Mathematics - 2024-08-18

leslie townes

00:05

jam: where do you get your expectation that the quotient ring would be isomorphic (as a ring) to ZxZxZ?

leslie townes

01:41

jam: math.stackexchange.com/questions/1065495/… indicates that Z[x]/(x^2 - 1) has no nontrivial idempotents (suggesting that it is not isomorphic as a ring to Z x Z, which has nontrivial idempotents like (1,0) and (0,1))

冥王 Hades

I love beige, bland food....

Xander Henderson

02:02

@冥王Hades sucks to be you. :P

Personally, I often like food that is so hot, it still hurts a day or two later.

zeta space

wow

Xander Henderson

On the other have, you might really enjoy the typical American Thanksgiving meal. Lots of beige.

zeta space

I like chicken (the good kind) and Euler characterstics

冥王 Hades

@XanderHenderson Yes I do, it's probably my favorite time of the year actually.

Xander Henderson

Though I tend to add more flavors. Paprika in the gravy, garlic and herbs in the potatoes, pineapple and maple syrup in the yams, etc.

冥王 Hades

02:05

A lot of bland meat, mashed potatoes, the whole 9 yards

zeta space

does bland mean lack of taste?

冥王 Hades

Not necessarily

zeta space

lack of flaovr?

Xander Henderson

Thanksgiving is, honestly, the only holiday I really celebrate in a year.

And I get to do it twice this year!

@zetaspace Fundamentally, yes.

zeta space

thanksgiving is nice

冥王 Hades

02:06

I like cranberry sauce too

zeta space

I don't like the meal very much

Xander Henderson

But also lack of texture. Lack of anything particularly interesting.

@冥王Hades I make a killer cranberry sauce.

Lots of orange zest, and ginger.

冥王 Hades

Also the pumpkin pie, though I usually have little room left for dessert after devouring a turkey

Xander Henderson

And I made a really good blue corn bread last year. It was cheesy and spicy (I put in a healthy amount of cayenne).

@冥王Hades pecan > pumpkin

But pumpkin is still good. Maybe my fourth or fifth favorite pie.

冥王 Hades

Yeah I've had pecan pie as well, pretty good.

zeta space

02:09

I won't be able to fit in my favorite pair of pants after thanksgiving

Xander Henderson

And I eat a lot of pies, all year long. Pie is the best. Cake is garbage.

zeta space

cake on thanksgiving is not right

冥王 Hades

@XanderHenderson The best cake on earth > The best pie on earth

Xander Henderson

Cake at any time is not right.

@冥王Hades WRONG!

Cake is garbage.

zeta space

what about strawberry rubarb pie hades?

冥王 Hades

02:11

@zetaspace Never tried that

Xander Henderson

@zetaspace That's a good pie.

Blueberry pie with coffee ice cream.

冥王 Hades

@XanderHenderson Have you ever seen one of those ludicrous wedding cakes that are taller than most people?

Xander Henderson

Grapefruit merengue pie.

@冥王Hades those are generally terrible.

They spend a lot of time going stale while people decorate them.

冥王 Hades

Yeah those are pretty stupid and taste terrible, but hey, good cake is good cake

Xander Henderson

And they have to be sturdy enough to be structural, which generally makes then less good to eat.

冥王 Hades

02:13

They're liked failed ice creams.

Xander Henderson

But even the best cakes are still awful.

冥王 Hades

Of course the best pie is, by a long shot, chocolate pie.

Xander Henderson

@冥王Hades Meh. Chocolate pies are okay, but they tend to be a little rich for my taste.

冥王 Hades

Of course, rich pies are for rich people such as myself.

Xander Henderson

You can have them.

Have you ever had chess pie?

zeta space

02:17

blackberry pie with real ripe black berries

Xander Henderson

@zetaspace I made a pie last week with berries I picked myself. Wild blackberries. So good.

冥王 Hades

@XanderHenderson Yes actually, back when I was in Texas

Xander Henderson

@冥王Hades but I thought you were rich? Chess pie is a Depression Era pie for poor people.

:P

冥王 Hades

@XanderHenderson Nice one. But it did taste nice. No reason rich folks can't enjoy "poor man's food"

I eat carrots, poor man's apple.

Xander Henderson

@冥王Hades haha!

Carrot cake is almost acceptable. It can be an honorary pie.

冥王 Hades

02:25

@XanderHenderson My girlfriend used to love that.

Still miss her

zeta space

Me too

I miss carrot cake

me mother used to make for us as kids

we were poor - sometimes the cake was missing the carrots

we lived in a shoebox on the side of the road

Four Yorkshiremen- Monty Python

►

leslie townes

02:44

jam: math.stackexchange.com/questions/2830085/… in qiaochu yuan's answer gives conditions on the ring R and positive integer n under which R[x]/(x^n - 1) will necessarily be isomorphic to R^n, suggesting at least some basis for an idea that this ought to be the case in general. but the answer also notes (in comments) that these rings are not isomorphic in the case R = Z and n = 2

Zac U.

03:17

Could a specific 2x2 matrix filled with 4 integers be useful as a parameter anywhere in higher math or physics?

Awhile ago I stumbled on a sort of potential coincidence which after four years of searching I think mustn't be developing in the way I originally thought it might and perhaps instead the entire set of four numbers must be the interesting component

one potato two potato

11:24

peep

nickbros123

11:51

If I have a cyclic group $G= \langle a \rangle:= \{a^z: z \in \mathbb{Z}\}$, then any subgroup would be cyclic. To prove this, I took an arbitrary subgroup $\{e,a^{j_1},a^{j_2},a^{j_3} \cdots \}$ where $j_1,j_2, \cdots$ is a sequence(possibly finite) of integers. Then 2 cases come up: if any pair $j_i, j_k$ has gcd$(j_i,j_k)=1$, then the subgroup becomes the main group itself. the other case is that every pair of indices have non unity gcd.

A subcase to this is, every index is a multiple of some common index. The other case is, there exists two pairs of indices $n_1,n_2$ and $k_1,k_2$ so that $(n_1,n_2) \neq 1$, $(k_1,k_2) \neq 1$ but $((n_1,n_2) ,(k_1,k_2) )=1$. In this case as well, the subgroup becomes the main group. Am I missing any cases?

also it appears quite powerful that if any two elements in the subgroup have indices that are coprimes, then the subgroup would necessarily have to be the main group

I hope im not making an y mistake here

psie

@Jakobian do you think you could explain what you mean by "a copy of $X$ in $\tilde X$"? $X$ is a set of points, $\tilde X$ a set of equivalence classes, i.e. a set of Cauchy sequences in $X$. Do you mean that we treat $X$ as a set of constant sequences, a sequence for each point?

zeta space

12:27

Can a math object possess local and global symmetries i.e. $(X,G,V)$ where $X$ is a surface which globally has discrete group symmetries i.e. $G$ but locally you can put a Killing field (vector field which preserves the metric)?

Ryder Rude

in formal math, "forall" actually means "for each", as e.g. $\forall x \exists y$ means the $y$ that exists can be different for each $x$

how can we construct "forall" in first order logic

nvm i got it

just say $\exists y \forall x (x+y=x)$

zeta space

I can't wait for the Banach people to show up

Soumik Mukherjee

13:01

@psie I don't know what the original question is but you can consider any point as the set of equivalent classes of cauchy sequences converging to that point. Not necessarily only constant sequences.

psie

ok 👍

Jakobian

13:20

@psie $e[X]$

$\tilde{X}$ is not a set of Cauchy sequences in $X$ but such sequences up to equivalence

$X$ is not treated as set of constant sequences but equivalence classes of such sequences

psie

ah ok, I was sloppy

Jakobian

@nickbros123 this is wrong approach.

nickbros123

@Jakobian could u tell me where im goign wrong

Jakobian

Everywhere

Start over

nickbros123

I do have another proof: I assume $a^j$ is in my subgroup for some $j \neq 0$. I look at the minimal $m$ so that $a^m \in $ subgroup. Then I argue that $a^m$ is the generator of subgroup

using euclids lemma

Jakobian

13:32

You mean that $m$ is the minimal positive such number?

nickbros123

hmm, yeah.

Jakobian

That's correct approach

nickbros123

yh so I can also assert that theres a positive $n$ so that $a^n$ where $n>m$ is in the subgroup, then $a^{mq+r}=a^n$ would force $r$ to be 0

Jakobian

You mean $n\geq m$?

Xander Henderson

@psie It helps to think of something more concrete. The real numbers can be constructed as a set of equivalence classes of of Cauchy sequences of rational numbers, right?

nickbros123

13:38

yes, but for the sake of a contradiction $n>m$

Xander Henderson

(Where equivalence is up to a null sequence.)

Jakobian

But any inequality on $n$ is irrelevant anyway

@nickbros123 what contradiction

Xander Henderson

It is then possible to embed the rational numbers in the real numbers by saying that a rational number $q$ is the real number with equivalence class represented by the constant sequence $(q)$.

That is, $\mathbb{Q} \ni q \mapsto [(q,q,q,q,\dotsc)] \in \mathbb{R}$.

Jakobian

You simply take an element $a^n$ in your subgroup and write $n = mq+r$ where $0\leq r < m$. Then you argue that by minimality of $m$, $r = 0$

nickbros123

@Jakobian we want to show that the elements in the sub group are of the form $(a^mj)$, and we take arbitrary $a^n$ so as to prove this. We already asserted m is the minimal positive integer so that $a^m$ is in the group. if the $n$ we take is equal to $m$, that wouldnt be helpful

anyways this is fine

Xander Henderson

13:41

Where the equivalence class $[(q)]$ is the set of all Cauchy sequences $x = (x_1, x_2, x_3, \dotsc)$ such that $(q-x_1, q-x_2, q-x_3, \dotsc)$ is a null sequence (i.e. $|q-x_j| \to 0$).

Jakobian

@nickbros123 it wouldn't not be helpful

It's additional noise

psie

What I don't understand about this completion business is the difference between (using my notation) $Y$ and the auxiliary complete space $\tilde X$, the set of all equivalence classes of Cauchy sequences. They both seem to be complete. Why introduce $Y$ at all?

Xander Henderson

@psie I haven't read all the way back.

Jakobian

@psie what did I tell you that you are essentially trying to show

psie

That they are equal up to isometry, or? I have forgotten what that means exactly.

Jakobian

13:44

That $Y$ is, up to unique isometry that preserves $X$, equal to $\tilde{X}$

In general you won't be constructing a completion, you might just be given one

And $Y$ represents this arbitrary completion.

The whole point is to see that the choice of completion is irrelevant

This is why you introduce $Y$, at all

Xander Henderson

@psie I am not going to read all the way back through the conversation to figure out what all of your notation means. The big idea in this part of analysis is that there are a lot of things which can be treated as the "same", so just do that---treat them as the same.

Jakobian

@XanderHenderson Essentially this is part of the proof of uniqueness of completion and $Y$ is some completion of $X$, for context

Xander Henderson

@Jakobian Yeah, I was kind of getting that idea. Where "uniqueness" is up to isometry.

Category theorists have a pretty diagram for explaining this. Something something universal property.

psie

ok, thanks. I feel like I am not done yet in my understanding of this, but I'll take a break here I think.

Xander Henderson

14:01

@psie Starting from first(ish) principles, the idea is that there are these things in the world called "complete metric spaces", which have the nice property that Cauchy sequences converge (the ur-example is $\mathbb{R}$). Then there are other metric spaces which are not so nice (such as $\mathbb{Q}$), in which Cauchy sequences may not converge.

As natural question to ask is "Given an incomplete metric space $X$, is it possible to find a metric space $Y$ such that $Y$ is complete and $X$ is "nicely" embedded into $Y$? If so, is there a smallest such space? If so, is this space unique in any way?"

psie

those are deep questions I think :)

Xander Henderson

So, start with an abstract definition: given a (possibly incomplete) space $X$, the completion of $X$ is a space $Y$ such that (1) there is an isometry $X \to Y$, and (2) if $Y'$ is any other space such that there exists an isometry $X \to Y'$, then there is an isometry $Y \to Y'$ (in this sense, $Y$ is the "smallest" space into which $X$ can be isometrically embedded).

(I'm doing this off the top of my head, so there may be errors in the above, but I think that is the basic idea).

Note that this is an abstract definition---nowhere in that definition do I claim that any such space $Y$ exists.

In principle, it might be possible to find several complete metric space into which $X$ embeds, but all of these spaces are incompatible, in the sense that none embeds into another. Or there may be spaces $X$ which cannot be embedded into a complete metric space.

So one has to show that this definition isn't useless.

Step 1 is usually to show that $X$ can always be embedded into at least one complete metric space. The standard "trick" is to take $\mathscr{C}(X)$ to be the set of all Cauchy sequences in $X$, then quotient out the null sequences $\mathscr{N}(X)$, giving the space $\tilde{X} := \mathscr{C}(X) / \mathscr{N}(X)$.

It is not too difficult to show that $\tilde{X}$ is a complete metric space, nor is it too difficult to show that $x \mapsto [(x)]$ is an embedding of $X$ into $\tilde{X}$.

psie

ok 👍

Xander Henderson

Step 2 is to show that $\tilde{X}$ is the completion of $X$, i.e. that if $X$ can be embedded into any other space $Y$, then $\tilde{X}$ can also be embedded into that space.

(oops---mashed the keyboard)

zeta space

14:26

$E$ is a space that locally (in region $A$) looks like 8 pairs of pant sewn together i.e a ball with 8 very erected cylinders.

Xander Henderson

@zetaspace 8 pair of pants would be 16 cylinders.

"Pair" means two.

Though "pear" means delicious fruit.

And "pare" means to remove the excess.

zeta space

@XanderHenderson right, i meants 4 pairs pants sewn together at the waists giving 8 cylinders

Consider $\Bbb R^3$ and associate to each unit $3$-cell i.e. $X=[0,1]^3$ four intersecting cylinders which collapse to points precisely at $\Bbb Z^3$ (one cylinder collapses at $(0,0,0)$ and $(1,1,1)$ another collapses at $(0,1,1)$ and $(1,0,0)$ etc.). One can delete the singular sets, amounting to $\chi=\Bbb R^3-\lbrace\Bbb Z^3\rbrace$ and then there some curves isotopic to $S^1$ contributing due to the four cylinder intersection in the interior of each $X$. Delete those getting $\beta=\chi-\lbrace S^1 \cup S^1 \cup\cdot\cdot\cdot \rbrace$.

pretty cool surface

And BY THE WAY - this is an abstract definition

the best way to describe it is a "neuron-network" because it looks like neurons in ones brains

Jakobian

15:05

@XanderHenderson a pair of pants is a cyllinder with a hole in it

Two holes if your zipper is open

zeta space

15:27

$$d(f_s, f_{s'}) = \left| \int_0^1 f_s(x) \, dx - \int_0^1 f_{s'}(x) \, dx \right|=\left| ||f_s||_M - ||f_{s'}||_M \right|$$

where

$$||f_s \cdot f_{s'}||_M \leq ||f_s||_M \cdot ||f_{s'}||_M$$

former is a metric

latter is a property that the norm satisfies

Xander Henderson

@zetaspace Gross. Use \lVert and \rVert for norms. Or \| if you are lazy, and don't care about spacing.

zeta space

$\lvert \lvert f \rvert \rvert$

looks the same lol

$$ \lVert f_s \rVert $$

I wonder where this comes $\lVert f_s \cdot f_{s'}\rVert_M \leq \lVert f_s \rVert_M \cdot \Vert f_{s'} \rVert_M$ $M$ here stands for multiplicative

oops I forgot to use lVert on that !

I wonder where that inequality comes into play

because usually I've seen additive norms

nickbros123

17:25

@Jakobian There seems to be also another way to prove this result (every subgroup of a cyclic group is cyclic), we can start by assuming $e, a^{j_1}, a^{j_2} \cdots $ are in your subgroup $H \leq G$ where $G=\langle a \rangle$. Case (1), There exists a finite subcollection of $j_1,j_2 \cdots $, call it $j_{n_1}, j_{n_2} \cdots j_{n_k}$ so that $gcd(j_{n_1},j_{n_2} \cdots j_{n_k})=1$.

From generalised Bezout's identity, we can find $x_1,x_2 \cdots x_k$ so that $1=x_1j_{n_1}+x_2j_{n_2} \cdots +x_kj_{n_k}$. This would mean that $H=G$ necessarily. Suppose then that for every finite subcollection of $j_1, j_2, \cdots$, the subcollection's gcd is not 1. This would mean that gcd(whole collection) != 1, which means that they all share a common factor. Hence, $a^{\text{that common factor}}$ would become the generator for this subgroup.

Jakobian

17:49

@nickbros123 why can't you just take $\gcd\{n : a^n\in H\}$ directly

but you're right that this is $m$ such that $m$ divides all $n$ with $a^n\in H$ and $a^m\in H$ (since $m\in \{n : a^n\in H\}$ from definition of $\gcd$)

and so $H$ is generated by $a^m$

I suppose there are cases to consider here but not the ones you mentioned

to assure $\gcd$ is well-defined one would need $H$ to be non-trivial

I suppose my approach was just reproving that $\gcd$ is the same as minimum

so this one is more direct, I agree it's better

user382894

20:27

Hello @XanderHenderson are you a moderator? If so, please note the answer in :mathoverflow.net/questions/477078/… . The account has made similar answers elsewhere

Xander Henderson

@user382894 I am not a moderator on MO. I would suggest that you flag the posts.

zeta space

21:06

3 people in chat

never seen that before

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