AlessandroCodenotti I'm not sure if $\dim A = \dim A_{\mathfrak{p}}$ is always true, when $\mathfrak{p}$ is a maximal ideal. I think stacks.math.columbia.edu/tag/02JE is a counterexample

but in most cases yes you're right

so, now i see where i was flawed: I was not checking in my proof that $f(1)$ should commute with every element of S, not just with f(R) ones.

Wait, the height of a maximal ideal can be lower than the Krull dimension?

@AlessandroCodenotti I think the link I attached seems to give an example? where the localisation of $B$ (in the link) at two different maximal ideals has different dimensions

I did not read the example carefully though

There is some discussion here on conditions guaranteeing that all maximal ideals have the same height

But it can indeed fail in general

oh of course - any non equidimensional scheme would do the trick! :p

It seems to work for nice enough rings though, which is reassuring (also I'm not surprised at all to find Nagata's name associated to yet another ugly counterexample :P)

But yeah I was still thinking of $\Bbb Z[i]$ where it works

This exercise wanted to know the dimension of the (co)tangent space in $(i+1)$ and while it can be computed explicitely I was trying to avoid doing so :P

Next it asks to determine $\pi_\ast\mathcal O_X$ and whether it is locally free, where $X=\mathrm{Spec}\Bbb Z[i]$ and $\pi_\ast$ is induced by $\Bbb Z\to\Bbb Z[i]$, but I think I got this, I'll write it down later

@AlessandroCodenotti most of the time if you're actually doing geometry it's going to work :)

I guess it's worth keeping in mind that dedekind domains = smooth curves

@loch What do you mean precisely?

i mean it's regular of dim 1

maybe i'll add "intuitively"

No, sorry, my doubt wasn't on smooth, but on "Dedekind domain implies smooth", what's the precise statement?

(we only defined smooth for a scheme $X$ over a field $k$ though where we defined it as locally of finite type, constant dimension $n$ and $\Omega_{X/k}$ locally free of constant rank $n$)

We also saw that if $k$ is perfect regular and smooth are equivalent, and Dedekind domains are regular, is that what you meant?

I see, that's actually something useful to keep in mind! I just wanted to make sure I wasn't missing any subtlety

I have to go now, thanks for your help!

np - it's important to be careful!

If $T: V\rightarrow W$ is an injective linear map does it follow that $imT=W$?

?

it would only if dimV=dimW, right?

@topologicalmagician You also need the spaces to be finite dimensional.

Consider the right shift operator from $\Bbb{R}^\omega$ to itself.

@user193319 Alright: So is the following statement true only when the two spaces are of the same dimension: A linear map from V to W (both finite dimensional) is invertible iff rankT=dimV?

The right shift operator takes $(a_1,a_2,a_3,...)$ and maps it to $(0,a_1,a_2,a_3,...)$

A linear map from V to W with what condition?

oops I meant to say an invertible linear map

I meant to say its invertible iff...

I believe what you say is correct. A $T : V \to W$ is a linear operator between vector spaces of the same dimension is invertible iff it is injective iff it is surjective. This essentially follows from the rank nullity theorem.

but is it true if they are not of the same dimension?

No.

that's what I thought

my lecturer didn't specify that they had to be of the same dimension

Every bijective linear map is invertible, right?

Yes, that is what it means for a linear operator to be invertible

I can't stand Hatcher and his rambling. Even my professor can't stand the book, yet he uses it because he feels somewhat obliged to do so.

how dare you disrespect Hatcher

Oh, don't get so butthurt over it.

lol

:P

an integral domain that has 1 prime has infinitely many primes?

Does anyone know why a symmetric bilinear form such as the metric tensor allows you to raise and lower indices in the summation convention?

because it gives you a map from the vector space to the dual

@LeakyNun Ah, I see. Thanks

@LeakyNun what about a field

@LeakyNun I can prove this for UFDs, but no idea in general

I can also prove it if you replace "prime" by irreducible and assume that R is Noetherian

Guys, what would you do if you wanted someone to check your proof but there isn't anyone to do it immediately?

wait.

or check it yourself.

i'm wondering, in group theory can there be a notion of continuity between groups?

There are topological groups

@loch it has no primes

like the group of rotational symmetries of a sphere

and if you have a topology you have a notion of continuity

@MatheinBoulomenos Noetherian? why should it matter at all lol

@AkivaWeinberger that’s... not just a topological group :p

Can you have a notion of continuity between fractals?

@LeakyNun in a Noetherian integral domain, every element is a product of irreducibles

By "between" do you mean "of a function between"?

yeah

@MatheinBoulomenos interesting

Sure, given any function from a subset of $\Bbb R^n$ to a subset of $\Bbb R^m$ you can define if it's continuous or not

and fractals are subsets of $\Bbb R^n$

unless I'm misunderstanding the question

ahh okay, I didn't know that.

Do you know the epsilon-delta definition of continuity?

yeah

It works when the domain is multidimensional, too

though instead of $|x-y|$ you have the distance between the points $x$ and $y$

yeah, I know that as well

So with two points on a fractal, if the fractal is in the plane, you can use the planar distance

I dunno what level you're at, sorry if I'm aiming too low

@AkivaWeinberger no lol its okay. I really haven't studied fractals yet, so I don't know anything about them. It was just a question I was curious about

Does the study of fractals and diff geo overlap at some point?

can you have a fractal where you can define differentiability?

@LeakyNun oh for some reason i read prime ideal oops

I don't know diff geo

@topologicalmagician people use geometric measure theory to study diff geo and to study fractals but the crowds doing one or the other don’t tend to be the same

(because differential geometers low key don’t care about fractals)

Hey everyone!

@Erico how sad, diffgeometers should get their priorities straight

@Daminark nah fam we good

:0

They study piecewise smoothness and then are unprepared for the roughness of real life

(which can be measured qualitatively through the Hausdorff dimension)

The thing is, you shouldn't really study fractals until you've studied the things they're made of

which is fractals

@AkivaWeinberger I guess i'll have to study fractals instead of fractals then

LOOL

Source

That plaque is amazing

What sound does duct tape make

Quact

huh? LOL

What's it called when you eat an alien

Astronomeat

knock knock

Who's there

yah

Yah who

.com

I have a great knock-knock joke but you have to start it

knock knock

Who's there

idk what to say

lol

(That's the joke)

lol

It's better in person

im sure

All my favorite jokes are anti-jokes

or at least it would be if I could keep an actual poker face for more than .2 seconds

What's red and bad for your teeth @topologicalmagician

@AkivaWeinberger same

what? @AkivaWeinberger

A brick

ahahahahahahahhahaa

I lol'd IRL

Whats the best way to brush your teeth? @AkivaWeinberger

How

use a toothbrush..

(I ping'd you specifically because I felt that Rithaniel probably knew it already)

@topologicalmagician Touché

While we're on stupid jokes

What's blue and smells like red paint

I knew it, but I had forgotten it until I read the punchline.

Blue paint

Have you heard of anti-anti jokes?

Is that something like a double-bluff

Does it rely on the audience being familiar with the relevant anti-joke

It starts up like it's gonna be an anti-joke and, at least on reddit, they just kinda become nonsense

reddit.com/r/AntiAntiJokes/comments/b0p0e4/…

@Daminark can I define a notion of continuity between the set of anti-anti jokes and anti jokes?

Like idk if there's anything to get

But like this cracked me up

There's also this mathematical joke

Let $\epsilon<0$

(It's somewhere between a funny joke, a bad pun, and heresy)

Heh

or let $\epsilon =0$

(joke) 10 people died in the bar

Lmaoooo

There are two types of people in the world

Those who can extrapolate from incomplete data

(joke) roses are red, violets are blue

I have something similar to that last thing but it's not actually a joke

There are four types of people in the world

(1) Those who like you for the right reasons

(2) Those who like you for the wrong reasons

(3) Those who hate you for the right reasons

(4) Those who hate you for the wrong reasons

The only type of people you should worry about is type 3

This is true

integrate this

what is $S^{-1} A \otimes_A T^{-1} A$?

feels like $(ST)^{-1} A$

But I don' wanna integrate that.

Oh it's pi day where I am

and in an hour-ish it'll be 3/14 1:59:26.53…

How many digits of pi does everyone know

but I probably won't be awake by then

@Ultradark all of them

you know pi digits of pi

@Ultradark 3.14159265358979323846264338327950288419

is as far as I know

and you'll just have to trust me that I did that from memory

I know six: 3.14159

7 is nex akiva

3.142

I basically assigned each digit to a note (1=C, 2=D, etc) and it makes a little melody

and then the rhythm you can make up

@AkivaWeinberger oh the page was not loading yesterday right

Ya

Though, how many digits of pi do you know in binary?

that's because it was on reddit... lol

but maybe you know already

11.01 @Rithaniel

does anybody know a quick way to calculate percents

Oh I'm wrong apparently

11.001001

Whoops

@Ultradark Depends on what it is

Yeah, I just plugged it into wolframalpha

Tip: x% of y is the same as y% of x

guys is it true that o(a)o(b)=o(ab) for ab in a group G?

where o is the order

(ex: 4% of 75 is the same as 75% of 4, which is 3)

@AkivaWeinberger 34% percent of 83

Not in general, magician.

36

'Bout a third-ish, so 27 and a bit? @Ultradark

28.22 apparently

a and b can both have order 2, but ab might have order 8675309 or something.

@topologicalmagician It's possible for $o(a)=o(b)=2$ and $o(ab)=\infty$

so when does o(a)o(b)=o(ab)?

Consider the symmetries of a line, where $a$ and $b$ are each reflections across a point

(different points for each)

$ab$ is then a translation (by twice the distance between the points)

@topologicalmagician if ab=ba then o(ab) = lcm(o(a), o(b))

Probably that only happens in very specific cases, magician, but I don't know a universal rule for every time it happens.

@Top

I might shorten your name to Top, Top

generalizing Akiva's example, in the dihedral group $D_{n,2n}$ you have $\rho \sigma$ which has order $2$, and $\sigma$ still of order $2$, but $(\rho\sigma)(\sigma) = \rho$ having order $n$

^Basically reflections of the circle across two different lines

where the composition is a rotation

@LeakyNun how would you prove your claim that if ab=ba then the lcm(o(a),o(b))=o(a)o(b)

?

Actually, take $o(a)=2$ then $o(a)o(a)=4$ but $o(aa)=1$, so you probably need to dictate that the elements are distinct.

@topologicalmagician think about it

@topologicalmagician Hint: if $ab=ba$ then $(ab)^n=a^nb^n$

@Rithaniel Oh yeah that's a point

and clearly $a$ commutes with $a$

Night

Gotta sleep sometime

G'night Akiva

oh yeah my thing isn't even true

goodnight @AkivaWeinberger

quick way to do percents. 39% of 521. $4 \times 5=20.$ add a zero

so $200$

I have a conceptual question about convex conjugates for example, in the equation in suvrit.de/teach/ee227a/lect4.pdf slide 24 of 103, what is a good interpretation of the variable "z" ?

precise answer is 203 ish

g2g guys, have a wonderful day!

thanks@LeakyNun @AkivaWeinberger @Rithaniel

cya magician, enjoy your day.

Integrate the integral

alternatively use monte carlo

$\int_0^k \pi(x)dx$

when is the sum of the number of primes less than or equal to $k$ prime?

Conjecture: the sum of primes less than or equal to $k$ is prime only when $k$ is prime

How do I show that $\sum_{n=1}^\infty \frac{1}{1+n^2x} \sim \frac{\pi}{2 \sqrt{x}}$ as $x \to 0^+$?

actually the conjecture applies to: $ \int_0^k \pi(x)\pi(k-x) dx$

jk

@user330477 looks like zeta(2)

conjecture: the sum of primes less than or equal to $k$ is prime or semiprime when $k$ is prime or semiprime