Mathematics - 2015-05-07 (page 1 of 6)

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Américo Tavares

12:43 AM

Is it possible to generate with the adequate query a list of my own questions and answers that received downvotes, detailing the number of downvotes per question/answer?
I think I have not many, but I would like to know, without having to visit all of them and see.

... an adequate query ...

That is: how to obtain automatically a list of my own questions and answers with downvotes?

@AméricoTavares I found this weird snippet, does it seem to do the right thing?

Ah, no, it's just numbers, not the posts themselves...

1:15 AM

Huh, data explorer is kind of fun. I modified a question to find your most downvoted tags, now I just need to figure out how to get the questions themselves...

@pjs36, I too am enjoying messing around with DataExplorer right now

haha, nice! I think this is my handiwork for downvoted tags...

mine are mostly proportional to the tags I'm active in

but wow, I've accrued a lot of downvotes

Hmm, something seems wrong @pjs36. Says I have 50 downvotes in Abstract-Algebra, but another thing told me minutes ago that I have only 17 downvotes total.

These are what you've downvoted.

1:18 AM

Ohh

user147690

Is a topological basis like a vector space basis, except instead of linear combinations via addition, we have unions of sets of the basis

it has one for me in class field theory, and I downvoted a question with that tag the other day, and I've definitely never posted in that tag

But yeah, mine are almost exactly proportional to my active tags

So the one I recently posted seems to be the tags that the input UserID has downvoted?

yes

@AlexClark: I guess I never thought of it that way, but yes. A basis for a topological space has the property that any open set can be written as a union of basis elements, and every union of basis elements is an open set.

1:20 AM

Apparently 1/3 of my posts get accepted. That's kinda cool.

user147690

@MikeMiller Thank you

But unlike a basis for a vector space, there is absolutely no guarantee of uniqueness.

hi

@MikeMiller, so the basis elements aren't necessarily $\in \tau_X$?

user147690

Hey @Karim

1:22 AM

@KajHansen: Yes they are.

The union of one set is just that set. :P

basis of topology ?

Oh I skipped over "every union of basis elements is open"

Reading too fast.

yeah that is instead of looking a topology of whole we look for the basis that make up the topology

Is there some easy way to invert the following system (I-T)x=b where T is a block\toeplitz matrix and Tx is computed with an FFT2? I got something with conjugate gradient, but I feel dirty.

2:06 AM

for any symmetric group is the generators always made up <r,m> where r and m are rotation and reflection ?

Yeah I think so like intuitively it makes sense that that is the case.

Do you mean any dihedral group? Symmetric groups are almost never generated by two elements.

@KarimMansour, the "symmetric" group of order $n$ refers to the group of permutations on $n$ elements. I think you're referring to the dihedral group - the group of symmetries of the $n$-gon.

yeah

However, the symmetric group on $3$ elements and the symmetry group of a triangle are the same :)

*group of symmetries * I meant to say

yeah but generally that isn't the case

2:09 AM

But yes, $D_n \cong \langle R, F \rangle$.

I see

alright thank you @KajHansen

hi @Kaj @Karim

Hi @TedShifrin

Hey @TedShifrin

hey, Kaj, did you get the book I sent you via Daniel?

2:12 AM

one I guess could argue why they aren't the same for n $\geq$ 4 is because you have $D_{2n}$ will have |$D_{2n}$| = 8 while for the symmetric group its 4! = 24 and same thing for elements bigger than 4 that factorial grows much bigger than multiplication by 2

Hi @TedShifrin

Are your final exams going well? Or just driving you to depression @TedShifrin ?

I was about to ask about you didn't see you for some time

@TedShifrin

No hello for me @Ted? ;)

I'm not very happy, no, @Kaj. I've spent the last two days doing nothing but grading and agonizing over grades, @Karim.

hi @AlexW :) Where's my cake?!

oh that sucks

2:13 AM

You didn't get it in the mail??

And unfortunately I haven't. I've been so busy the past few days that I haven't seen Daniel since last week sometime. I'll see him at the end of May when we meet up for the REU though @TedShifrin

nope, @AlexW

Oh, ok, @Kaj ... We thought you might like Dugundji's Topology book. It's a challenging classic.

Ugh @ grading. Couldn't your students at least send you off on a happy note, @TedShifrin? How rude.

No, @pjs, and some of them are going to be arrogant enough to complain that I've "made" them not graduate.

Perhaps one of your students ate it in zealous grief after your diff geo final, @Ted. :)

2:13 AM

oh, nice timing

morning folks

Oh, thanks! I actually did enjoy quite a bit of that course in retrospect. It's given me a lot to think about @Ted

goodnight, @MikeM

Why are you agonizing over grades, you just have to give them...

Howdy, @Mike

I almost said good morning back =P

@Ted: Why are the odd Betti numbers of a Kahler manifold even?

2:14 AM

LOL @AlexW ...

$h^{p,q}=h^{q,p}$, @Mike.

compact Kähler

and Hodge decomposition, of course.

Ah, right. Thanks!

I should teach you some complex manifold stuff sometime ... oh wait, I've forgotten.

@Alex: I wrote the first part of the talk as a blog post

Hippa will have to make a new meme for me.

Excellent, @Mike. I'll be reading that now. :)

Will you regularly make posts, do you think?

2:16 AM

I've forgotten everything I knew about complex manifolds too, @Ted

well, you have little to forget, @MikeM ... I've taught courses.

@AlexW Probably. I need to write the sequel about h-cobordism, and I have a couple ideas for posts.

One more day, eh, @AlexW? :)

@TedShifrin, do you know whether we'll have to wait for the actual deadline for professors to post our grades to see our grades? It seems that's how Athena is set up now, which is rather annoying. In the past I could check Oasis whenever I wanted :(

Cool! I'll be keeping an eye on things, then.

You got it, @Ted. :) Tomorrow is going to be a good day.

2:17 AM

Athena/Banner is not configured that way, @Kaj. They'll just get loaded when grades close Monday afternoon. I don't like it, either.

Of course, @Kaj, most faculty now procrastinate even more.

@Ted: Well, I tried to make you feel better. :P

It keeps me in suspense for much too long!

I'd rather leave town before some of my grades are posted, @Kaj :P

Have you found a place in California yet?

No, no, I don't go until 5/27, @Kaj.

2:20 AM

memegenerator.net/Ted-Shifrin @TedShifrin Looks like hippa made more than I expected!

Those are some strange memes.

@Paul: Hippa did, or you're helping?

Not me

The page disappears after a second.

Those are all very mean-spirited.

2:22 AM

yes, that's why I call Hippa badly behaved.

I don't think I'll help him any more ...

The notes from my students yesterday were much nicer :P

@Mike: silly typo which you may not care about. In your statement of the h-cobordism theorem, you write "...h-coboridsm".

maybe it was a Freudian slip, @AlexW, as he was borid.

2

I do care! I'll fix that when I get home.

@Ted LOL

@Ted: I remembered I had a question about differential forms for you. :P

2:26 AM

BTW, @MikeM, for reasons I completely do not understand, I can stay logged into chat on my iPhone, but still lose the connection on my iPad after minutes of being gone.

@AlexW: You shouldn't encourage me :D

Too late :B

What question, @Mike?

I have a 2-form. No other assumptions. Why is $\tau([F,G],H)-\tau([G,H],F)+\tau([H,F],G)=0$? I think I got my signs right.

that sort of stuff shows up in symplectic geometry ...

Hence my interest. My algebra just sucks.

2:29 AM

yeah, but there you know closedness.

so you can use identities for $d\omega$.

I don't for this problem.

It's about the failure of the Poisson bracket to be a Lie bracket when my 2-form isn't closed. I would have the desired formula if I could kill the above thing.

well, everything is tensorial, so if we show that expression is $C^\infty$ linear, it follows immediately.

My 2-form is still nondegenerate but I can't see how that would help.

I don't see why, @Ted. Sorry. :P

Because then you can check on coordinate vector fields, where everything's dumb.

The point being that the Lie bracket is zero there. OK, I see.

Thanks.

2:32 AM

Yup.

I think that's gonna work ...

Well, I guess my work here is done.

@Ted: perhaps you'd be interested in helping me? :)

I am doing an exercise, but I think I have found a counter example, and want to make sure I am not screwing up. The problem is if $X$ is a topological space, $A$ is a closed subset and $B$ is a proper subset of $A$. Let $X'=X \setminus B$ and $A' = A \setminus B$, then $ X / A $ is homeomorphic to $X'/A'$. I have shown that there is a bijection, but I think there may be a counter example to the problem (which I will share in a bit).

Does anyone know off the top of their head if the problem is wrong?

@Mike: another small thing. I don't think this is a mistake, per se, but just an odd bit: "Milnor was able to write down a few exact sequences that help us with your problem".

The counter example, or at least I am pretty sure is a counter example. Would be $X= \Bbb R$, $A= [0,1]$ and $B=\{0,1\}$.

@AlexW: I'll be home in about 15-30, so ping me with this then to remind me.

2:46 AM

Ok, sounds good. Really nice stuff. I don't have the topology to appreciate strong portions of this, but I'm getting a good general picture.

Thanks! Feel free to ask questions - in about half an hour. :P

Haha, sounds good.

This is because the "obvious" function that you would think would be a homeomorphism doesn't seem to work. Basically the point $\{A'\}=\{(0,1)\}$ in $X'/A'$ is open, since $(0,1)$ is open in $\Bbb R \setminus \{0,1\}$. But the "obvious" bijection $ \Bbb R/ A \to (\Bbb R \setminus \{0,1\}) / (0,1)$ would have the preimage of $\{(0,1)\}$ to be $\{[0,1] \}$, which is not open, since $[0,1]$ isn't open in $\Bbb R$

The "obvious" bijection by the way is $x \mapsto x$ when $x \neq A$ and $A \mapsto A'$

You're correct that the problem is wrong. Your example shows this. The correct assumption is that $B$ is in the interior of $X$. You might need to assume $B$ is open but I don't think so.

interior of $A$

First year I didn't ask for german chocolate cake for birthday. My mother decided to be silly and adventurous, having come across some maple-bacon frosting in the store, so I've got a batch of maple-bacon cupcakes (on a spicecake base) this year.

Sweet&salty. mmm

2:54 AM

@MikeMiller Thanks, I will see if I can get it to work if $B$ is in the interior.

the joys of bacon-flavoured everything

what a weird phenomenon

well, maple-bacon donuts are my favorite variety, with oldfashioned and chocolatefrosted-cake close behind

ok, I'm home @AlexW

Cool @Mike. Do you need me to resend you that message?

No, I remembered it. Fixed.

3:04 AM

Excellent, thanks!

@AlexW: Help with what?

Happy birthdsy, @JMoravitz

2

Hehe, nothing important @Ted. :)

Tease ...

Haha, I don't mean to be a tease! It concerns this question: math.stackexchange.com/questions/1259312/quotients-of-mathbbzi/…

Hi!!!!!!!@TedShifrin

3:11 AM

I just finished reading this very nice indeed !www1.spms.ntu.edu.sg/~frederique/groupsymmetryws.pdf

Hi @Rememberme

The approach outlined in the question is very nice. I had posted an answer, which was wrong, which suggested using the fact that $\mathbb{Z}[i]$ is a Euclidean domain. You can apply Euclidean division by any element $z$, so that the quotient set is at most elements with norm less than $N(z)$.

hi @Rememberme

Happy Birthday @JMoravitz

Hi@KarimMansour how's it going?

Unfortunately, this isn't strong enough. $N(1+i) = 2$, for example, but there are five elements with norm strictly less than $2$. I was wondering if there was a way to salvage the approach, @Ted.

3:13 AM

Ah, @KarimMansour, you should try and get ahold of Conway's Symmetries of Things, you'd enjoy it.

Well, one quadrant's worth of those ...

good how about you

okay thank you for the suggestion @pjs36

Still in with linear transformations and isomorphisms

But $1$ and $i$ are associate in $\mathbb{Z}[i]/(1+i)$. Don't they belong to the same quadrant?

I think there's some approach with fundamental domains here, but it's just out of reach to me at the moment.

Right, you have to think about the square, @AlexW.

I have this stuff in my akgebra book.

3:16 AM

@AlexWertheim Define quadrant to be, like, things with argument in $[0,\pi/2)$. Don't include both axes.

You want the square with sides $z$ and $iz$.

you can use units to push something into the top right quadrant excluding imaginary axis; go from there

@KarimMansour is I have a linear transformation from $\Bbb{R^3} \to \Bbb{R^2}$ can I say that the linear transformation is not injective

no

Over what fields?

3:17 AM

oh

linear map

never mind

Uh huh. Reading helps :)

Some fields F

Ahhh, I see now. D'oh. Thanks @Ted, @Mike.

it will be not surjective and not injective

for example

@AlexW: did you have questions about the post? You sounded like you did.

3:19 AM

(3,2,3) will go to (3,2) what about (3,2,4)

hm

1 second

for sure it will be not surjective

Hence the inverse doesn't exist.....there we go my proof is finished I think (3,2)

You shouldn't make assumptions about the map, @Karim

I do @Mike, but I'm not high functioning enough atm to parse the answers. If you're willing to be annoyed tomorrow, I'd happily ask then. :)

yeah I was overthinking that @TedShifrin

Sure. I asked because if not, I was gonna take a nap. :P

3:20 AM

it depends on the map @Rememberme

to test it out see if kernel(T) = {0}

No map given...... :(

Have a good last day, Alex.

hm

Haha, good then. I'm actually going to run to bed, so that works particularly well.

I don't know kernel's yet

3:21 AM

Thank you, @Ted! :)

let's analyze it carefully

@Rememberme Is this an exercise in the book?

Night all.

So we have a map from $R^3$ $\rightarrow$ $R^2$

night!

3:22 AM

I have to show that the inverse doesn't exist of a transformation UT where T is the map who h I told and U is a linear transformation from R^2 to R^3@KarimMansour

that means we sends the elements of $R^3$ which has more elements to $R^2$ so for sure it will be surjective however injectivity will fail

well

that is easy @Rememberme

since

Well, it won't necessarily be surjective

dim($R^2$) = 2 and dim($R^3$) = 3 so they have different dimensions so they can never be isomorphic

No isomorphisms yet

in particular inverse don't exist

3:24 AM

if there's an injective map from $\mathbb{R}^3$ into $\mathbb{R}^2$ then $\mathbb{R}^3$ is isomorphic to a subspace of $\mathbb{R}^2$, so that $\mathbb{R}^3$ has dimension 0, 1, or 2

But can we go this way...the injective surjective one

@SamuelYusim He just said it was over some fields, not over $\Bbb R$

yeah also it won't be necessarily surjective aswell it depends on the map too as @pjs36 said

oh sure

well even over any arbitrarily field my argument still work

3:25 AM

@KarimMansour No it doesn't

why ?

$\mathbb{R}^3$ has the same "number" of elements as $\mathbb{R}^2$

Can I somehow show that the inverse doesn't exist

If I can then UT is not invertible

inverse exist if it is isomorphic double implication

Thats what my first instinct was

3:27 AM

Have you learned this theorem, @Rememberme: A linear map $T: V \to W$ is injective if and only if is kernel (null space) is $\{0\}$?

that is if T is isomorphic then inverse exists and if inverse exist then it is isomorphic

@pjs36 no kernels @KarimMansour no isomorphisms

but you have that the dimensions of your fields you know them right?

In general no you can't. Although since I know you are working in Hoffman and Kunze I think it is safe to say you are working over the field $\Bbb R$

dim($F^3$) = 3

3:28 AM

I have learnt nonsingular stuff if I am right is that what you call a kernel?

@Rememberme In H&K kernel is the same thing as null space

Then I would just think about what would happen to a basis of $\Bbb R^3$. Could it get sent to a linearly independent set?

for example

fix a basis for $R^3$

@PaulPlummer bit they haven't mention any fields and they also haven't mention that if if there is no mention of any fields what should I take

B = {(1,0,0),(0,1,0),(0,0,1)}

3:30 AM

Standard basis vectors

can it ever be sent to independent set by your linear transformation ?

@PaulPlummer how come if we take any arbitrarily fields my argument wouldn't work ?

@KarimMansour map is not given..

But if I assume that it maps to the standard basis vectors in R^2

@KarimMansour Because over the field $\Bbb Q$ they have the same dimension, which is $\mathfrak{c}=2^{\aleph_0}$, so they are isomorphic as vector spaces over $\Bbb Q$ (assuming some axiom of choice or zorns lemma)

@pjs36 I have learnt that theorem

well suppose that it sends (1,0,0),(0,1,0) to (1,0) and (0,1) what about the third vector

(0,0,1) @Rememberme

@PaulPlummer oh I see

3:34 AM

It is not sent to anything .....

Because I think we can only sent it to two vectors....@KarimMansour

AbstractionOfMe

Anyone here tried a low-carb diet?

but that is in the domain so it must act on the third vector aswell @Rememberme

because

Yes.....

because we want that

So the 0 vector

3:36 AM

since it is a bijection

isomorphisms are bijections aswell as homomorphisms

since linear transformation is by definition homomorphic so we don't need to check that

So??? conclusion

all we need to do is check for bijection and what is bijection is that it must be surjective aswell as bijective, so it can't "not" act on (0,0,1) so this will force your linear transformation to be not bijective.

so in essence inverse doesn't exist

So for any map the following linear transformation for T will always not be bijective

In fact @KarimMansour the dimension thing is useful because then you can show that $(\mathbb{R},+)$ is isomorphic as a group to $(\mathbb{R}^2,+)$

Groups??

3:40 AM

well yea

Huh

How come here

because a field if your restrict it to the addition operation its by definition a group @Rememberme

remember what is the definition of a field

But isn't a field also a group if I take the nonzero elements under multiplication @KarimMansour

abelian group that satisfies operations that is related to multiplication and distribution law

So field is abelian under addition and multiplication if I take the non zero elements @KarimMansour

3:43 AM

ye

no

wait

ye if you remove your zero

The rise of @Incurrence

@Incurrence hi

@PaulPlummer Typed it in by mistake lmao

haha

So my question was: The discrete topology - what is this?

3:44 AM

where is discipline of barney didn't see him for long time

@Incurrence How did all that homework go?

oh I see :D

that is why :D

@PaulPlummer Semi-decently

@Incurrence The discrete topology is basically this: every set is open

every subset

@PaulPlummer Presentation is tomorrow now(or whenever I get to talk)

3:45 AM

@KarimMansour Yah, I am Disciple of Barney

Are you ready to present?

:D

Yeah I was going to present yesterday, but we didn't have time

Did you manage to show that it doesn't split?

I will change my talk though I think, since it seems to be about 15 min long in its current form

Ah, makes sense

3:47 AM

@PaulPlummer Why is a central extension of H_3(Z) important in your opinion haha

Wait one min, getting a coffee from the coffee thing 20 meters away

One of the important things about extentions is that you break a group into simpler components, and sometimes saying things about the simpler components is easier, but leads back to the full group @Incurrence

Woah, you've got a coffee thing 20 meters away from you, @Incurrence? Sweet gig!

So for example We have $1 \to \Bbb Z \to H \to \Bbb Z^2 \to 1$ is "the" extention. Now it isn't necessarily obvious that $H$ is nilpotent or solvable, but if you know a thing or two about extentions and nilpotent groups, as soon as you split $H$ up like that you know that it is nilpotent

@pjs36 Haha at uni atm, so there are coffee shops all over the place

@PaulPlummer I will try to incorporate this when I rewrite it tonight. It was really rough as it was

@Incurrence taking summer classes or that is still winter semester?

3:57 AM

@KarimMansour Haha I'm in Australia :P

Okay, Did you show it didn't split?

oh I see

lool

@KarimMansour We are going into winter soon :)

yeah so I guess your at the end of your fall semester :D ?

@KarimMansour Fall = Autumn?

Well we have ~17 weeks semesters, only two a year

3:59 AM

yeah

what !!?

@PaulPlummer Not explicitly perhaps

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