Mathematics - 2020-05-05 (page 1 of 2)

Ted Shifrin

00:04

@Edward: Right, that seems to be what it's saying. Didn't you forget to write that the horizontal lines are exact?

(I haven't thought about this in a while.)

Edward Evans

Yeah I did, thanks

That's very confusing to me lol

Ted Shifrin

So they're saying you can put a $0\to$ on the left of the bottom and a $\to 0$ on the right of the top. If it holds then, it holds in the general exact situation.

I recall this is a (not difficult) diagram chase.

loch

my assumption would be something like the case r injective, j surjective is covered in class, and it's asking you to use this to deduce the general situation yourself

geocalc33

@anakhro conjecture: Given two transversal $\Bbb R^{1,1}\cap \Bbb R^{1,1},$ the multiplication of any pairs of lines of constant time and space yields a minimal surface and of course geodesic curve, whose length encodes the information-entropy

anakhro

What is R^{1,1} again?

Edward Evans

00:09

The version I wrote above (without the assumptions on $r$ and $j$) was covered in class

I'll stick a few zeros on the outsides and see what happens when I stare at it for long enough

Ted Shifrin

Wait, @Edward, so they did the general case already?

They're just trying to get you to deduce the general case from the special case?

Edward Evans

Right

Akiva Weinberger

What is the population density of me

Ted Shifrin

Gotcha.

Heya DogAteMy.

Akiva Weinberger

in terms of people per square mile

Heyup

Ted Shifrin

00:12

I just heard from your math prof ... I was sad to hear he skipped the best part of the course.

Edward Evans

Hi @Akiva

Akiva Weinberger

Who, Patrick?

Ted Shifrin

Yup.

Akiva Weinberger

What part is that?

Also what got you talking?

Ted Shifrin

Differential forms through Stokes's Theorem.

Akiva Weinberger

00:13

Ah

Ted Shifrin

He emailed me ages ago to ask permission to use my videos when school went remote.

Akiva Weinberger

Ah, makes sense

And you said?

Ted Shifrin

Given his being a combinatorist, I'm not sure why he loves teaching this course ... students should see forms and Stokes for sure.

Well, of course I said no problem. They're on YouTube, after all.

Akiva Weinberger

Mhm

Ted Shifrin

@Edward: This seems easy to me. For example, replace $N_1$ with $N_1/\ker r$.

Akiva Weinberger

00:16

Been thinking about how all computably enumerable sets are Diophantine

Ted Shifrin

Are you turning into a computational logician? :D

Akiva Weinberger

The Fibonacci numbers have a relatively simple Diophantine expression: $\{n\in\mathbb N:\exists x\in\mathbb N,n^2-nx-x^2=\pm1\}$

Apparently so does the complement of the Fibonacci numbers ($y^2-xy-x^2=\pm 1$ is satisfied only by pairs of adjacent Fibonacci numbers so you can express a non-Fibonacci number as fitting in between solutions to that hyperbola)

@TedShifrin Just been doing some reading

Most of the Diophantine equations you find in these papers are horrible

And I couldn't find a reference for a Diophantine expression of the powers of 2

Ted Shifrin

I know absolutely nothing thereabout.

Akiva Weinberger

(and I don't know how to find one myself)

$\{\text{powers of 2}\}$ equals $\{n\in\mathbb N:\exists x_1,\dots,x_k,P(x_1,\dots,x_k,n)=0\}$ for some polynomial $P$

(It doesn't matter if the $x_i$ range over the integers or naturals)

Ted Shifrin

I don't begin to understand/believe this.

Akiva Weinberger

00:21

In all likelihood $P$ is much more horrible than the one for the Fibonacci numbers (only assuming because if it was simple I feel like someone would've written a paper on it)

Though, there exists a paper with an explicit polynomial that gives you the set of primes

Ted Shifrin

how can it be that you get only powers of $2$?

Akiva Weinberger

Why is that less plausible than the Fibonacci one?

Ted Shifrin

Because the Fibonacci one was an explicit quadratic.

Akiva Weinberger

The big theorem is that all computably enumerable sets can be written like this. To quote Wikipedia: "This is far from obvious, however, and represented the culmination of some decades of work."

(Computably enumerable means: you can write a program that outputs all the elements of the set, though not necessarily in order)

Balarka Sen

This is Matiyasevich's theorem

Famous

Akiva Weinberger

00:26

Yes

(so if $n\in S$ is true then you'll know eventually, but if it's false then you might never know)

Ted Shifrin

Yeah, this stuff is totally beyond me.

Akiva Weinberger

(There are computably enumerable sets whose complement is not computably enumerable)

loch

ah Hilbert's tenth problem

Ted Shifrin

Any interesting homework on that topology set, a?

Balarka Sen

None

It's droll check

Ted Shifrin

00:28

Is that like droll call?

Balarka Sen

@Alessandro Are perfect maps Cartesian-closed?

I think so

Oh apparently it's a theorem in Engelking. OK

Just bunk Munkres what is even the point of doing these kind of point set topology

One should do proper point set topology

Ted Shifrin

I think it's fair to say that Munkres has (more than) what every serious mathematician needs to do analysis, topology, or whatever.

Nah, I totally disagree.

Balarka Sen

I am being extremely facetious :)

Ted Shifrin

OK, since our narcissistic emperor started employing (his version of) sarcasm, I no longer have a sense of humor.

Balarka Sen

Hahah

@TedShifrin The last exercise is to check $\Bbb R^2_\ell$ is not normal, so that's a little interesting.

Mostly fine though

Ok, gotta finish writing

Ciao

Ted Shifrin

00:35

Ciao. I am not sure I have ever written that proof out. I know I haven't taught it.

manooooh

Hello all!

Ted Shifrin

Hi manoo

manooooh

I need to translate this problem into propositional logic

Ted Shifrin

Ugh.

You need to get back to math.

manooooh

"In naturals, if $a\mid c$ and $a\mid(c+2)$, then $c=1$ or $c=2$". I translated it into the following: $$\forall a,c\in\Bbb{N}(a\mid c\;\land\;a\mid(c+2)\to c=1\lor c=2).$$ Is that correct?

Thank you!!

Of course it is false, but I am asking for the translation

Ted Shifrin

00:41

Yeah, it's right.

manooooh

Thanks @TedShifrin!! I love your videos btw

Ted Shifrin

I personally don't think of it that way, though. I think of it as "if for some $c$, we know blah for all $a$, then blah."

So I would not mark it wrong if someone wrote IF there exists $c$ such that for all $a$ blah, THEN $c=1\vee c=2$.

I prefer it that way, actually.

manooooh

So you would write it as $$\exists c\in\Bbb{N}\forall a\in\Bbb{N}(a\mid c\;\land\;a\mid(c+2)\to c=1\lor c=2)$$

Ted Shifrin

No, not quite.

manooooh

@TedShifrin I understand it from your message, sorry. How would you write it, though? Sorry for the silly symbols, I have to...

Ted Shifrin

00:46

$(\exists c \forall a, a\mid c \wedge a\mid (c+2)) \to c=1\vee c=2$.

manooooh

Ohh you have changed the scope

Ted Shifrin

Note I had the if before the "for some $c$".

I dunno.

manooooh

01:06

@TedShifrin I have asked the professor and he told me he thinks both statements are equivalent to each other. Thank you Ted again, your suggestion was awesome

Now I am asking him for a proof that both are equivalent... I think it's not so simple to prove it, at least in the course we are teaching haha

Ted Shifrin

01:28

LET me know if I fail. :)

EnjoysMath

01:52

Got a good type theory question: math.stackexchange.com/questions/3659221/…

Not twin primes :)

Why isn't there a category of types

I googled that and didn't find what I wanted

Balarka Sen

03:38

@TedShifrin I ended up learning something actually

If $f : X \to Y$ is a perfect mapping, $X$ is second countable then $Y$ is also second countable. Perfect mappings are closed, surjective, continuous maps with compact fibers.

Seems like a useful fact.

Ted Shifrin

03:56

So proper closed maps.

Balarka Sen

Weaker than proper, just fibers of singletons are compact

But in particular, yeah

(if Y is T1 :P)

Ted Shifrin

05:17

@manooo: I do fail. The statement is a. “whenever”. statement, not an “If for some” statement. Not equivalent.

manooooh

05:34

@TedShifrin I don't see any difference

How should the statement be translated if we pick your interpretation? What should change?

Ted Shifrin

Suppose we make a simpler version: $\forall x (p(x)\implies q(x))$ versus $(\exists x p(x))\implies q(x)$

What if it holds for one $x$ and fails for another?

manooooh

@TedShifrin the first expression would be false and the second one would be true?

Ted Shifrin

Right.

I still prefer English words.

manooooh

05:54

@TedShifrin I do know

@TedShifrin thanks, will consider it

Hawk

Sylow 2nd theorem says if $H \leq G$ and $|H| = p^j$ for some $j$ then $H$ is contained in some Sylow p-subgroup of $G$. Now Sylow p-subgroups are the maximal subgroups with order $p^k$ right? So if $|H| = p^j$ doesn't that mean $p^j$ divide $p^K$? The books proof uses orbit-sabilier to prove this but I want to know why my reasoning is wrong.

Ted Shifrin

You don't have any reasoning.

Hawk

oh its because i haven't shown the existence of this maximal subgroup that's the problem

which is the point of the theorem.

Ted Shifrin

You're presumably using the converse of Lagrange's Theorem or some such.

Hawk

I think I actually answered my own question...lol

Ted Shifrin

06:01

Even if you had the one of maximal order, it still doesn't follow without justification.

Hawk

oh i see where this is going

yeah that maximal group may contain a subgroup of order $p^j$ but it might not be my $H$

Ted Shifrin

Right.

Hawk

but they should all be conjugate that's the point right

Ted Shifrin

Of course, generally there's no unique maximal one. But part of the theorem is that they're all conjugate.

Thorgott

10:01

In $\omega_1+1$, the closure of $[0,\omega_1)$ is $[0,\omega_1]$, but $\omega_1$ is not a limit point of any sequence in $[0,\omega_1)$, right?

order topology, of course

Balarka Sen

Correct

Thorgott

is there an easier example of a limit point of a set that's not the limit of any sequence in the set?

Balarka Sen

I don't think so. Fundamentally you need that point to not have a countable neighborhood basis.

Alessandro Codenotti

It's also very easy to show that no sequence converges to $\omega_1$ in that case

Leaky Nun

do you need choice

Balarka Sen

10:06

Countable choice, at least, @LeakyNun

Alessandro Codenotti

It's just that countable union of countable ordinals is a countable ordinal

Balarka Sen

Oh, you meant for Alessandro's statement? Clearly not

Every countable set in $\omega_1$ is bounded above

Thorgott

it sure is convenient that I learned ordinals right before this topology course, so that I can come up with cool counterexamples

Leaky Nun

why don't you need choice for that

Alessandro Codenotti

@BalarkaSen $\mathrm{cof}(\omega_1)=\omega$ is consistent with $\mathsf{ZF}$

Balarka Sen

10:10

@LeakyNun Concretely: Let $A$ be a countable set, look at union of the sections $\{x \in \omega_1 : x \leq \alpha\}$ for $\alpha \in A$. This is countable, so the complement contains a point by uncountability of $\omega_1$. Every point in $A$ is less than that by construct.

It's just definition-chase, I don't see where choice is used.

Alessandro Codenotti

countable union of countable sets is countable uses (very little) choice

Balarka Sen

lol does it

You win

Leaky Nun

because for each set you need to choose a bijection

Balarka Sen

Got it.

Alessandro Codenotti

yeah, there are models of $\mathsf{ZF}$ in which the reals are a countable union of countable sets

I don't actually know how to build one though. Maybe I should learn that

Balarka Sen

10:13

Nuts

Alessandro Codenotti

building models of ZF is always finicky

Thorgott

oh god

I don't wanna live in a world where the countable union of countable sets isn't countable

Alessandro Codenotti

Because if you force over a model of ZFC you cannot break choice, and forcing over models of ZF is painful

The only construction I've ever seen was a model of ZF in which the reals are not well orderable and it was done by first forcing over a nice model of ZFC and then looking at an inner model of the forcing extension

Balarka Sen

@Alessandro Help me define sheaves over simplicial complexes

Alessandro Codenotti

Why

Balarka Sen

10:17

I need this language I think

A simplicial complex naturally gives rise to a category, where the objects are faces and morphisms are inclusions

A presheaf on this is a functor from the opposite category to AbGrp say

What is the correct gluing axiom? Should it just be that given a colimit diagram, if the colimit exists, the functor sends it to a limit?

Colimit exists only if it's an arrangement of faces in a simplex or something

This looks subtly different from gluing as I would understand it but I don't think this is an issue

Alessandro Codenotti

Uhm not sure. It looks like gluing to me though

Balarka Sen

Like, I would understand gluing to mean glue some stuff over two different simplices such that along the faces they intersect these stuff are the same. But colimit exists only for diagrams like, $\Delta$ is a simplex, and you have an arrangement of faces $\Delta_1, \Delta_2, \cdots$ of $\Delta$

Doesn't matter right?

Alessandro Codenotti

Urgh I don't know, this looks awful

Balarka Sen

No, it's very weird. I am confused

learning_mathematician

hello everyone

Balarka Sen

10:23

Colimits never exist unless you have a face inside a face inside a face

In which case the notion is trivial

Garbage

learning_mathematician

does any of you have some measure theory experience?

Alessandro Codenotti

@Thorgott This is a nice example if you are an analyst math.stackexchange.com/a/2344090/136041

Balarka Sen

Eh, actually I don't need a sheaf. A presheaf is fine with me, that's enough to take a Cech cohomology

Alessandro Codenotti

@learning_mathematician Yes but we can't tell whether it's enough to answer your question if you don't ask it, so go ahead

learning_mathematician

@AlessandroCodenotti haha ok

In a measure space (X,μ), let f ∈L1 and f≥0. for every measurable set E let μf=∫Ef = ∫XfχE

1) show μf is a measure do I have to show that f induces a measure on E of X?

so to show μf is a measure on E, I must prove the criteria, μf(empty set)=0, pairwise disjointness of En's and union=sum

Alessandro Codenotti

10:30

Right

Where are you stuck?

learning_mathematician

so to show μf(emptyset)=0

should I just let E=0 so that ∫fχ∅=∫f0=0?

Alessandro Codenotti

Yes, integral of any f over the empty set is 0 pretty much by definition

learning_mathematician

or should I take the μf(EU∅)=μf(E)+μf(∅)

I know this is a basice question but I don't like having doubts

Alessandro Codenotti

The former, μf(A) means the value of the integral of f over A, so μf(emptyset) is the integral of f over the empty set

learning_mathematician

I see so μf(∅)=∫∅ f which clearly is 0 since the integral of f over nothing is 0

Alessandro Codenotti

10:41

indeed

robjohn

using the bookmarklet listed at the upper right, you can use MathJax here.

$\int f(x)\chi_{\emptyset}(x)\,\mathrm{d}x$

Thorgott

my functional analysis is too weak, but that looks cool

learning_mathematician

for the second part for the indicator function of the union of En that is X_(⋃En)=∑χEn.

should I just show what the indicator/characteristic function of the union of En looks like?

robjohn

@robjohn for example

Balarka Sen

@Thorgott My functional analysis is weak* but it should get better if I find a compact book on it to study from

learning_mathematician

10:50

@robjohn bookmarklet?

Thorgott

lol

robjohn

@learning_mathematician Does your browser have a bookmark bar along the top?

learning_mathematician

@robjohn yes

robjohn

@learning_mathematician Then go to this page and drag the link for "start ChatJax" to that bar. That should add the bookmarklet there.

clicking on that bookmarklet should start rendering the MathJax on this page. Each time you refresh the page will stop the bookmarklet from running, so you will need to click it after each refresh.

learning_mathematician

11:12

done

thanks robjohn

let's try $\mu$

it works

for the second part for the indicator function of the union of $E_n$ that is $X_{⋃En}$=∑χEn.
should I just show what the indicator/characteristic function of the union of $E_n$ looks like?

so $\chi_{\cup E_n}$

Anonymous

11:39

Could someone explain to me the part in bold: $$1 \to G \to E \to Q \to 1$$

Given a normal subgroup $G$ of a group $E$ with quotient $Q$ there exists a natural homomorphism $\varphi: Q \to \mathrm{Out}(G)$, where $\mathrm{Out}(G)$ is the group of outer automorphisms of $G$ defined as $\mathrm{Aut}(G)/\mathrm{Inn}(G)$, where $\mathrm{Aut}(G)$ is the group of all automorphisms of $G$ and $\mathrm{Inn}(G)$ is the normal subgroup of inner automorphisms of $G$. The mapping $\varphi: Q \to \mathrm{Out}(G)$ is defined by choosing an arbitrary lift $\hat a$ of $a \in Q$ in $E$, and using the autom…

Anonymous

Why is the automorphism (considered as an element of $\mathrm{Out}(G)$) independent of the choice of the lift of $a$?

Balarka Sen

@Triskelion Because you quotiented by $\text{Inn}(G)$. Write down the map $Q \to \text{Out}(G)$ for me.

Anonymous

@BalarkaSen I guess $a \mapsto (g \mapsto {\hat a}g{\hat a}^{-1})$?

Balarka Sen

Change $\hat{a}$ to $\hat{a}'$ such that both are lifts of $a$. How are $\hat{a}$ and $\hat{a}'$ related?

Anonymous

@BalarkaSen Umm, I'm not sure how $\hat{a}$ and $\hat{a}'$ are related apart from the fact that they are arbitrary elements of $E$

Anonymous

11:51

The map corresponding to $\hat{a}'$would be $a \mapsto (g \mapsto {\hat a'}g{\hat a'}^{-1})$

Balarka Sen

They are clearly not arbitrary elements of $E$, because they are both lifts of the same element $a \in Q$

What does that mean, being lifts of $a$?

Anonymous

@BalarkaSen By "lift", do they mean something like a section $s: Q \to E$ (one-to-one homomorphism)? That is, the order of $a \in Q$ and $\hat a \in E$ has to be the same?

Balarka Sen

No I just want you to write down what "$\hat{a}$ is a lift of $a$" means, mathematically :)

Anonymous

@BalarkaSen Well, I don't know what "lift" mathematically means in this context :P I never came across the term before

Balarka Sen

Then your confusion seems to be before the sentence you put in bold, because they mention the word "lift" in the line right before the one you bolded.

Am I accurate in understanding that

Anonymous

11:56

@BalarkaSen True, yes

Balarka Sen

Given a surjective homomorphism $f : G \to H$ a lift of an element $h \in H$ is an element $\hat{h} \in G$ such that $f(\hat{h}) = h$

Anonymous

@BalarkaSen Oh, I see. That sounds similar to the concept of a section (splitting map) though

Anonymous

That is, if I have a section $s: H \to G$ that maps $h \mapsto \hat h$, then $f \circ s = \mathrm{id}|_H$

Balarka Sen

A section is choice of a lift for every element of $H$, and in a consistent fashion (aka $s$ is a homomorphism)

Way stronger than what is needed

You just want a single lift of a single element

Anonymous

Oh, right. I understand

Anonymous

12:01

Okay, so now we have to see why the part in bold is true

Balarka Sen

Tell me, yes.

Now you know the definition of a lift, so you should be able to figure it out

Anonymous

$$1 \to G \overset{g}{\to} E \overset{f}{\to} Q \to 1$$

Anonymous

I think I'd require $f(\hat a) = f(\hat a')$

Anonymous

Where $\hat a$ and $\hat a'$ are two different lifts of $a \in Q$

Anonymous

Is that right?

Balarka Sen

12:05

Yes.

Anonymous

What do we need to show now? That $g \mapsto {\hat a}g{\hat a}^{-1}$ and $g \mapsto {\hat a'}g{\hat a'}^{-1}$ are identical automorphisms in $\mathrm{Out}(G)$?

Balarka Sen

Correct.

Anonymous

Well, one thing is that $G$ is a normal subgroup of $E$

Anonymous

So if I can show that ${\hat a}g{\hat a}^{-1} = {\hat a'}g{\hat a'}^{-1}$ we'd be done?

Balarka Sen

That's false.

So you cannot show it

Anonymous

12:13

Oh, we don't have to show equality elementwise, just for the whole group $G$. So rather ${\hat a}G{\hat a}^{-1} = {\hat a'} G{\hat a'}^{-1}$ I think

Anonymous

I'm not sure at this point

Sam

Hello

Anonymous

What's the condition for equality of two automorphisms in $\mathrm{Out}(G)$?

Balarka Sen

Use definition. What does it mean for two elements of $G$ to give the same element in $G/H$ upto quotienting by $H$?

Sam

If someone would ask you what the opposite of X is (talking generally), what would be the correct answer? -x, y, or z? Or isn't there any?

Balarka Sen

12:16

Apply to the case $\text{Aut}(G)/\text{Inn}(G)$ accordingly.

You should revisit some fundamentals of group theory before attempting to understand short exact sequences, @Triskelion

Sam

X is 90 degrees from y and z, but 180 degrees from -x. Does -x make that the opposite?

Because don't opposites even out? so 1 + -1 will make 0

Im confused

AMC

I'm trying to prove Kunneth formula for cohomology on $\mathbb{Z}$

i tried a "theoretical" approach

but to me it seems wrong and most importantly "too" obscure as i am not understanding what is going on

Sam

Jumping back to my previous thing, it all relies on perspective

If you see it and you know it you know that x is 90 deg from y and y is 90deg from z and z is 180deg from x

but if you see it from another side

no

z is not 180deg what am i saying

they are all 90deg

how did i go with 90deg

180*

Mats Granvik

Life is an open book exam.

Sam

anyway

so what is it? all is 90deg away from eachother

x is 180deg from -x

forward is 180deg from backward

forward is the opposite of backward

x is opposite of -x?

Anonymous

12:24

@BalarkaSen Well, I think there is the theorem that two elements $g_1, g_2 \in G$ lie in the same coset of $G/H$ if their "difference" $g_1g_2^{-1} \in H$?

Balarka Sen

That's nearly the definition of quotient of groups, not a theorem :)

I think it'd useful at this point to go back to foundations instead of pondering on short exact sequences.

Anonymous

@BalarkaSen Yeah, I'll do that soon

Anonymous

But I think we're close regarding my original question

Mike Miller

@AMC I'm not sure what a non-theoretical approach to Kunneth is. in the end there is a Tor term and that's given by some homological algebra nonsense. One has to do some homological algebra nonsense somewhere.

Anonymous

I guess we need to show that $g \mapsto {\hat a}g{\hat a}^{-1}$ composed with the inverse of the automorphism of $g \mapsto {\hat a'}g{\hat a'}^{-1}$ lies in $\mathrm{Inn}(G)$

Balarka Sen

12:29

That is correct.

Anonymous

Which is $g \mapsto (\hat{a} \hat{a}'^{-1}) g (\hat{a}' \hat{a}^{-1})$

Anonymous

For now we know $\hat{a}\hat{a}'^{-1}$ and $\hat{a}'\hat{a}^{-1}$ are both elements of $E$

Balarka Sen

More is true about those elements

Anonymous

But for an inner automorphism, we need to show that there exists an element $\gamma \in G$ s.t. $(\hat{a} \hat{a}'^{-1}) g (\hat{a}' \hat{a}^{-1}) = \gamma g \gamma^{-1}$

Anonymous

@BalarkaSen Like?

Balarka Sen

12:36

I claim $\gamma = \hat{a} \hat{a}'^{-1}$ works

Anonymous

@BalarkaSen Uh, but for that we need $\hat{a} \hat{a}'^{-1}$ to lie in $G$ rather than just $E$. Do the $f(\hat a) = a$ and $f(\hat a') = a$ conditions come to rescue somehow?

Balarka Sen

Indeed.

Anonymous

For one, $f$ is homomorphism

Anonymous

So $f(1) = f(\hat a'^{-1} \hat a') = f(\hat a'^{-1})f(\hat a') = 1 \implies f(\hat a'^{-1}) = a^{-1}$

Anonymous

Also, $f(\hat a) = a$

Anonymous

12:43

So $f(\hat a \hat a'^{-1}) = 1$

Anonymous

But $G$ is the kernel of $f$ and so $\hat a \hat a'^{-1} \in G$

Anonymous

Looks okay?

Balarka Sen

Correct

Anonymous

Gotcha, thanks!

Stupid Kid

13:07

hola

Ola amigos

I need help with 3d curl

I know everything

What I am feeling buggy is about the normal vector used in formal definition

Why we get a vector and get unit normal vector and find curl?

Why normal instead of using 2d curl definition and add I j k component

though I j k curl might be tedious

I can't uunderstand why we take arbitrary point and get some unit normal vector??!!

oops i understand

It is like a plane with a stick perpendicular and we can move the stick.

thanks for intution!

Sha Vuklia

14:36

Could someone clarify wou the inclusion of $T_{e_i}S^3$ into $\mathbb R^4$ is the space $e_2,e_3,e_4$?

I guess I can see it visually for $S^2$ into $\mathbb R^3$

maybe I have to write out how the differential looks

Astyx

Do you see it in $3D$ ?

Sha Vuklia

yep

Astyx

It's exactly the same in $4D$

The sphere is the set of $x$ that satisfy $\langle x, x\rangle = 1$

Since this is a bilinear application its differential at $x$ is just $h\to 2\langle x, h\rangle$

So the tangent is the kernel of that, ie vectors that are orthogonal to $x$

Sha Vuklia

ah, that's a nice approach

thanks

Astyx

you're welcome

Sha Vuklia

15:01

@Astyx btw, I am slowly but steadily getting used to various identifications in diffgeo, and that is making life a lot easier. But do I keep checking manually that identifications are justified, as I don't want to be hesitant about them anymore

also, olas Ted

Anonymous

15:17

Is a direct product like $K \times H$ of any two groups $H$ and $K$ (even non-abelian ones) necessarily a central extension of $K$ by $H$ or $H$ by $K$? I'm a bit confused as I hear people say that direct products are central split extensions. But non-abelian subgroups can't be in the center of any group!

Astyx

16:14

@ShaVuklia Honestly you probably know more differential geometry than I do

Sha Vuklia

16:25

I doubt it:p

More Anonymous

Anyone here good at measure theory?

Thorgott

17:12

Is it obvious what the prime ideals in $k[X,Y]$, $k$ field, look like?

Tobias Kildetoft

@Thorgott Not really. Even when the field is algebraically closed, you get some weird ones

Thorgott

ok, thanks

Tobias Kildetoft

Not sure if there is some nice description of them

geocalc33

17:27

$X_1$ and $X_2$ are transversal lorentzian submanifolds s.t. the convolution $X_1\star X_2=X_3$ is a manifold. Is $X_3$ always a geodesic manifold w.r.t. $X_1$ and $X_2?$ That is, is any path $l,$ in $X_3,$ a geodesic through $X_1 \cap X_2?$ Equivalently, is any path $l$ in $X_3$ a minimal surface through $X_1 \cap X_2?$

loch

well, you should think of prime ideals as corresponding to irreducible (reduced) (closed) varieties.

in this case, let's take k to be alg. closed. Then your maximal ideals are closed points (0-diml varieties). They look like (x-a,y-b)

Your prime ideals which are not maximal or 0 are necessarily going to have height 1 (if i rmb my commutative algebra correctly), which just means geometrically that they correspond to 1-dimensional varieties in \C^2.

But of course that means that they are hypersurfaces - you can show that in particular they're given by the zero locus of an irred polynomial.…

Leaky Nun

nobody said $k$ is alg. closed

loch

17:43

i did

Thorgott

I think that answers the "is it obvious" question affirmatively with "no"

wait, that's not what affirmatively means

Ted Shifrin

Hmm, I had no idea I was here.

Edward Evans

E v e n i n g

Ted Shifrin

Yes, @Astyx's argument is correct. The tangent space of a sphere (centered at the origin) is always the orthogonal complement of the position vector. Mathematically it is clear, but it is also physically clear: If your velocity had any radial component, your distance from the center would change.

Tobias Kildetoft

@TedShifrin Hi

Ted Shifrin

17:56

Hi @Tobias

Tobias Kildetoft

@TedShifrin So, how severe is the American lockdown at the moment?

Ted Shifrin

Varies radically from state to state — this is what happens with the worst possible US government ever.

Still no testing.

In CA things have been more responsibly handled since the beginning, but ...

Tobias Kildetoft

So I assume by no testing you mean apart from people with symptoms?

Ted Shifrin

The GA governor has opened everything up despite all advice not to.

Yes.

Tobias Kildetoft

Denmark is slowly opening again. Next stage of that will be on the 10th, though nobody knows what that will entail yet

T_01

17:59

If $f$ is a vector space automorphism on $V$ and $U$ a subspace of $V$ such that $f(U) \subseteq U$ - why then $f^{-1}(U) \subseteq U$ follows?

Ted Shifrin

The question is how bad the re-surges will be ...

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$Mathematics$ Mathematics