The usual. Writing a paper :P

Making good progress on it?

Also have you finished with grading?

Yup, all resits I have to grade are over

About 60 students have now filled out the poll I made about the speed of my lectures. They are generally considered too fast, though not by a huge amount. Plenty of good comments that are mainly constructive.

First paper is finished, supervisor made some adjustments and now it's in the hands of the co-author... Hopefully he'll have time to read it soon

Second paper, well, a first draft should hopefully done by the end of the week

Nice, to both of you

Also almost done with the paper I refer to in my project description (or at least a version with those things mentioned in that project)

Sick. You've got a bit over a week left, right?

Yeah, deadline is Friday next week

Nice, good luck

thanks

@TobiasKildetoft where, exactly, do they want you to slow down?

(just curious :^)

@skullpatrol The poll seems to indicate that they feel the speed is about equally too fast for all parts. Some of the comments point to it being me speaking too fast and it would help if I just took a short break occationally and summarized.

Also, many of them would like me to include more examples and go into more details with those

Hmm...

...summarization will give them a chance to catch-up.

One thing which might help with respect to fast talking, I was told this by one of the TAs because the first lecture I gave for the summer camp was just way too fast, for many incomprehensible.

He told me that there's a prof who also speaks quite fast, but what he often does is give a sort of meta-justification for what's happening. I've seen this from some of his lectures myself, he'd both give a proof as well as the reason why he expects the statement to be proven in that way

Yeah, hearing it the second-time definitely helps :)

nice idea

Lies!

lol

Also, you should totally draw a picture before writing down the proof

^

Topologist: "Here is the proof" draws incomprehensible picture of wiggling spaghetti

Now that's accurate :P

When Emerton did some basic Fourier on finite groups, the structure theorem came up and he was telling us "Well, so you have a tension, if you try to mod out by larger subgroups, you're more likely to have a cyclic quotient but also you might mod out by the element you're looking for. There's the reverse problem if you mod out by small ones. What saves us is that the group is abelian, so we have all the normal subgroups we need to hit the right middle ground"

Chef: hands you a pudding

The proof is in eating it?

Ding

:D

Or it was something to that effect. But yeah it wasn't pictorial, though finite Fourier analysis and the structure theorem for finite abelian groups isn't that pictorial anyway. Still, I think this sort of intuition is good for algebra. Better than all of Hatcher's handwaviness inb4 Balarka raeg

throws wiggling hot spaghetti at @Daminark

@Balarka I just tabbed in and saw that and now idk what to do with my life

FOOD FIGHT!!!

nullhomotopes Balarka

::makes popcorn::

nullhomotopes the popcorn

Popcorn: I have higher homotopy groups, b*tches!

:O

@Eric We should take the silliness to the trash room

don't want to bring the Darude Trashstorm here

Lmao

@TobiasKildetoft examples are a great idea, wherever possible :-)

@skullpatrol Yeah, we will have a lot more now (we would anyway, as the topic we get to now is groups which lends itself way more to examples than the elementary number theory we have done so far)

does anyone know of a lie algebra which is semi simple but has a non trivial ideal?

I feel it's simple, but i'm just forgetting my examples

@mdave16 Take the product of two simple ones

ffs, i just saw i even wrote that down, but still continued being confused

it is in some sense the only example (apart from taking products of more than two simples)

yeah, i was getting confused with the direct product of vector spaces being a vector space

which is a bit embarrassing, but it happens

i had written sl(2,C)^2, and then was trying to work out what the bracket would be on it exactly, and the basis, and whether this was legal, but it is, because it's still a vector space!

hi chat

Yo

@Semiclassical Sup semi

not much atm

I have a proof of <a^k > = < a^gcd(n,k) > , can you tell me if it is true or not?

I mean like if I took wrong step or stuff like that =p

Go ahead!

all righty

what's n

the ord (a) = n

sorry forgot to mention that

Yo @PVAL

hi

if we pick an element of < a^k> , for example , (a^k)^i = a^ki+nl for some integer l , whitch belongs to < a^(gcd (n,k) >

thus <a^k> is a subset of <a^gcd(n,k) > and <a^gcd(n,k) > is a subset of <a^k >

they are equal

I used ofc that a^n = e

@KasmirKhaan Looks fine.

Yup

@TobiasKildetoft @Daminark thanks guys

Dammit I'm getting sniped by Tobias now?

the book proved it by starting that d | n and d| k

n =qd and k = pd

*Correction: I meant I was even able to prove that $R^*$ must be commutative

It's the goal of the world to snipe Demonark.

So it seems

@Helios: What do you mean by $R^*$ trivial?

Group of units of $R$ only contains 1, I think?

Clearly false. Try upper triangular real matrices.

What are the units in that?

@TedShifrin, in case of the real matrices all but zero is a unit. Did you correctly interpret the story, as I am trying to prove that commutative implies only trivial units (id est 1), hence why the reals not work

First, you're wrong in the first sentence?

Oops, ignore the first part, talking about the matrices themselves it is ofcourse all the matrices with det not zero, which implies no zero on diagonal.

Right.

I misread that myself yes, I thought you were talking about the real numbers

OK, so we're assuming the units in $T$ form a commutative group.

No, we assume that $T^*$ is commutative

I just said that?

Oops, sorry sorry, yes

So commutativity forces the diagonal elements to be equal. And the only way to guarantee that is to have only one nonzero choice. Agreed.

Why do they have to be equal?

Work out what it means for two upper triangular matrices to commute.

I do not see it, unfortunately. So what I did was literally working it out, see: latex2png.com/output//… .

From here, it is clear that the diagonal elements should commute, but then I do not see why they have to be equal

OK. So how are the upper-right entries equal?

I know that is the thing where I should conclude it from yes, but how?

*that that

Elementary algebra. :)

What equation must hold?

Equality between both entries

So write down the equation for me.

ae + bf = db + ec

And this must hold for all units $a,c,d,f$ and arbitrary $b,e$?

I realise that yes

So can you manipulate the equation to get something useful?

I am sorry if I am being silly at this point, but that's the point I have been stuck at for quite some time

Note there are two things with $b$ and two things with $e$.

I see, but the problem to me is that bf and db both have a b of course, but not at the same side

Aha ... OK, I was ignoring this issue because I thought we'd already decided the things on the diagonal had to be in the center.

In the center of $R^*$ yes, but not necessarily of $R$, right?

I see. Good point.

Maybe it suffices to use $b$ and $e$ already units, then.

What if you do that?

But is that general?

No. But maybe it's enough to give you what you need?

Try it!

Learn to do some mathematics by wishful thinking :)

Wait, is it misunderstanding or are you saying I should assume that $b$ and $e$ are units?

That's what I'm saying. See what happens if you do that.

That might work out, but that is not as it is stated in my textbook. It does make things easier yes ;)

We're trying to prove that $a=c$ and $d=f$. If we can do this by choosing certain $b,e$, we are fine.

I do not agree with this, we would only prove it in the case of certain $b,e$

Not in the general case

You're giving up before you think.

As we want to eventually use that $a=c$ and $d=f$ to say something about $R^*$ in the general case

You're not getting the logic.

If using $b,e$ units forces $a=c$ already, then it must hold that $a=c$.

I may be wrong, but I don't believe I am.

I get the idea behind your point, but I wish to carefully think about this, if that is okay

Of course it's OK.

But before you think about this, make sure you see that I'm right to get the conclusion I want.

Only thing I can get to is $e(a-c)=b(d-f)$

OK, perfect. Now try to be clever. :)

Hmm.. The only thing I can come up with is that it is allowed to take $b=e$, such that $a-c=d-f$

Not clever enough. :)

Wait, but what if i note that it must also hold for $d=f$?

Aha. Forget $b=e$. Go on.

Oh that's slick

So for $d=f$ it implies that $a-c=0 \implies a=c$

The demonic peanut gallery speaks :)

Because you're using that $e$ is a unit, note!!!

I do get now that it is indeed allowed to assume $b,e$ to be unit!

*units

Note that this is a good technique to prove things. Sometimes you can assume a general thing is specific and use that to draw general conclusions.

Yes, I know that, I used it before. But at this moment it just did not cross my mind

You happier now? :)

Ofcourse I am! Thank you very much, this does really help with starting up my algebra proving skills again

Cool :)

@Alessandro I summon you

If you're there lmao

You think Alessandro is a genie in a bottle?

are you gonna talk about measure theory again

It's worth a shot, call his name 3 times and he appears or smth

Not exactly

Borel sets

yeah

no

too much analysis on chat these days

There's this proof from analysis which to this day everyone is clueless about

Hush, Balarka. Anything is better than category theory ****.

Oh, the genie appeared.

You see?

:P

@Ted but i get left out nah, higher topos theory is the real mathematics

Then it's time for me to disappear from the room forever.

i am going to learn it and converse with myself on higher topoi all day in chat

so that none of you will understand it

hence will not be able to participate in conversations

Then you too shall be ignored, just like Chris'ssis.

But yeah the thing is to prove that the Borel hierarchy terminates in $\omega_1$ steps. Proving that in that time it'll be done is alright, because if you take countably many elements then there's something after it in an $\omega_1$ list (vague but I'll take that for now)

@Ted I am Ramanujan after all

@Dami why hast thou summoned me?

To help me understand a step in the proof that the Borel hierarchy does not terminate in countably many steps

And I think this'll be up your alley

So, $A\subset \mathbb{R}^2$ is said to be universal wrt a property $P$ if it satisfies $P$ and if any subset of $\mathbb{R}$ satisfying it can be given as a horizontal section of $A$

Hmm, I'll check Folland to get that down better

I'm a bit busy right now, I'll be back in a moment, but the linked question should be relevant

Lattices in dimensions 8n?

But the proof I had in mind is to construct a universal set for the property of membership in a given Borel class that was achieved in countably many steps, but then it turns out (for reasons I'm clueless about) that for such a universal set $\mathcal{g}$, $B = \{x:(x,x)\in \mathcal{g}\}$ is in that class while its complement isn't

But yeah, thanks!

I'll put this on hold, @Balarka okay so now here's your thing

So a vector bundle is locally trivial, meaning a section can be locally given by a map from the base space to the fiber, yeah?

A local section?

@BalarkaSen eh, that doesn't seem ambitious enough. Learn IUT :p

Hi, DogAteMy.

@Daminark Well, yes, and local triviality is equivalent to having n (rank of v.b.) independent local sections from the base.

I think he means a global section coincides with a map from the base space to the local fibers but I'm not sure

Global section corresponds to local sections that are mutually compatible somehow.

@Semiclassical Good idea.

Welcome to sheaf cohomology land. :P

@Daminark Why do you say "to the fibers"? A (global) section is a map $s : B \to E$, to the total space; however over each $p \in B$, $s(p)$ belongs to the fiber above $p$.

I'd jokingly suggest that I'd become a category theorist, but um

I guess that is why you are saying it.

He means the fiber. Vector-valued function locally.

No just no

Oh, he's thinking about that picture. Sure.

Then you have to consider a global section as a "patching up" of the maps $s_\alpha : U_\alpha \to \Bbb R^n$ according to the transition functions, like Ted said.

@Ted I have almost never used this picture; I suppose it's useful in algebraic geometric lingo.

And differential geometry for sure. That's the whole point of the principal bundle viewpoint of connections.

Remember that we talked about tensoring with a line bundle as twisting? (What's $L\otimes L$ if $L$ is the tautological line bundle on $\Bbb RP^1$, otherwise known as the Möbius bundle?)

Mhm.

I am not sure what you mean by the principal bundle viewpoint of connections by the way; a connection on a G-bundle to me is an Ehresmann connection on the associated vector bundle which is G-equivariant

Oh wait okay I think I get things now

The transformation rule under the $G$ action for the $\mathfrak g$-valued $1$-form comes from change of frame formulas :)

@TedShifrin Oh I see. Agreed.

You're not squeaking your way out of differential forms :)

heheh

I still remember you hated multivariable calc when we first started chatting.

Basically the thing I was wondering was that Atiyah put a vector space structure on the set of sections, but why is he able to do it globally?

Though I guess you can handle local things separately and patch together

How do you multiply a section by a scalar or add sections?

No, you don't need to localize.

I am not sure if Atiyah uses the local patch-up definition or the global definition I wrote

ie which one is Daminark using?

I can't read Demonark's mind. He is demonic.

Well, his thing was that a local section is given by a vector-valued map on the base space

Ok, and a global section?

See that's the jump I'm wondering

I suggest thinking about the global definition first. A section of $E/B$ is nothing but a map $s : B \to E$ such that $s(p)$ belongs to the fiber over $p$ in $E$ :)

If $U$ is a small local patch on $B$ over which $E$ trivializes, then you can write $E|_U$ (restriction of the vb) as $U \times \Bbb R^n$ (upto isomorphism).

Now a map $s_U : U \to U \times \Bbb R^n$ such that $s(p)$ belongs to $\{p\} \times \Bbb R^n$ is nothing but a map $s_U : U \to \Bbb R^n$ by projection to 2nd factor.

So local sections over $U$ are the same as maps from $U$ to $\Bbb R^n$; precisely because there is a canonical identification of the nearby fibers over $U$ to $\Bbb R^n$.

@Daminark Is that good?

Hi nice folk

Oh good. I am not included.

can anyone give me a good example of kernel of a group ?

like explain it to me in cool example

@TedShifrin haha why not ? :D

<---- mean ogre.

@TedShifrin you are very nice too !

You need a question that makes sense.

well I just heard about the kernel of a group

no, kernel of a group HOMOMORPHISM

Yeah that's not a thing

sniped

end my life

We did not do anything about it at class , but I want to know a little about it

I think am asking to much ><

Ill return other time with better Q:s

o/

I'm gonna stay up for Federer - DelPo, Ted :D

Hey, this is a very nice talk.

(non-mathematical)

Sorry my mom called me for... quite some time

But yeah so I got that much down, so global sections manifest locally as maps into a vector space

yup

so can you see why they form a vector space? (eg, what's scalar multiplication or vector addition?)

Yeah, pointwise

That much I'm happy about already

Right, OK. In fact $\Gamma(E)$ is a $C^\infty(B)$-module.

Okay so now I wonder why we can make a global jump

That's my concern here, for each region on which you're trivial you can handle it, but imagine you have, say a space with two connected components and you're considering product bundles on each with different dimensions?

Okay that example might be trivial but in general

Can you explain what you mean by "making the global jump"?

So on our set $U$ where it's trivial, it's just a set of continuous maps to $\mathbb{R}^n$

But on two different sets $U$ and $V$, if the set of local sections on each is a vector space, how do we know that when we can do it to both at the same time and preserve continuity?

Okay I guess I have the right way to say it now

So let's say we're on a trivial bundle

If we take a function $f:U\to \mathbb{R}^n$, that should give us a section by saying $s(x) = (x,f(x))$, right?

Mhm.

Thus, any sum of sections is also a section

In this context

So you see why $\Gamma(U)$ is a vector space for an open subset $U$ of the base over which $E$ trivializes, but not why $\Gamma(E)$ is a vector space? Is this what you mean?

Yeah

If you have two sections you can pass to the trivial patches and add/scale the sections restricted to each patch, but how do you know that you can now take the resulting collection of localized sections and turn that back into a global section?

Technically, you don't need to localize to a trivial patch to "add" two sections, is the point.

If $s_1, s_2 : B \to E$ are two sections of $E/B$, it makes sense to write $s_1(p) + s_2(p)$ for any $p \in B$, because $s_1(p), s_2(p)$ are two vectors in the same vector space (namely, $F_p$, fiber over $p$ in $E$).

But then is their sum still going to be a continuous map to E?

The thing is there is a well-defined notion of vector addition on each fiber; exploit that to pointwise add two sections by adding over each fiber.

@Daminark Yeah, because it's locally continuous, by just what you said.

Locally continuous means globally continuous.

Oh really?

Wait yeah

For any point $p \in B$, choose a trivial patch $U$ around $p$ and then you have $(s_1 + s_2)|_U$ as a section of the trivial bundle over that patch, which is continuous because you can write that as a map to $\Bbb R^n$.

@Daminark Yup

Okay that question was retarded, sorry

No need to apologize. It's a cute "woah" argument.

Useful to know.

I should actually learn point-set at some point

And of course similar for scalar multiplication there too

@Daminark You know a lot. Just think of $B$ as a metric space.

Nobody cares

hi chat

Yo

@orlp Yep, pretty funny to read through. :D

Nice schemes, though.

@Wildcard I'm a bit sad that there's so much elaborate answers that give wildly different answers, and it also appears that their authors do not intend to improve/correct/clarify them

as it stands I still have no idea what the correct answer is