Mathematics - 2022-12-03 (page 1 of 2)

12:04 AM

hey @Ted, I'm thinking about an exercise right now that I think you probably like. it's "classify all conics (meaning hypersurface of degree $2$) in $\mathbb{P}^2$".

I'm getting either a projective line or the union of two projective lines meeting in a point, that sound reasonable?

(up to abstract isomorphism, the ambient classification should be finer, I think)

Ted Shifrin

Over what field?

Not that I like asking that.

Thorgott

algebraically closed field

of characteristic not 2

(though I've not used the latter assumption anywhere, so there's probably an oversight in my thinking)

Ted Shifrin

Char 2 is crazy, I think. Ah.

Jakobian

@leslietownes MSE existed when I was integral hungry, but I didn't know that it does

Ted Shifrin

You are missing a double line.

Thorgott

12:08 AM

what's that?

Ted Shifrin

The square of a linear, rather than product of two different linears.

Thorgott

but that gives the same variety as the linear that is being squared

perhaps there's a difference scheme-theoretically

Ted Shifrin

Not perhaps. Very different for all sorts of reasons, not the least of which is Bezout.

Intersect with a line. You need to get 2 points.

leslie townes

be-who?

Thorgott

Bezout holds for reducibles too?

Ted Shifrin

12:16 AM

Yup.

leslie townes

bezout holds for rezoutables zout.

Thorgott

then it makes sense you have to count differently. we'll hopefully cover Bezout soon in lecture.

but, in terms of varieties, it should just be those two cases, right?

and I would suspect if we classify up to ambient isomorphism, there should be three cases? one is two lines meeting in a point, one is a linear embedding of a projective line and the other is the degree 2 veronese embedding of a projective line.

Ted Shifrin

To me, you want a closed set … so limits of families should be included, hence the double line. But I don’t know your conventions. If you say reduced, then OK.

No, you said degree 2 to start with, so you can’t say a line.

I win.

I hope robjohn’s OK.

Thorgott

my convention is (pre-)varieties, so everything is a locally ringed space locally isomorphic to some (classical) affine variety. so, in terms of schemes, yes, everything is reduced.

@TedShifrin $V_p(xy-z^2)$ is certainly a line and degree 2, no?

Ted Shifrin

Huh? That’s a smooth conic.

Thorgott

12:24 AM

I don't know what smooth means, but yes

oh, I see, my terminology is very bad

by "line" I meant "isomorphic to $\mathbb{P}^1$", but probably "line" should be reserved for things like $V_p(ax+by+cz)$

Ted Shifrin

Indeed. Say rational.

Thorgott

this subject has so many adjectives

Ted Shifrin

Doesn’t every subject?

Or object, for that matter.

Thorgott

fair, I guess it's just familiarity at this point

I know all the separation axioms by heart, so I shouldn't judge

Ted Shifrin

Uh huh.

Thorgott

12:32 AM

so I think what I have already shown is that every closed embedding $\mathbb{P}^1\rightarrow\mathbb{P}^2$ is, up to post-composition with a projective automorphism, given as $[x\colon y]\mapsto[x\colon y\colon0]$ or $[x\colon y]\mapsto[x^2\colon xy\colon y^2]$. I suspect these are not equal up to projective automorphism.
the first type is a line and that's clearly preserved under projective automorphism. the latter has image $V_p(xz-y^2)$. this cannot equal a line $V_p(ax+by+cz)$, for otherwise $\sqrt{xz-y^2}=\sqrt{ax+by+cz}$ by the projective Nullstellensatz, but both polynomials are prime…

Ted Shifrin

Where are you using char $\ne 2$?

Thorgott

has to be in the "I have already shown" part

Ted Shifrin

Note $x^2+y^2+z^2=(x+y+z)^2$.

The erstwhile-mentioned double line.

Thorgott

hmm, I actually don't see where in my proof I have used the characteristic =/=2 assumption

so I must be making an error somewhere in any case

leslie townes

according to the search function, 'erstwhile' has been used only seven times in this chat (not counting this one), and each time by ted shifrin.

Ted Shifrin

12:47 AM

I am far eruditer!

D.C. the III

only one of us with an expansive vernacular

see even uses more fancy words than vernacular

Thorgott

1:05 AM

ok, so there's this claim that the vanishing locus of a degree 2 irreducible in $\mathbb{A}^2$ is, up to affine-linear automorphism, either $V_a(x^2-y)$ or $V_a(xy-1)$. does this need characteristic =/=2?

Ted Shifrin

Yes, surely.

I prefer to think homogeneously, but whatever.

Thorgott

ahh, I have a "proof" not using the assumption and can't figure out for the life of me where it's going wrong

Ted Shifrin

1:21 AM

Well, test out my example above of the generic quadratic form.

Who says the affine curve has to be irreducible? In char 2 is there an irreducible?

Thorgott

the issue with that example is that it's reducible

nobody says affine curves have to be irreducible (I think), but I'm classifying the irreducible ones

the closures of these in projective space correspond to irreducible conics (and the reducible ones are unions of two lines, either the "double line" or two lines meeting in a single point)

Ted Shifrin

In char 2 is $xy=1$ irreducible?

Thorgott

yeah, I think so

but even if it wasn't, that just mean even less possibilities, so the classification would at worst get smaller

Ted Shifrin

Well, did you prove those are irreducible or not? Why not others?

Thorgott

1:40 AM

an arbitrary degree 2 polynomial over an algebraically closed field has the form $(x-ay)(x-by)+cx+dy+e$ (up to scaling, which doesn't change the locus). if $a=b$, we substitute $x\mapsto x+ay$ and obtain $x^2+cx+d'y+e$. now, $d'\neq0$ as otherwise the polynomial is irreducible (which is preserved under affine linear automorphisms), so we can substitute $y\mapsto-\frac{1}{d'}(cx+y+e)$ and obtain $x^2-y$.
if $a\neq b\neq0$, i can substitute $x\mapsto x-\frac{a}{b}y$ and $y\mapsto\frac{1}{b}(x-y)$ to obtain $(1-\frac{a}{b})^2xy+c'x+d'y+e$. the coefficient is non-zero, so I can remove it by res…

robjohn

4:07 AM

@TedShifrin I am okay, but some testing being done. Sorry, I have been quiescent here for a while.

3

Ted Shifrin

4:21 AM

@robjohn just thinking of you!

robjohn

Nice to be remembered

@Ted how are things with you?

Ted Shifrin

A bit hoarse from going to a gathering with too many loud people — hopefully, nothing worser.

robjohn

I was worried about going to UCLA to proctor an exam 2 weeks ago. It has been about 3 years since I've been to UCLA and sometimes after going there, I get some cold/flu symptoms

leslie townes

4:38 AM

i have some in long beach if you'd like more. fresh from day care!

Ted Shifrin

This is today’s life. With the weather going up and down and with allergies, I always have nose/throat issues.

leslie townes

we've had one RSV, one pneumonia (probably post-RSV), and one covid report in the last week.

Ted Shifrin

Is that all?

leslie townes

well, no obligation for day care to report to us if there's no testing in the first instance, or if it's done but not reported to the day care.

we had got sniffles and sore throats and no appetites here, but tested negative for covid.

Ted Shifrin

Sometimes the home tests take a week to register positive.

leslie townes

4:48 AM

could be any day now. i still test if i have to go out. it is hard to home test a 4-year-old.

one potato two potato

5:36 AM

Let $T^n$ be an $n$-torus $S^1\times\cdots\times S^1$. A closed covector field $\omega$ on $T^n$ is exact iff $\int_{\gamma_j}\omega =0$ where $\gamma_j:[0,1]\to T^n$, $\gamma_j(t) = (1,\ldots,e^{2\pi it},\ldots,1)$.

Let $\iota_j:S^1\hookrightarrow S^1\times\cdots\times S^1$ and let $\omega_j = \iota_j^*\omega$ and $\tilde{\gamma}_j(t) = \pi_j\circ\gamma_j$. Then $\int_{\gamma_j}\omega = \int_{\tilde{\gamma}_j}\iota_j^*\omega =0$ so $H^n_{dR}(T^n)\simeq H^1_{dR}(S^1)\otimes\cdots\otimes H^1_{dR}(S^1)\simeq\Bbb R$ via integration, so the statement follows.

(Roughly) the correct argument?

Ted Shifrin

5:54 AM

What statement follows?

The first statement was about $1$-forms.

one potato two potato

Ah,,,, I see.

Ted Shifrin

6:11 AM

The first statement is basically multivariable calculus and line integrals. How do you define a well-defined potential function?

Daniel Donnelly

9:02 AM

YAPOPT - yet another proof of prime twins:

1

Q: Can we take advantage of the slackness $\sqrt{x + 1}..(x-2)$ in the Sieve of Eratosthenes for twin prime averages $x$?

Let $X$ be the set of twin prime averages, then for all $x \in X$ we have that $x \neq \pm 1 \pmod {p_i}$ for all $i = 1..f(x)$ where $\sqrt{x + 1} \leq f(x) \leq x - 2$ for all $x \geq 1$. Conversely, if $x$ is not a twin prime average, then $x = \pm1$ for some $i = 1..f(x)$. This will be gen...

But my previous proofs had errors, so this one probably does as well.

Mad

10:17 AM

Trying to prove: $\lVert Ax\rVert \leq \lVert A\rVert\cdot \lVert x\rVert$
Write $Ax = a_i * x$ as a dot product between row vector and x then
$ \lVert Ax\rVert = \sqrt(\sum_i a_i *x )^2) \leq \sqrt(a_1x)^2 + \cdots \sqrt(a_n * x)^2 = \sum_i \lvert a_i x \rvert \leq \text{by Cauchy inequality} \sum_i \lVert a_i \rVert \lVert x \rVert $

What am i doing wrong?

one potato two potato

10:57 AM

@TedShifrin $\int_{\gamma_j}\omega = 0$ condition will imply the potential function $F(x) := \int_{\gamma_x}\omega$ for $\gamma_x:[0,1]\to T^n$ a path from $0$ to $x = (x_1,...,x_n)$ obtained by concatenating $(0,...,0)\mapsto (x_1,0,...,0)$, $(x_1,0,...,0)\mapsto (x_1,x_2,0,...,0)$... well-defined.

Xander Henderson

12:19 PM

@Jakobian I also mistyped it... gah! Same mistake, twice. I read "high school" as "kindergarten."

I thought you were saying that you were excited about integration when you were in kindergarten.

(And then when I explained it, I had already mentally corrected "kindergarten" to "high school", and wrote something that makes no sense).

Jakobian

ah. Funny

Xander Henderson

12:35 PM

@Jakobian Doubly funny because my phd advisor used to joke about being taught integration in kindergarten. :)

Odestheory12

1:10 PM

Hey

If $x_n$ is a convergent sequence, does it imply that $x_m$, where $m=\frac{1}{n}$, is convergent?

Does that even make sense? ^^

Xander Henderson

That does not make sense.

What is $x_{1/n}$?

When you say that "$x_n$ is a convergent sequence...", I assume that you mean that $(x_n)_{n=1}^{\infty}$ is a convergent sequence, i.e. you have a sequence of terms $(x_1, x_2, x_3, x_4, \dotsc)$.

You haven't defined terms with indices that aren't natural numbers.

Odestheory12

Yeah ^^

Xander Henderson

So what is $x_{1/n}$?

Odestheory12

It doesn't make sense :D

I was trying to prove $x_n=n\sin{\dfrac{\pi}{n}}$ is cauchy using the definition

So I thought in doing $m = \frac{1}{n}$

I was getting $\left |x_n-x_m \right |=\left | n\sin{\dfrac{\pi}{n}} - m\sin{\dfrac{\pi}{m}} \right | \leq | n - m | \leq n+m < 2n$

Which gave me the idea of the substitution but meh

Xander Henderson

1:27 PM

Are you familiar with the small angle approximation for the sine function?

I would expect that to play a role.

Daniel Donnelly

-1

Q: There exist infinitely many $2k$-separated prime pair averages. Twin primes is $k = 1$. An elementary proof exists.

Let $X$ be the set of $2k$-prime pair averages for fixed $k\geq 1$. I.e. $X = \{ x \in \Bbb{N} : x \pm 2k \in \Bbb{P}\}$. When $k = 1$, $X$ is the set of twin prime pair averages. Equivalently, by the standard Sieve-of-Eratosthenes arguments, we have that $X = \{ x \in \Bbb{N}: x \neq \pm k \pm...

elementary-number-theory solution-verification prime-numbers prime-gaps prime-twins

@shintuku

Okay, I removed that

Xander Henderson

Please also do not use comments below a question to complain about downvotes.

Daniel Donnelly

please though check it out. The proof appears flawless

@XanderHenderson okay, I removed that too

one potato two potato

Only the downvoter knows the reason for downvoting

Xander Henderson

@DanielDonnelly No, you didn't. I did.

Daniel Donnelly

1:32 PM

No, I did, lol I beat you to it

Maybe we both did and the system decided

Xander Henderson

@DanielDonnelly You didn't.

But whatever.

Daniel Donnelly

I did though

I right clicked and deleted

Xander Henderson

Daniel Donnelly

I mean the dropdown

Okay, you won on that one but I removed the one here

I did both actually

:D

Xander Henderson

In any event, based on the title of your question, I doubt that it is really appropriate for this forum. It seems that you claim to have resolved the twin prime conjecture. Such a resolution should be submitted to a peer-reviewed journal, where people with the appropriate expertise can judge whether it is correct or not.

Math SE is simply not equipped to provide that kind of peer review.

Daniel Donnelly

1:35 PM

I disagree

Odestheory12

And shouldn't be a question to MathOverflow rather than Math SE?

Daniel Donnelly

It's the first place one should start

MO banned me just like most of us

auto-ban for whatever reason

Xander Henderson

@Odestheory12 MO specifically disallows announcements of new results.

Daniel Donnelly

This isn't a new result. It's an elementary proof, I'm not showing any new math, I'm just analyzing and applying old math

Odestheory12

@XanderHenderson Oh I see, interesting. What is the purpose of MO? (I guess I can google this question :P)

512

Q: Differences between mathoverflow and math.stackexchange.

What are the differences between MathOverflow.net and Math.stackexchange.com? Why two communities for Mathematics? Wasn't one enough?

discussion faq mathoverflow

Yeah :D

Daniel Donnelly

1:37 PM

Also there's the proof-verification tag which I used

I'm wondering if I made some serious error

Most likely yes

I do suss out my own errors a lot of times, but this one has run through my checking procedure, and passed

Xander Henderson

@DanielDonnelly Yes, and? If you follow the meta conversation about the proof-verification tag, you might understand that this is a controversial tag, and that it is really intended to tag questions where a student has written a solution to a problem, and is looking to know whether their proof is correct.

It is not intended as a tag for new research results and peer review.

Daniel Donnelly

It's not new research

It's not peer review

It's pure math

Xander Henderson

Do you disagree with my assessment of the nature of your question, i.e. "It seems that you claim to have resolved the twin prime conjecture"?

Daniel Donnelly

No, hasn't been resolved yet until it's peer reviewed in a journal about 20 years from now

Xander Henderson

...

The title of your question is "There exist infinitely many 2𝑘-separated prime pair averages. Twin primes is 𝑘=1. An elementary proof exists."

Daniel Donnelly

1:42 PM

They set the bar so high and impossible for amateurs to publish that it's probably not even possible.

Xander Henderson

To me, it seems as though you are claiming to have found an elementary proof of the twin prime conjecture.

If that is not your claim, I would suggest that you choose a better title.

Daniel Donnelly

I always title differently from what the content is, to make the post interesting

The question is at the bottom

Xander Henderson

So you have intentionally titled your question in such a way as to mislead potential readers?

(-1)

Daniel Donnelly

Okay, I retitled it

Xander Henderson

Also, "Where in the above argument is the error if any?" is not a good question for Math SE. A good question on Math SE should highlight a specific place (or places) where the asker feels there is likely a problem. Asking the community to sift through an entire question to spot an error is not really a good fit for the site.

That isn't the kind of question Math SE is designed around.

Daniel Donnelly

1:46 PM

Most of the post is just background info to get the reader up to speed, otherwise they wouldn't know what the argument entailed.

Notice, that I assumed the reader knows about the Eratosthenes sieve, so I didn't overload it too much

Xander Henderson

@DanielDonnelly In reading your post, how is a typical reader supposed to know that? Have you highlighted any part of the argument where you believe that there is a flaw?

Also, now that I have actually skimmed through the question, you end with "Therefore, there exist infinitely many 2𝑘-prime pair averages for any 𝑘≥1. In particular there are infinitely many twin primes. QED" So, in essence, you are asking the Math SE community to vet a purported proof of the twin prime conjecture (which is exactly what your original title suggested). This is off-topic for Math SE.

nickbros123

Hi guys, I'm new to math chat, I'll fire away: i realised every problem on Spivak links a general result from another previous problem, with a concept. Concept is fine, but does one remember all obscure formulas that come up in previous problems ?

Xander Henderson

@nickbros123 Maybe?

Daniel Donnelly

I noticed there is one flaw: you need existence of at least one $2k$-pair. I noted in the comments

Xander Henderson

Generally speaking, I don't bother to memorize a bunch of formulae. If a formula is important, I will tend to internalize it (for example, when I am working on my own research projects, there are a few infinite series who sums I get to know pretty well), but for things I don't use that often, I either remember that there exists some formula that I can look up, or I rederive it when I need it.

(For example, when I have been away from research for a while, I always have to rederive the values of geometric series, because if I rely on my memory, I am going to include an off-by-one error).

shintuku

1:59 PM

@DanielDonnelly hwat?

Xander Henderson

@leslietownes I see this pinned on the starboard. I am not even going to look at the context. It needs another star.

Wow... super pretty sunrise right now:

The camera phone doesn't quite do it justice. The whole neighborhood is orange.

There are perks to living in bf-nowhere Arizona. :D

one potato two potato

2:26 PM

Is there any formula that relates pushforward and pullback between smooth manifolds?

robjohn

@XanderHenderson Lovely! If you up the saturation a bit:

don't know if that's much better

Here is the sunset we had from Nov 4:

I did not enhance the saturation there.

nickbros123

@XanderHenderson sir, i generally agree with your take. Internalising the in and outs of a results and it's relations is better than memorising formulas, but in this case (Spivak) almost all problems' shortest proof requires using a result from another question. If I'm not writing the shortest proof or learning the shortest I feel like I'm missing out which makes this book extra bit harder for me

robjohn

2:50 PM

@nickbros123 However, rather than reproving something, it is often easier to cite a previous result. It can improve the readability also.

But I agree, that for understanding, knowing the ideas behind a result is important.

Xander Henderson

@robjohn To make it match what I am seeing more better, the contrast needs to be upped. The saturation is also a problem, but that blue should be much darker. I could make it look nicer, but didn't have the energy (I was on my way out the door to go grab some things from my office).

user4539917

@nickbros123 Welcome to the chat!

nickbros123

@robjohn sir, what you're saying is true, quoting another result shortens the proof and makes it easy to digest, but some of the relations are a bit obscure in Spivak, thinking in that direction doesn't come obvious

@user4539917 thank you sir

user4539917

3:08 PM

In all honesty, something is "obscure" only until you've memorized/internalized it. How much "obscurity" you're comfortable with depends on how deeply you want to understand the ideas behind a result.

Thorgott

3:44 PM

@TedShifrin characteristic =/=2 was unnecessary after all

the classification of conics is the same

Koro

4:33 PM

Suppose that A takes a tensor F to $\sum_{\sigma} sgn (\sigma) F^{\sigma} $. Given AF=0. Why is $A(F\otimes G) =0$?

I see. It has complicated proof.

Thorgott

4:46 PM

what is your definition of "tensor", what does your notation mean, what is G

this is very incomplete

Koro

By tensor (k-tensor) here, I mean a linear functional on $V^k$, where V is a vector space.

The question I asked above is required in proving associativity of the wedge product.

Thorgott

I assume you'd rather want multilinear

Koro

Ohh yes. Multilinear map on V^k taking real values.

Thorgott

5:08 PM

I still don't understand what you're trying to conclude

this is probably not complicated, but requires precision in actually writing down what you want

robjohn

5:44 PM

@XanderHenderson I took the dog to the park, but I tried a bit more processing:

Koro

6:15 PM

Hi@robjohn!!

PNDas

In LaTeX how to interchange \epsilon and \varepsilon?

Koro

@Thorgott I'm trying to conclude $A(F\otimes G)=0$ given $AF=0$. A is as given earlier. $F$ may be assumed a multilinear real valued map on $V^k$, $G$ on $V^l$.

Thorgott

I think it's probably easier to just prove more generally that $A(F\otimes G)=A(AF\otimes AG)$

perhaps up to some factor depending on whether your conventions are good or not

Koro

One would write: Given $v\in V^{k+l}$, $A(F\otimes G)v = \sum_\sigma sgn (\sigma) (F\otimes G)^\sigma v=\sum_\sigma sgn(\sigma) F(v_{\sigma(1)},\cdots, v_{\sigma(k)})g( v_{\sigma(k+1)},\cdots, v_{\sigma(k+1)})$

I have no idea how to show this to be zero now.

notation: $F^\tau (v_1,v_2):= F(v_{\tau (1)},v_{\tau(2)})$.

ILikeMathematics

When you are told to make a negative angle of elevation of a line positive, the convention is to add 180°, right? Not 360°

Ted Shifrin

6:27 PM

@PNDas Just make your own nickname with \def

Koro

Spivak's proof is unexplainable.

Ted Shifrin

@Thorgott The issue is that the generic conic in characteristic 2 is reducible.

Koro

Munkres proof is also not clear.

I also checked Tu's book. There also the proof is unexplainable.

PNDas

@TedShifrin Hmm I got that idea after asking that.

Koro

Checked it on mse, the post used that result as a lemma without giving any proof.

Thorgott

6:29 PM

@TedShifrin I don't know what that means. All I'm saying is that the isomorphism classification is the same.

Tu, Lee, they will all have reasonable proofs of this fact

Ted Shifrin

@nickbros123 I taught a year-long course out of Spivak (Calculus, I assume you're talking about) 15 times, and I'm not sure what you're complaining about. What obscure formulas are so necessary?

PNDas

I also got this '\let\tmp\phi \let\phi\varphi \let\varphi\tmp' it interchanges them.

Ted Shifrin

@Koro WTF does "unexplainable" mean?

Koro

I mean the proof has no head and tail.

Ted Shifrin

This looks like a standard cosets of symmetric group proof.

@Koro This is just as worthless a comment, seriously.

Koro

6:33 PM

:(

Ted Shifrin

Koro, sometimes you are too quick to criticize well-respected authors instead of expending effort thinking.

Thorgott

just checked, this is Lemma 3.24 in Tu

Ted Shifrin

Fix $\sigma(k+1),\dots,\sigma(n)$ and consider the sum over the permutations over $1,\dots,k$.

Thorgott

and what do you know, it's the same as I suggested :P

Koro

Munkres': $A(F\otimes G)v = \sum_\sigma sgn (\sigma) (F\otimes G)^\sigma v=\sum_\sigma sgn(\sigma) F(v_{\sigma(1)},\cdots, v_{\sigma(k)})g( v_{\sigma(k+1)},\cdots, v_{\sigma(k+1)})$. Then collecting those terms where g(...) is the same, we get $sgn(\sigma)\color{red}{[\sum_\tau sgn(\tau)F(v_{\sigma(\tau 1)},\cdots, v_{\sigma(\tau k)})]} g( v_{\sigma(k+1)},\cdots, v_{\sigma(k+1)})$. I actually don't understand the red part.

Ted Shifrin

6:38 PM

It's exactly what I said. Consider cosets of $S_k$ in $S_n$.

Koro

I'm not criticising anyone.

Ted Shifrin

Every permutation fixing $k+1,\dots,n$ is a permutation of $1,\dots,k$. You can account for all of them by picking one and then composing with elements of $S_k$.

You're using adjectives to describe the proofs of the authors rather than saying "I cannot understand."

This is not the first time.

Koro

Oh sorry. I didn't mean it that way. I do/did not understand is what is/was intended.

Ted Shifrin

Anyhow, did my additional explanation allow you to figure it out?

This group theory "trick" is quite standard in mathematics and is quite important.

Koro

Did you mean fixing $\sigma (k+1),\sigma(k+2),\cdots, \sigma(k+l)$ instead?

Ted Shifrin

6:47 PM

Sure. I was making the key idea clear. Now you can fix it.

Koro

yes you did.

Jakobian

@Koro I wanted to write something like that before you said the proof is complicated

tbh

Ted Shifrin

If you want a fancy explanation, it's a variant of Fubini. Summing (integrating) over $G$ can be done by summing over $G/H$ the sum over $H$.

Koro

it actually is (to me).

Ted Shifrin

The fancy terminology is "integration over the fiber." :D

Koro

6:50 PM

why is that $sgn (\sigma)$ out of the $\sum$? Why is it $\sigma(\tau 1)$ instead of $\tau(\sigma 1)$?

Ted Shifrin

There's a sum outside that whole thing. The $\tau$ is inside because you're permuting $1,\dots,k$ and then acting by $\sigma$.

$\tau\in S_k$ does not even know how to act on $\sigma(1),\dots,\sigma(k)$.

Sometimes you should write out an explicit example (with small values of $k$, $n$) and do it yourself.

Koro

7:45 PM

what is the mathjax for wedge product?

Lukas Heger

$v\wedge w$

\wedge

Koro

thanks :-)

Thorgott

8:01 PM

one of math's biggest atrocities is that the wedge sum is denoted by $\vee$

Jakobian

8:19 PM

wouldn't it be funny if to read a book about books

Koro

8:30 PM

I understand it now.

Spivak's proof makes sense to me :-).

Ted Shifrin

Yippee!

Koro

Let $G=\{\sigma\in S_{k+l}: \sigma \text{ fixes k+1, k+2, ..., k+l}\}$. So we break the sum in two parts: 1) summation over G, 2) over those $\sigma$'s which are not in G. 1) gives 0. 2) We take a $\sigma_0\notin G$ and define $G\sigma_0:=\{g\sigma_0: g\in G\}$ and we take summation over this and it is zero.

Now the idea is to note that G and G \sigma_0 have empty intersection. We continue like this (i.e., take another \sigma_0' not in G and consider G \sigma_0' to get sum over it as 0) and get a partition of $S_{k+l}$. Over each of the sets in the partition, we get the summation as $0$.

This is so nice.

Ted Shifrin

By the way, when I have taught this proof, I've written it this way: $$\sum_{\bar\rho\in S_{k+\ell}/S_k}\sum_{\tau\in S_k} \text{sgn}(\rho\tau) (S\otimes T)^{\rho\tau} = \sum_{\bar\rho}\text{sgn}(\rho)\left(\sum_\tau \text{sign}(\tau)(S\otimes T)^{\tau}\right)^\rho = 0.$$

Koro

I typed a very long post. While typing it, I understood it.

Magnus Alexander

Hello everyone! Just a quick question on Fourier series. I have a task to prove the formula: $\sum\limits_{n=1}^{\infty} \dfrac{sinnx}{n}=\begin{cases} \dfrac{-x-\pi}{2}, x\in [-pi, 0) \\ \dfrac{-x+\pi}{2}, x\in (0, pi] \\ 0, x=0 end{cases}$ by decomposing the right side into Fourier series. Am I right that I have to do it with each part of the function? For example, do I need to take $\dfrac{-x-\pi}{2}$ and consider x to be from $[-\pi,0]$?

Ted Shifrin

8:36 PM

You really do want to think of it in terms of cosets.

Koro

nah, looks complicated.

Magnus Alexander

Meaning that when I will calculate the Fourier coefficients I'll get integrals from -pi to 0? I calculated them, but it doesnt work out the way it should be, so I'm curious if I misunderstood anything.

Ted Shifrin

No, your integrals are always from $-\pi$ to $\pi$.

Magnus Alexander

Hm, so I will get $a_n=\dfrac{1}{\pi}(\int\limits_{-\pi}^{0} \dfrac{-x-\pi}{2} + \int\limits_{0}^{\pi} \dfrac{-x+\pi}{2})$, for example?

Uh, sorry, forgot to add $cosnnx$

Ted Shifrin

You need backslashes with sin and cos.

Jakobian

8:40 PM

It's funny how some people are so much smarter than you but you can't really tell from the surface level

Koro

why is it funny?

Ted Shifrin

@Magnus Something looks suspicious. If the function has only $\sin nx$ terms, then it needs to be odd.

Jakobian

@Koro I meant that, it's interesting

Magnus Alexander

@TedShifrin, thank you! I was just taking wrong integrals, now I will try to make a correct one and it should be fine

Koro

it's also interesting that if you tell someone- "make eye contact with me for 18 seconds", then both burst into laughter.

Magnus Alexander

8:45 PM

@Jakobian, I learned that you should never compare yourself to others, most of the time it's depressing, meaningless and mathematics is really not about that at all

Jakobian

@Koro I think that only happens if one of the parties attempts to make funny faces

Koro

oh, irregardless of that.

I don't know why it happens though. haha

Ted Shifrin

@Magnus Well said. I studied with some of the most brilliant mathematicians of the 20th century, and no way I was going to be comparable to them :P

Magnus Alexander

@TedShifrin I mean if Grothendieck compared himself to other mathematicians and took it serious, he would've probably given up on maths immediately ahaha

Jakobian

@MagnusAlexander I don't think so, you just need to be realistic about it

Rome wasn't built in a day

Magnus Alexander

8:50 PM

That's actually what he wrote in his book somewhere. He was talking about his years in university, and how advanced everyone were compared to him, etc

@Jakobian What do you mean by being realistic? I mean, ok, Pontryagin solved Hilbert's problem when he was 20 years old, Ramanujan read like 1 maths book and shook the world with his resuts, and Terrence Tao completely demolished his opponents at maths olympiads who were significantly older than him. We can compare ourselves to those people and someone else, but I don't see the point

Jakobian

comparing yourself to someone doesn't have to entail striving to be like them

one needs to understand they can only be who they are, of course

Magnus Alexander

So, what's the goal of it?

Jakobian

I think it's dangerous to focus on the meaning of things

Magnus Alexander

Ahaha I'm sorry I don't really understand what you mean

Jakobian

I just find that someone is wise to be admirable, and I find it inspiring

I didn't want to go for the whole comparing yourself to others discussion, I know people do that and that's a problem

Magnus Alexander

9:10 PM

I'm not sure we are comparing when we are admiring someone

Jakobian

oh. Maybe I messed up something here

Balarka Sen

I tend to admire mathematicians who are good expositors, helpful teachers, ... people who are generally nice. Mathematics in itself is a useless endeavor, so by extension I don't consider mathematicians to be worthy of admiration just in virtue of knowing a lot of math and proving a lot of theorems

robjohn

@Koro hey there. I’m out of town right now, so on my phone. I still dislike using mobile here.

Jakobian

knowledge is admirable. Being a decent human being is also admirable

Balarka Sen

that is a subjective thing

Jakobian

9:23 PM

I suppose. It's admirable to me

Balarka Sen

interesting

why?

Jakobian

I agree that a lot of mathematics is useless, but even so, I'm not studying it because it's useful. I suppose you could say I admire the art itself

Balarka Sen

why do you study it

Jakobian

It's like an adventure, no?

leslie townes

jakobian is in a court-ordered program where he agreed to do math instead of jail. it serves the main function of jail, i.e., keeping him off of the streets where he gets into trouble, without the expense to the public.

Balarka Sen

9:28 PM

@Jakobian if the romantic viewpoint is a sustainable source of interest, then sure

Avv

Hello Guys,

Any idea why we need to create new columns with quadratic terms for all the relevant prediction variables before we do backward model to find statistically significant variables in SPSS please?

Just looking for hints if possible

Magnus Alexander

I really don't understand why some of you refer to mathematics as to something "useless", do you mean most of it is not applied anywhere? First of all, not applied yet, and second -- do you consider art useless?

Jakobian

You have a point. Saying it's not useful anywhere else would be more correct

user4539917

There are critics of art for art's sake.

Balarka Sen

art gives you insight into history of human culture; i can arrange the political history of the last century by just listing down the stylistic shifts in modern painting. what insight does a infinite dimensional frechet Lie group give me

"most of math is not applied anywhere" is correct. if you claim heegaard-floer homology will be used to make machines of the future some fifty years down the line, i will gladly take this wager

most of math will never be applied anywhere, is my claim :)

i find the comparison between art and math very tiresome. generally people who do not understand art make this comparison

user4539917

9:38 PM

how does one "understand" art?

Balarka Sen

historically

or culturally, same thing

Jakobian

Isn't it that when one thinks of art, they think of consuming it, and of the artist, their endeavor to make it

Balarka Sen

maybe. but what would that imply

amWhy

@robjohn planning on the "same ole" Santa Hat? (It's a classic, so go for it.) But, another possibility I was thinking maybe you donning the green-mean-grinch-with-Santa-cap? Eagerly awaiting your Christmas identicon. (I think you might have previously worn just the same.)

@TedShifrin I know better now, but I can't help but interpret your identicon, when small and in front of your comments, as a unique sea-shell!

Jakobian

That comparison between art and mathematics is a valid comparison.
And, if we want to understand art historically, we can do that with mathematics as well. For example, we can look at Soviet mathematics

user4539917

9:49 PM

and French

leave Physics to the Germans :-)

Balarka Sen

you misunderstand me. what i mean is when you tell me "expressionism" i think of post-WW1 germany and the incoming weimar republic

arranging math historically gives you no greater insight into history of human culture

it doesn't evoke any imagery that can help you arrange history

fundamentally, its because math is detached from reality

user4539917

detached by extension

Balarka Sen

"That comparison between art and mathematics is a valid comparison." how? all sorts of commodities are produced by some and consumed by others. does that mean they are comparable?

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$Mathematics$ Mathematics