Mathematics - 2017-07-12 (page 4 of 4)

Now, that's $1+2+4+3+\dotsb$ and then it repeats @Danu

8:27 PM

It's still too small to read, Perturbative.

but I cannot remember what an isomorphism is

@AkivaWeinberger OK

and its partial sums are $1,3,2,0$

Yes, so?

@TedShifrin just click on it. It opens a new window. Then just zoom in with your browser.

8:28 PM

That is, everything in $\Bbb Z_5$ other than $1/(1-2)=4$.

@TedShifrin The webpage shrinks things so the width fits in chat

That pattern works whenever we have a generating element of a $\Bbb Z_p$

@AkivaWeinberger Proof?

Left as an exercise

So what's your question, Perturbative. This looks like a Guillemin & Pollack exercise.

8:29 PM

Like for $x=3$ in $\Bbb Z_7$, the series is $1+3+2+6+4+5+\dotsb$ and the partial sums are $1,4,6,5,2,0$ @Danu

and it hits everything other than $1/(1-3)=3$

So yeah I can leave that as an exercise 'cause it's fairly easy when looked at the right way I guess, but $x$ does need to be a generating element

@TedShifrin It is, I'm just having some trouble with the highlighted part, I'll give you a general form of my question now

So like $x=2$ in $\Bbb Z_7$ doesn't work, because that gives you $1+2+4+1+2+4+\dotsb$ whose partial sums are $1,3,0,1,3,0$

though maybe there's another pattern that works for all $x$ that I haven't noticed yet

$2\mapsto 2(2)+1=5\mapsto 2(5)+1=4\mapsto 2(4)+1=2$

@AkivaWeinberger Ah yes, I think I see why it happens

It's because you have to fill out all of the elements before returning to the original one

That's why it cancels out

So there's an orbit $2\mapsto 5\mapsto 4\mapsto 2$ in addition to $1\mapsto 3\mapsto 0\mapsto 1$

8:33 PM

@Danu What do you mean?

and I guess $\sum_{i=0}^{p-1}i \equiv 0 \mod p$

@Semiclassical Oh whoa cool

That's everything except for $1/(1-2)=6$

which is in its own orbit

@AkivaWeinberger So for some power, you come back to the original number. The point of a generating element is that it hits everything else first.

Right

Equivalently, $6=-1$

8:34 PM

And then this thing about the sum

And the question is, why do the partial sums hit everything (even zero) other than $1/(1-x)$

so $-1\to 2(-1)+1=-1$

idk

ain't nobody got time for this

I have to present on my thesis in 6 days

^

back to reality :(

8:35 PM

That's a pretty good reason.

Tobias Kildetoft

@Danu That's plenty of time to prepare

No no no

More generally, one needs $(1-x)^{-1}$ to map to itself

@TobiasKildetoft Not when the results aren't even done

Go go now now now

Tobias Kildetoft

8:35 PM

@Danu So you are supposed to present it in 6 days, but it has not been written yet?

Avantgarde

good luck Danu

$(1-x)^{-1}\mapsto a(1-x)^{-1}+b=(1-x)^{-1}$

And it's also my only (remaining) chance to get a PhD position in the near future... "fun" times.

@Semiclassical Your map is "multiply by $x$ and add one"?

@TobiasKildetoft It's not the thesis defense. It's a PhD interview.

8:36 PM

Clearly $1+x+x^2+\dotsb$ is a loop under that. @Semiclassical

Tobias Kildetoft

@Danu Ahh, I see

@Danu It's basically pigeonholing. Multiply all those partial sums by $1 - x$ so you end up with $1 - x^{n+1}$. Now as you say $x^{n+1}$ hits everything else except $0$, so you just miss $1 - 0 = 1$

@akiva it would be if that were finite.

that's the same as missing $1/(1 - x)$ before

but presumably you mean something slightly different.

I'm actually annoyed that you write it with $x$.

8:37 PM

@Semiclassical Yeah but it acts the same as $1/(1-x)$

Also, is today your birthday? Congrats

So if$X$ and $Y$ are smooth manifolds of dimension $k$ and $l$ respectively, and $x \in X$ and $y \in Y$ and we are given parameterizations $\phi : U \to \phi[U]$ and $\psi : V \to \psi[V]$ of neighbourhoods $\phi[U]$ of $x$ and $\psi[V]$ of $y$ with $psi(v) =y$ and $\phi(u) = x$, then $d\phi_u : \mathbb{R}^k \to TX_x$ is an isomorphism and $d\psi_v : \mathbb{R}^l \to TY_y$ is an isomorphism, and $d(\phi \times \psi)_{(u, v)} : \mathbb{R}^{k+l} \to TX_x \times TY_y$ is an isomorphism

@Semiclassical Oh, 'cause it looks like this should be $\Bbb Z[x]$?

Because it means that I have to write it as $y\mapsto xy+1$ rather than $x\mapsto ax+1$.

@BalarkaSen 1.5 hours to go.

8:37 PM

Ohh.

Sure, @Perturbative.

(it's already my birthday in India, I guess)

close enough :)

I'da used like $a$ or something

other than $y$

Except we usually write $T_xX$, etc.

8:38 PM

Maybe $t$, $~t$ is nice this time of year

well you weren't born in india

so that can't be right :p

Mostly I want to have $a,b$ as parameters and $x$ as the independent variable.

Anyways. If I take my mapping to be $x\mapsto ax+b$

So why does ${d(\phi \times \psi)_{(u, v)}}^{-1} = {d\phi_u}^{-1} \times {d\psi_v}^{-1}$?

then the condition for an element to map to itself is $ax+b=x$

If $a=1$ mod $p$ and $b\neq 0$, this is impossible.

If $a=1,b=0$ then it's just the identity map and therefore trivial.

@Perturbative I don't know, but I saw something about block matrices, so maybe $\begin{bmatrix}A&\\&B\end{bmatrix}{}^{-1}= \begin{bmatrix}A^{-1}&\\&B^{-1}\end{bmatrix}$ is relevant

8:40 PM

If $a\neq 1$, though, then this becomes $x=b(1-a)^{-1}$

You have to interpret things appropriately, @Perturbative. We have $d\phi_u\colon\Bbb R^k\to\Bbb R^n$, so it's not invertible. But it is an isomorphism to its image, $TxX$, and then there's a smooth map $g$ defined on a neighborhood of $X$ extending $\phi^{-1}$ as a map from a neighborhood of $x$ to $\Bbb R^k$. By definition, $d(\phi^{-1})_x$ is the restriction of $dg_x$ to $T_xX$.

For the case of $b=1$ this means that $(1-a)^{-1}\mapsto (1-a)^{-1}$ under $x\mapsto ax+1$.

DogAteMy, I think he's worried about manifold issues, not linear algebra.

So $(1-a)^{-1}$ is indeed an orbit unto itself.

@Perturbative: Note that G&P define a map on an arbitrary subset to be smooth if at each point it has a local smooth extension.

8:41 PM

I forget what the name for that is, though.

a singleton orbit?

oh. a fixed point.

lol

So $(1-a)^{-1}$ is the unique fixed point of $x\mapsto ax+1$ so long as $a\neq 1$.

@Semiclassical But I think the "obviously $1+a+a^2+\dotsb$ is fixed under that, and it equals $1/(1+a)^{-1}$ if you handwave enough" argument should work

Looking back at it, maaybe not

lol

I dunno, should be fixable somehow waves hands

8:44 PM

I think the way one would do it formally is this.

By Fermat's little theorem, $a^{p-1}=1$ mod $p$.

So one really should have $1+a+\cdots+a^{p-2}$.

Hm, I wonder if you could make a topology on $\Bbb Z_p$ where geometric series converge

That's $0$, Semiclassic

I said that an hour ago.

(to a unique limit)

@TedShifrin Read again?

That's what p-adics are for, DogAteMy.

But it's a finite set…

8:46 PM

Oh, never mind.

and then $a(1+a+\cdots+a^{p-2})+1=a+a^2+\cdots+a^{p-1}=1+a+\cdots+a^{p-2}$.

Alessandro Codenotti

@AkivaWeinberger yes, but the trivial topology is boring

Oh this is like the "sum of roots of unity are zero" trick

Multiply it by a generator and they just rotate around

Right.

right.

8:48 PM

@AlessandroCodenotti sniped kinda

But I don't see why one should identify $1+a+a^2+\cdots$ with $(1-a)^{-1}$ in this context.

The only formal content it has is that $a(1-a)^{-1}+1=(1-a)^{-1}$

and this follows more simply just by writing $a=1-(1-a)$.

@TedShifrin I understand that, but I'm not seeing exactly how the fact the inverse of the derivative of a product of maps equals the product of the derivative of maps (if that makes some sense)

Alessandro Codenotti

@AkivaWeinberger woops

Oh, so it is what DogAteMy said, @Perturbative. The inverse of a product mapping is the product of the inverses. Just look at the block matrix.

$(A\oplus B)^{-1}=A^{-1}\oplus B^{-1}$ where that's the matrix direct sum.

8:51 PM

@Perturbative: Inverses of linear maps are unique, so just check it works.

Oh that's a cool notation

Makes sense I guess

Ah okay thanks @TedShifrin and @AkivaWeinberger!

@Perturbative: I know that book backwards, backwards, and forwards.

I'll come to you then, Ted if I have any more problems :D

Is that in the order of ways you've tried to read it

8:57 PM

Who reads, DogAteMy?

@TedShifrin So, tell me, what's meroeht fpoH-eracnioP

You worked hard at that, didn't you, Balarka'?

Oh is that the one with the secidni

How's the Spanish going, DogAteMy?

Bien

Ahora entiendo la palabra "se" mejor

8:58 PM

Múy bien?

read Tlon Uqbar Orbis Tertius in Spanish

porque no sabía que era la voz pasiva 90% del tiempo

…Tlon?

But in English we're taught to avoid the passive voice.

preferably while smoking weed

@Akiva Tl\"on to be precise

@TedShifrin Did you just…

9:00 PM

That's Borges, isn't it?

yeah

("Pero en inglés se enseñamos evitar la voz pasiva")

(to translate your sentence)

I should be heading to bed now

@LeakyNun Can you Spanish my me check

Night, Balarka'.

9:08 PM

Black hole. noun: avoid