Mathematics - 2023-08-08 (page 1 of 2)

CroCo

00:22

no one is willing to help?

Axoren

00:36

@CroCo I think you might have abused notation. You do mean $M = \frac 1 2 \left( A + A^T \right) + \frac 1 2 \left( A - A^T \right)$?

No, actually, I think you may not have, looking at it again.

Yeah, that is true for $M = A$.

robjohn

@sunny Can you show that it is not uniformly convergent?

CroCo

This is my answer and I hope you give me your feedback
https://math.stackexchange.com/questions/4748751/when-is-a-a-top-psd-even-though-a-is-not-psd/4749269#4749269

mr_e_man

02:36

Algebraic number theory question: Are the fields $\mathbb Q(\cos(\pi/n))$ all distinct, for $n>3$?

In other words: We know that the cyclotomic fields $\mathbb Q(e^{i\pi/n})$ are all distinct (see Theorem 4.1 in Keith Conrad's notes). Are their real subfields also distinct?

Ted Shifrin

03:31

What is the quadratic extension that gives the complex field over the real?

Semiclassical

based on the stuff i've been rambling about lately: suppose i have some spherical quadrilateral on the unit sphere. if i swap two adjacent side lengths, is this also a valid spherical quadrilateral?

Ted Shifrin

No clue. Keeping the other sides fixed?

Semiclassical

right

fixed in length, that is

(I think you can do it even if you leave the other two sides unchanged entirely)

Ted Shifrin

What about Euclidean?

Semiclassical

for Euclidean you definitely can, and i think a similar proof would work for spherical?

Ted Shifrin

03:35

No clue.

Semiclassical

take the quadrilateral ABCD, and suppose you want to swap AB with BC. then draw the diagonal AC and its perpendicular bisector. i can then reflect ABC across said bisector to get A',B',C' with A'=C and C'=A by construction

so the quadrilateral AB'CD now has lengths |AB'|=|BC|, |B'C|=|AB|, |CD|,|AD|

i think that should also work for spherical, with an appropriate great circle serving as bisector

Ted Shifrin

Yeah, seems right.

Semiclassical

this feels like one of those old facts about spherical geometry that people knew a hundred years ago and just stopped caring about

mr_e_man

@TedShifrin - The extension is $\mathbb Q(e^{i\pi/n})=\mathbb Q(\cos(\pi/n))(\sqrt{\cos^2(\pi/n)-1})$.

Ted Shifrin

Write down the quadratic equation.

It should be $x^2-2\alpha x+1=0$, right?

So you want to consider $\beta \in \Bbb Q(\alpha)$. Should be decidable.

Hmm, I dunno. Maybe it’s harder.

leslie townes

04:07

not to be 'that guy' but don't keith conrad's notes imply that the fields Q(e^(i pi/n)) are not all distinct? theorem 4.1 says that there's a tiny amount of nondistinctness, in that you get the same thing for k and 2k, whenever k is odd.

i have nothing to contribute to the actual discussion.

mr_e_man

@leslietownes - Note that $\pi/n=2\pi/(2n)$, so we're actually dealing with even numbers.

leslie townes

or is that somehow wrapped up in what the definition of what mu_n is.

oh, i see.

Dannyu NDos

Is it decidable, given positive integers $m$ and $n$, whether $\pi_m(S^n)$ has Betti number zero?

Axoren

What was the format for \parray?

I just need a column vector.

The matrix notation seems heavyhanded.

Semiclassical

04:23

@Axoren pmatrix is what i always use for bigger columnvectors

sometimes i get lazy and use \binom for two-vectors

Axoren

I remember there was a shorthand in ChatJax for vertical, but I can't seem to remember it well enough

I thought it was \parray

$\parray{1 // 2 // 3}$

I'm clearly wrong of course, but I can't shake that there was a simple way

Ted Shifrin

You need \\, not //. But I do not know this syntax.

Axoren

$\array{1 \\ 2 \\ 3}$ Testing

It's just \array. Thanks Ted!

Ted Shifrin

no simpler than using bmatrix or pmatrix. …

Axoren

I didn't like the begin/end tags.

But it looks like parray and barray came from packages I was importing that chatjax might not have

leslie townes

04:41

@DannyuNDos this seems to be completely answered in the 'finiteness and torsion' section of en.wikipedia.org/wiki/Homotopy_groups_of_spheres

Dannyu NDos

Dang. So it wasn't worth for a challenge for Code Golf SE.

mr_e_man

As a special case of my question, can we prove that $\mathbb Q(\cos(\pi/7))\neq\mathbb Q(\cos(\pi/9))$? That seems like it should be easy.

leslie townes

well, i doubt that anything undecidable would be a good code golf challenge. although i somehow thought the homotopy groups of spheres were more along the 'really complicated' side of hard to compute than the 'undecidable' side, so maybe there is something worth asking about them.

PM 2Ring

@AkivaWeinberger Yes, exactly 100. Here are the complete faultline count stats. (h, v) means h horizontal faultlines, v vertical faultline.

(0, 1) : 50
(1, 0) : 50
(0, 2) : 258
(1, 1) : 600
(2, 0) : 258
(0, 3) : 186
(1, 2) : 1086
(2, 1) : 1086
(3, 0) : 186
(0, 4) : 5
(1, 3) : 400
(2, 2) : 1362
(3, 1) : 400
(4, 0) : 5
(1, 4) : 20
(2, 3) : 342
(3, 2) : 342
(4, 1) : 20
(2, 4) : 35
(4, 2) : 35
(2, 5) : 1
(5, 2) : 1

one potato two potato

betti number of homotopy group is not a big deal. One can change the problem to homology groups using Hurewicz map

Akiva Weinberger

04:53

@PM2Ring Ha! Way fewer than I would have thought. (That's worth posting as a comment, at least.)

What do the 5s and 1s look like?

PM 2Ring

@AkivaWeinberger I'll probably post an answer with various stats, and my scripts. Even if it's not a direct answer, it kind of qualifies as useful related material. ;)

I want to do stats for the actual faultline positions. But I got side-tracked...

Axoren

Is there a cleaner way to write out this definition of $\bold u$ and $\bold v$?

$$
\bold u, \bold v &\in \mathbb R^4 \\
\bold u &= \text{Rot}_\text{arb}^{(4\times4)} \bold v \\
\bold e_4 ^T \bold v &= 0 \\
\bold e_4 ^T\bold u &\geq 0 \\
||\bold v||_2 &= 1 \\
||\bold u||_2 &= 1
$$

u and v are both 4-dimensional real vectors, they differ by some arbitrary rotation, v has 0 as its 4th coordinate, whereas u has a non-negative as its 4th coordinate. They are unit vectors. It's been so long since I've done any math, I've forgotten how to notate things correctly.

PM 2Ring

@AkivaWeinberger What do you mean? There aren't any solutions with (1, 5) or (5, 1).

Axoren

Well, clearly I've screwed up.

Picture.

leslie townes

axoren: most of that looks fine to me but "differ by some arbitrary rotation" is not the kind of thing that one would usually convey only with notation. i'm also not sure that it has substance here, at least if "rotations" get to come from the usual place (i.e. orientation preserving isometries of R^4, where any two vectors of the same length would differ by some such thing)

Axoren

05:04

I guess that does follow from the equal norms.

PM 2Ring

@Axoren I can't see any obvious errors. But sometimes ChatJax clashes with Markdown, because Markdown uses underscore & asterisk.

leslie townes

if you are doing something weird where R^4 is sometimes thought of as "R^3 + 1" or something and only certain isometries are "rotations," maybe it has content.

Axoren

Is it understood that $u_4$ is the 4th coordinate of $\textbf u$ usually?

Because I don't like the multiplication by canonical vectors.

Dannyu NDos

$\mathbb{R}^4$ admits double rotation, no?

Axoren

Double rotation? I don't need a uniqueness condition.

leslie townes

05:07

that is up to you to decide. it is not uncommon to denote the usual basis for R^n by e_1, ..., e_n. although i might not do that if you planned to be considering different n's at the same time. or indeed, if you wanted to give a kind of uniform meaning to subscripts.

Axoren

I could always break out the monster $\ ^a_c \textbf v ^b_d$.

Don't know where I'd put a 5th script

Dannyu NDos

At least in layman's terms, "double rotation" is a rotation done in two independent axes. I thought it was relevant?

leslie townes

axoren: on the vertices of an inverted pentagram around v, of course.

Axoren

$\ ^a_e\left(\ ^b_f \textbf v ^c_g \right)^d_h$

Dannyu NDos

$\mathbb{R}^4$ and $\mathbb{R}^5$ admit double rotation, $\mathbb{R}^6$ and $\mathbb{R}^7$ admit triple rotation, $\mathbb{R}^8$ and $\mathbb{R}^9$ admit quadruple rotation, and so on; at least that's what I've known.

Axoren

05:10

For when you want to publish something to the trash.

It might not even get accepted, there.

Dannyu NDos

As for the notation for $\mathbb{R}^4$, Einstein summation convention is quite standard, if that's relevant for you.

Axoren

$$
\textbf u , \textbf v \in \mathbb R ^ 4 \\
u_4 \geq v_4 = 0 \\
||\textbf u||_2 = ||\textbf v||_2 = 1
$$

Might be for when I start actually applying transformations.

Is there a notation for an almost-identity matrix? (one of the 1s is missing)
Would it be something like this?
$$
\begin{bmatrix}
I^{(3\times3)} & 0 \\
0 & 0
\end{bmatrix}
$$
?

Or $I_3$ instead?

leslie townes

there are no universally adopted notational conventions at this level. people do use variations on that, including that, usually after explaining what their convention is. that would be a fine one.

if the dimension will not be varying you might suppress the 3x3 and just have an understanding that I means I of some fixed size.

similar thing with the euclidean norm - you might not decorate it with the subscript 2 if it is the only norm you will be using. to "declutter" the landscape a little.

alternatively, sometimes when it matters, people would clutter all of that with subscripts or superscripts. or at least a 0 and 0^T in upper right and lower left to indicate that those 0s are different 0s from the 0 in the lower right, which is the number 0.

people who do that often fix a 'style' of vector as unadorned (here maybe column vectors) and then adorn row vectors with the ^T. i find it distracting to do that with things like the 0 vector, but some people live by those kinds of rules.

Axoren

I'm seeing $\text{diag}(1, 1, 1, 0)$, which I might prefer.

leslie townes

05:26

you will see a lot of stuff, just keep in mind that none of it is universal, even if all of it is very close to self-explanatory.

Thomas Finley

0

Q: Show that the possible ideals of a ring of quaternions over integers modulo an odd prime is either the $(0)$ or the set itself.

Using the ring of real quaternions as a model, we define the quaternions over the integers $\text{mod}$ $p,$ $p$ an odd prime number, in exactly the same way; however, now considering all symbols of the form $a_0 + a_1i + a_2j + a_3k,$ where $a_0,a_1,a_2,a_3$ are integers mod $p.$ (a) Prove that ...

Help needed?

Trying for 2 days now...

leslie townes

thomas, i mentioned an issue with the attempt as currently written the other day. in a finite field, a being nonzero does not imply that the sum of squares of the components of a is also nonzero. that sort of inference is specific to real numbers or at least ordered fields, which finite fields are not.

i also renew my recommendation to think carefully about eric's hint (top comment).

in particular, you should abandon your program of attempting to prove that nonzero elements of this ring are necessarily invertible. that is a nice idea (and would establish the desired result if it were true) but this ring does not have that property. that intuition is specific to commutative rings.

ted, no ducks today, but two rabbits. did not catch either of them.

Axoren

$\text{diag}(I_3, 0)$ is probably the best I'll get that isn't confounded with something else obvious.

leslie townes

diag(, , , , ) with matrices inside for a block diagonal matrix is not something i have seen before. that is wild. you, sir, are a madman.

i like it.

Thomas Finley

05:48

@leslietownes well, but that's absurd. A zero element is, 0+0i+0j+0k. So, if a is an element of the ring and is non-zero then, all of it's component is not zero and as the components are integers modulo p i.e reals so, the sum of squares of components is definitely non-zero

@leslietownes Any idea about what's $i^{-1}$ ?

Axoren

The inverse element of $i$.

Thomas Finley

@Axoren wrt what?

Dannyu NDos

...which is $-i$.

Axoren

In a modulo space? There is another representation.

Thomas Finley

I know the multiplication table of quaternion units. But to speak about the inverse of a quaternion unit, we have to think the set of eight quaternion units as group under multiplication , @Axoren

Isn't it?

Axoren

05:54

For example, $2 \mod 5$ has an inverse that isn't $-2$, it's $3$.

So, the question to ask in regards to Eric's post is whether or not he meant the additive inverse or the multiplicative one.

Thomas Finley

@Axoren I think $i^{-1}$ in all probability means the multiplicative inverse of i, when the set of eight quaternion units are considered a group under multiplication with identity element 1

Axoren

The only structure you've assigned to them is that they are a Ring, which means the multiplicative operation does not necessarily have an inverse. I don't think you need to elevate the set to Octonions if that's what you mean.

Oh, you just meant the 8 quaternion unit vectors.

Thomas Finley

@Axoren yes

@Axoren that's the problem.

And this is where Eric's hint is nonsensical

I only have the info that R is a ring. Nothing more and nothing less

But then, no one reveals their work for some strange reason. Dunno but did it stumble everyone?

leslie townes

it's not a nonsensical hint. some elements of your ring have two-sided inverses, including i, and ^(-1) makes sense for those elements.

i asked before about what happens when you compute (i + j + k)^2 mod 3. i see you still haven't done that.

it's a lot easier to attempt to harangue people into helping you in the way that you want to be helped, than it is to listen to what people who were, in fact, trying to help have already suggested.

PM 2Ring

06:12

@AkivaWeinberger Here are the position stats for the faultlines. gist.github.com/PM2Ring/… Eg, (0, 1, 2, 3, 4) (1, 3) : 1 means there's one tiling with all 5 horizontal faultlines, and vertical faultlines at 1 & 3, i.e., the tiling with all dominoes horizontal.

Axoren

06:22

Okay, special relativity makes sense. That means I've been awake too long.

Good luck, all

Unknown x

07:14

@Unknownx So what is $\alpha(s)$? — Ted Shifrin Aug 3 at 18:43

$ $\alpha(s)$ is the integral of $T(s)$. Right?

for become a helix, what is the condition?

one potato two potato

I can't access the above Chat guidelines and Latex in chat sites

leslie townes

one: hm, is it tinyurl on the fritz? try math.meta.stackexchange.com/questions/3890/… and math.ucla.edu/~robjohn/math/mathjax.html

one potato two potato

Thanks for the links. I remember I could access long before.

leslie townes

08:38

haha, someone asked a day or two ago why MSE has all of these questions about inequalities in multiple variables that are symmetric in the variables, and now its all i can see. people can't stop posting them. e.g. math.stackexchange.com/questions/4749426/…

one potato two potato

08:58

arxiv.org/abs/math/0110143

Hmm

Dannyu NDos

Didn't actually read that, but is it like "nuking the ant house"?

one potato two potato

I didn't read that either

leslie townes

09:14

that would be a good stage act, whatever it is

Sourav Ghosh

Hi :)

Soumik Mukherjee

Hi

Seeing you after a long time

leslie townes

welcome back. seems like lot of people were 'on vacation' lately :)

冥王 Hades

10:08

Physics and engineering should just rename themselves to Differential Equations

Dannyu NDos

10:18

But Quantum Mechanics do that on Hilbert spaces.

sunny

10:39

@robjohn that's challenging, I guess one would have to resort to the negation of uniform convergence. Do you have quick solution? Otherwise, don't bother I'd say.

What I really want to show is $$\frac{d}{dx}\sum_{k=1}^\infty \log\left(1+\frac{1}{k^2x}\right)=-\frac{1}{x}\sum_{k=1}^\infty \frac1{1+k^2x},\quad x>0.$$

In other words, according to a certain theorem, I need to show the left-hand series converges pointwise (this follows from $\log(x)\leq x-1$ for all $x>0$) and that the right-hand series converges uniformly on $x>0$.

The challenging part is the uniform convergence of the right-hand series.

Akiva Weinberger

@PM2Ring No, I mean the 5s and 1s in the rightmost column of your table. (0,4) and (4,0) have 5 solutions and (2,5) and (5,2) have 1 solution.

robjohn

12:04

@sunny If that's all your interested in, then you just need to show uniform convergence where you have. Then the statement you want is true for $x\ge a$ for any $a\gt 0$. You don't need uniform convergence on $(0,\infty)$.

Which is good, because the convergence is not uniform on $(0,\infty)$.

sunny

Ok, I see.

Thorgott

12:41

@Jakobian how did you arrive at $\Phi$ defined on all $R$? must be a Zorn's lemma argument, but I'm not sure what the conditions on $\Phi$ are

Thomas Finley

13:06

@leslietownes I don't understand why you are going so agitated in a simple online discussion. No one's haranguing anyone, I just wrote what I felt. Even Eric didn't take it that way the way you did!

That's were things get funny, you know.

Anyways, let's just end the topic here.

I dont understand how this claim follows from density theorem? Am I missing anything?

Jakobian

13:24

@Thorgott I didn't use Zorn lemma there

All $a\in R$ are algebraic over $F$ so they are roots of some minimal polynomial

from Sturm's lemma we know that the minimal polynomial has the same amount of roots both in $R$ and in $R'$

so $\Phi$ is defined on all of $R$

Thomas Finley

I found this in a real analysis book by Bartle-Sherbert

Jakobian

@ThomasFinley agitated? Leslie?

Axoren

@ThomasFinley Provide 2.4.8, not everyone has that book.

冥王 Hades

Any ideas yet? Anyone?

Thomas Finley

13:44

@Axoren This is it.

@Jakobian Every human has the right to get agitated! Isn't it(?)

Anyways

Jakobian

yes but when you say that it makes me think you're trying to play some psychological games here

Thomas Finley

This is the density theorem 2.4.8 in the book

@Jakobian what kind of "psychological games" ?

Axoren

@ThomasFinley Picking any point in $x \in I$, and any neighborhood around $x$ in $A_5$, what do we notice about those neighborhoods?

Thomas Finley

Look, friend @Jakobian I don't know him personally and neither do I know you. The point is that this is an online forum, please don't make simple matters complicated. I think I have not said something offending.

@Axoren the neighborhood around x in A_5 contains only rational points present in the neighborhood of x in I

Axoren

@ThomasFinley How many?

Thomas Finley

13:52

@Axoren infinitely many?

Axoren

Then, what is the definition of a cluster point?

Thomas Finley

@Axoren A point $x$ in $X\subset R$ is a cluster point if every epsilon neighborhood of $x$ intersects I in some point other than $x$

Axoren

If you're still not getting it after seeing all those pieces in front of you, reformulate the Density Theorem in terms of $x = u, y = u+\varepsilon$ where $u, \varepsilon \in \mathbb Q$. In that way, the Density Theorem is a direct claim about neighborhoods around points $u$.

Thomas Finley

Oh, I see what you mean. The point is, that if $x\in [0,1]$ then, any epsilon neighborhood of $x$ contains rational points not equal to $x$ such that those rational points are in $[0,1]$ and there we have the thing, i.e the epsilon neighborhood intersects the set $[0,1]\cap Q$ at some point not equal to $x$

@Axoren This is the thing, right?

Axoren

Even simpler. For every $u$, there is $r$ such that $u < r < u+\varepsilon$ in some epsilon neighborhood that intersects $I$. This is the definition of a cluster point.

Thomas Finley

14:01

@Axoren Yes, but is my comment, i.e

3 mins ago, by Thomas Finley

Oh, I see what you mean. The point is, that if $x\in [0,1]$ then, any epsilon neighborhood of $x$ contains rational points not equal to $x$ such that those rational points are in $[0,1]$ and there we have the thing, i.e the epsilon neighborhood intersects the set $[0,1]\cap Q$ at some point not equal to $x$

Rigorous enough?

Axoren

You used too many unnecessary words. You can be rigorous and brief.

Thomas Finley

@Axoren So, you mean those unnecessary words, made it non-rigorous?

Axoren

No, just that I didn't want to read it because it was too long. :P

"Even simpler" doesn't mean wrong. It means simpler, that's all.

Thomas Finley

@Axoren But my final question will be: After (if) you have read it, I wanna know precisely if my comment be passed as a rigorous explanation to the assertion?

I feel proof writing is often an important thing

I myself dont have the right to judge it. So, I am nagging it this way.

Axoren

I can sympathize with the sentiment of trying to be as formal as possible. However, the more you write, the more opportunities you have to be wrong. I'll give it a proper read once I finish my breakfast.

Thomas Finley

14:07

@Axoren Ok, happy breakfast time ! :)

Cheers!

Thorgott

@Jakobian I don't follow, you haven't defined $\Phi$. Sure, you can extend $\Phi$ to any intermediate extension generated by an $a\in R$, but there's no guarantee these choices are compatible for different $a$, no?

Axoren

@ThomasFinley "every epsilon neighborhood" instead of "any epsilon neighborhood" is better for formal proofs. If you were even just a slight bit careless with other parts of your sentence, you would have implied that only a single neighborhood needs to include a rational number to meet the criteria. $q$ could then be a cluster point of $\{ q, r, q+\varepsilon \}$, if you're not careful.

Jakobian

14:22

@Thorgott I did define $\Phi$, wdym

$\Phi$ maps roots of minimal polynomials to roots of minimal polynomials in increasing order

see the edit to my answer maybe it's more clear now

Thomas Finley

14:47

@Axoren So, my so-called statement is valid?

Axoren

It could be misunderstood, and it's a bit messy in presentation. Communication is two-fold: Both parties must make an effort to understand as well as be understood. If you don't make an effort to balance your signal-to-noise ratio, you cannot expect people to parse your signal from your noise.

Thomas Finley

@Axoren understood.

Jakobian

@ThomasFinley this is the definition of acting like a victim

you're avoiding criticism by saying things like "this person is just mad" etc.

Axoren

People can be mad and have a point. I am regularly both, lately.

Jakobian

doesn't invalidate my point

Axoren

14:57

I'm not disagreeing with you. I'm emphasizing your point.

Even if they are mad, that doesn't mean there is no substance to them.

Jakobian

ah, I misunderstood you then

Ted Shifrin

Mad = crazy. Pointless is better :)

Thomas Finley

@Jakobian Oh, now will you speak some sense, please! I never used the word, "mad". Neither did I say, "this person is mad". You spend too much time on phone, due to which you're hallucinating brother, get some sleep. This is an online community and a networking site, but I think you're getting too serious about this. Please don't try to pick up a fight with me.

Jakobian

do you ever read what you write

Thomas Finley

No

Also, it's impolite to speak between two persons. You are the third person here

The conversation was between me and leslie. Not u and me

Ted Shifrin

15:00

Thomas, you are not the king of the room. Hush.

Axoren

Come on, now. Now you're just being semantically pedantic. Also, this is a public chatroom, not a private room.

Thomas Finley

@TedShifrin I know, but he's just bringing up the same issue over and over. I will just move on and suggest him the same

Jakobian

I prefer they/them

robjohn

@ThomasFinley The directed messages are so that more than one conversation at a time can be conducted. This room gets busy, and it would be hard to only hold one conversation at a time.

geocalc33

15:18

$$ \lambda(L) = \min_{v \in L- {\mathbf \lbrace{0}\rbrace}} ||v||_N $$

Shortest vector problem $N$ is the Euclidean norm $L^2$

$v$ is for vector

$L$ is for lattice

$\lambda(L)$ denotes the shortest non-zero vector in the lattice.

Ted Shifrin

So ?

Axoren

What is the question here?

Ted Shifrin

G'morning, @robjohn. You and the pup doing well? :)

Axoren

@geocalc33 Whatever it is you're trying to accomplish, it doesn't even seem like it's guaranteed to have a unique minimum for an arbitrary lattice. For example, $L = \text{vertices of a unit square}$. There are two vertices with the minimum $L_2$ norm: $(0, 1)$ and $(1, 0)$. Whatever you're trying to do might be ill-posed.

Ted Shifrin

@Axoren You're missing the point. There is of course a shortest length (as long as the lattice is discrete).

Axoren

15:32

Ahhh, mistook it for the argmin. I'm on suboptimal sleep this week.

There's a much easier formulation then, if it's discrete. Just meet all the points together and take the norm of that. $||\bigwedge \left(L\setminus\{0\}\right)||_2$.

Jakobian

it's a different kind of lattice

I'm pretty sure

Ted Shifrin

I have no idea what you just said, Axoren.

Jakobian

I think Axoren mistook the group-theoretic lattice for the order-theoretic one

Axoren

The meet is the infimum between two points. Lattices are closed with respect to infimum, aren't they?

Oh, probably.

Ted Shifrin

To me that is the opposite of a "much easier formulation."

Jakobian

15:36

Lattice (group)

In geometry and group theory, a lattice in the real coordinate space R n {\displaystyle \mathbb {R} ^{n}} is an infinite set of points in this space with the properties that coordinate-wise addition or subtraction of two points in the lattice produces another lattice point, that the lattice points are all separated by some minimum distance, and that every point in the space is within some maximum distance of a lattice point. Closure under addition and subtraction means that a...

Xander Henderson

@冥王Hades I fixed this image for you, and you continue to use the broken one. :(

Ted Shifrin

I thought Hades's image was beyond repair.

Sine of the Time

hi

Ted Shifrin

Hi, Sine.

Semiclassical

@leslietownes these questions feel like contest math to me

they're not literally so, of course

Sine of the Time

15:43

how are you, Ted?

Semiclassical

but they seem like they show up mostly as journal challenge problems

Jakobian

@Semiclassical most of them probably are

robjohn

@TedShifrin G'day. We are. We've been out for a walk with one of Rosie's and Holly's friends.

Ted Shifrin

@Semiclassic The interesting thing (as I've mentioned here before) is that a symmetric function may have non-symmetric critical points. People are used to max/min problems with the answers only with all variables equal.

Doing fine, thanks, Sine. You?

Semiclassical

yeah

Ted Shifrin

15:45

@robjohn Rosie's made friends already?

geocalc33

Yes I'm back I was typing about the lattice. I'm using the definition of a lattice found in the wiki article Linked directly above

Sine of the Time

Good, I'm preparing a couple of exams

Semiclassical

something-something spontaneously-broken symmetry

(it's not really that, but it reminds me of it)

Ted Shifrin

@geocalc But is $\Bbb Q$ a lattice in $\Bbb R$?

@Semiclassic I always assigned such a function to my multivariable math class. Sadly, I discovered the phenomenon after the book was published.

robjohn

@TedShifrin Rosie's had friends for a long time. Holly has met most of them and plays with them.

Ted Shifrin

15:46

Oh, oops, I reversed the names of the dogs, @robjohn. Sorry.

Semiclassical

@TedShifrin actually, what's the simplest nontrivial example of that? i can sorta mentally imagine one but i don't know the simplest example

Sine of the Time

@TedShifrin Commutative property is not always valid

robjohn

Holly is now over 14.5 lbs.

Xander Henderson

@robjohn That's a big cat.

robjohn

We used to have a cat that big. I think our heaviest cat now is around 12 lbs

Ted Shifrin

15:47

@Semiclassic $f(x,y)=2(x+y)+xy(1+x+y)$

Semiclassical

hmm, okay

Ted Shifrin

My cat is about 6.6 lbs.

Xander Henderson

@robjohn It's not a cat? Are you sure you didn't misplace a decimal point, then?

Surely you mean 145 lbs?

:P

geocalc33

@TedShifrin Yes precisely

robjohn

@XanderHenderson no, Holly is about 11 weeks old. If she were 145 lbs, I would worry.

Sine of the Time

15:49

why do you use lbs :((((

Semiclassical

the example i was cooking up was $(x-\sqrt{3}/2)^2+(y+1/2)^2+(x+\sqrt{3}/2)^2+(y+1/2)^2+(x-1)^2$

geocalc33

oops I meant to reply to the other message

Xander Henderson

@robjohn Ah, it is a baby dog.

Ted Shifrin

To me, a lattice is a discrete subgroup. Is that in their definition?

geocalc33

@TedShifrin Yes precisely

Xander Henderson

15:49

It'll eventually get to a reasonable size.

@SineoftheTime Because we are Americans, and we use freedom units.

geocalc33

@TedShifrin Yes

Ted Shifrin

@Semiclassic You have a typo, I think, but that doesn't have $f(x,y)=f(y,x)$.

Xander Henderson

None of these poncy French "SI" units for us!

'MERIKA!

Jakobian

@XanderHenderson which are defined using SI units

Ted Shifrin

@geocalc33 Oh, so then the answer to my first question is no.

Semiclassical

15:50

yeah, i was thinking of 3-fold rotational symmetry. but i should have gone all the way to 6-fold

Ted Shifrin

So what is the question?

Xander Henderson

@Jakobian Frech propoganda.

Semiclassical

tho i guess 6-fold is overkill

Xander Henderson

And lies.

Ted Shifrin

2-fold suffices to understand it.

Semiclassical

15:51

hmm

Jakobian

@XanderHenderson :P

Semiclassical

my examples are all quadratic in $x,y$ so i think they won't suffice

geocalc33

Oh right, I see why the answer to the first question is no @TedShifrin

Sine of the Time

@XanderHenderson 😒

Semiclassical

(whereas the example above was cubic)

geocalc33

15:53

I want to know if it makes sense to minimize $H$ over the $L^2$ norm here

$$\lambda(H) = \min_{v \in H- {\mathbf\lbrace{1}\rbrace}} ||v||_{L^2}$$

$H = w_1^{\mathbb Z} \odot w_2^{\mathbb Z},$ where $w_1=(a,1)$ and $w_2=(1,b)$ for some $a,b>0; a,b\neq 1,$ and $(x_1,y_1)\odot(x_2,y_2)=(x_1 x_2, y_1 y_2)$ and $w^{\mathbb Z} = \lbrace w^n \mid n \in \mathbb{Z} \rbrace.$

Ted Shifrin

Right. My point was for calculus students, not physics students. And calculus teachers. I suspect a lot of teachers tell their students that if $f$ is symmetric about the diagonal, then every critical point will be found on the diagonal.

Semiclassical

i note that this particular example doesn't have a global maximum/minimum

ZaWarudo

If I'm able to prove that the double indexes sequence $(a_{n,m})_{n,m\in\mathbb{N}}$ is decreasing in $n$ for each $m \in\mathbb{N}$, that is $a_{n+1,m} \le a_{n,m}$ for each $n,m \in\mathbb{N}$, is it correct that $\sup_{n,m\in\mathbb{N}} (a_{n,m})=\sup_{m \in\mathbb{N}} a_{0,m}$?

Ted Shifrin

True.

geocalc33

Where I've defined $H$ as an image of $L$ where $L$ is a lattice

Semiclassical

15:55

presumably that's a matter of coefficients etc

hmm, or not. for large $x,y$ you're dominated by $x y(x+y)$

Ted Shifrin

Well, the coefficients, as I recall, were a delicate matter.

But, yeah, I think it's the nature of cubics.

@geocalc Offhand, I don't see why there is a min in this case. You're switching additive questions to multiplicative ones.

geocalc33

@TedShifrin I'm switching addititve questions to multplicative ones but measuring the multplicative thing with an additive scale

Ted Shifrin

Yeah, I don't think that's going to work. The inf can certainly be $0$, can it not?

Semiclassical

@TedShifrin i think there's no way to get (non-symmetric) global maxima/minima without going to quartic

Ted Shifrin

Well, and of course you cannot get both. Probably you can get local of both.

Semiclassical

16:01

right

geocalc33

@TedShifrin yeah that is true, good point. So it sort of becomes trivial I suppose. Wondering if I can make it more interesting

Maybe I could choose $\mathrm{max}$

Semiclassical

i was hoping i could do something like $((x-1)^2+(y+1)^2)^2+((x+1)^2+(y-1)^2)^2$ but that has $(0,0)$ as global min

i guess i can subtract off something like $(x^2+y^2)^2$ with sufficiently large coefficient to "penalize" the origin

Ted Shifrin

@geocalc33 Nope.

Semiclassical

ack, but if i do that then i just make the origin a global minimum...sigh

oh. no, this does work

geocalc33

Okay I figured it out just have to do closest vector problem instead of shortest vector problem

Lattice problem

In computer science, lattice problems are a class of optimization problems related to mathematical objects called lattices. The conjectured intractability of such problems is central to the construction of secure lattice-based cryptosystems: Lattice problems are an example of NP-hard problems which have been shown to be average-case hard, providing a test case for the security of cryptographic algorithms. In addition, some lattice problems which are worst-case hard can be used as a basis for extremely secure cryptographic schemes. The use of worst-case hardness in such schemes makes them among...

If my heuristics are correct, it should get very difficult to determine the CV

Semiclassical

16:13

@TedShifrin the example you see a lot in physics is $V(x,y)=(x^2+y^2)^2-a (x^2+y^2)$

where for large enough $a$ you get a circular ring of local minima

oops, for any $a>0$

Ted Shifrin

Yeah, I consider that degenerate.

Semiclassical

yeah

to get non-degenerate examples you have to perturb that somehow

e.g. $(x^2-y^2)^2$

i guess perturbing with $xy$ is simpler

Ted Shifrin

I confess that I used Mathematica to create my example :)

Semiclassical

lol, fair enough

i'm using mathematica to plot mine

Ted Shifrin

I could have done it by hand, but why bother once I’ve spent the money to buy Mathematica.

Semiclassical

16:21

i guess $f(x,y)=(x^2+y^2)^2-a(x^2+y^2)+bx y$ is the simplest "physical" example i know

huh, maybe i don't even need the second term

yeah, $f(x,y)=(x^2+y^2)+8xy$ has critical points at $(x,y)=(1,-1),(-1,1)$

Ted Shifrin

But saddles?

Semiclassical

there's a saddle point at the origin, sure

i guess this still has a critical point at $(0,0)$ too

Ted Shifrin

What about the others?

Saddle surfaces are saddle surfaces.

Semiclassical

oh, blah. dumb typo

$f(x,y)=(x^2+y^2)^2+8xy$

missed the square on the first term, bleh

Ted Shifrin

Ohhhh

Now I need to compute ….

冥王 Hades

16:37

Shouldn’t a wireless mouse be called a hamster?

Ted Shifrin

@Semiclassic Nice. Your function appears to have two (global) minima and one saddle.

Semiclassical

yeah

Axoren

@冥王Hades Is that the definition of hamster? Tail-less rodent?

Semiclassical

in polar coordinates i can write that second term as $4r^2 \sin(2\theta)$, so there's obvious generalizations if you want to get more complicated examples

Axoren

Does that mean hamsters are bounded distributions?

Semiclassical

16:49

@Axoren booooo

Xander Henderson

@冥王Hades Etymological fallacy. The origin of a word does not necessarily tell you anything about how the word is used now.

Etymological fallacy

In classical Aristotelian logic, an etymological fallacy is committed when an argument makes a claim about the present meaning of a word based exclusively on that word's etymology. It is a genetic fallacy that holds a word's historical meaning to be its sole valid meaning and that its present-day meaning is invalid. This is a linguistic misconception. The inverse negative form of the fallacy treats the current meaning as the sole true meaning, requiring negation of the etymology from which the current meaning was derived. == History == Ancient Greeks taught that a word's meaning could be tracked...

Ted Shifrin

Soon this will be the etymology & epistemology room.

Axoren

I remember back when you could get a Kentucky Fried Man (see: Diogenes)

Jakobian

@XanderHenderson I believe, it was, a joke

Axoren

@Jakobian I believe Xander was also joking.

Ted Shifrin

16:57

Now we need a “Are you joking?” flag.

Axoren

Or better yet, /hj

The least useful tone indicator: half-joking.

Jakobian

@Axoren I don't know, but I did learn a new word

Ted Shifrin

Or, better yet, a John McEnroe “You can’t be serious” yell.

Axoren

"I am being stochastically facetious."

冥王 Hades

No I call it trolling

Transcript for

$Mathematics$ Mathematics