Mathematics - 2014-02-18 (page 1 of 5)

agent154

00:02

@PedroTamaroff I'm not quite sure how to continue. Normally, the inverse of the matrix (working just in $\mathbb{R}$) is

$$\frac{1}{-15}\begin{pmatrix}1 & -1\\-19 & 4\end{pmatrix}$$ but now I'm left with $\frac{1}{-15}$ that makes no sense mod 26. Do I need to find the inverse of 11 mod 26?

Pedro Tamaroff

@agent154 Yes, it does make sense!

Karl Kronenfeld

Since 15 is relatively prime to 26, you can invert it.

agent154

well, $-15+26=11$ so I'd be inverting 11 right?

Karl Kronenfeld

Sure you could have inverted 15 then negated as well. Either option works..

agent154

OK, thanks.

Pedro Tamaroff

00:10

@agent154 This is the fancy way to invert stuff.

agent154

I'm wondering if I did something wrong. The result I got fits in with the first equation but not the second

Pedro Tamaroff

@agent154 What do you mean?

agent154

I think I found the wrong inverse.

Karl Kronenfeld

It's easy to check, just multiply your purported inverse by 11=-15

agent154

The resulting vector I got was

$$\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}3\\15\end{pmatrix}$$

Nevermind, it's right... I made a stupid arithmetic error when verifying

Thanks for the help!

Pedro Tamaroff

00:24

@KarlKronenfeld

You suggested showing any homomorphism $Z_2\otimes Z_2\to S_3$ is trivial, yes?

Fernando Martin

That's not true @Pedro

Pedro Tamaroff

30 mins ago, by Karl Kronenfeld

@PedroTamaroff If you want to do it formally, just show that a direct summand collapses for any map $(\mathbb Z/2\mathbb Z)^{\oplus2}\to S_3$.

Fernando Martin

Just a direct summand collapses, not the whole group

Pedro Tamaroff

Provided it goes well with inclusions, I take it.

Oh.

So $0\times Z_2$ and $Z_2\times 0$ will be in the kernel?

Fernando Martin

one or the other

usukidoll

00:27

BUNNY!

Pedro Tamaroff

@FernandoMartin OK, suppose I have $\eta_1:Z_2\to S_3$ given by $0\to 1$ and $1\to (12)$, and $\eta_2:Z_2\to S_3$ with $1\to (13)$.

Fernando Martin

For instance, projecting to $\mathbb{Z}/2\mathbb{Z}$ is non-trivial; then compose the projection with a non-trivial map $\mathbb{Z}/2\mathbb{Z}\rightarrow S_3$ and you have a non-trivial map from the product to $S_3$.

ok, you can factor that through the product

Pedro Tamaroff

I want to show no homomorphism $\eta:Z_2\times Z_2\to S_3$ exists for which $\eta\iota_1=\eta_1$ and $\eta\iota_2=\eta_2$.

usukidoll

bunny!!! playbites @PedroTamaroff 's ear

Fernando Martin

ok, but those $\eta_1$ and $\eta_2$ won't work for this, since you can factor them through the product

choose $\eta_1$ to be non-trivial as well

Pedro Tamaroff

00:32

@FernandoMartin Isn't $\eta_1$ nontrivial...?

@usukidoll Hello, hello.

Fernando Martin

Sorry, I totally misread what you said

You're right

usukidoll

yayayyaaayy! plays with @PedroTamaroff

Fernando Martin

Yes, what you said is right

for some reason I read what you wrote as $\eta_1$ being trivial and $\eta_2$ mapping $1\rightarrow (12)$

Mike

01:05

@TedShifrin You might like this

user4140

@Mike what year is it?

Pedro Tamaroff

@Mike That's cool.

Mike

@robjohn Answer the ones that take a single line to answer, that way your effort:point ratio is super small

Fernando Martin

@Mike That's really neat

robjohn

@Mike yeah, but there are those people who like to downvote high rep users who answer easy questions.

Mike

01:12

Ah, good point

Answer only hard questions, but answer them snarkily

5space

@Mike cool geometry link

user4140

what do you mean snarkily?

Mike

with snark

like so

Making a square in 8 steps is killing me

user4140

lol noob

you need to cut back on a line

there was a question about that problem on se about a year ago

haha, found it.

15

Q: How can I construct a square using a compass and straight edge in only 8 moves?

I'm playing this addictive little compass and straight edge game: http://www.sciencevsmagic.net/geo/ I've been able to beat most of the challenges, but I can't construct a square in 8 moves. To clarify a move is: Drawing a line Drawing a circle Extending a line is not a move. Lines can onl...

geometric-construction

@Mike

Mike

01:28

oh, i was just whining

i want to figure it out myself

thanks though

user4140

what are examples of mathematical objects coming from biology?

Pedro Tamaroff

@user4140 Exponential growth.

@Mike YAY. I did the circle three pack.

The six pack should be the same but with an hexagon.

Mike

@PedroTamaroff I did that one in 10 moves... time to figure out what move I wasted

user4140

@PedroTamaroff lol, I meant the names

Fernando Martin

I don't get how you can do the circle pack 2 in under 5 moves

just drawing 3 circles takes up 3 moves

user4140

01:32

same as the square

Pedro Tamaroff

@user4140 cellular blah

?

Fernando Martin

cellular homology

haha

I doubt the name comes from biology though

What about amoebas?

ams.org/notices/200208/what-is.pdf

Mike

@FernandoMartin Took me 4 moves to make 3 circles, and then after that the fifth made my big circle

Fernando Martin

hmm

Mike

Takes longer to do it inside your starting circle, that took me 8 moves

Fernando Martin

01:34

Ahhh, I see

It isn't updating though

(the scoreboard, that is)

Mike

weird, it updates for me the moment i get it in that many moves

are you resetting after you finish a construction?

Fernando Martin

yes

Anthony

@PedroTamaroff !!!!

I'm still confused :(

Pedro Tamaroff

WAT.

Mike

now to do a circle pack 3 inside of the original circle...

Anthony

01:39

With the rhat/r^2 thing.

I know that spherical isn't well defined

And I know you're dividing by zero

But I feel like it's something that shows up elsewhere too.

I don't know why you can't just take the divergence of rhat/r^2

Pedro Tamaroff

Paste the problem here.

Anthony

The one my confusion stemmed from?

Pedro Tamaroff

Yes.

user129566

Hello!

Anthony

Sorry about the terrible quality

Pedro Tamaroff

01:49

What is $e^{-r/\lambda}$ suppose to mean?

Is $r$ a vector?

@Mike

Mike

@@

Pedro Tamaroff

I want to prove that if $G$ is abelian with $G\simeq \sum_{i\in I}G_i$ then $mG\simeq \sum_{i\in I}mG_i$.

For any integer $m$.

So I have an isomorphism $\eta:G\to\sum_{i\in I}G_i$.

Anthony

@PedroTamaroff It's just the radial potential

Pedro Tamaroff

@Anthony I don't know what that is. Is it a scalar function?

Anthony

Yeah

Sorry

So grad of potential should give e field

Pedro Tamaroff

01:54

Dunno man. Physics elude me.

Anthony

and del of e field gives charge

Aight, but my point is, at some point, you get del of 1/r^2 I think

And using the spherical divergence formula, you get the wrong answer. I just don't know why you can't take the del of 1/r^2 using that formula!

I mean I know that everyone is saying spherical isn't defined at the origin, and that you divide by zero, but those don't seem like they are the reasons... I don't see why you can't take the divergence, just because it diverges!

You can take derivatives of things like ln(x) which diverge at zero.

Pedro Tamaroff

@Mike So, this is what I have in mind.

Mike

@PedroTamaroff Isn't it sufficient ot just restrict your isomorphism?

Ah, nevermind, I see.

Go on.

Pedro Tamaroff

We have the inclusion maps $\iota_i:G_i\to \sum G_i$. I want to define $\psi_i:mG\to mG_i$ by $mg\mapsto m \psi \iota_i(g)$.

Then since $\sum G_i$ is a coproduct, I can extend this to a map $\psi:m G\to \sum mG_i$.

Mike

That seems like the way to go.

You can, of course, also identify $G_i$ with a subgroup of $G$

which might help

But I think you're almost done with your approach

Fernando Martin

02:03

Don't coproducts work the other way around?

Mike

You're right

You can factor morphisms to groups through their product, and you can factor morphisms from groups through the coproduct

Pedro Tamaroff

@FernandoMartin Sorry, should've put the guy in the other place.

The idea still holds.

Fernando Martin

yup

Pedro Tamaroff

I should have said $$\eta:\sum_{i\in I}G_i\to G$$

And now define $\psi_i:mG_i\to mG$ as follows.

Mike

Yeah

I need to get some reading done... after I draw this stupid circle 3-pack

Pedro Tamaroff

02:06

$$\psi_i(m g_i)=m\; \eta\circ \iota_i (g_i)$$

Here $g_i\in G_i$ is a generic element.

Fernando Martin

good old "multiply by m"

Pedro Tamaroff

Then I have a unique map $\psi :\sum_{i\in I}mG_i\to mG$ such that $\psi \iota_i=\psi_i$. Feeding an element $mg_i$, I get that $$\psi \circ \iota_i(mg_i)=m\eta\circ \iota_i(g_i)$$

@FernandoMartin So, that looks good yes?

Fernando Martin

Yes

What you said holds in $R$-modules as well

Pedro Tamaroff

@FernandoMartin OK.

Pedro Tamaroff

02:25

@FernandoMartin I am being stoopid I think,

$mG_i\leqslant G_i$ so $$\sum mG_i\leqslant \sum G_i$$ If $f:\sum G_i\to G$ is an isomorphism, define $f_m:\sum mG_i\to mG$ via $\{ma_i\}\mapsto m f(\{a_i\})$. @Mike

Then I claim ${\rm im}\; f_m=m{\rm im}\; f$ and $\ker f_m=\ker f$.

@anon

Fernando Martin

Why do you say you're being stupid?

Pedro Tamaroff

I was overcomplicating things.

►

That video is hilarious.

Fernando Martin

Not really, your first approach is categorical

Pedro Tamaroff

That sounds badass. Categorical!

Well, let me finish one of the approaches at least. =P

Mike

I've got a funny video, but I don't think this is the appropriate place for it

Pedro Tamaroff

02:38

irony?

Mike

deal

Pedro Tamaroff

@FernandoMartin I want to finish that approach.

I have show $\psi$ is an isomorphism.

user129566

Mike, is that a Fermat joke?

Fernando Martin

Ok

Mike

no

Pedro Tamaroff

02:55

@FernandoMartin I think I have an idea?

@mixedmath Hello.

mixedmath

@PedroTamaroff Hiya

It's been a bit

Pedro Tamaroff

Yeah.

I was guessing you could help me with what Fernando was suggesting. That is, giving a categorical proof that if I have an isomorphism $\eta:\sum G_i\to G$ of abelian groups I also have an isomorphism $\psi:\sum mG_i\to mG$ for any nonzero integer $m$.

I constructed $\psi$ already.

Now I have to show it is an iso.

I used $\sum G_i$ is a coproduct in $\bf Ab$.

I can repeat it for you: define first maps $\psi_i:mG_i\to mG$ by $$\psi_i(mg)=m \eta(\iota_i(g))$$

Where $\iota_i:G_i\to\sum G_i$ is the inclusion that comes with the coproduct.

Then I get a unique morphism $\psi :\sum mG_i\to mG$ such that $\psi\iota_i=\psi_i$.

I want to show this is an isomorphism.

Fernando Martin

Sorry, I wasn't around

Let's see

Pedro Tamaroff

@FernandoMartin I wrote it all there.

Fernando Martin

Try to write an inverse for it

I don't think I can find a categorical way to describe the inverse though

Pedro Tamaroff

03:04

Yeah. I was trying to do that =P

Ryker

Hey everyone, I have a quick question. Tangent spaces are manifolds, right?

Mike

sure, since a tangent space is just a vector space

but that's not very interesting

if you mean to ask whether tangent bundles are manifolds, yes

Ryker

Yeah, I know they are.

But I wanted to check tangent spaces themselves.

I tried googling it and was puzzled by the lack of yes/no answers.

I mean, I was pretty sure they are manifolds, but needed to check, just to be sure.

Thanks!

Mike

sure, every $\Bbb R^m$ is a manifold

they're locally euclidean (duh!) and second countable :)

Ryker

Is there a way for me to see LaTeX code rendered here?

Mike

03:08

see the "LaTeX in chat" link on the rigt

Ryker

I don't have that link.

Oh, I do.

Mike

:)

Ryker

start ChatJax

Mike

you have to bookmark that link and then click the bookmark

Ryker

Yeah, just did :)

Awesome, I was wondering how everyone here got along with all the $'s and LaTeX code, but nothing showing up haha.

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$Mathematics$ Mathematics