Mathematics - 2017-04-24 (page 1 of 3)

BAYMAX

00:57

Cantors function is not absolutely continuous function

as the map maps from [0,1] onto [0,1]

?

Daminark

01:29

I mean it maps the Cantor set (measure 0) into [0,1]

That's the problem

ALannister

01:48

A whole lot of grief.

@MikeMiller eew did that beeeyatc&^ win?

1

Q: Existence of the least common multiple in a Unique Factorization Domain

Note: The other questions that have thus far been posted (by other people) relating to this topic have not asked for differences in the proof for the case when either $x$ or $y$ is zero or a unit element (I have been told these cases are trivialities, but sometimes seeing trivialities completely ...

The beeyatc&^ being Le Pen...eeew

Anybody around who can take a look at my question?

BAYMAX

02:14

yes

Loosely absolutely continuous function maps a set of zero measure to a set of zero measure

@Daminark

ALannister

Hey Betamax ;)

BAYMAX

hello Gangsta

ALannister

;)

BAYMAX

(' . ') :)

ALannister

What is the lcm of a unit element and a non unit element?

in a unique factorization domain?

BAYMAX

02:21

sorry don't know.I hope somebody will help

ok,will be back!

leo

@ALannister What is the $lcm(3,1)$?

ALannister

3

but that's different...in every ring, the units aren't just 1

leo

Then lcms are up to unit multiples

ALannister

So, the lcm will be the product of the non-unit element and the unit element?

leo

It would be the non-unit element up to unit multiples

ALannister

02:28

nice!

Thanks @leo

leo

Meaning, any multiple of a unit $u$ a non-unit $x$ would be divisible by unit multiples of $x$

Space Otter

10:33

Is there a limit to how high a power of e can be

Alessandro Codenotti

no, $\lim\limits_{x\to+\infty}e^x=+\infty$

Rajesh Dachiraju

11:38

4

Q: A minimization problem in function fitting setup

Let $\Omega$ be a convex, closed, compact set in $\mathbb{R}^d$ with a smooth boundary. Given a data $(x_i,d_i)$, $x_i \in \Omega$,$d_i \in \mathbb{R}$, $i = 1,2,3...N$, $N>d$ and $\sum\limits_{i=1}^N d_i = 0$ I want to find a continuous function $f:\Omega \to \mathbb{R}$, such that, $\int_...

functional-analysis real-analysis convex-optimization

Rajesh Dachiraju

11:53

Anyone interested in commenting on the above problem?

user123733

12:56

1

Q: Function with 3d geometry

i assume the functions as 5x,2x and 3x respectively . Solving through this I got option c as answer . but the answer is given as BCD

functions 3d

anyone answer this

Fawad

13:34

wolframalpha.com/input/…

I think I should know how to solve it^ but Iam not getting how to solve.

Poptart

Good morning all!

I have a question about Galois theory: If $\mathbb{Q}$ is the rationals and $\zeta_7$ is a primitive seventh root, I know that the Galois group for the extension $\mathbb{Q}(\zeta_7)$ (adjoining the primitive seventh root) is $\mathbb{Z}_6$, the cyclic group of order $6$ since we can observe that $\zeta_7$ can be mapped to any other primitive seventh root and any one of those mappings generates all the other possible automorphisms.

What I'm not uncomfortable with for some reason, (I'm not sure why I'm uncomfortable), is that if I view this extension as a vector field ove…

Semiclassical

14:15

Am I missing something silly in my comment here? math.stackexchange.com/q/2249692/137524

I can't see how one avoids the determinant vanishing regardless of $n$.

Mike Miller

you're right, since the rows literally repeat after 3

Semiclassical

Yeah.

idgi

Tim The Enchanter

Err, I happen to have seen this exact problem before, It doesn't say $|p|^2 \ne 0$, it says $P^2 \ne 0$

Semiclassical

The only thing that comes to mind is that one wants the matrix whose entries are the squares of P.

Tim The Enchanter

So I'm pretty sure they're talking about the zero matrix.

Its the same, right down to the options, I've got it in front of me right now.

Semiclassical

14:25

Hmm.

That'd make considerably more sense, yes.

Tim The Enchanter

I'm pretty sure it is the zero matrix whenever n is a multiple of 3

Semiclassical

The person asking it seems to be confused on that point, though. (As does the answerer, for that matter).

Tim The Enchanter

Because then you've got triples of the cube roots of unity.

Semiclassical

Right.

Tim The Enchanter

@Semiclassical Yeah, I think ill leave a comment

Semiclassical

14:32

@TimTheEnchanter The remaining issue is that it seems like there's more than one right answer.

Tim The Enchanter

@Semiclassical Yeah this is one of the multiple answer ones (sadistic I agree).

Semiclassical

If it were [P]^2=0 then that'd be fine.

Oh. So more than one can be right.

That's icky.

For an explicit approach, one can factorize $P=\Omega u v^T$ where $\Omega$ is diagonal and $u,v$ are appropriate column vectors. Then the point is that $u^T v=0$ iff $3$ divides $n$.

I guess I could just say $P=uv^T$ instead, though. Especially since in this case $u=v=(\omega,\omega^2,\ldots,\omega^n)^T$.

Tim The Enchanter

Oh yeah that would work.

Never thought of it that way, I just bruteforced it.

Semiclassical

The point remains to show that $\sum_{j=1}^n \omega^{2j}=0$ iff $n$ is a multiple of 3.

But that's not hard, since $\omega^{2}$ is itself a cube root of unity.

Tim The Enchanter

Because you just get a repeating sum of $\omega ^2 + \omega + 1$ until the tailing terms.

ALannister

15:08

What is the lcm of two unit elements?

Arrow

Let $r>0$. Suppose we center at each point of $\mathbb Z^2\subset \mathbb R^2$ a circle of radius $r$. Consider the line $y=ax$ with $a\in \mathbb R$. If $a\in \mathbb Q$ it's clear the line intersects some circle (in fact even some lattice point). But if $a$ is irrational? How can I prove it intersects some circle?

ALannister

15:25

0

Q: Least common multiple of two unit elements

If $R$ is a commutative ring with unity that is also a Unique Factorization Domain, I was wondering what the least common multiple of two unit elements is? I know that normally, the least common multiple is unique up to multiplication by a unit element. Also, in the integers, I know that every e...

abstract-algebra ring-theory unique-factorization-dom least-common-multiple

Okay, I deleted that question b/c it had something in it that wasn't true.

But, is the lcm of two unit elements simply the product of the unit elements?

Or since the lcm is unique up to multiplication by units, the lcm could be either unit element?

Balarka Sen

Hi @PaulPlummer

Paul Plummer

@BalarkaSen Hey

Balarka Sen

15:42

Gonna try and learn some more foliations right now

Paul Plummer

Nice. Going to try to learn more about this hierarchically hyperbolic stuff

Balarka Sen

sounds fun

Semiclassical

This question makes me lol. (See my comment and then Dietrich's answer.) math.stackexchange.com/q/2249877/137524

Smit

hello

i need help with graph-theory can anyone assit me?

say wether it is possible to construct a simple graph with 10 vettices and 9 edges which (i) is not a tree, (ii) a tree

Paul Plummer

The humor seems to be lost on me

Tim The Enchanter

15:47

@Smit A connected graph is a tree iff the number of vertices = 1 + the number of edges

Semiclassical

Both "if" and "only if" being important here.

Smit

In this case, it is a tree right? i have explained in a concept of such that if there a n vertices than there should be n-1 edges for a graph to be a tree(connected and no cycles)

Tim The Enchanter

Yep

Semiclassical

For (ii) you can also just draw said graph.

Smit

but we know a tree is a graph, thus we can contruct graph for (i) and (ii) right?

Tim The Enchanter

15:50

@Smit We know it must be a tree so (i) is impossible

Semiclassical

^

I don't recall how one proves that, alas, but if that's a theorem then (i) is indeed impossible.

Smit

Oh! Makes sense! I intrepreated the part i wrongly. thanks a ton @TimTheEnchanter and @Semiclassical

Semiclassical

np.

Tim The Enchanter

^

Smit

I need a one more help to see if my answer is correct. Exams coming soon!

Suppose that we form 6-digit numbers by rearranging the digits of the number 123456 (i) How many dufferent numbers are possible? soln: 6! (No repetation, order matters) (ii) How many numbers contain the digit dequence 1234? soln: 3! (iii) How many numbers contain at least one of the two digit sequences 12 and 23? : soln 5! + 5! + 3! such that 5! for (12,3,4,5,6) another 5! for (23,1,4,5,6) and 3! for (1234,5,6)

is this correct

Semiclassical

15:58

How did you get 3! for (ii)? I also get 3 but not from 3!.

Balarka Sen

(1234)56

1234 is considered as a single digit

Semiclassical

Oh, yep.

Balarka Sen

@Smit Are you not overcounting at (iii)?

(1234, 5, 6) can be a case for both of those 5! cases.

Smit

oh @BalarkaSen is inclusion exclusion thingy?

Tim The Enchanter

imo should be 5!+5!-4!

Balarka Sen

16:00

Right, by inclusion-exclusion.

A = numbers containing 12

B = numbers containing 23

Then you are counting #(A $\cup$ B)

Semiclassical

oh, on an entirely egoistic note

woo just passed 10k

Balarka Sen

congrats

Tim The Enchanter

woot woot

Semiclassical

(huzzah for meaningless numerical goals)

Smit

congrats @Semiclassical

@BalarkaSen thanks a ton! it was great break down inclusion exclusion.

Balarka Sen

16:04

No problem. I personally don't know combinatorics; what I do know is set theory and that's what I try to reduce everything down to :P

Paul Plummer

Your first step in becoming Tony Montana @Semiclassical

Semiclassical

oh lawd

Imago

16:16

Are there such functions f, g such that: f o g is strictly increasing and g o f strictly decreasing? From my intuition that shouldn't be possible.

Felix.C

16:26

If a real matrix has a determinant of 1 then we now it's at least invertible but are there other things we can get out of it?

SteamyRoot

16:46

It belongs to the special linear group

BAYMAX

any help on argument principle in complex analysis

?

Semiclassical

If it's composed entirely in terms of integers, one furthermore has that it's inverse is also entirely integer.

(There's a proof of that on MSE somewhere)

BAYMAX

$I = \frac{1}{2\pi i}\int _{C} \frac{f^{'}}{f} dz$ ,where $C$ is $|z - 1 -i| = 2$ , so what is the value of the integral when $f = \frac{z^2 - 9}{z^2 + 1}$ ?

SteamyRoot

Just write the inverse in terms of the adjugate

BAYMAX

Argument principle states $I = N - P$

Where $N$ is number of zeros inside $C$

And $P$ is the number of Poles inside $C$

$N$,$P$ are zeros and poles of $f$ respectively.

I get $N = 0$

and $P =2 $ as $i , -i$

is the number of $P$ correct ?

Semiclassical

16:56

Well, doing that integral numerically in Mathematica gives me -1 not -2.

So that doesn't bode well.

Actually, here's the issue: The distance between $-i$ and $1+i$ is $|1+2i|=\sqrt{5}>2$. @BAYMAX

So the pole at $-i$ isn't inside $C$.

BAYMAX

Yola

Thanks @Semiclassical

Sha Vuklia

17:18

guys, can we apply Gram-Schmidt on complex vectors?

in order to do that, I'd need to know how the projection of complex vectors works

however, I'm not sure about the order of the hermitian product

SteamyRoot

Any inner product space works, no?

Sha Vuklia

I'll give an example

Say we want to orthogonalised $v,w\in V$

where $V$ is a complex inner product space

Then we have two options:

$$
v^{\perp}=v-\frac{\langle v,w\rangle}{\Vert w\Vert^2}w
$$
or
$$
v^{\perp}=v-\frac{\langle w,v\rangle}{\Vert w\Vert^2}w.
$$

and those are not the same

while they are for real inner products

I could try both out and see which one is orthogonal

maybe they both are

and then work backwards

SteamyRoot

It's always the second one

Sha Vuklia

are you assuming linearity in the first or second component?

SteamyRoot

First

Sha Vuklia

17:21

and why is it the second option?

今天春天

Hi!
Just a short question. I want to compute probability $P(A+B = 2 | B + C = 1)$ if I know that $p(A=0) = p(A=1) = p(B=0) = p(B=1) = p(C=0) = p(C=1) = 1/2$. I just use the definition: $P(A+B = 2, B+C=1) / P(B+C=1) = (1/2)^3 / (1/2)^2 = 1/2$. Is this correct?

SteamyRoot

I'm not sure what you mean with that question?

Usually, the second option is just how we define projection

Sha Vuklia

Shouldn't $P(B+C=1)$ be: $P(B=1,C=0)+P(B=0,C=1)=1/4+1/4+1/2$? Instead of $(1/2)^2$

@SteamyRoot and oh okay

well if it's a definition, i'm good

SteamyRoot

Perhaps it helps if you think of this

If you have an ONB $e_1, \dots, e_n$

you know you can write any vector $v$ as a linear combination, say $v = v_1e_1 + \cdots + v_ne_n$

Sha Vuklia

yes

(hi @Ted btw!)

SteamyRoot

17:26

You want the projection of $v$ on $e_i$ to be $v_i$

Sha Vuklia

yes

so we get: $P_{e_i}(v)=\langle v_1e_1+\dots+v_ne_n,e_i\rangle$ ?

or $P_{e_i}(v)=\langle e_i,v_1e_1+\dots+v_ne_n\rangle$

今天春天

@ShaVuklia you're right thank you!

SteamyRoot

Well, which one is it? :P

Akiva Weinberger

This really old answer of mine keeps getting upvotes for some reason

Sha Vuklia

the first, because we are linear in the first component!

:)

Akiva Weinberger

17:29

It has like 19 upvotes now and I'm not sure why or how

SteamyRoot

Yup

Sha Vuklia

so you put the vector that is going to be projected in the component that is linear, right.

Akiva Weinberger

I wrote it in Nov '14 and got an upvote on it today for some reason

Sha Vuklia

right. = I got it :P

which one @Akiva?

nv

got it

:P

well it's a popular question, that makes sense!

i upvote everything that helps me, even if it is ancient

Akiva Weinberger

This one

Daminark

17:32

Hey everyone!

Sha Vuklia

yo @Daminark

have you had your test already?

also hi @Astyx

Daminark

Yes, and I didn't die, :D

Sha Vuklia

hahaha! Good to hear!

did they ask the questions you predicted?

Astyx

Hi chat @Sha @Akiva @Daminark

Daminark

Might lose a few points for sloppiness in a couple questions, but not too many, and the idea was there for every question

One of them

Sha Vuklia

17:34

well sounds good, I think!

Daminark

Proving that there's no finitely additive, translation invariant measure on $\ell^2$ for which the measure of every ball is positive and finite

Well, how are things going with you?

And what's up @Astyx?

Astyx

One last exam tomorrow morning until I move on to another group of school on the next day :p And you ?

Sha Vuklia

nothing much here, just started with my homework:d

Daminark

Well, I've got my manifolds midterm, though our professor did say it was more of a test to make sure we were following, as opposed to a "gg get rekt" type of thing, you know?

Astyx

That's nice

SteamyRoot

17:39

I now get the urge to write "gg get rekt" instead of "good luck" on future exams...

Not sure if students will appreciate, even if they realise it's a joke

Astyx

I would laugh so hard if I read that coming from my prof

Daminark

One of my friends TA'd linear algebra first quarter, and whenever someone clearly put no effort on homework, he'd put a "Good job!" sticker

Professor felt it was a bit antagonizing and asked him to stop, but still, lol

And yeah I'd crack up

Ted Shifrin

17:52

A grad student grader when I took Guillemin & Pollack made all sorts of sarcastic remarks and I did not appreciate them as an undergrad who'd put in a dozen or more hours of work. ... In hindsight, his comments were correct, but still very annoying. I tried to be less caustic grading as a professor.

Sha Vuklia

@Steamy! I've shown it generally. Say we have $v=v^\perp+v^\parallel$, and we use the "wrong" inner product first:
$$
\langle w,v\rangle=\langle w,v^\perp+v^\parallel\rangle=\overline{\langle v^\perp+v^\parallel,w\rangle}=\overline{\langle v^\parallel,w\rangle}=\overline c\Vert w\Vert.
$$
This yields to:
$$
\overline c=\frac{\langle w,v\rangle}{\Vert w\Vert}\iff c=\frac{\overline{\langle v,w\rangle}}{\Vert w\Vert}=\frac{\langle v,w\rangle}{\Vert w\Vert},
$$
so in the end, the vector that is being projected will always be in the linear component.

Ted Shifrin

Hi @Astyx

DogAteMy: I continue to get upvotes for a throw-away answer on trace of a matrix. I'm up to 85 now. It's ridiculous.

Daminark

l m a o

I still think my best rep answer is the fake proof of Cayley-Hamilton

SteamyRoot

My best answer is probably still mentioning Lothar Collatz as a mathematician who's famous only for a conjecture and not for any actual results of theirs...

Sha Vuklia

(shit i had to add a square)

Ted Shifrin

17:57

But you excel at fake-i-tude, Demonark.

Daminark

And yeah I imagine. We never had any of that when he TA'd my class first quarter, though we did actually try on it, and if someone gets a perfect score he usually makes some "purr-fect" pun and puts a cat sticker (he's very much a cat person)

gasps

Eric

Hi chat

Daminark

Hey @Eric!

Eric

Yo when is your manifolds midterm

Daminark

Thursday

Eric

17:59

Ah ok my diff geo is then too

Ted Shifrin

heya @Eric

Eric

Hey @Ted how's it going

Astyx

Salut @Ted

@Daminark What's this fake proof ?

Daminark

Ah, makes sense, you guys are immediately after us

The proof is that the characteristic polynomial is $p_A(t) = \det(tI - A)$, so plug $t=A$

Astyx

Lol

Ted Shifrin

18:02

This reminds me of my horrendous not-proof of Lagrange multipliers in a course my freshman year. And I knew better, too. I guess I've learned a little since then.

Can I help confuse you about any geometry, @Eric? :D

Demonark: I approve of cat persons :)

Felix.C

If we have a submanifold in $\mathbb{R}^{nxn}$ constructed by a function $F:\mathbb{R}^{nxn} \mapsto \mathbb{R}$ over a regular value, what is the dimension of this submanifold? Would say that it has dimension (n*n-1)?

Ted Shifrin

Sure, when you have a regular value, @Felix, the codimension is given by the dimension of the range. So codimension $1$ here, meaning dimension $n^2-1$.

Daminark

How does that proof go @Ted? And lol definitely, and yes @Felix

Ted Shifrin

As with linear algebra, each constraint equation cuts down the dimension by $1$.

Demonark: You pretend it's a regular critical point as opposed to a constrained critical point.

Dodsy

Hey

So I got a 99% on Unit two of chem.

so I've sent my transcripts

and my average is 88%

we'll see what happens

just thought I'd update you Ted.

Ted Shifrin

18:10

Well, that's encouraging, Nate. Thanks for letting me know. :)

Dodsy

Yeah well I'm pretty stressed out.

But, it's out of my hands at the moment.

Daminark

Also @Eric talked to Fefferman briefly about life, he said that if 209 goes well, then the bootcamp would give the go ahead for grad analysis, and he said that if there are space considerations, it may be worthwhile to pick up point-set independently and do algebraic topology

Ted Shifrin

your sentence confused me, Demonark. I wish you'd put commas around Eric. I thought you were saying Eric talked to Fefferman. ... You definitely need some point-set but not inordinate amounts for algebraic topology.

But you need some serious algebra (like familiarity with structure theory of finitely generated modules and more), so ...

Daminark

This'll be undergrad atop, so it won't assume algebra at a higher level than what I'll be doing, I think

Ted Shifrin

Ohhh .. undergrad.

Daminark

18:13

But yeah I guess I'll read Hatcher's notes on point-set

Ted Shifrin

And don't forget to learn some more about forms and integration on manifolds!

Daminark

Fefferman's point in not going straight to grad atop was that some professors (e.g. Calegari) who teach it do assume background in it

Ah, yeah definitely

Ted Shifrin

And actually compute a few integrals, dammit, Demonark.

Daminark

I've got a pretty significant block of time between the end of this year and the bootcamp, so during that time I'll hopefully pick up some of that then

computes the integral of the 0 function and calls it a day

Lol jk

Ted Shifrin

Cool. Of course, I won't be around for the month of June to torture you, so you'll probably get away with ...

Daminark

18:17

Haha, perhaps

Ted Shifrin

mumbles annoyedly

Daminark

But yeah, it definitely seems like there are a lot of things to be picked up on the side. I don't know how much harmonic analysis we're doing in 209, but I think not much, so that'll probably be one thing, as well as forms, categories, point-set, etc

Variational calc also seems to be a reasonably important thing that won't really come up in any class so

Ted Shifrin

Well, Demonark, you can't learn everything. Save some things for your more advanced age.

@Sha: You deleting yourself?

Daminark

Lol @Sha this happens to me a lot, I can't figure why things aren't working but once I say it to someone else I'm like "Wait hold on a second..."

Ted Shifrin

Demonark: This is a good one for you to ponder.

Sha Vuklia

18:24

@Ted yes, because I saw the mistake, so I didn't want to bother you guys unnecessarily!

@Daminark it's called the "duck" effect, right?

in programming at least

something like that

where programmers tell their code to a duck :P

"rubber duck debugging", apparently

Daminark

This is pretty neat @Ted, and yeah some stuff will have to be relegated to another time. The main thing that I will be making a directed effort to figure out separately is point-set, also categories because I'm in a group and we're doing that

Forms if we don't do it in difftop (which I still am not sure about)

Eric

LOL @Ted for now I'm confused enough trying to work through the problem sets you sent me

Dominykas Mostauskis

I'm implementing a clustering model, and it uses exp(-d) as a neighbourhood func tion, where d is index denoting order by distance. The idea is, if d is 0, you get 1, and as d increases its influence decreases fast, so it can be used to se lect neighbourhoods.
This works when, in d, 0 means closest and biggest one means farthest; however, problem is, I have to use indexes that are semantically inverse: 0 means farthe st, while biggest means closest. What can I use, instead of exp(-d), that would have similar neighbourhood-type outputs?

Eric

currently thinking about the Backlund transform one, was this in your diff geo notes? I seem to recall this exact thing but my memory is fuzzy

@Daminark idt Neves is big on forms

Daminark

So it seems, he mentioned integration as one of a couple topics to be done (though it was like, in one week he'd cover it among other topics, so not really), but he's going a bit slower than originally intended so I wouldn't be surprised if he cuts that out

Eric

18:31

weird if he doesn't cover integration considering the formal title of the course is smooth manifolds and integration

although my version of the course didn't cover it and it was weird

Daminark

I mean in my experience it seems like course titles are treated as loose recommendations

LeGrandDODOM

@AlessandroCodenotti do you know what the following means ? It's written in a deep learning book "out of all possible probability distributions with the same variance, the normal distribution encodes the maximum amount of uncertainty over the real numbers. We can thus think of the normal distribution as being the one that inserts the least amount of prior knowledge into a model."

especially the first sentence

Eric

lol @Daminark or are completely ignored and transmuted by Greg Lawler

Daminark

Yeah... though I think the probability of his including some amount of complex analysis in there is reasonably high :P

Alessandro Codenotti

No idea @LeGrandDODOM

Daminark

18:35

Still amusing that he changed the title of the class

Though many do that, I think

Eric

I think Lawler is the only person ive seen do it

Daminark

Well, few people add on stuff for sure, it's mostly shortening

Are you guys covering parts of diffgeo that are not Riemannian with Neves?

Eric

I don't think so

he said we'd do all of do Carmo, and maybe more

idk if the more includes stuff that isn't Riemannian

Daminark

I see

Well, Neves did retitle our class "Differential topology", which seems to be the basis for mostly (and maybe completely?) ignoring stuff like integration, forms, de Rham cohomology, etc

Eric

we're like pretty far in do Carmo now so it seems like it'll be on track

Daminark

18:40

We've been going a bit slow-ish

Ted Shifrin

@Eric: Yeah, the Bäcklund transform was the last exercise in the undergrad notes section 3.3.

hi @Alessandro, @LeGrandDODO

LeGrandDODOM

hi Ted!

Daminark

So far in the quarter it's basically been just chapters 1, 2, and now 4 of Milnor

Ted Shifrin

@Eric Demonark: It still upsets me that so many people teach a year-long graduate geometry course and mention forms only when they have to. But oh well ...

Daminark

@Ted I think the second quarter of the sequence is a lot on differential forms

Eric

18:43

I think the reluctance to do forms has to do with not wanting to teach the algebraic preliminaries

kind of lame because you end up with a powerful tool as a trade off

Daminark

Like I know when Calegari taught grad differential topology he basically spent a full 3 weeks on forms

Ted Shifrin

@Eric: It does take a few days to go through all the tensor, alt stuff — or one can avoid it all, as I do in my multivariable course. But for diff top and grad diff geo I would do it all right, for sure. In particular, in the grad course I want students to see tensor products of bundles more generally (especially if they're going to do complex geometry).

Daminark

And I think Wilkinson did it this year as well?

Eric

Yeah Neves said at the beginning of the course that he really didn't wanna talk about all the tensor algebra and the looking at vector bundles other than the tangent bundle cause he wanted everything to be as concrete as possible

Mike Miller

vector bundles are extremely concrete imo.

Balarka Sen

18:47

morning @MikeMiller

Daminark

Hey @Mike and @Balarka!

Balarka Sen

hi hi

Ted Shifrin

To me universal bundles on $\Bbb P^n$ (and their various tensor powers) and $G(k,n)$ are very concrete. It's really a matter of what style geometry one does. Modern day Riemannian geometers don't think about that stuff nearly as much as I did for my research and interests.

Hi @MikeM and @Balarka.

Daminark

Lol I don't think we're doing vector bundles much, just some particular stuff like the tangent bundle, normal bundle

Balarka Sen

Hi @Ted.

LeGrandDODOM

18:49

@AlessandroCodenotti seems like normal distribution has the smallest entropy among all distributions with fixed variance

Ted Shifrin

Demonark: I'm talking about the grad course. In your course I followed G&P and just did tangent and normal bundles. I'm fine with that. There's a big step up in the grad stuff.

Eric

I only saw some bundle stuff when I did a reading course on difftop with farb

Daminark

Oh, I see

Balarka Sen

Tangent and normal bundle are very essential for differential topology.

Eric

purely intro though, we didn't really do anything with them

Balarka Sen

18:49

A smooth map is literally a bundle-morphism of tangent bundles

As in, that's the coordinate-free definition

Ted Shifrin

Eric, I'm talking about grad stuff, for sure. For example, it's highly significant and important that if you have $M^n\subset\Bbb R^N$ (or $\Bbb P^N$ or whatever), then the Gauss map pulls back the tautological $n$-plane bundle on $G(n,N)$ to the tangent bundle of $M$.

Well, @Balarka, that's getting a bit carried away. Next you'll be deluged in categories.

Balarka Sen

nah

Ted Shifrin

@Eric: Granted it's a bit more algebraic, but it can also be important to understand the tangent bundle of symmetric spaces in terms of the algebraic structure. I'm always amazed at how diff geo books don't make that clear.

Daminark

@Balarka Digitally join our category theory reading group this summer! Come to the dark side

Ted Shifrin

puts Demonark on permanent ignore

Eric

18:52

right

@daminark don't let Schlag know youre doing that

Balarka Sen

i am not joining you until you do topos theory

Eric

if he hears you say the word category he will unleash his fury

Balarka Sen

that's the real stuff

Ted Shifrin

adds Balarka to ignore list

Daminark

Robby told me about topos theory and it seems wild but really cool

Balarka Sen

18:53

@TedShifrin Well a manifold is a topoi!

Mike Miller

@Balarka why is a bundle map of TM a smooth map?

Balarka Sen

intentionally annoying grammar there

Eric

Does anyone know why a manifold is called a manifold?

Mike Miller

you ever seen Boy's surface?

Daminark

@Eric There was this one time when we were doing a bit of ODEs on manifolds, and Schlag asked us about the image of a vector field, we were pondering it and I'm thinking, oh this is the tangent bundle thingamabob that Alex talks about

Got quite some sass for that...

Ted Shifrin

18:55

@Eric: I was confused in my high school days about whether it should be intake or exhaust.

@MikeM: So your point is that we have immersed submanifolds?

Daminark

Probably deservedly so, but yeah I could see why sufficiently many incidences of the word "category" or "functor" would make him displeased.

Balarka Sen

@MikeMiller hmm, I suppose I messed that up?

His point is that Boy's surface has lots of folds

Ted Shifrin

@Eric: for "many dimensions" ...

Balarka Sen

I think

Eric

but the sphere has no folds

@Ted ohhhhhhh that makes total sense

Ted Shifrin

18:57

The meaning of the word has nothing to do with folds in singularity theory.

confuzled by the intermingling discussions

@Eric: I already recommended this question to Demonark, but I'll point it out to you, too.

Eric

Neves made a joke that if you give something a bunch of folds you expect it to be called a manifold but it isnt and we call them orbifolds or something

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