Mathematics - 2018-12-09

1:05 AM

Let $M$ be a surface in $\mathbb{R}^3$. Then if $\zeta$ is a unit normal vector field on $M$ and $v$ is a unit tangent to $M$ at $p$ and $v$ is an eigenvector of $A_\zeta:T_pM\to T_pM$ say $A_\zeta(v)=\lambda v$, then, if we intersect the normal plane (generated by $\zeta$ and $v$) with $M$, we get a curve with planar curvature exactly $\lambda$ (at $p$). Does this generalizes for hypersurfaces in higher dimensional $\mathbb R^n$?

The problem is that in higher dimensions, the normal plane intersected with the hypersurface gives us something at least 2D, right? So, this does not determine a curve to look at...

Ted Shifrin

@AndersonFelipeViveiros: First, the planar curvature is $|\lambda|$. Next, you still have to slice with normal $2$-planes in higher dimensions.

Mike Miller

1:28 AM

I consistently forget that planar curvature is non-negative.

I know I shouldn't.

Ted Shifrin

No, all curvature is non-negative. In the plane you can use signed curvature.

But the signs may not agree; it depends on choice of normal vector.

IF you use the normal vector to orient the plane, then things should agree.

Mike Miller

Right, and you will never have curvature of a curve that changes sign, yes?

Ted Shifrin

No, signed curvature does change sign at inflection points.

Mike Miller

Ok. I feel that I had this discussion recently with someone else and they told me otherwise. I took them on faith.

My badley, Chadley.

Ted Shifrin

I believe I'm qualified on this one. I may have misspoken, but I'm still qualified :P

Mike Miller

1:35 AM

Yes, I take you as the higher authority than a grad student reviewing his calculus lecture.

I asked precisely whether that was true (existence of signed curvature and whether it changes at inflection points).

Because it seemed clear.

Ted Shifrin

well, this doesn't belong in calculus

Mike Miller

multivar 1 has planar curvature here

Ted Shifrin

OK, that's reasonable.

Secret

Suddenly think of a strange question related to surface. Is there a quantity that characterise how "locally twisted" a point on a surface is?

Ted Shifrin

I have no idea what that means.

Secret

1:44 AM

for example, the point produced when you place your finger on a towel and then twist you finger so the towel become crumbled around the finger. Is there a mathematical characterisation of that point where the twist converges to?

Ted Shifrin

So you're talking about singular points.

You can talk about cone angles.

Secret

Ted Shifrin

Singularities in dimension higher than 1 are very subtle.

Balarka Sen

ear perks up at the mention of singularities

Ted Shifrin

LOL, Balarka is back from unsleeping.

Ultradark

1:52 AM

are we talking about diffeomorphisms

oh, singular points...interesting

Secret

hmm...

I don't know... What I do know is that if those three ridges you saw in that photo is more curved, then there is more twisting in the neighbourhood of that singular point, but I am not sure if cone deficit angles is enough to capture the complexity of that

Ted Shifrin

No, it's not, Secret. I'm not sure what is.

Secret

What is the 1D example you are thinking of that involve cone angles?, do you mean the apex of a cone or something more subtle?

Ted Shifrin

No, I never said anything about 1D and cone angles. That was for the surface.

Ultradark

What are some questions to ask about regular square lattices and diffeomorphisms of regular square lattices

I guess I'm talking about non-rigid transformations

mrnovice

2:14 AM

anyone here know anything about mixture distributions

and rejection sampling?

Ted Shifrin

Nope.

mrnovice

:(

Ted Shifrin

That's truly statistics. Is there a Statistics SE?

Ultradark

Cross-validated

anyone know about modular flow

Ultradark

2:42 AM

or horocyclic flow

CaptainAmerica16

3:17 AM

@Michael.P I lost the Spivak room. Send the link again?

BAYMAX

3:31 AM

for nay integer n, 5 does not divide n^2 - 2

i thought of doing two cases of

n even

and n od

if n is even then n^2 multiple of 4

but after that?

Balarka Sen

4:05 AM

Well, it seems people also have the power to flip me over still.

Mike Miller

Fudgesicle, @BalarkaSen. That sounds like bull-sugar.

Balarka Sen

Well shove up you backstabbing addendum

Mike Miller

Hi @Aloizio

Semiclassical

@BAYMAX typically the casework you'd do for divisibility by m is to consider n having remainder 0 mod m, remainder 1 mod m, ..., and remainder m-1 mod m

Fargle

This is a [REDACTED] server, no expletives (such as [REDACTED]) are allowed

Balarka Sen

4:11 AM

@Fargle Yeah we can [EXPUNGED] see that, captain obvious

Semiclassical

obviously that involves a lot of cases if m is large, but for small m you can just test them all

Mike Miller

Wow it's like the lamest possible SCP

Fargle

Finally, a promotion from my rank as Lieutenant Obvious

Semiclassical

yet-another-polytope-problem

Balarka Sen

Stop

No more polytopes

Semiclassical

4:13 AM

aww

Mike Miller

@AlessandroCodenotti So I guess you're going to be getting into algebraic group theory

Curiously, that is the theory of algebraic groups, which is not the algebraic theory of groups, because an algebraic group is not a group

Balarka Sen

I should write an answer to this using small cancellation theory

BAYMAX

if n is even n^2 is even and n^2 - 2 is also even, so when its of the form 2k , so does 5 divide 4k^2 - 2 ? suppose 5 diivides 4k^2 - 2, then 5 must divide 4k^2 and 5 divide 2 but 5 doesnot divide 2 hence it cannot divide 4k^2 - 2 ----- will this work ?? @Semiclassical

Mike Miller

is that something you use for simple presentations by free groups? so you're cancelling stuff in words?

Balarka Sen

Yeah, you bound the length of the word you cancel when multiplying two relators

Semiclassical

4:17 AM

@BAYMAX "suppose 5 divides 4k^2-2, then 5 must divide 4k^2 and 5 divide 2" is definitely not sound

BAYMAX

oh ok

Balarka Sen

If it's less than 1/6-th the length of the relators, the group automatically shows hyperbolicity

Semiclassical

5 doesn't divide 13 and 5 doesn't divide 2, but 5 does divide 15

BAYMAX

i m thinking how how

Mike Miller

Is this your C'(1/6) thing?

BAYMAX

4:18 AM

to do thius

Balarka Sen

Ya

Mike Miller

But this is a GGT thing, not an algebra thing

Semiclassical

like i said, the usual approach by cases is to consider one case for each possible remainder modulo your divisor

Balarka Sen

It's basically combinatorics that C'(1/6) groups have cyclic centralizers

Not quite GGT

but my partial motivation while writing an answer is to learn some GGT

Mike Miller

I guess what i mean to say is "Is it accessible to OP (who will never read it)?"

Semiclassical

4:19 AM

I'm dubious that thinking in terms of even/odd will be very helpful, since even/odd numbers can be divisible by 5

Balarka Sen

I will do an expository!

Mike Miller

Sick

I am excited to read

Balarka Sen

Like all my answers, which are quality as you know

I have started using MSE as a blog basically :p

Semiclassical

and yet, does anyone actually use the MSE blog?

Balarka Sen

No lol

Mike Miller

4:21 AM

They actually killed the whole blogSE

@BalarkaSen "You have 7 days"

Semiclassical

last one is from 2015

math.blogoverflow.com

BAYMAX

hm

Semiclassical

@BAYMAX to get a sense of why the remainders approach is handy: Suppose you write down the first few instances of n^2-5 and see what their remainders mod 5 are

say, write down n^2-2 mod 5 for n=1 through 10

Mike Miller

BAYMAX

@Semiclassical 2,3,4

Semiclassical

4:27 AM

that's the first three. keep going.

BAYMAX

did for n=3 to n=7

Semiclassical

be more precise. What's the list you get?

BAYMAX

i GOT THE LIST OF REMAINDERS WHEN $N^2 -2$ IS DIVIDED BY 5 IS 2,3 AND 4

SORRY FOR UPPER CPS

Caps

Balarka Sen

@MikeMiller 7 days for what

BAYMAX

and the remainders keep repeating though, like where they should be repeating!!

@Semiclassical

Semiclassical

4:31 AM

Right. More precisely, you're getting 4,2,2,4,3,4,2,2,4,3,...

BAYMAX

left remainders are 0, 1

Semiclassical

so it's repeating every fifth term

by comparison, if you look at the remainders mod 5 for n=1 on up, you get

Mike Miller

@BalarkaSen

Semiclassical

1,2,3,4,0,1,2,3,4,0,...

Balarka Sen

Oh

Shite

Now I gotta be fast

BAYMAX

4:33 AM

Yup, @Semiclassical

Semiclassical

so it has the same periodicity, and the remainder of n mod 5 identifies what the remainder of n^2-2 mod 5 will be

if you can prove that, then the fact that n^2-2 is never divisible by 5 follows from what you've just observed

BAYMAX

yup

when we dive n by 5

it leaves remainder 0,1,2,3,4

now when we do n^2

it should leave remainders 0,1,4,4,1

Semiclassical

right

BAYMAX

when we do -2

leaves us with -2,-1,2,2,-1

and when we do mod 5

3,4,2,2,4

so never with remainder 0

nor 1

Semiclassical

there's a whole realm of number theory in this vein: Which numbers occur when you take n^2 mod p?

As you've seen, you get 0,1,4 for squares mod 5 but not 2,3

BAYMAX

4:37 AM

nice!!

Semiclassical

So 0,1,4 are said to be quadratic residues mod 5

and 2,3 are quadratic nonresidues mod 5

looots of stuff known about that: en.wikipedia.org/wiki/Quadratic_residue

BAYMAX

why you included 4 in quadratic residues?

Semiclassical

(nothing I'm terribly an expert in myself)

Because I'm talking about n^2 mod 5, not n^2-2

BAYMAX

ohh cool!

i had a question

For every $a,n \in \Bbb{N}$ with n greater than equalto 2, there exists k,l in N such that $n $ divides $a^k -a^l$

Semiclassical

You're trying to show that?

BAYMAX

4:42 AM

prove or disprove that -

1 min ago, by BAYMAX

For every $a,n \in \Bbb{N}$ with n greater than equalto 2, there exists k,l in N such that $n $ divides $a^k -a^l$

Semiclassical

Ah.

Where does N start for you, at zero or one?

BAYMAX

starts from 0

Semiclassical

okay

in terms of modular arithmetic, you're wanting to show that you can always find k,l such that a^k = a^l mod n

(so long as n>=2)

Simplest proof I can think of is the pigeonhole principle: a^k mod n is some value from 0 to n-1

Actually. Hmm. I think one could make that work but I'm not totally sure it's valid come to think of it

BAYMAX

hmmm

Semiclassical

i mean, the basic observation is this: Suppose you start writing the various values for a^k mod n

If you ever see the same value twice in that list, then take those indices as k,l to get an example

And the fact that the values you get mod n must be between 0 and n-1 means that you eventually have to repeat something in that list

you can't write down an infinite list of numbers using 0,1,2,...n-1 without eventually repeating yourself

The pigeonhole principle is the relevant concept from combinatorics.

BAYMAX

4:51 AM

0

Q: Divisibility of the difference of powers

Consider the following theorem: For any $a, b \in \mathbb{Z}^+$, there exist $m, n \in \mathbb{Z}$ such that $m > n$ and $a\ |\ b^m - b^n$. What's the best way to prove it? I have an idea (and I know it's true because of that idea), but I don't know how rigorous it is to constitute a proof.

elementary-number-theory

Semiclassical

oh hey

that looks pretty much what I was sketching

BAYMAX

says it is enough to ocnsider $0 \leq n,m \leq a$

Semiclassical

for reference:

Pigeonhole principle

In mathematics, the pigeonhole principle states that if n {\displaystyle n} items are put into m {\displaystyle m} containers, with n > m {\displaystyle n>m} , then at least one container must contain more than one item. This theorem is exemplified in real life by truisms like "in any group of three gloves there must be at least two left gloves or at least two right gloves". It is an example of a counting argument. This seemingly obvious statement can...

BAYMAX

but for m ei ht equestion it has only given distinct k and l

so i think if we add the condition

$0 \leq k,l \leq n$

then only the statekment would be true?

Semiclassical

You can assume without loss of generality that $k>l$

BAYMAX

4:56 AM

okay but i was interpreting this case for my problem -

3 mins ago, by BAYMAX

says it is enough to ocnsider $0 \leq n,m \leq a$

for my problem this condiiton would be $0 \leq k,l \leq n$

Semiclassical

Sure.

They're arguing that it's sufficient to consider that.

BAYMAX

but i was saying that since in my question it says that distinct k,l but has never givien such conditon

so it should be false right

Semiclassical

No.

BAYMAX

oh ok got it!

Semiclassical

Their point is that you can find $k,l$ in that range, then you can certainly find them without that condition.

Amit

4:59 AM

@MikeMiller Thanks @MikeMiller

BAYMAX

oh ok

so we need remainder(a^k / n) = remainder(a^l / n)

Semiclassical

So for instance, if n=100, you only need to write down a^k mod 100 for k=0 to 100. You're guaranteed to repeat remainders mod 100 somewhere along that list.

You can of course find examples with k,l larger than 100. but you only need one such example, so you don't need to look farter than that

BAYMAX

but seems to be much experimenting, like no rigorus proof?

seems would need to code a program

Semiclassical

Nah. The pigeonhole principle makes it precise.

Make a set of n boxes, one for each possible remainder mod n.

Then start writing down powers of a, and putting them into boxes depending on which remainder you get.

Suppose you do that a total of n times (so a^k for k=0 through n-1) so that you've inserted n possible numbers into those boxes.

BAYMAX

ok

Semiclassical

5:07 AM

At this point, it's conceivable that no box has more than one number in it: one number for each of the n boxes.

BAYMAX

yup

Semiclassical

however, if I then write down even one more power, then i must put it in an already occupied box

and therefore I'll have my repeat.

So I only need to consider n+1 different powers. Therefore, it suffices to consider k=0 through n.

BAYMAX

yup

Semiclassical

Hence I must repeat a box eventually. Since whichever powers $a^k,a^l$ which I put into that same box have the same remainder mod $n$, we have $a^k=a^l$ mod $n$

i.e. $n$ divides $a^k-a^l$.

This is essentially the idea which is in the earlier answer, just with the pigeonhole principle being used explicitly

Mike Miller

@BalarkaSen What do you know about spectral sequences?

BAYMAX

5:11 AM

I see, thanks@Semiclassical

Balarka Sen

I have done one or two computations

But I am not at all comfortable with it

Mike Miller

Do you want to do the Serre SS computation from that post with me?

Balarka Sen

I know the Serre spectral sequence and for simple enough bundles I can get the E^infty page

Erm not right now though

Mike Miller

Okie doke

Balarka Sen

6:33 AM

@MikeMiller So $\Bbb Z_2$ acts freely on $S^2 \times S^2$ and quotient is $X$. I take the classifying map $X \to B\Bbb Z_2$

Homotopy fiber is $S^2 \times S^2$

Mike Miller

@BalarkaSen Start by telling me what SS starts from and where it goes to

In general

Balarka Sen

First of all, does the base act trivially on cohomology of fiber? $\pi_1 B\Bbb Z_2 \cong \Bbb Z_2$ and the action of $1$ is just by $f_*$ where $f$ is the self-map of $S^2 \times S^2$ given by $f(x, y) = (-x, -y)$, I guess?

That's not trivial on $H_2$... what am I messing up?

Mike Miller

You're correct that's not trivial on $H_2$

What's the statement of SSS you know?

Daminark

By the end of this conversation it'll be "So we compute SSSSSSSS..."

Silent

I want to know Fundamental theorem of calculus better, hence I was staring at this seductive pic. What I can't understand is in $\int_a^b f(x)\,dx=F(b)-F(a)$, why do we need to subtract $F(a)?$ Can I see that in the linked pic?

Mike Miller

6:38 AM

I am just inconsistent between Serre Spectralsequence and Serre Spectral Sequence

The latter is better because I do not believe it abbreviates to a well-known nazi organization

Balarka Sen

I have a fiber bundle $F \to X \to B$ such that $\pi_1 B$ acts trivially on $H_* F$. Then there's a spectral sequence with $E^2$ page $H_q(B; H_p(F))$, differentials going diagonally, such that there is a filtration of $H_n(X)$ with successive quotients being $E^\infty_{k, n-k}$.

Homology of $E^i$ is $E^{i+1}$, $E^\infty$ is whatever survives at all pages

Mike Miller

Ahhh. Well, if you follow through the proof, there is no difficulty removing that nasty assumption.

$\pi_1 B$ does not need to act trivially on $H_* F$. Rather, that becomes a local system on $B$. The SSS still starts from $H_q(B; H_p F)$.

No change at all in the proof, just epsilon more work identifying the $E_2$ page (which you can still do by hand).

Balarka Sen

Ahh, I know that one

I have encountered it while going through the proof in de Rham context.

Mike Miller

This is the version I want to use today

Balarka Sen

OK, I don't know how to compute cohomology with local coefficients though, but I will learn that today I suppose. Go slow!

Mike Miller

6:44 AM

I am going to cheat.

So, what is the $E_2$ page in this case?

Not explicitly yet, just abstractly.

Balarka Sen

OK, $E^2_{p, q} = H_p(\Bbb{RP}^\infty; H_q(S^2 \times S^2))$.

For $q = 1$ and $q > 4$ it's all 0

Mike Miller

1) I guess we should probably know the action, huh?

2) is there another name for this?

Roll up and smoke Adjoint

Alpha Testers Please:

zoomspace.bubbleapps.io

ZoomSpace my app

Desktop app

No viruses

:D

@Daminark

@BalarkaSen

Balarka Sen

Before thinking about those two, it's also easy for $q = 0, 3, 4$, right? $H_3 = 0$ by PD and the action on $H_0$ because nothing happens to connected components when we flip, and $H_4 \cong H^0$ (or, nothing happens to fundamental class/orientation when we flip)

So $q = 2$ is indeed the only problem, where we have to compute actual twisted homology instead of standard vanilla singular homology.

Roll up and smoke Adjoint

@BalarkaSen nice avatar

@Daminark nice avatar, mspaint.exe?

Mike Miller

6:55 AM

@BalarkaSen Yeah, I agree.

Balarka Sen

So for $q = 2$, $\underline{\Bbb Z \oplus \Bbb Z}$ is the sheaf on $\Bbb{RP}^\infty$ you get when replacing the fibers of $S^\infty \to \Bbb{RP}^\infty$ by $\Bbb Z \oplus \Bbb Z$.

Kind of

I don't know how to say it.

Mike Miller

I know what you mean, but you're going slightly off game here.

What's special about the base?

(Also, very precisely what is the action of $\pi_1 B$ on $H_2$? I'd like to see this presented as a standard abelian group with an endomorphism,)

Balarka Sen

(I think I meant that the pullback of the sheaf to $S^\infty$ is the constant sheaf with values in $\Bbb Z^2$) The action of $\Bbb Z_2$ on $\Bbb Z^2$ is $1 \cdot (x, y) = (-x, -y)$ - the diagonal action of $\Bbb Z_2$ by the only nontrivial automorphism of $\Bbb Z$.

Do you want me to say that the sheaf trivializes away from $\Bbb{RP}^1 \subset \Bbb{RP}^\infty$? I guess it's not clicking with me.

I guess I have a homomorphism $H_*(S^\infty; \Bbb Z^2) \to H_*(\Bbb{RP}^\infty; \underline{\Bbb Z^2})$ by the pullback thing I said, by functoriality of sheaf cohomology.

Should I try to understand this homomorphism?

Mike Miller

7:13 AM

No

I want you to tell me what you know about RP^infty

(Don't think cohomology)

(Homology of $S^infty$ is zero)

Balarka Sen

@MikeMiller True, dumb of me.

Even for singular homology that hom gives no information

(Well, by Bockstein it does, but whatever)

Mike Miller

RP^infty is very special and we will exploit this

Balarka Sen

Do you want me to do the explicit computation, making $\Bbb Z^2$ a $\Bbb Z[\Bbb Z_2]$-module using the action, and taking tensor product with the cellular chain complex of $S^\infty$?

Oh, the local homology in this case is a certain group homology, I suppose.

Mike Miller

That's what I wanted!

Balarka Sen

It's $H_*(\Bbb Z^2; \Bbb Z[\Bbb Z_2])$ isn't it?

Mike Miller

7:23 AM

A local system over K(G,1) is a G-module A, and the local system (co)homology is group (co)homology with coefficients in A.

Balarka Sen

Right. Nice!!

(A is like a bundle over G, so it makes sense)

Mike Miller

Yup. Now, the way you just suggested is the way to do the computation explicitly - use cell structure of $S^\infty$.

Right, a G-module is a bundle over BG

But you could also just use a link I was about to provide that gives group cohomology of cyclic groups.

9

Q: Group cohomology of the cyclic group

It is well known how to compute cohomology of a finite cyclic group $C_m=\langle \sigma \rangle$, just using the periodic resolution, $\require{AMScd}$ \begin{CD} \cdots @>N>> \mathbb Z C_m @> \sigma -1>> \mathbb Z C_m @>N>> \mathbb Z C_m @> \sigma -1>> \mathbb Z C_m @> >> \mathbb Z \en...

finite-groups cohomology group-cohomology

Tbh I just go there whenever I need one of these calculations.

Balarka Sen

Cool. It all makes sense.

Should I write down $E^2$ explicitly now?

Mike Miller

So now it's time to use that to pencil in your diagram

Yup

For fun, do it with Z and also Z/2 coefficients

Actually F where F is not characteristic 2 is also fun

Balarka Sen

I'm heading for lunch now; I'll try to write it down. Computing the group homology will take me time for sure since I have forgotten all algebra

Mike Miller

7:35 AM

I was suggesting just using the computation there

Balarka Sen

But I see that the rest is not hard; the nonzero rows of the $E^2$ page alternates, so $E^2 = E^\infty$, isn't it?

@MikeMiller Erm I don't see how the link computes $H_*(\Bbb Z^2; \Bbb Z[\Bbb Z_2])$

Mike Miller

I am doing cohomology and it does it with all coefficients

Also, thYs not the correct Z/2 action

Balarka Sen

Oh, I have the homology groups wrong.

Mike Miller

Also the thing you're thinking of is that if some page is concentrated in even bidegree (2p, 2q), all differentials are zero - but there could be d_3 or d_5 in our picture

Balarka Sen

Oh, $E^2 = E^3$.

Mike Miller

7:39 AM

There must be some, in fact. It converges to the cohomology of a 4-manifold!

Yup!

Balarka Sen

9:13 AM

@MikeMiller I don't get it. $H_n(\Bbb{RP}^\infty; \underline{\Bbb Z^2})$ is the cohomology of $C_*(S^\infty) \otimes_{\Bbb Z[\Bbb Z_2]} \Bbb Z^2$. In the standard cellulation of $S^\infty$, $C_*(S^\infty)$ is $\cdots \to \Bbb Z^2 \to \Bbb Z^2 \to \Bbb Z^2$, where every map $\Bbb Z^2 \to \Bbb Z^2$ sends $(1, 0)$ and $(0, 1)$ both to $(1, 1)$.

Where standard cellulation of $S^\infty$ is lift of the cellulation of $\Bbb{RP}^\infty$ to $S^\infty$, so in each dimension I have two cells.

So when I tensor with $\otimes_{\Bbb Z[\Bbb Z_2]} \Bbb Z^2$, doesn't it remain the same chain complex, and homology is zero??

I should take the augmented $C_*(S^\infty)$, which is $\cdots \to \Bbb Z^2 \to \Bbb Z^2 \to \Bbb Z \to 0$.

And the group homology is $H_*(\Bbb Z[\Bbb Z_2]; \Bbb Z^2)$, right?

Er, $H_*(\Bbb Z_2; \Bbb Z^2)$ I mean

No, in $\Bbb Z^2 \otimes_{\Bbb Z[\Bbb Z_2]} \Bbb Z^2$, $\Bbb Z_2$ acts on the two copies of $\Bbb Z^2$ in very different ways. In the first copy, $1 \cdot (x, y) = (y, x)$. In the second copy $1 \cdot(x, y) = (-x, -y)$

I'll try this computation out later today, I need sleep before this. Too many different actions, too many Z's, too many 2's

But it's $H_*(\Bbb Z_2; \Bbb Z^2)$ we are computing, that's for sure.

user10354138

9:43 AM

...

@Haran here?

Secret

> ... THIS SEEMS PLAINLY ABSURD; BUT WHOEVER WISHES TO BECOME A
PHILOSOPHER MUST LEARN NOT TO BE FRIGHTENED BY ABSURDITIES.

-- BERTRAND RUSSELL

In fact, a good philosopher embraces absurdity without losing sense

Mike Miller

@BalarkaSen Yeah. The action is by negation.

Mats Granvik

11:18 AM

.

Leaky Nun

,

Secret

Creation retold under ZFC. 2nd attempt:

In the beginning, there is God (some kind of meta thing that ensures the existence of mathematics itself)

Day 1, God created classical logic

Day 2, God created nothing $\varnothing$

Day 3, classical logic defined induction and the Peano axioms

Day 4, God created the union of sets

Day 5, God created unordered pairs

Day 6, From nothing, Union and unordered pairs, by induction, comes something, the finite ordinals and finite sets

Day 7, God created subsets and powersets

Day 8, God banish sets that contain itself

Day 9, God made all sets unique

Day 10, God ensures all definable functions maps to subsets

Secret

12:01 PM

Day 11, God created the first infinity, $\omega$

On the final day, God choose to well order all his creations. And he then go to rest

Holo

2:01 PM

@Secret ZFC=>PA, but to create ZFC we don't need them

Also, unordered pairs and union of unordered pairs is meaningless(does not add anything) if you have union

Secret

Well, I am trying to say axiom of pairing when I mentioned unordered pairs

user 1357113

2:28 PM

math.ucla.edu/news/ciprian-manolescu-receive-2019-moore-prize

5

Holo

3:09 PM

@Secret Oh

Dude156

3:57 PM

hey guys

Hey I have a question, why are we allowed to cancel common factors in division?

user21820

4:12 PM

@Dude156 Because a/b = c means ( a = b·c and b ≠ 0 ), so you can see what happens with common factors.

newUser

4:35 PM

Hi. Do you know if the result is correct or not? math.stackexchange.com/questions/3032571/question-on-forms

Tobias Kildetoft

5:24 PM

Am I missing something, or is the claim asked about in math.stackexchange.com/questions/3032421/… wrong? It seems to me that there should be an embedding of rings, since one is the algebraic closure of the field of fractions of the other (assuming choice).

Leaky Nun

@TobiasKildetoft sure there is an embedding of rings but not R-algebras

Tobias Kildetoft

@LeakyNun Right, that was my conclusion (the embedding of rings is just weird of course)

Leaky Nun

@TobiasKildetoft I'm trying to formalize the statement $R[S][T] = R[S \cup T]$

and I'm not sure how to state it

$i:R \to A$ is an $R$-algebra, and $S, T \subseteq A$

so naturally $R[S \cup T]$ is an $R$-subalgebra of $i$

what is $R[S][T]$?

Alessandro Codenotti

Funny problem, at first I thought I need a trascendental over $\Bbb R$ in $\Bbb C$ to play the role of $x$, which is impossible, but there actually is such an embedding, with a bit of choice of course

Leaky Nun

if we want $R[S][T]$ then first we need to make an algebra $i^S : R[S] \to A$

which is possible because $R[S]$ is a subring of $A$ (but in doing so we forgot that $i_S : R \to R[S]$ is a subalgebra of $i$)

and at this point I am way too hungry to think about all this nonsense

user193319

5:37 PM

Problem: Given spaces $X,Y$, let $[X,Y]$ denote the homotopy classes of maps of $X$ into $Y$. If $X=I = [0,1]$ and $Y$ is path-connected, prove that $[X,Y]$ has a single element. Attempt: Let $f : I \to Y$ be some continuous map. It suffices to show $f$ is homotopic to the constant path from $y_0 \in Y$ to $y_0$. For every $s \in I$, there is a path $p_s$ from $f(s)$ to $y_0$. Consider $H : I \times I \to Y$ defined by $H(s,t) = p_s(t)$. Then $H(s,0) = p_s(0) = f(s)$ and $H(s,1) = p_s(1) = y_0$.

But it isn't obvious that $H$ is continuous.

How do I show that $H$ is continuous?

Leaky Nun

1. $i^R_A : R \to A$
2. $j^R_S : \operatorname{sub}(i^R_A) : R \to R[S]$
3. $i^R_S : R \to R[S]$
4. $R[S] \subseteq A$
5. $i^S_A : R[S] \to A$
6. $j^S_T : \operatorname{sub}(i^S_A) : R[S] \to R[S][T]$
7. $(i^R_S)^{-1} (j^S_T) : \operatorname{sub}(i^R_A) : R \to R[S][T]$

so somehow I need to create pullback of subalgebras

Alessandro Codenotti

Sanity check: The product in $\mathsf{Sch}/S$ is the fibered product over $S$ in $\mathsf{Sch}$, right?

Leaky Nun

@AlessandroCodenotti yes

depending on the definition of "is"

you have a forgetful functor Sch/S -> Sch

that sends product to fibred product over S

Alessandro Codenotti

Yes, yes, modulo technicalities

Leaky Nun

sorry I've been thinking about technicalities too much

Alessandro Codenotti

5:48 PM

Because if I draw the diagram for a product and then add the two bonus arrows toward S I immediately get that the product in Sch/S has the same universal product as the fibered product in Sch

Leaky Nun

right

see Proposition 3.6

blue_eyed_...

6:11 PM

0

Q: Solve the following system of equations using Gaussian Elimination Method

Solve the following system of equations using Gaussian Elimination Method $$x+2y+3z=2$$ $$x+y-z=1$$ $$2x+3y+2z=3$$. My Attempt : $$x+2y+3z=2………(1)$$ $$x+y-z=1…………(2)$$ $$2x+3y+2z=3………(3)$$ Subtracting equation $(1)$ from equation $(2)$, we have $$y+4z=1………(4)$$ Multiplying equation $(1)$ by $2...

systems-of-equations gaussian-elimination

Logarithmic Derivative

7:12 PM

I've kind of done the first one, but I'm a bit confused over the second one. Any ideas?

Akiva Weinberger

7:43 PM

What does it mean when a question is "protected"

Oh I see

user10478

8:11 PM

Hello, is the matrix A here (youtube.com/watch?v=oaiiyIsbNdI) always the Hessian matrix of the quadratic form?

s.harp

8:48 PM

@user10478 im not watching your link, but any symmetric matrix $A$ is the hessian of the quadratic form $x\mapsto x^t A x /2$

Evinda

Hello

Let $f: \mathbb{R} to \mathbb{R}$ with the properties: $f(x)>0, \forall x \geq 0$, $f$ is decreasing and $f'(0)=0$. I want to prove that $f''(x)=0$ for some $x>0$.

Could you give me a hint?

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