Mathematics - 2016-04-18 (page 1 of 5)

12:21 AM

@Anthony Followup — what about $\eta+\eta$, where $\eta$ is that of the rationals? Unless that was done in class, I guess

I found a book on Google Books a while back on order types. It was interesting, and completely free as far as I could tell (via the "look inside" feature — normally it only lets you view so many but I think that didn't happen for this one), but I didn't finish it. I didn't get very far in at all, actually

and I completely forgot about it until now

Anthony

12:37 AM

@AkivaWeinberger Hmm.

I don't expect it to be equal to $\eta$... but how to show it.

Hmmmmmmm.

Akiva Weinberger

Why are you learning about order types, anyway? What class has these?

(I saw your edit. Don't hide your feelings!)

Anthony

I'm taking a set theory class for a breadth requirement.

(:P)

Akiva Weinberger

Ah.

Oh, is it because of the relation to ordinals?

Anthony

Yah.

Akiva Weinberger

It's so weird how you can have a well order of well orderings, but no natural total order of total orderings (that I know of).

Anthony

12:45 AM

lol

Akiva Weinberger

They're so weird. Consider $\sum_{q\in\Bbb Q}\operatorname{denom}(q)$, where $\operatorname{denom}(q)$ is the denominator of $q$ in lowest terms.

Consider it for no reason other than the fact that it's weird-looking.

Still embeds in $\Bbb Q$, though

Anthony

What is that? You're summing over the denominators?

Akiva Weinberger

I'm adding order types.

Anthony

Oh

Akiva Weinberger

So, like, $\eta=\sum_{q\in\Bbb Q}1$

where $1$ is the order type of one element

Anthony

12:48 AM

oh boy

my mind isn't ready for this

but so is $\eta+\eta = \eta$?

Akiva Weinberger

@Anthony The sigma-notation-for-order-types thing or the $\eta+\eta$ thing?

Anthony

the sigma-notation

lol

Akiva Weinberger

@Anthony Yes. :)

Anthony

And how do you show that?

Akiva Weinberger

For better notation, let's write $(a,b)$ instead of $\Bbb Q\cap(a,b)$

So $\Bbb Q=\dotsb\cup(-1,0]\cup(0,1]\cup(1,2]\cup\dotsb$

Anthony

12:53 AM

This is true

bolbteppa

How would you even motivate the definition of a tensor operator as one which preserves expected values en.wikipedia.org/wiki/… if you didn't know any physics?

Pallas

how do i solve lambda = 2x - mu; lambda = 2y - 2 mu; lambda = 2z - 3 mu?

ive been trying for an hour now

i ccant solve it

Akiva Weinberger

It's also equal to$$\dotsb\cup(1,2]\cup(2,3]\cup(3,3.1]\cup(3.1,3.14] \cup\dotsb\\ \cup \\ \dotsb\cup(3.15,3.2] \cup(3.2,4]\cup(4,5]\cup(5,6]\cup\dotsb$$

Pallas

i cant get x in terms of y or whatever

Akiva Weinberger

@Anthony

Pallas

12:55 AM

it doesnt work

ive tried every combination i could think of

im starting to think its not possible

Anthony

Sure.

Akiva Weinberger

So 2x-μ = 2y-2μ = 2z-3μ = λ ?

Pallas

i cant figure out how to do this. is this even possible?

yeah

nothing cancels out

i cant get rid of all the mus

Akiva Weinberger

@Anthony Each term of the top row can be mapped linearly to an interval of $\Bbb Q$ as written in the comment before

Anthony

how quaint.

Akiva Weinberger

12:58 AM

I approach pi!

Essentially, I break $\Bbb Q$ into two pieces at $\pi$ @Anthony

and each piece has the order type of $\Bbb Q$

Anthony

And why doesn't this work for $\mathbb{R}$?

Pallas

is this possible?

maybe my equation is wrong

Akiva Weinberger

You can't split $\Bbb R$ into two open sets like $\Bbb Q$, @Anthony

Pallas

i dont think this is possible, because the mu cant be eliminated

Anthony

Oh, I see.

Pallas

1:00 AM

pleaes ive been trying to do this for an hour now and its not even lagrange which is what im trying to study

Akiva Weinberger

@Pallas Possible solution, μ = 2, x = 1, y = 2, z = 3, λ = 0

But there's more than one solution, I think.

Pallas

Akiva but how do you get that?

i just need one variable in terms of another

like x in terms of y, or y in terms of z

or something that doesnt include mu or lambda

Akiva Weinberger

I just set x, y, and z to 1, 2, and 3 to see what happens. It's not the general solution, though, so it's not what you're looking for

Pallas

it doesnt work though because theres an uneven amount of mu that wont cancel out

it i dont know why it doesn towk

i can get all of the equations in terms of each other but thats not useful

Akiva Weinberger

If you double one of the equations, you get 4x-2μ=2λ, and this can be subtracted from 2y-2μ=λ to cancel out the mus

Pallas

1:02 AM

i dont get what i have to do

Akiva Weinberger

Probably easier in terms of linear algebra, though.

Pallas

but when you subtract that, don't you get a lambda?

i cant use linear algebra. i didnt learn it yet

Akiva Weinberger

I do. Maybe it's not the best

Pallas

you get lambda = 4x - 2y

how does this help though because there is still lambda

Akiva Weinberger

I don't think you can get an equation with just two variables

Ted Shifrin

1:04 AM

@Pallas: Don't forget that you also have a constraint equation to use.

Akiva Weinberger

I think the least you can get is three, like you have here

Pallas

so its impossible?

but its not possible

because im trying to solve lagrange

Akiva Weinberger

@TedShifrin He does?

Ted Shifrin

If this is a Lagrange multipliers problem, yes.

In fact, there are two constraint equations if there are $\lambda$ and $\mu$.

Pallas

f(x,y,z) = x^2 + y^2 + z^2 subjected to contraints: g(x) = x + y + z = 1 and h(x) = x + 2y + 3z = 6

and yes

i know

i get <2x, 2y, 2z> = lambda<1,1,1> + mu <1,2,3>

Akiva Weinberger

1:06 AM

10 mins ago, by Pallas

how do i solve lambda = 2x - mu; lambda = 2y - 2 mu; lambda = 2z - 3 mu?

Ted Shifrin

Yeah, better to drop the $2$'s, but yes.

But you also know that $x+y+z=1$ and $x+2y+3z=6$, so you have to use that information.

Pallas

yeah, those are the equations i get from the above

Akiva Weinberger

I completely forgot everything about Lagrange

Ted Shifrin

Read my book, DogAteMy :D

Pallas

ok, so z = 1 - x- y

Akiva Weinberger

1:07 AM

By the way, I just learned that my homology book (not Hatcher) was \$50.

Ted Shifrin

Better, substitute your expressions for $x,y,z$ in terms of $\lambda$ and $\mu$ into the constraint equations, @Pallas.

Akiva Weinberger

(Relevant because I was shocked at how much your thing was)

Ted Shifrin

Then you'll get two linear equations in $\lambda$ and $\mu$.

Akiva Weinberger

(but I guess my dad likes throwing lots of money at my love of math)

Pallas

@TedShifrin but i thought to solve lagrange, i need some variable in terms of another

i find that variable in the first constraint

and i find the third variable in the second constraint

Ted Shifrin

1:08 AM

Trust me, @Pallas.

Pallas

in my textbook, they always eliminate mu and lambda

ok

Ted Shifrin

Yes, in my book, when there's one Lagrange multiplier, I eliminate it always. But with two you cannot always do that. You often have to solve for them. It depends on the problem.

Pallas

so right x,y,z in terms of lambda and mu then substitute them back

Ted Shifrin

DogAteMy: Your dad would much rather nurture your love for math than nurture a love for, say, drugs.

Yes, @Pallas.

Akiva Weinberger

How do I pronounce "Pallas"? With a Spanish ll, so "payas," or like the word "palace"

Pallas

1:10 AM

i have x = (1/2)(lambda + mu)

Ted Shifrin

or it could be French, DogAteMy :P

Pallas

probably like 'palace', it was greek

Akiva Weinberger

@TedShifrin But little does he know…! (jk)

Anthony

@TedShifrin What was your motivation for writing your text? (And also HI!) ((And also, I mean your calc book. Have you written two books?))

Akiva Weinberger

Well, there was a lot of throwing money at my drugs. But, like, medicinally prescribed drugs. Medical problems and all that.

Ted Shifrin

1:12 AM

Four @Anthony, but who's counting?

And hi :)

Anthony

FOUR!?!?!?!?!?!?

Wow.

Pallas

i have 3 lambda + 6 mu = 2

Ted Shifrin

I didn't mean that sort, DogAteMy, and you know it.

Pallas

6 lambda + 14 mu = 12

Akiva Weinberger

It should be stated that having to give oneself an injection every night for a few years does wonders for getting rid of the fear of needles

Ted Shifrin

1:13 AM

So simplify the arithmetic and solve for $\lambda$ and $\mu$ and then you're almost done, @Pallas.

Pallas

i have mu = 10?

Ted Shifrin

I never had a fear of needles, although the last time I was in the hospital for surgery, the nurse who needed to redo my IV was terrible and stabbed me about 10 times. Finally, one of my friends asked her to find someone competent.

Pallas

lambda = -58/3

Akiva Weinberger

Oof

Ted Shifrin

No, @Pallas. Try again.

Pallas

1:14 AM

:(

Akiva Weinberger

How did you get mu=10?

I think you forgot to multiply the right-hand side?

Pallas

my equations were 3 lambda + 6 mu = 2

6 lambda + 14 mu = 12

oh i forgot to divide rhs

mu = 4?

Ted Shifrin

Yup.

Pallas

lambda = -22/3?

Ted Shifrin

Yeah, yucky, but that's right.

Pallas

1:17 AM

cool :)

Akiva Weinberger

Looks right to me

Pallas

then i substitute this back into the other equations

Ted Shifrin

Yup.

It depends on the particular problem. Sometimes with two constraints you can be more clever, but here this is the way to go.

Mike Miller

Hi @Ted.

Ted Shifrin

Rehi @MikeM. On the road?

Mike Miller

1:19 AM

Off the road at SLO for a bit.

Ted Shifrin

Ah.

Mike Miller

Here is a question you could answer authoritatively.

Pallas

thanks so much for explaining. i was really confused with this

Ted Shifrin

Taking 101 all the way, @MikeM?

Just keep an open mind about different ways to approach problems, @Pallas. You'll be fine.

Pallas

hmm, i got the wrong answer unfortunately

i got x = -5/3, y = 1/3, z = 7/3

Akiva Weinberger

1:21 AM

Shouldn't they add to 1?

Or something

Pallas

oh wait nvm, i reading off the wrong chart

got 25/3 :D

thanks again

Ted Shifrin

Um, they do, DogAteMy :D

Not sure I want to get messed up in there, @MikeM. I doubt very many graduate students in "geometry" know much of that stuff. I could recommend he read Pedoe's wonderful book.

Akiva Weinberger

Oh, whoops

You're right, they do

Ted Shifrin

nods ... Some mathematicians can do simple arithmetic :D

Akiva Weinberger

…57

Ted Shifrin

1:25 AM

Heinz?

SAWblade

What's everyone up to? :0

Akiva Weinberger

With regards to geometry,

let me suggest this gem that I really need to finish one of these days

Pallas

How do i find the point lying on the intersection of x + 1/2 y + 1/4 z = 0 and the sphere x^2 + y^2 + z^2 = 9 with the largest z coordinate? I'm confused with how this is set up

Ted Shifrin

What do you want to minimize/maximize? @Pallas

Akiva Weinberger

Does affine, projective, hyperbolic, and elliptic geometries

Ted Shifrin

1:28 AM

I don't know that book. Do take a look at Pedoe's book (Dover, so cheap). Fabulous. Student and co-author of the very famous Hodge (famous for algebraic geometry and Hodge theory).

Akiva Weinberger

Approx 500 pages

Also, I somehow ended up in possession of this?

Several decades old

Klein's lectures, I believe

Pallas

I want to maximize the z

i think

i dont know

Ted Shifrin

OK, so $f(x,y,z)=z$ is the function you want to maximize. What are the constraints?

Akiva Weinberger

I also have another on analysis

Pallas

it needs to be on the intersection of x + 1/2 y + 1/4 z = 0 and the sphere x^2 + y^2 + z^2 = 9

does interseection just mean it needs to satisfy both equations?

Ted Shifrin

1:32 AM

So what are the constraint functions/equations? YES.

Pallas

h(x,y,z) = x + 1/2 y + 1/4 z

g(x,y,z) = x^2 + y^2 + z^2 = 9

Ted Shifrin

OK, so go for it.

Pallas

<0,0,1> = lambda < 1, 1/2, 1/4> + mu <2x, 2y, 2z>?

Ted Shifrin

Yup.

Pallas

but this is weird because the euqations = 0

Ted Shifrin

1:34 AM

What do you mean?

Pallas

i guess that helps

Ted Shifrin

Right, it helps.

Pallas

just like 0 = lambda + 2x mu

Ted Shifrin

Again, you have to figure out how to use (at least) one of your constraint equations.

Or note that the two equations with 0 in them tell you x and y are related immediately.

Pallas

ah, x = 2y

Ted Shifrin

1:36 AM

Yuppers.

That's all you need, actually, along with the constraint equations.

Pallas

so in my textbook, they substitute this into one of the constraint equations to get like z in terms of y or whatever

should i do that here also?

Ted Shifrin

Or substitute into both constraint equations, yeah.

Pallas

oh ok

i got y = sqrt(9/105)

Ted Shifrin

I'm not working on it, @Pallas. You're in charge.

Mike Miller

@Ted: That's the PCH? We plan to.

Pallas

1:39 AM

ok

Ted Shifrin

No, I think PCH is a lot of Rte 1, actually, @MikeM. Not sure.

Pallas

i think it worked :) except the answers were all negative, but i got the write general answer

Mike Miller

I think the question could benefit from an answer, but I agree it's a little "controversial", say.

I can't weite an andwer because I don't have an "authoritative voice" here.

Ok, well we're going to drive that. I'm feeling a little sick.

Ted Shifrin

@MikeM: No point taking Rte 1 in the dark. Just do 101. It's still like 3 hours or so.

Mike Miller

Not my choice ... I'm not driving :)

Was trying to read a good book in the car but the curviness got to me.

Pallas

1:50 AM

how do i set this lagrange problem up? The cylinder x^2 + y^2 = 1 intersects the plane x + z = 1 in an ellipse. Find the point on the ellipse furthest from the origin?

so I said f(x) = x^2 + y^2 + z^2

h(x) = x^2 + y^2 = 1

g(x) = x + z = 1

Is this the correct setup?

anon

2:46 AM

you mean h(x,y,z) = x^2+y^2-1 and g(x,y,z)=x+z-1 ?

Simeon

If $f$ is continuous with at least one root, is it true that for any point $x$, there is a root of $f$ that is closest to $x$?

anon

or more than one

Pallas

anon, no

i mean it doesnt even matter because the gradient would be the same anyway

idk why its so wrong

mathb.in/55842

my attempt in nicer format

i cant figure out why its wrong

i think its to do with the cylinder

idk why though

the cylinder equation is definitely wrong

because there's no z value even though z can be anything

what is the gradient for the cylinder?

x^2 + y^2 = 1

is a circle and thats why its wrong

and we are in R3 it acts like we are not in R3

thats why its wrong

because its a plane

i dont know how to fix it though

does anyone know?

no one knows how to do this

no matter what lambda = 1

i ask everyone

anon

it's not really correct to write things like h = x^2+^2 = 1 I don't think, but I see what you're doing is equivalent to what I know

in any case you differentiated slightly wrong I think

Pallas

where?

Forever Mozart

2:59 AM

the cylinder is not the graph of a function of $\leq 2$ variables

anon

@Pallas wait nevermind

Pallas

what?

anon

@ForeverMozart not relevant

Pallas

idk whats wrong

anon

you can't just make up lambda=1 can you?

Pallas

3:00 AM

no

because the answer suggests the y = 0

so it would be dividing by 0

this problem is not even worth doing

i know how to lagrange. this problem isnt worth all this time is pent on it

anon

ah, you divided by y to get lambda=1. I guess you have to do cases with y=0 and not 0

Pallas

you can didivide by 0

i thought the premise of lagrange was thaat the values are non 0

Forever Mozart

what happened

Pallas

idk though

i cant figure this out

anon

@Pallas what values? lambda and mu?

Pallas

3:02 AM

its such a stupid problem anyway

mathb.in/55842

my answer is wrong

i odnt know why

this isnt worth doing though

this is insane that i care so much

Forever Mozart

yes insanity

Pallas

i hate this so much. fuck this

anon

been awhile, J.

you assumed y was not 0 to get lambda=1. assume y=0 and see what it gets you.

Pallas

i still dont get an answer

i cant do this

Forever Mozart

don't be so pessimistic

user147690

3:06 AM

Nothing is impossible! Just DO IT!

Forever Mozart

90% of things you try in mathematics will fail

you get maybe 1 good idea out of 100 if you're lucky

Pallas

i am so pessimistic

im so behind in math. im going to fail exam

user147690

If you're sick of starting over, STOP GIVING UP! Just DO it!

anon

@Pallas You've done 99% of the problem. When you assumed y was not 0, you got the local solution (1,0,0). This is locally a optimal solution. Then you need to assume y=0 and see what happens.

Pallas

why would i know to assume y = 0 though?

Forever Mozart

3:09 AM

Shia LaBeouf "Just Do It" Motivational Speech (Original Video)

►

anon

When you assume y=, from x^2+y^2=1 you get x=1 or x=-1. If x=1, you get the solution (1,0,0) again, that's nothing new. If x=-1 you get (-1,0,2). It remains to compare the distances of (1,0,0) and (-1,0,2) from the origin. The latter is the maximum distance, the former is I believe the minimum.

@Pallas You already had the intuition to assume y was not 0 to see where it got you (from the equation 2y=2y lambda). If you explored one case where y was not 0, you need to explore the other case where y is 0.

Pallas

when i assume y = 0, then i get the two equations 2z = mu and 2x = 2x lambda + mu

anon

@ForeverMozart, @AlexClark We've been at this for literally years. I'm not sure of what help platitudes or speeches will be.

Pallas

what do i do with them though

anon

I already told you what to do when you assume y=0.

Forever Mozart

3:11 AM

you have been working on this problem for years?

Pallas

yes i have

for like two hours

anon

@ForeverMozart the pessimism problem, that is

user147690

Wait this is the famous one

user147690

The one from Jonas era?

anon

yes

user147690

3:11 AM

Oh my

Pallas

x lambda + mu = 1

anon

3 mins ago, by anon

When you assume y=, from x^2+y^2=1 you get x=1 or x=-1. If x=1, you get the solution (1,0,0) again, that's nothing new. If x=-1 you get (-1,0,2). It remains to compare the distances of (1,0,0) and (-1,0,2) from the origin. The latter is the maximum distance, the former is I believe the minimum.

(should say when you assume y=0)

please delete your comment, no names unless volunteered

Pallas

anon, i dont understand what to do

how do you get that solution

i assume y = 0

but it changes nothing

i still cant find x in terms of z

anon

I told you what to do

read what I wrote

Pallas

oh, you use the constraint equations

oh wow it works

anon

3:14 AM

from y=0, the equation x^2+y^2=1 reduces to x^2=1, which means x=1 or x=-1. If x=1, then filling in the missing part of (1,0,?) from the constraint yields (1,0,0). If x=-1, then filling in the missing part of (-1,0,?) from the constraint yields (-1,0,2).

Pallas

yeah, got (-1,0,2) now.

:D

:D

Thanks!

no worries.

this was giving me a minor panic attack

Forever Mozart

3:18 AM

panic !#)!#$@!#$@!@$~@!#@$*@!~

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

4:49 AM

@user1618033 thanks for the update :-)

Interesting stackoverflow link @Agawa001 thanks

Balarka Sen

5:36 AM

@Anthony "Anthony" is just too much to write.

Sie

Game dev SE chat is dead. I got lonely so I found this place.

Balarka Sen

Oh no, a game developer.

Sie

Kind of a stretch to call me that :P. More like mash spaghetti code together until it works then cry when it breaks again.

user147690

@Sie You can stay if you write scripts for me

user147690

Or not scripts, more like programs

Tobias Kildetoft

5:39 AM

@AlexClark Do you happen to have access to a computer that is decently fast and can run something for a week or so?

user147690

@TobiasKildetoft Yes

Tobias Kildetoft

@AlexClark Does it also have GAP installed?

user147690

No, but I can install it

Fargle

Hello, chat.

Balarka Sen

Hi Fargle.

user147690

5:41 AM

@TobiasKildetoft Did you want me to run something?

Sie

@AlexClark Behold. pastebin.com/6XuxampS

Tobias Kildetoft

I would like to test if a certain type of group exists of order 768 (I have tested all other orders up to 1000), but it takes too long and I need to more my laptop to go between home and work

user147690

Sure, I can do that

Tobias Kildetoft

@AlexClark Great, let me just put together the code in a single file

Sie

Also fyi in case it wasn't obvious that will cause a stack overflow exception so if you run it on a total toaster then don't blame me if it locks up.

Balarka Sen

5:46 AM

@MikeMiller I went to sleep yesterday, but I haven't been able to see things quite as much as geometrically I'd like to see yet. All I have is this: $\Bbb P^2$ blown up at a point $p$ lives in $\Bbb P^2 \times \Bbb P^1$. That blown up at a point $q$ (outside the exceptional subvariety say) lives in $\Bbb P^2 \times \Bbb P^1 \times \Bbb P^1$. Call the resulting object $Z$.

I can restrict projection $\Bbb P^2 \times \Bbb P^1\times \Bbb P^1 \to \Bbb P^1 \times \Bbb P^1$ to $Z$. I am pretty sure this is itself a blowup, with exceptional subvariety the line passing through $p \times \Bbb P^1$ and $q \times \Bbb P^1$.

(the picture is that if I pinch the one unique line passing through $p \times \Bbb P^1$ and $q \times \Bbb P^1$, I wedge $p \times \Bbb P^1$ and $q \times \Bbb P^1$ togather. And now families of lines passing though points of each of these should give the family of meridians and longitudes in $\Bbb P^1 \times \Bbb P^1$)

I think a bit of formula writing should easily prove this but that's not something I want to do.

Tobias Kildetoft

@AlexClark It is dropbox.com/s/n2d130kbsmglqaw/tobiastest.g?dl=0 so if you Read that into GAP then run TobiasTest(768.768);

user147690

Sweet, will get it going very shortly, and it'll probably take 2 weeks?

Tobias Kildetoft

(it is probably horribly inefficiently written)

user147690

Haha np

Tobias Kildetoft

No idea actually, I just know that 48 hours was not enough for my computer to finish it

If it returns "fail" then that is a good thing, as that means it was not able to find a counter example to a conjecture of mine

Balarka Sen

5:52 AM

irrelevant to the discussion: I wish I had a couple more examples of moduli spaces

Tobias Kildetoft

@AlexClark A reviewer asked me to expand on why order 768 was problematic and I decided to try once more to actually do the calculation

@AlexClark Woops, looks like I had a typo in what I told you to run. It should be TobiasTest(768,768); (a comma instead of a dot)

Mike Miller

I gave you all I can, @BalarkaSen

PVAL

@MikeMiller Hi Mike

Arkansas was fun

Balarka Sen

@MikeMiller I am not sure what you are referring to. You mean to say the thing about blowup? Yeah, what you said (about it being a version of the real case) was very illuminating, but it's not obvious to me how I would carry the proof out in similar vein.

I'm just being dumb.

Hi @PVAL.

Mike Miller

@PVAL Was it really?

@Balarka I am not saying you're dumb. I'm just saying I don't have anything more I can say. I don't know how to prove that those two things are biholomorphic.

PVAL

6:04 AM

@MikeMiller Ya the grad student workshop was really boring, but the talks were all pretty interesting.

Balarka Sen

@MikeMiller Alright. Thanks.

Mike Miller

Good to hear. I didn't know anything was happening in Arkansas.

I am getting sick of travel. I'm going to Toronto in a few weeks and then I'm not going to go anywhere this summer unless I have to.

Balarka Sen

I got to get back to work. See you all later.

bolbteppa

@BalarkaSen some baby moduli spaces:
Lines https://www.quora.com/What-is-an-intuitive-explanation-of-a-moduli-space
Triangles http://www.austms.org.au/Gazette/2008/May08/Do.pdf
Ellipses and more http://www.ams.org/journals/bull/2004-41-03/S0273-0979-04-01024-9/S0273-0979-04-01024-9.pdf
all I've really looked at so far sadly

user147690

@TobiasKildetoft TobiasTest(768,768); or TobiasTest(1,768); ?

PVAL

6:08 AM

@MikeMiller I am now beyond convinced that in the contact/symplectic world, in high dimensions topology \ne algebra.

Tobias Kildetoft

@AlexClark 768,768 (otherwise it also tests all numbers up to 767 which I already did years ago)

Mike Miller

@PVAL That's cool. Any good examples?

PVAL

@MikeMiller The whole Legendrian front calculus for higher dimensional Weinstein manifolds. I don't really understand how much of this is worked out.

@MikeMiller Lots of interesting questions about embedding contact 3-manifolds in contact 5-manifolds some of which reduce to looking at Lefschetz fibrations.

user147690

@TobiasKildetoft I seem to be able to run your code for most $(n,n)$ where $n\not\in \{768,1536\}$

user147690

Maybe I need a library

user147690

6:19 AM

Error, AllSmallGroups / OneGroup: groups of order 768 not available called from
SelectSmallGroups( arg, false, false ) called from
OneSmallGroup( Size, i, IsSolvableGroup, true, IsMonomialGroup, false,
erTaketaGr, true, erNyGr, false ) called from
<function "TobiasTest">( <arguments> )

Tobias Kildetoft

@AlexClark Ahh, yeah, you need the full SmallGroups library for order 768

(it works for order 1024 because that is a prime power, so it doesn't actually have to test that due to various results)

user147690

@TobiasKildetoft Sure np

Fargle

@TobiasKildetoft What type of group are you looking for?

Tobias Kildetoft

@Fargle One thst is a Taketa group but not a PR-group (I conjecture that no such groups exist). The definitions are in my paper arxiv.org/abs/1311.1383

Fargle

@Tobias ...I know some of these words.

Better brush up on my group theory, that's what I just learned.

user147690

6:34 AM

@TobiasKildetoft It's begun. It's running on a uni PC that'll never be bothered

Tobias Kildetoft

@AlexClark Cool, thanks

datalava

I'm looking for a hint on this analysis problem. Clearly I want to define some bounded linear function to help me out, but I'm not sure what my function should be.

The problem: $X$ is a Banach space and $Y$ and $Z$ are closed subspaces such that $X=\{y+z:y\in Y, z\in Z\}$ (i.e. $X=Y+Z$). Show that for every $x \in X$ there are $y,z$ such that $x=y+z$ and $||y||+||z|| \leq M||x||$ for some positive $M$.

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$Mathematics$ Mathematics