Mathematics - 2017-07-05 (page 6 of 10)

Akiva Weinberger

18:01

Proof for the equation: Note that $(A-2)^2+(B-2)^2=1$ gives a circle in the $AB$-plane that doesn't intersect either the $A$- or $B$-axes.

So one $S^1$ factor comes from choosing a point on that circle, essentially

gxyd

@EricStucky and that is because the transcendence degree of $\pi$ is $\infty$?

Akiva Weinberger

and the other two come from $w^2+x^2=A$ and $y^2+z^2=B$ (note that $A$ and $B$ are always positive).

Oh, crap. Typo. I meant $(w^2-x^2-2)^2+(y^2+z^2-2)^2=1$

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

@BalarkaSen In any case: the inverse image of the equator in $\pi_4(S^2)$ is gonna look more-or-less like that, modulo an inverse stereographic projection to actually get it on the sphere.

The inverse image of the two points on the equator are 2-tori, twisted in such a way that they link with each other.

Balarka Sen

Ya

Akiva Weinberger

The linking prevents the map from being contractible, similar to the Hopf map, I'm guessing.

You saw my conjecture for how the 2-tori live on the 3-torus?

Eric Stucky

Yes, gxyd

Akiva Weinberger

18:06

Essentially, both perpendicular to a space diagonal, when we view it as a cube with opposite sides identified.

Balarka Sen

@AkivaWeinberger No, I don't think so

Akiva Weinberger

You know how, on the Hopf map, when we unroll the torus (preimage of equator) into a square, the preimages of the points are diagonal lines?

Tobias Kildetoft

@EricStucky In what way is the trancendence degree of $\pi$ infinite?

Balarka Sen

Ya

Akiva Weinberger

So I'm thinking this is essentially the same but up one dimension

You know how you can slice a cube to get a hexagon

or a triangle

(those two are parallel slices)

So, I think the 2-tori would be like that.

Balarka Sen

18:08

Oh yeah so these 2-torii inside 3-torii should live like an m/n "surface" in T^3

for a generic point

That is, take an m/n curve in T^2, and product it with S^1

For the things lying on the equator m = n = 1; diagonal curves

Akiva Weinberger

You should need three numbers to specify a torus there, I think.

Think of it as the coordinates of the normal vector.

I'm thinking of (1,1,1) tori.

Balarka Sen

Ah

I suppose you are indeed right

Akiva Weinberger

I just wanted to specify that I do not want a (1,1,0) torus (which kind of looks like a slanted parallelogram I guess)

Hence the verbosity

Mike Miller

I am so damn sleepy

Akiva Weinberger

Do you want a drugs

Alessandro Codenotti

18:16

isn't $\Bbb Q (\pi)$ isomorphic to $\Bbb Q (x)$?

Akiva Weinberger

@BalarkaSen You said you have a visual reason why this is order 2?

@AlessandroCodenotti Yes, I believe so

Tobias Kildetoft

@AlessandroCodenotti Yes

Akiva Weinberger

Why?

I think you'd even call $\Bbb Q(x)$ a transcendental extension of $\Bbb Q$

which makes sense if it's isomorphic to $\Bbb Q(a)$ for all transcendental numbers $a$

Astyx

By definition of transcendental numbers no ?

Balarka Sen

@AkivaW So our map is $S^4 \stackrel{\Sigma h}{\to} S^3 \stackrel{h}{\to} S^2$, right?

Akiva Weinberger

18:19

Yeah

Astyx

They are the $a$ such that $\Bbb Q[a]$ is isomorphic to $\Bbb Q[X]$

Akiva Weinberger

I guess you'd only need to show the first one is order 2 (if that's true)

Balarka Sen

Does it not suffice to prove $\Sigma h$ has order 2?

Damn sniped

Akiva Weinberger

@Astyx Hm, yeah, I suppose so

Then do the quotienting

Balarka Sen

There is a Thom-Pontryagin argument I can tell you

but I think I knew a more visual one, hmm

Alessandro Codenotti

18:21

@Astyx yeah, that follows fom the first isomorphism theorem

hi @Ted

Astyx

ohi Ted

Ted Shifrin

hi @Alessandro — smashed into any cars lately? :)

ohi @Astyx

Akiva Weinberger

@AlessandroCodenotti Why do you ask?

Alessandro Codenotti

nope, but I'll have adriving lesson tomorrow morning so I might soon

Astyx

What's up ?

Akiva Weinberger

18:22

Hey Ted, how is you?

Balarka Sen

Oh, I actually know something else

Alessandro Codenotti

@AkivaWeinberger someone asked how to show that a basis of $\Bbb Q(\pi)$ over $\Bbb Q$ is infinite above

Ted Shifrin

hi DogAteMy, Balarka

I assume we get to assume $\pi$ is transcendental?

Alessandro Codenotti

hopefully

Tobias Kildetoft

@AlessandroCodenotti Actually, I asked in what way the trancendence degree was infinite since that was what was asked and confirmed by Eric

Akiva Weinberger

18:23

So it's the same as $\Bbb Q(x)$, as we sanity-checked for him

Balarka Sen

Suppose you glue a $D^4$ to $S^3$ along $\Sigma h$. Then the resulting space is the mapping cone of $\Sigma h$. But $C(\Sigma h) = \Sigma Ch$, and if you glue a $D^3$ to $S^2$ along $h$ you precisely get $\Bbb{CP}^2$.

So this space is the same as $\Sigma \Bbb{CP}^2$

Alessandro Codenotti

@TobiasKildetoft I don't know what the trascendence degree is

Akiva Weinberger

Cones commute with suspensions? I guess that's probably true

Balarka Sen

I think this can be used to prove non-null-homotopic-ness (...?) of $\Sigma h$. If it was nullhomotopic, $\Sigma {\Bbb{CP}^2}$ would be homotopy equivalent to $S^4 \vee S^2$

I am sure there is a cohomology argument for doing this.

Tobias Kildetoft

@AlessandroCodenotti That's the thing. Trancendence degree is usually something a field extension has, and in this case the extension by a single trancendental element just has degree $1$.

Balarka Sen

18:26

@AkivaWeinberger Yeah

it does

Ted Shifrin

Hmm, no one remarked on yesterday's Jumble, so maybe I'll stop posting them.

Akiva Weinberger

Oh I think I see it I guess @BalarkaSen

Mike Miller

@Bal You're trying to see why the Hopf map survives to stable homotopy?

Alessandro Codenotti

just looked up the definition, the trascendence degree looks crazy hard to compute even for reasonable extension ($\Bbb Q(\pi,e)$)

Ted Shifrin

Well, $\pi$ and $e$ have no algebraic relation ...

Mike Miller

18:27

This (like you say) is equivalent to $\Bbb{CP}^2$ not admitting some stable splitting, and that's probably more or less equivalent to knowing Steenrod squares

@Ted As far as you know!

Ted Shifrin

Is that not known? Surely ...

Akiva Weinberger

@TedShifrin No known algebrsniped

Mike Miller

No, transcendental stuff is very hard.

Balarka Sen

Openopenopenopen

Ted Shifrin

oh well — who cares :P

Balarka Sen

18:28

@MikeMiller Ah ok

Alessandro Codenotti

@TedShifrin that's open apparently

Mike Miller

That's whyu Balarka just does topology now.

Akiva Weinberger

$\pi+e$ might even be rational

Ted Shifrin

LOL

Tobias Kildetoft

@TedShifrin We only know that one of $\pi e$ or $\pi + e$ is trancendental. The other might not be for all we know

Balarka Sen

18:29

my tiny brain cannot take number theory

Mike Miller

@BalarkaSen This is a common trick. Stabilizing forgets the ring structure but remembers Steenrod squares, and the Steenrod square is easy to compute on Cp2

Balarka Sen

if i am going to get the fields medal, i am going to get one for topology, not transcendental number theory

so better to focus on that

Mike Miller

Axiomatically Sq^2 alpha = alpha cup alpha is nonzero

So the same is true after you suspend

Akiva Weinberger

Do they show any favoritism to field theory, I wonder?

Mike Miller

but it's now a higher dim class so no relation to relate it to cup product

Alessandro Codenotti

18:31

@AkivaWeinberger someone made a joke in chat that if your research has no multiplicative inverses you get a rings medal at most

Balarka Sen

I actually know literally nothing about Steenrod squares, but it makes sense. They are higher order cohomology operations; whereas cup product (with a specific class) is a low-grade cohomology operation which distinguishes $\Bbb{CP}^2$ and $S^3 \vee S^2$

Ted Shifrin

smacks Alessandro

your language is inaccurate, Balarka

not classes, operations

Mike Miller

I wonder if one can prove that Steenrod squares exist algorithmically (prove you can generate every space using operations the axioms respect)

Thanks @Ted

Ted Shifrin

now if only Balarka could learn apostrophes so easily ...

Balarka Sen

@TedShifrin Thank's

Ted Shifrin

18:32

Uh huh

Your welcome

Mike Miller

I have to go do some algebra I think.

Tobias Kildetoft

@BalarkaSen Thank is what?

Balarka Sen

You're very kind

Mike Miller

See you later.

Astyx

@BalarkaSen Do you know about Bohr's model for the atom ?

Balarka Sen

18:33

@MikeMiller I am sorry for you.

Mike Miller

@BalarkaSen Are you trolling him by using the correct you're to make him think you were trying to use an incorrect one, but weren't wrong?

Ted Shifrin

gets dizzy and collapses

Balarka Sen

Your' correct

Eric Silva

hi chat

Ted Shifrin

hi @EricS

Eric Silva

18:35

how is everyone

Krijn

@MikeMiller This is good.

Mike Miller

doing homological algebra

so, you know how it is

Akiva Weinberger

I can put apostrophes wherever I wan't

Tobias Kildetoft

@MikeMiller From a topological point of view?

Ted Shifrin

good thing DogAteMy will be self-banishing for 5 weeks

Balarka Sen

18:36

Eric Silva

oh @Ted i wanted to ask you if one could generalize crofton's formula to work for $m$-dim submanifolds of $\mathbb{R}^{n}$

Astyx

I live right next to the lawful good then

Krijn

@BalarkaSen I have no idea why these things would correspond to these texts

Ted Shifrin

of course, @EricS ... there are all sorts of generalizations to more interesting geometry than volume too. These are the kinematic formulas. Chern did them in $\Bbb R^n$, I did them for complex submanifolds in $\Bbb P^n$.

Akiva Weinberger

@TedShifrin Spanish has no apostrophes

It makes me so sad

Ted Shifrin

18:37

But you have all those accents to play with, DogAteMy (fewer than French, admittedly).

Balarka Sen

@Krijn It's an algebraist-proof joke

You're not supposed to get it

Eric Silva

interesting

Tobias Kildetoft

@BalarkaSen I am not sure where to place myself on that chart

Mike Miller

@TobiasKildetoft nope

i just need to do it

Akiva Weinberger

@TedShifrin Does French have a ü?

Eric Silva

18:38

I was wondering if you could use it to define an $m$-dim measure that differs from the hausdorff measure

Ted Shifrin

If you want to see a classic overview, look up Santaló's book on Integral Geometry, @EricS.

Eric Silva

although it might just be the same thing

Tobias Kildetoft

@MikeMiller And I assume also not from the point of view of abelian categories in general?

Mike Miller

Someone is doing volume geometry? Poor choice

Eric Silva

i read somewhere that there was no unique measure that gave $m$-dim surface area somewhere though

Ted Shifrin

18:38

um, no, that's German, DogAteMy, although French has an occasional ë and ï.

Balarka Sen

@Astyx Bourbaki...

Mike Miller

@TobiasKildetoft That would be actively unhelpful in this case

Akiva Weinberger

I'm just saying, Spanish has it (albeit rarely)

You have us beat on the è_é front, though

Ted Shifrin

@EricS: You mean on the ambient space? No. Unless you're doing Kähler geometry. That's one of the amazing things about complex geometry — you have a universal form on $\Bbb P^n$, for example, who gives $2k$-dimensional volume for $k$-dimensional complex submanifolds.

Tobias Kildetoft

@MikeMiller Yeah, probably

Eric Silva

18:40

oh wow @Ted, that's actually very surprising to me

Mike Miller

@TobiasKildetoft I have structure in this situation that story doesn't respect

Astyx

@BalarkaSen The leader of our pantheon

Ted Shifrin

It's one of the reasons complex geometry/topology are so taut.

Akiva Weinberger

Any language that has more vowel sounds than vowel letters is stupid, though

(That includes English)

Eric Silva

@Ted is the book you have in mind called introduction to integral geometry or integral geometry and geometric probability

Tobias Kildetoft

18:40

@MikeMiller What sort of context are you working in?

Mike Miller

I just need to do a lot of work, frankly

Ted Shifrin

But you don't need a global ambient form to get Crofton or generalizations, @EricS.

Mike Miller

@EricSilva It's the definition of Kahler form

Tobias Kildetoft

@AkivaWeinberger Don't most languages have that?

Krijn

@AkivaWeinberger Dutch combines vowel letters for new vowel sounds

Ted Shifrin

18:41

Probably the latter, @EricS. I've forgotten the difference.

Alessandro Codenotti

@Balarka What do you call someone who reads papers on cathegory theory? A coauthor

Akiva Weinberger

@Krijn Exactly. Stupid.

Eric Silva

right

Krijn

@AkivaWeinberger No, brilliant

Mike Miller

@Tobias I want to work out the basics of equivariant cohomology for compact Lie groups in an algebraic setting (for chain complexes with G-action)

Akiva Weinberger

18:41

@TobiasKildetoft Not Spanish, is my point

Krijn

Need a new sound? Combine o and e for oe

Akiva Weinberger

@Krijn Make up new letters!

Mike Miller

These have four different flavor of cohomology all satisfying certain properties

Eric Silva

@Mike i still dont know any Kahler stuff unfortunately...

Mike Miller

So, gotta get it

Akiva Weinberger

18:42

Don't use an alphabet with insufficient vocalic capacity

Ted Shifrin

OK, I'm gone for a bit. Bye.

Mike Miller

@EricSilva memorize everything in Bessie

Balarka Sen

See ya

@Alessandro I like that joke

Akiva Weinberger

Vai

Tobias Kildetoft

@MikeMiller Hmm, that sounds close to the sort of thing I ought to learn at some point

Eric Silva

18:42

im very behind on stuff i wanted to do this summer, what with moving into a new place and stuff

lol besse is open on my laptop rn

Mike Miller

@TobiasKildetoft I am a very slow writer but I can send you the appropriate section when I have it done

Alessandro Codenotti

is someone here familiar with probability?

Akiva Weinberger

Probably

Eric Silva

someone is probably familiar with probability

Akiva Weinberger

Almost surely

Eric Silva

18:45

sniped

RIP me

Tobias Kildetoft

@MikeMiller Sure. I am still trying to figure out what these new geometric ideas really are that seem to be coming up everywhere. There are a lot of words that I keep mixing up (perverse sheaves, parity sheaves, intersection cohomology).

Mike Miller

I suspect if I stay in academia I will end up one of those people constantly apologizing for how late I am on papers

Alessandro Codenotti

@AkivaWeinberger ...

Mike Miller

Ah, I care about those. A friend of mine is trying to do Lagrangian intersection theory of hokomorphic Lagrangians so he learned all that perverse stuff

Tobias Kildetoft

@MikeMiller Everyone is always slower to write than they expect (well, everyone I have worked with except Mazorchuk who seems to do everything at insane speeds)

Mike Miller

18:47

I am about half a year late on this paper

I had most of my current results then

Alessandro Codenotti

@SoumyoB are you around here by any chance?

Semiclassical

Hrm:

1

Q: I'm a neuroscientist, what book should I read to improve my maths?

I'm a neuroscientist, but I don't know much maths. I like to learn the functions, and what they mean exactly and how/when I should use them. I'm specifically interested in knowing what mathematical functions actually mean and what I'm supposed to understand from them. I want to read a little bi...

education

Akiva Weinberger

Someone knows something of value, I expect @AlessandroCodenotti

Semiclassical

I'm a bit torn with this question. On the one hand, I applaud the intention of the question.

On the other hand: it's broad, it's low on context, and it's strongly opinion-oriented.

Balarka Sen

perverse sheaf? dayum, and them algebraists call the pictorial people perverts

Akiva Weinberger

18:51

(I've run out of probability puns)

Semiclassical

[something something dessin d'enfants] @BalarkaSen

Krijn

@BalarkaSen Grothendieck was very much against the use of that word

"What an idea to give such a name to a mathematical thing! Or to any other thing or living being, except in sternness towards a person—for it is evident that of all the ‘things’ in the universe, we humans are the only ones to whom this term could ever apply.”

Mike Miller

It's a complex of constructive sheaves

Complex because chain complexes are good

Constructive means it's made out of cozy little pieces

Semiclassical

and that's perverse? someone clearly had too much imagination.

Mike Miller

I dont know the origin

Krijn

18:54

@Semiclassical That "someone" is Goresky

35

Q: What is the etymology of the term "perverse sheaf"?

Grothendieck famously objected to the term "perverse sheaf" in Récoltes et Semailles, writing "What an idea to give such a name to a mathematical thing! Or to any other thing or living being, except in sternness towards a person—for it is evident that of all the ‘things’ in the universe, we huma...

soft-question terminology reference-request

Here he answers himself how he came up with it

Balarka Sen

+1 to Goresky man

algebra is totally perverted stuff

Goresky is a nice man

Krijn

That is also the only thing he ever did on MO, which is quite funny

Like Tate

Semiclassical

so it all comes down to, in his words, the 'badness' of the relevant objects.

in english one perhaps would use the word 'defect' instead...in which case the scenario is even worse, since they'd now be defective sheaves.

Alessandro Codenotti

So I have a sequence of independent random variables $\{X_i\}$, each which an exponential distribution of parameter $1$. We define $M_n$ as $\text{Max}\{X_k,k\le n\}$ and I'm asked to show that $M_n-\log n$ converges in distribution, $\Bbb P(M_n-\log n<y)\to\exp(-e^{-y})$

Mike Miller

yeah, so they're sheaves that measure how perverse things can be

perversity-controlling sheaves

Semiclassical

18:59

heh. to pun on the terminology I just gave above, perhaps they should then have been "detective sheaves"

Transcript for

$Mathematics$ Mathematics