Mathematics - 2016-02-17 (page 6 of 6)

Ted Shifrin

22:00

MyDogAte: So I'm alleging that you can prove that when $R$ is any quotient of an integral domain (a ring with no zero-divisors), so it'll hold in any $\Bbb Z/n$.

Akiva Weinberger

Also, $(xy-1)a=(xy-1)b=0$

Semiclassical

whereas $D$ definitely isn't. (the momentum operator $p=-i D$ is, but then that gives $xp-px=i$)

Ted Shifrin

yes, I get the point, @Semiclassic. Cool.

Semiclassical

one also should have that $a^{\dagger}a$ serves as a 'number operator', namely $(a^{\dagger}a)a=a^\dagger a-1$

nope, i'm still wrong

Ted Shifrin

Not quite sure what that means.

Mike Miller

22:03

@PvAL: Still that's a nice start.

Ted Shifrin

You're doing well so far, MyDogAte.

Balarka Sen

Hmm, I think I understand implicit and inverse function theorem now. Will do harder problems tomorrow morning.

Semiclassical

the main reason $a^\dagger a$ matters is that, if $a^\dagger a \psi = n \psi$ for some state $\psi$, then $a^\dagger a(a^\dagger \psi) = (n+1)\psi$

so if we have a ground state $\psi$ such that $a\psi=0$, we immediately have a tower of eigenstates $\psi, a^\dagger \psi,(a^\dagger)^2 \psi,\cdots$ with integer eigenvalues

and if one uses the representation I said above, then $\psi = e^{-x^2/2}$ is such an example

Ted Shifrin

Ah, there's that eigenfunction for the Fourier transform again ...

Semiclassical

in which case one finds $(a^\dagger)^n \psi = H_n(x)\psi$ where $H_n(x)$ are the Hermite polynomials

Ted Shifrin

22:08

Ah, right.

Semiclassical

@ted yep. note that $a^\dagger a = \frac{1}{2}(x+D)(x-D) = -\frac{1}{2}D^2+\frac{1}{2}x^2 +\frac{1}{2}$

Ted Shifrin

Yup, I can do Algebra I :D

Semiclassical

which maps to itself under Fourier transform

Evinda

Suppose that we have two parallel vectors and want to find all the unit vectors that are orthogonal to these vectors.
Then it suffices to find all the unit vectors that are orthogonal to one of these vectors, let d.
If we find two non-collinear vectors, a and b, that are orthogonal to d , does it hold that all the vectors that are orthogonal to the initial two vectors are $t \cdot a+s \cdot b, t,s \in \mathbb{R}$? If so, why does this hold?

Ted Shifrin

What dimension are you working in @evinda?

Evinda

22:11

$\mathbb{R}^3$ @TedShifrin

Semiclassical

in terms of $x,y$ as earlier, one has identites like $(yx)x=x(yx-1)$ and $(yx)y=y(yx+1)$

Ted Shifrin

So draw some pictures. If you fix a nonzero vector in $\Bbb R^3$, what do all the vectors orthogonal to it look like?

Semiclassical

"drawing some pictures" is really the first step one should always do in a problem like that

Ted Shifrin

One reason my books are full of pictures ... even the algebra book. :)

Semiclassical

yep

going back to my stuff, one also can consider the eigenvalue problem $a\psi = \alpha \psi$

Ted Shifrin

22:14

Typographical nightmare.

Semiclassical

in which case one gets the so-called coherent states, which are a bunch of fun

maybe, but it's exactly the kind of nightmare you'd see :P

Evinda

@TedShifrin They will be parallel to it

Ted Shifrin

I prefer $T$ and $\lambda$, which don't look the same :P

What does orthogonal mean, @evinda?

Semiclassical

probably for these purposes I'd do $x\psi = \lambda \psi$

Ted Shifrin

No, $x$ is already the variable.

Balarka Sen

22:15

@TedShifrin Say I have a sequence $f_k : B \to B$ of contraction mappings, $B \subset \Bbb R^n$ a compact ball, converging pointwise to a contraction mapping $f : B \to B$. When can I say that the fixed points $x_k$ of $f_k$ converge to a fixed point $x$ of $f$?

Ted Shifrin

Hi @Khallil

Balarka Sen

Semiclassical

true. i was trying to honor the fact that $xy-yx=1$ was used earlier

Danu

Sigh, I'm stuck on a "trivial" question again :P

Evinda

Oh sorry , it means perpendicular @TedShifrin
But what can we say geometrically?

Ted Shifrin

22:16

@Balarka: A priori, $x_k$ needn't even converge.

Balarka Sen

But I do not know if this is true for arbitrary $\{f_k\}$ converging to $f$.

Ted Shifrin

So what are the things perpendicular to the vector $(1,0,0)$, @evinda?

Semiclassical

but $T\psi = \lambda \psi$ with $T S-S T=1$ is fine.

Balarka Sen

@TedShifrin I am not sure of that. Can you give me an example?

Evinda

$(x,y,z) \cdot (1,0,0)=0 \Rightarrow x=0$ @TedShifrin

Semiclassical

22:18

one is typically interested in the time evolution generated by $\exp(-i \tau ST)$

Ted Shifrin

Right ... So you get all vectors $(0,y,z)$ for any $y,z$.

What does that look like geometrically, @evinda?

Danu

Doing some analysis, @BalarkaSen? :P

Ted Shifrin

Oh, you're assuming contraction mappings.

Balarka Sen

Oh no, I am on the perfect path to becoming a real analyst. Nightmare.

Khallil

Hey, @TedShifrin!

Balarka Sen

22:18

@Ted Right.

Khallil

Long time no see, @BalarkaSen too! ^_^

Ted Shifrin

I was thinking of nonunique fixed points. Can't happen, then.

Balarka Sen

Hi @Khallil.

Evinda

That there is no intersection point? @TedShifrin

Balarka Sen

@TedShifrin Cool. Can you give me a proof? I need this thing to prove something I want to prove.

@Danu Exactly.

Ted Shifrin

22:19

Not yet I can't, @Balarka.

@evinda: Draw pictures.

Danu

4 hours ago, by Danu

Please don't make people solve your problems for you

(just kidding)

Semiclassical

going back to differential forms: I'm forgetting if there's a tidy expression for $d(fg)$ for scalar $f,g$. is it just $f\,dg+g\,df$?

Danu

PRODUCT* rule?

Semiclassical

yeah, i guess in the scalar case there's no need to get complicated

Danu

In fact that's kind of what one aims for in the general case, too

Balarka Sen

22:21

@Danu That's a strange sense of humor.

Ted Shifrin

This is typically a situation where one applies the inverse/implicit function theorems, @Balarka: If you have a family of functions depending on parameters, and you want to study the solutions to some equations, do they vary continuously/smoothly with the parameters?

Danu

You can define $d$ as the thing that extends this kind of thing

Ted Shifrin

Yes, @Semiclassic. It's called the product rule :)

Semiclassical

hrm

Balarka Sen

Right, I came up with this trying to generalize a portion of the proof of inverse function theorem.

Evinda

22:22

Can all the orthogonal vectors to (1,0,0) be written as follows: $s(0,2,0)+t(0,0,3)$ ? @TedShifrin

Danu

^But also in a simpler way

Semiclassical

how does it look for $d(\alpha\wedge \beta)$ with $\alpha,\beta$ being $n$-,$m$-forms respectively?

Ted Shifrin

Sure, @evinda, but no need to put $2$ or $3$. $1$'s will do.

Danu

@Semiclassical Product rule up to a factor $(-1)^{\operatorname{deg}\alpha}$ on the second term

Semiclassical

i seem to recall that it's just $\alpha \wedge d\beta$ and $d\alpha\wedge \beta$ with appropriate signs

Ted Shifrin

22:23

Never mind. Danu answered.

Danu

(as is easily seen in coordinates)

Semiclassical

there we go

Balarka Sen

@TedShifrin But note that applying IFT means I need C^1 assumptions. I only have C^0.

Ted Shifrin

I know, @Balarka. Yes, this is a standard uniform convergence exercise. What's an example of a sequence of contraction mappings that does not converge uniformly?

Evinda

So if we have a vector and two non-collinear vectors that are orthogonal to the vector, then any linear combination of these two non-collinear vectors will be perpendicular to the initial vector since the three vectors span $\mathbb{R}^3$ ? @TedShifrin

Balarka Sen

22:26

$f_n : I \to I$, $f_n(x) = x^n$ works, I believe.

It converges pointwise to the char function at $1$, which is obviously not continuous, so doesn't converge uniformly at all.

Ted Shifrin

The limit function isn't a contraction mapping.

Are you assuming convergence is to a contraction map?

Balarka Sen

Oh, you want the limit function to be C^0 and a contraction. Hmm.

Ted Shifrin

Well, I'm asking.

Balarka Sen

OK, so a sequence of contractions converging to a contraction ptwise but not uniformly.

Let me think.

Semiclassical

okay, then. so from my above i have $$d(\omega F)=F\,d\omega+\omega\, dF=\star d(\omega G)=G(\star d\omega)+\omega(\star dG)$$

Danu

22:29

What is $F$?

Ted Shifrin

@Semiclassic. Most of us use $\omega$ as a differential form, so this is very confusing.

Danu

and $G$

Oh, $\omega$ is not a form? Okay, I guess I'll shut up :P

Semiclassical

some scalar functions such that $dF\wedge \star dG =0$

Balarka Sen

Physicists have weird taste for notations, of course.

ducks twice

Semiclassical

not sure what else to use for an integrating factor :/

pffffft

Ted Shifrin

22:30

Um, $\lambda$ is nice. So is $u$.

Semiclassical

i'll go with $u$.

Ted Shifrin

@Balarka: I have an example on $\Bbb R$, I believe. I won't ruin your fun.

Semiclassical

$$d(uF)=F\,du+u\, dF=\star d(u G)=G(\star du)+u(\star dG)$$

better?

Balarka Sen

@TedShifrin At least now I know the thing I am looking for exists. I'll keep on thinking.

I should probably read about Lipschitz continuity following @MikeMiller's advice on becoming the world's leading expert on Lipschitz manifolds. Analysis is good stuff.

Danu

What is $u$? And aren't you omitting some wedges?

Ted Shifrin

22:33

Well, maybe I do. Maybe not. BTW, do we consider the zero function a contraction?

Danu

Why wouldn't you?

Semiclassical

anyways, i can wedge those expressions with $dF$ and get $F dF\wedge du = G (dF\wedge \star du)$

Balarka Sen

Well, the Lipschitz constant of the zero function is not in $(0, 1)$.

Semiclassical

well, all the forms in there are 1-forms

Ted Shifrin

But some people allow it to be in $[0,1)$.

Danu

22:34

Ah, you take it to be in the open interval, fair enough

Ted Shifrin

I think in my book I didn't do that, though.

Semiclassical

so i don't see how including wedges would matter since the only products are with 0-forms

Balarka Sen

Ah, alright.

@TedShifrin Nope!

Ted Shifrin

So, I don't have an example yet.

Balarka Sen

:)

Semiclassical

22:35

doing the same with $\star dG$ gives $F du\wedge \star dG = G(\star du)\wedge (\star dG)$

Mike Miller

@Balarka: In particular, of course, the definition of a form on a Lipschitz manifold.

Semiclassical

which is starting to hurt my head a bit

i think i've done something nonsensical back there :/

Balarka Sen

So I need to learn chapter 8 either ways. Good for me!

Ted Shifrin

Huh? @MikeM

Semiclassical

no, wait, i think i'm fine. i keep forgetting that $\star$ maps 1-forms to 1-forms here

Mike Miller

22:38

@Ted: Mm?

Ted Shifrin

What do Lipschitz manifolds have to do with contraction mapping constants?

Balarka Sen

8 mins ago, by Balarka Sen

I should probably read about Lipschitz continuity following @MikeMiller's advice on becoming the world's leading expert on Lipschitz manifolds. Analysis is good stuff.

Ted Shifrin

Ohhhhh. Sorry.

Balarka Sen

Side-comments can be confusing, I agree.

Ted Shifrin

LOL @ "Analysis is good stuff."

Akiva Weinberger

22:41

@BalarkaSen It looks like you have enough problems to keep you busy, at the moment, but I'd like to remind you of that topology problem I gave you a while ago.

Ted Shifrin

MyDogAte: You still have my (revised) algebra problem :)

Balarka Sen

@AkivaWeinberger I remember it :) I'll do it as soon as I get free.

Ted Shifrin

I think I put a false problem on a topology qual some years ago. I was about to put it on my students' topology midterm, and I think it's not quite right.

Akiva Weinberger

Namely, if $A$, $B$, and $C$ are closed subsets of $S^2$, then at least one of $A$, $B$, $C$, and $S^2\setminus(A\cup B\cup C)$ contains a pair of antipodes.

Wow, that took a while to type.

Semiclassical

different forms question: $\star(\alpha\wedge\beta)=?$

Mike Miller

22:43

@Ted: My officemate does quasicomformal manifolds. Those are cool.

Akiva Weinberger

@TedShifrin Yeah, I know, still thinking about it

Ted Shifrin

Suppose $f: X\to Y$ is a surjective local homeomorphism. If $X$ is compact and $Y$ is Hausdorff, prove that $f$ is a covering map.

Balarka Sen

@AkivaWeinberger Speaking of, did you ever prove that S^\infty is contractible?

Ted Shifrin

Ah, Lipman Bers stuff, @MikeM.

Akiva Weinberger

@BalarkaSen Nah, I looked it up. Sorry.

Semiclassical

22:43

(in the 2D setting i'm working with, it looks like $\star\alpha \wedge \star\beta = -\alpha\wedge \beta$)

Akiva Weinberger

But I don't know how to do it using only the CW structure (i.e. without coordinates)

misheekoh

Could someone help me with this latex? $$P(1) = \frac{4\choose2*5\choose1}{9 \choose 3}$$

Ted Shifrin

Are you sure about the minus, @Semiclassic?

Balarka Sen

@TedShifrin That seems true to me.

Semiclassical

no, i'm not. geometrically it seems wrong

Balarka Sen

22:45

@AkivaWeinberger OK, here's a geometric way.

Ted Shifrin

You don't want an asterisk, @misheek. I don't use \choose. I use \binom.

Semiclassical

rotating two planar vectors by ninety degrees each shouldn't change their cross product

Ted Shifrin

\frac{\binom{4}{2}\binom{5}{1}}{binom{9}{3}}

misheekoh

@TedShifrin thanks!

Balarka Sen

$S^\infty$ is built inductively from gluing two $n$-cells to $S^{n-1}$, correct? You get $S^n$, and then you glue 2 $n+1$-cells, and so on and so forth.

Akiva Weinberger

22:46

Yeah

Ted Shifrin

It seemed true to me, too, @Balarka, but I seem to be wrong most of the time.

Danu

@Ted Would you know any book that could give me a very basic introduction to sheaves and related things (in a concrete fashion)? Would Forster's book on Riemann surfaces be good?

Mike Miller

@Ted: Donaldson and Sullivan did work on these in dim 4, which people abandoned soon after since they're not as smart as Donaldson and Sullivan.

Balarka Sen

So each $S^n$ is realized as a subcomplex of $S^{n+1}$ as the equator.

Ted Shifrin

I like Forster. The introductory stuff on sheaf theory in Griffiths/Harris is good, too.

Bott/Tu is a great source, too, @Danu.

Semiclassical

22:47

but when i do $\star \alpha =\star(\alpha_x dx+\alpha_ydy) = \alpha_y dx-\alpha_y dy$ and wedge it with $\star\beta$

Balarka Sen

Note that I can nullhomotope the equator $S^n$ inside $S^{n+1}$ by pushing it through one of the attached disks (equators).

Danu

@Ted I told you Forster is lecturing on his book here next semester, right? :)

Ted Shifrin

Try $\alpha = dx$ and $\beta = dy$, @Semiclassic. Don't be so complicated.

Akiva Weinberger

@BalarkaSen Small typo when I reminded you of that topology problem: we need $A\cap B\cap C=\varnothing$

Semiclassical

fair enough

Akiva Weinberger

22:47

Forgot to mention that

Ted Shifrin

Yes, @Danu. I've lectured out of his book several times, but I'm sure he'll do it much better :D

Akiva Weinberger

Not that it's immediately relevant

Danu

@TedShifrin In the introduction he says it's material for ~3 semesters. I assume that means he'll skip a lot of stuff.

Balarka Sen

So just push each $S^k$ through the equator of $S^{k+1}$, at each level. This gives you a contraction :)

Semiclassical

in that case, the minus sign goes away. so now i should figure out what i'm doing wrong

Balarka Sen

22:48

@AkivaWeinberger Ok, gotcha.

Ted Shifrin

I did a bit more geometric stuff with curves than he does, in fact. But I did just one quarter. So I only covered about half the book.

Danu

Which makes me sad, and also makes it harder for me to read up on it before the course start (which I wanted to do, since I'll be taking too many courses next semester)

Semiclassical

bah, caught it. yeah, $\star\alpha\wedge \star\beta = \alpha\wedge \beta$

Ted Shifrin

The right question, @Semiclassic, is: What is $\star\star$? It depends on $n$ (dimension) and $k$ (degree of form).

Akiva Weinberger

@BalarkaSen Why is that continuous at $t=1$?

At the very end of the contraction

Balarka Sen

22:49

Exactly the question I expected from you :D

Akiva Weinberger

Yay?

Balarka Sen

You need to do it with care. Then you get continuous.

Semiclassical

should be $-1$ in this case, since $\star$ rotates $\vec{A}$ in $\alpha=\vec{A}\cdot \vec{dx}$ by ninety degrees in the plane.

Akiva Weinberger

*waves back*

Balarka Sen

More precisely, call the nullhomotopy of $S^n$ as $H_n$.

Akiva Weinberger

22:50

That sounds handwavy, is what I mean

Ted Shifrin

OK, @Semiclassic.

Balarka Sen

You need to glue the $H_n$'s by a $[1/2^k, 1/2^{k+1}]$ trick.

Akiva Weinberger

Remind me what nullhomotopy is?

Mike Miller

@Akiva: Then do the work to make it not.

Balarka Sen

^

Akiva Weinberger

22:51

Challenge possibly accepted

Semiclassical

i'm more used to the case of 3space, i suspect

Balarka Sen

@AkivaWeinberger Homotopy between the map you have and the constant map.

Akiva Weinberger

Ah

Mike Miller

@Ted: Here's a question I luke to ask students. What happens to * when you change the metric?

Akiva Weinberger

OK, gotcha

Still have a nasty knot at $1$ for me to untangle.

Balarka Sen

22:52

You're doing the nullhomotopy quicker and quicker as you get closer and closer to the vertex point, intuitively speaking.

That's why you don't mess up.

Ted Shifrin

Sometimes, MyDogAte, it's ok to wait a few years, as I've suggested before :P

Akiva Weinberger

But it sounds like I may get a $\sin(1/x)$-type problem

Where it also gets faster and faster until continuity breaks

Balarka Sen

Um?

Yeah, well, check that you don't.

Ted Shifrin

You have to consider what the topology on $S^\infty$ actually is, too.

Speaking of topology, @Balarka, what's your proof that that's a covering map?

Akiva Weinberger

Right. *mutters something about the inverse image of neighborhoods of $1$*

Balarka Sen

22:54

The subspace topology from R^\infty is nasty.

The CW topology is cool.

(I don't think they are the same, but it's contractible in both topologies anyway)

@TedShifrin Give me a few minutes.

Mary Star

I want to show that $\text{Aut}(\mathbb{Z}_p)$ is isomorphic to $\mathbb{Z}_{p-1}$.

The group $\mathbb{Z}_p$ is cylcic and is generated by one element. The possible generators are $1,2,\dots ,p-1$.

Each automorphism maps each of the element of the set $\{1,2,\dots ,p-1\}$ to one of the element $\{1,2,\dots ,p-1\}$, right?

But how can we show that that the automorphism group of $\mathbb{Z}_p$ is isomorphic to $\mathbb{Z}_{p-1}$ ?

Ted Shifrin

OK, @Balarka, but soon I have to go help 4th graders at the library.

Akiva Weinberger

@MaryStar Only for $p$ prime, right?

Aren't the automorphisms $z\mapsto az$ (or $z^a$ if you're using multiplicative notation, which would be weird)?

For $a\ne0$

Balarka Sen

@TedShifrin Yeah, probably now is not the best time. I am going to drool off soon.

Akiva Weinberger

(That doesn't completely solve it, though)

@BalarkaSen What time is it where you live?

Balarka Sen

22:59

4:30 AM.

Semiclassical

okay, i've entirely lost the thread of what i was trying to do

Akiva Weinberger

*Googles "time in Kolkata"*

Oh, wow

Ted Shifrin

OK, @Balarka, let me know tomorrow. I think it's wrong.

Akiva Weinberger

Don't you have school to go to…?

Ted Shifrin

As I said, I'm mostly wrong these days :)

Mary Star

22:59

@AkivaWeinberger It isn't specified but I think so.

Balarka Sen

Talk to you tomorrow. I don't think it's wrong, but I'll have a look carefully.

Akiva Weinberger

How are you even mathing competently…

Ted Shifrin

Yes, $p$ has to be prime.

It is ridiculous, isn't it, MyDogAte?

Mary Star

@AkivaWeinberger Why are these the automorphism?

Ted Shifrin

What is special about cyclic groups, @MaryStar?

Akiva Weinberger

23:01

'Cause $f(x+y)=a(x+y)=ax+ay=f(x)+f(y)$

Balarka Sen

@AkivaWeinberger For the first question, school's off. And I was about to ask the same question to you. For the second question, I have learnt to be nocturnal. I think @Ted and @Mike has noted that I am saying garbage a lot less than usual.

Akiva Weinberger

And similarly for inverses

Danu

@TedShifrin Do you know anything about mirror symmetry, and why mathematicians care about it?

Akiva Weinberger

It's 6pm here, school's over

Ted Shifrin

Not any more, @Danu. I knew some a long time ago. It's about beautiful algebraic geometry stuff.

Akiva Weinberger

23:02

Technically I have choir but I'm too sick to sing

(Cold)

Ted Shifrin

I think this room should just be left to the 15- and 16-year olds.

Danu

@TedShifrin Okay.

@TedShifrin Pfft

Balarka Sen

@Danu You want to see yourself asymmetrical when you look in the mirror?

Akiva Weinberger

Other than me and Balarka, who… @TedShifrin

Balarka Sen

Observation 1: Physicists say "pfft" a lot more than normal standards.

Mary Star

23:03

@TedShifrin Each of the elements $\{1,2,\dots ,p-1\}$ generate the group, right?

Ted Shifrin

There were lots of others ... Soham and Julian and ...

Balarka Sen

Explanations: String theory.

Forever Mozart

hi

Ted Shifrin

Not always, @MaryStar, but $1$ generates the group always.

Akiva Weinberger

@BalarkaSen Maybe you're not saying "pfft" enough.

Danu

23:03

I'm interested to learn about the mathematical ideas around mirror symmetry, since I'm supposed to start working on it from the physics side at some point soon :P

Ted Shifrin

hi @Forever.

Then you need to learn a ton of math, @Danu.

Mike Miller

@Danu: What are you looking for?

Mary Star

@TedShifrin Why not always? Do you mean in general or for this specific group?

Danu

There is this huge (~950 pages I think) book on mirror symmetry but it really doesn't do justice to the math, even though it pretends to

@MikeMiller Some mathematics books that could somehow provide a "way in"

Akiva Weinberger

@MaryStar Does $2$ generate $\Bbb Z_8$, for instance?

Mike Miller

23:04

To what?

Danu

To mirror symmetry

Ted Shifrin

@MaryStar: Try $\Bbb Z_4$. Does $2$ generate it?

Danu

I don't need to be an expert but I'd like to have some kind of idea

Mike Miller

Note that classical mirror symmetry and homological mirror symmetry look a lot different.

Danu

Both are probably interesting to me.

Mike Miller

23:05

The latter is now known to imply the former, I think, by work of Perutz.

Danu

(Though I think the physical one is the classical one, but I'm not too sure)

If you know any references that'd be highly appreciated.

Mike Miller

To see why symplectic geometers might care look up the review of Seidel's book on Fukaya categories.

Balarka Sen

@Danu "Through the Looking Glass" - Louise Carroll.

Semiclassical

@MikeMiller is that the one i linked you?

Danu

Did my joke earlier piss you off a little @Balarka? :P

Mary Star

23:06

@AkivaWeinberger No the generators of $\mathbb{Z}_8$ are $1,3,5,7$, right?

Mike Miller

Yes!

Balarka Sen

It did, thus I am trying my best to piss you off with my jokes.

Semiclassical

neat

Mary Star

@TedShifrin No, the generators are $1$ and $3$, right?

Danu

@BalarkaSen -1/10

You're way too obvious

Ted Shifrin

23:08

Yes, @MaryStar. But $1$ always generates $\Bbb Z_n$ for any $n$.

Good night, @Balarka.

Danu

Also it isn't "Louise" Carroll :P

Mike Miller

@Danu: If you want to understand homological mirror symmetry, you need to start by knowing what the Fukaya category is. Auroux has an intro to this that I like.

Danu

@MikeMiller Okay, let's take about 23 steps back from there. Is there an "intro to the intro"?

Semiclassical

I like that i linked that review to you because of the first half of the title, not the second :)

Balarka Sen

@Danu auto-correct.

Ted Shifrin

23:09

Poor @Danu.

Danu

@TedShifrin I feel motivated, if anything.

Mike Miller

No.

Danu

So, for clarity, I know beginner-level (2 courses) diffgeo and some stuff about Seifert-Van Kampen. Everything else that is above undergrad is probably unknown to me.

I don't expect to know anything about mirror symmetry by the end of next year, but maybe in 3?

Mike Miller

Learn some symplectic geometry is a good start.

Balarka Sen

I'm off to bed. See you all later.

Ted Shifrin

23:12

Yeah, and some stuff about complex algebraic geometry.

Danu

@MikeMiller Okay, there's a course on that next semester.

Balarka Sen

Or maybe I'll die tomorrow and won't see you all later again.

Danu

@TedShifrin I'll be taking two courses on alg.geom. next academic year.

@BalarkaSen err, okay.

Mike Miller

I don't know anything about the complez geometry side.

Danu

For now, the planning for next semester is (1) complex geometry (2) Riemann surfaces (3) math. gauge theory (intro; very basic as Mike already heard) (4) alg. top. 2 on (co)homology

Mike Miller

23:13

@Balarka: See you on the flip side.

Danu

+physics courses

Mike Miller

You're insane.

Danu

...but there's still the course on symplectic & bi-lagrangian structures

:\

@MikeMiller I think it doesn't have to be so terrible. I've got about 2 months of holidays now to start preparing, too.

I've got the notes of last year's complex geometry, and the book on Riemann surfaces that we'll be using

Math. gauge theory will probably overlap with a lot of stuff I already know (as we discussed)

I'm worried most about the physics courses and schedule clashes :(

PVAL

einfach-zusammenhangend translates as easy-coherent.

Danu

Lol

PVAL

23:19

It took a while to see it meant simply-connected

Danu

No, as simply connected

PVAL

Ya I thought it was quasi-coherent

Danu

quasi in English -> quasi in German

PVAL

for like a minute and had no idea what was going on.

Danu

@MikeMiller Are you still in the stages of taking any courses, or have you already progressed beyond that?

Akiva Weinberger

23:22

@TedShifrin The truth is, I'm terrified of learning a math concept for the first time in an actual math class

L33ter

hi @TedShifrin

@BalarkaSen

Mike Miller

I will probably sporadically take whatever Inferestig courses we have until I leave.

Danu

@AkivaWeinberger It'll be nice: Someone actually trying to make you get it ;D

Akiva Weinberger

It's so much easier to have at least a general idea of the concept before actually having to do homework about it

Danu

...that's what the lecture is for!

Akiva Weinberger

23:23

@Danu But I'm trying to make me get it!

L33ter

you know @TedShifrin so I didn't study anthropology or even attended classes for that class. So, one day I discovered we had an exam for that class and so I had to do it the exam people were finished after 30 min(it was multiple choice), but I took whole class since I was analyzing the questions to see which answer makes sense. So, I ended up getting 95 % in the exam haha.

that is very risky though

Akiva Weinberger

@Danu But what if I don't get it after the lecture? If it's not my first time seeing the concept, I understand the lecture much more

L33ter

I can't bring myself to get motivated on art electives

Danu

@AkivaWeinberger There will also be a book or lecture notes, don't stress!

L33ter

@Danu your a master student?

Danu

23:25

Yes

L33ter

that is insane semester !!!

Danu

@L33ter Sorry, I don't understand what you mean

L33ter

Are you taking those subjects in 1 semester?

1)complex geometry 2)Riemann surfaces 3)gauge theory 4)alg top ?

the one you mentioned above

Danu

I plan to try!

I'll probably fail to keep it up and end up having to drop one; I've also got physics courses to take care of.

L33ter

I took last semester grad algebra,grad mechanics,grad topology,complex analysis, and English.

It was hell

I did ok, but nonetheless, it was something I am never doing again.

graduate courses take a lot of time to understand

Danu

23:30

Well, I'm a grad student so I should be taking grad courses!

L33ter

I don't know how it is in your country, but here it is actually 6 courses + a thesis for masters.

Danu

Here we need 8, I think.

I'm taking a lot more than that, though.

L33ter

that is nice

Danu

I'm spending an extra year here to do more courses

L33ter

when I enter masters next year I will also attend some undergrad courses to brush up on some stuff

Mary Star

23:35

@TedShifrin Ah ok... How can we use this fact to show the isomorphism?

Danu

So what information do you need to specify to determine an automorphism of a cyclic group?

L33ter

you just need to show where generators go

the generator of your cyclic group capture all its information.

HowDoIMath

I'm not a native English speaker. Would you say "logarithmically increasing"? It is the "logarithmically" part I'm concerned about.

I certainly wouldn't be able to say it ten times in a row.

L33ter

brb

Danu

@L33ter You're spoiling the god damn exercise!

Mary Star

23:43

The automorphisms map the generators to generators, or not? @Danu @L33ter

Danu

You should be able to finish the proof on your own now

Mary Star

All the possible generators of $\mathbb{Z}_p$ are the elements $\{1,2,\dots , p-1\}$. The automorphisms map each of the elements $\{1,2,\dots , p-1\}$ to one of the element of $\{1,2,\dots , p-1\}$.
That means that $\text{Aut}(\mathbb{Z}_p)$ contains $p-1$ automorphisms, right?
But how do we continue to show the isomorphism?
Could you give me a hint? @Danu

Forever Mozart

@BalarkaSen I got a $T_1$ counterexample now

MickLH

@HowDoIMath I would totally say this... unless I could go a step further and say that it's "increasing at a rate proportional to the reciprocal of time"

Forever Mozart

hom

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$Mathematics$ Mathematics