Mathematics - 2016-06-02 (page 2 of 2)

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Mike Miller

21:00

yeah, i understand

arctic tern

my guess is that it's equivalent to the following: put a deflated balloon on a soccer ball, start trying to wrap it around the soccer ball (so there are two layers of rubber) until the crease contracts to a single point on the other side of the ball. at that point in time, twist the inner layer of rubber around 360 degrees. then do the original spreading-out thing in reverse.

Mike Miller

i've got a picture, let me think to see if that agrees with what i'm seeing

i'm having trouble visualizing what you're doing... it doesn't sound like you start with image a point

oh, i see

arctic tern

imagine you deflated the ballon to a single point to begin with :P

Mike Miller

i think that works.

here's one way of checking whether it does. a smooth map $S^3 \to S^2$ has Hopf invariant $d$ if, given regular values $x,y$ with connected preimages, the preimages have linking number $d$

so... find out if that's true

Semiclassical

21:34

@MikeMiller just to check if I'm forgetting something

if $M$ were a manifold without boundary, then wouldn't this formula be trivial? math.stackexchange.com/q/1810136/137524

i.e. it's only an interesting question if $M$ is a manifold with boundary

Akiva Weinberger

22:01

I think my mind was just blown

$\bigcap\emptyset=V$

This is not OK

arctic tern

heh

Mike Miller

@SemiC yup

arctic tern

22:44

send stickers? eh?

you can use ascii symbols like normal. you can still type the text-based smileys. chatrooms don't generally have picture emoticons. you can of course upload images in chat though. and in this one we have $\LaTeX$ we can use.

images using the "upload" button. you'll just have to link to pdfs you upload elsewhere.

right next to the send button, which is right next to the textbox I type in

probably.

dunno. why?

sure

see room description

Mary Star

23:07

Hello!!

We have that $R$ is a commutative ring.

Suppose that $0\rightarrow A\rightarrow B\overset{f}{\rightarrow} C\rightarrow 0$ and $0\rightarrow C\overset{g}\rightarrow D\rightarrow E\rightarrow 0$ are exact sequences.

Then $0\rightarrow A\rightarrow B\overset{gf}{\rightarrow} D\rightarrow E\rightarrow 0$ is also exact.

I want to show that each exact sequence can be arised by short exact sequences as above.

Can we show that by induction?

quid

@Liam no.

I replied to a particular message of yours. Obviously I answer the question in that one;

plus you could follow the link.

quid

23:23

I do not share you assessment, but have no interested in further discussion of MO.

But if others want to discuss MO. More power to them.

Not really.

George Smyridis

Hey guys. Hope i am not interupting! I have a quick question. When solving a limit in MVC can δ(ε) have the form of δ'+c ?

Evinda

Hi @GeorgeSmyridis
Are you from Greece?

George Smyridis

Yeah, is it obvious ?

Evinda

Because of your name :D @GeorgeSmyridis

George Smyridis

LoL

Evinda

23:30

From which city? @GeorgeSmyridis

George Smyridis

I thought you had to be greek to recognise it :P

So you are greek ha?

Evinda

Who says that I am not :P @GeorgeSmyridis

George Smyridis

Thessaloniki.

Evinda

And you also study there?

George Smyridis

Yeah! This is the first year i study physics!

Evinda

23:31

Nice

George Smyridis

Evinda? That doesn't seem greek

Evinda

It's not my real name

George Smyridis

So from which city are you?

Evinda

I am from Crete

George Smyridis

Do you study there?

Evinda

23:34

Yes

George Smyridis

Maths?

Evinda

Yes...

quid

@Liam mainly I said two things: a) offensive content should best be handled discreetly. b) most of the time there is no need to follow up on unhandled flags with chat-messages to moderators.

Evinda

So... I will go to sleep now... See you... Kalhnuxta :) @GeorgeSmyridis

George Smyridis

So, can anyone answer my question?

Kalhnyxta :D

quid

23:36

'Too bad ST had not read it...

In this room it likely will take you a bit longer.

But you sure are making a good start.

It was to cheer up the idle summer time.

Akiva Weinberger