Mathematics - 2012-04-16 (page 3 of 3)

t.b.

23:00

Conversely, $(g_i)_{i \in I} = \Phi(\Psi((g_i)_{i \in I}))$ by how $g$ is determined...

Benjamin Lim

hey theo

t.b.

yes, ben?

Benjamin Lim

you see that map $g$ was determined by just one $g_i$ right

wait

t.b.

No, there was a for all $i \in I$.

Benjamin Lim

so the point $(g_1, g_2, \ldots....)$ was just collapsed to $g$

sorry I am not assuming $I$ is countable or anything just wanna show

that the sequence of $g's$ is collapsed to $g$

t.b.

23:02

I see what you mean.

After all, we want to prove an equality $\Hom(\ldots) = \prod \Hom(\ldots)$

Benjamin Lim

yes

t.b.

on the left hand side you have just one $\Hom$ thing, and on the right hand side a product of $\Hom$-things. So that's how it should be :)

N3buchadnezzar

Heya, mind anyone help me find a post in the site?

Benjamin Lim

yes

N3buchadnezzar

I am looking for a closed form of a integral

Benjamin Lim

23:03

that's what I was going to note

Jonas Teuwen

@N3buchadnezzar Again?

Benjamin Lim

One thing on the left was sent to a product of things on the right

okook

N3buchadnezzar

$$ \int_0^\infty \frac{x^{n-1}}{1+x^m} \, \mathrm{d}x$$

I know it is lurking around on the site, but I can not remember where. Nor how to search for it. I know anon posted an answer ..

ops

Benjamin Lim

@tb Right

Here's my go at why $f = \Psi(\Phi(f))$

Jonas Teuwen

@N3buchadnezzar That's just a hypergeometric function.

t.b.

23:07

I'm here.

Benjamin Lim

By the universal property of the direct sum we have a map $g: \bigoplus M_i to N$ such that $g \circ j_i = f \circ j_i$

By injectivity of the $j_i$ this means that $g = f$

N3buchadnezzar

@JonasTeuwen Yes, but it has a closed form. Something with 1/sin(m \pi) or something

Jonas Teuwen

That is closed.

t.b.

@BenjaminLim No, the universal property tells you that there's a unique $g$ such that blah blah, and $f$ already does that blah blah.

Benjamin Lim

ah crap

yes

there is a unique map $g$ from $\bigoplus M_i$ to $N$

but then we already have $f : \bigoplus M_i \to N$ that makes the diagram commute

so $ f = g$

t.b.

23:10

yes: you start with $\Phi(f) = \left( f \circ j_i\right)_{i \in I}$, get a unique $g$ such that $g \circ j_i = f \circ j_i$ for all $i$, so $g = f$.

Benjamin Lim

yes yes yes yes

t.b.

Now in the other direction: you start with $(g_i)_{i \in I} \in \prod_{i \in I} \Hom{(M_i,N)}$.

Benjamin Lim

that gets sent to the unique linear map $g : \bigoplus M_i \longrightarrow N$

t.b.

the universal property gives you a unique $g \in \Hom{(\bigoplus_{i \in I} M_i, N)}$

Benjamin Lim

and this $g$ under $\Phi$ gets sent to $(g \circ j_i)_{i\in I}$

t.b.

23:12

such that $\left(g \circ j_i\right)_{i \in I} = (g_i)_{i \in I}$.

and $\Phi(g)$ is by definition ^

(yes)

Hi, Ilya!

Benjamin Lim

yeah so that was easy

@Ilya pryvyet

N3buchadnezzar

@Ilya: Hey

Ilya

@tb: good morning

@Ben: privet

@N3: hi

Benjamin Lim

@tb Because $g_i$ by definition is $g \circ j_i$

Ilya

decreasing number of letters

t.b.

23:14

@BenjaminLim yes.

Benjamin Lim

@tb So

That will not help me in that exercise because at the end you want to use Yoneda right??

You want to say that $\hom(M,-) \cong \hom(M',-) \implies M \cong M'$

N3buchadnezzar

@Ilya Remember anything about posts in the site regarding $ \int_0^\infty x^{n-1}/\left( 1 + x^m\right)\,\mathrm{d}x$ by any chance ? =)'

t.b.

@BenjaminLim actually we're applying $\Hom{({-},N)}$

Ilya

@N3 reminds of Chebyshev integrals a bit, but I rarely deal with such functions

Benjamin Lim

ok

@tb You know what

all these things, we never really used that $M_i$ were modules right?

just some "object"

t.b.

23:17

No, it's a purely categorical argument.

Benjamin Lim

yeah

like I said man my view of mathematics has changed very rapidly

@tb Look at this guy employing thermonuclear weapons here: math.stackexchange.com/a/132327/5783

Ilya

@N3 $m,n$ are integers? because in this case the anti-derivative can be found

t.b.

@BenjaminLim I saw that. The edit from divisible to torsion free was due to a comment of mine :)

Benjamin Lim

oh ok

@tb Proof that $\Bbb{Z}$ is a PID with topology???????????????????????????

t.b.

Next thing to understand: $\Hom{(M \otimes M', N)} = \Hom{(M,\Hom{(M',N)})}$

Benjamin Lim

23:21

that's a proposition from atiyah macdonald

Ilya

@N3: ah, that would be not enough. Anyway, it may be worth taking a look on the last part here

t.b.

@BenjaminLim Probably. Thing is both correspond to bilinear maps $M \times M' \to N$ via universal property.

Benjamin Lim

@tb I think it is obtained from tensoring some exact sequence with some stuff

and the fact that if you have an exact sequence $0 \to M' \to M \to M'' \to M$

Ilya

@tb: I dare to interrupt your conversation - how is Kannappan doing?

Benjamin Lim

Then applying hom or something like that to this one is exact too

@Ilya He seems to be freaking out a little over the nilpotent elements question

t.b.

23:23

@Ilya dunno, seems still annoyed and pops up occasionally.

(very briefly)

Benjamin Lim

disease?????????????

Jonas Teuwen

@Ilya Hi 8-).

Ilya

@Jonas: hi

t.b.

@Ilya I don't know. I haven't really talked to him about that. Last time he mentioned it was quite a while ago.

Just around when you left.

Benjamin Lim

@tb What does he have? Can you tell me then delete the comment?

Ilya

23:25

@Jonas: how are you? I'm leaving for a breakfast in 5 minutes

Jonas Teuwen

@Ilya Breakfast? I'm okay!

You?

Ilya

@JonasTeuwen me too, thanks! :) yesterday we enjoyed Chinese cuisine

Benjamin Lim

I can't find it in the transcript

ok

you can delete the comment

Jonas Teuwen

@Ilya Nice 8-).

Ilya

my colleagues also went to see The Great Wall and The Forbidden City but I couldn't joint them since I was in Italy

but I will grab some photos they've taken

Jonas Teuwen

23:29

@Ilya :-).

Good.

I need to sleep!

Ilya

what are the news in Holland? Quine is ok?

t.b.

@BenjaminLim back to

8 mins ago, by t.b.

Next thing to understand: $\Hom{(M \otimes M', N)} = \Hom{(M,\Hom{(M',N)})}$

Benjamin Lim

@tb yes

Jonas Teuwen

@Ilya Nothing.

Benjamin Lim

Ok

Now I only need to understand why you had that equality between two direct products

Ilya

23:30

@Jonas: sleep well

Jonas Teuwen

@Ilya Thank you :-).

Good night guys.

t.b.

good night, Jonas

@BenjaminLim no, let's think about this thing again, it's not hard!

Benjamin Lim

ok

t.b.

The left hand side corresponds to bilinear maps $M \times M' \to N$ by the universal property of the tensor product.

(using the map $M \times M' \to M \otimes M'$)

Benjamin Lim

yes

because any map from the tensor product to $N$ is completely determined by the map from $M \times M' $ to $N$

t.b.

23:32

and conversely the map $M \times M' \to M \otimes M'$ gives you a bilinear map $M \times M' \to N$ for every linear map $M \otimes M' \to N$ (by pre-composition).

Benjamin Lim

yes

t.b.

On the other hand, a bilinear map $b: M \times M' \to N$ gives you a map $M \to \Hom{(M',N)}$

Benjamin Lim

so @tb This is kinda saying that given any two sides of the triangle we can complete the third side

t.b.

exactly.

(and uniquely!)

Benjamin Lim

@tb Yes you do that trick of fixing some element in $M$

t.b.

23:34

yes, $m \mapsto b(m, \cdot)$.

Benjamin Lim

yes

t.b.

conversely a linear map $B: M \to \Hom{(M',N)}$ gives you a bilinear map $b(m,m') = B(m)(m')$.

and again those are mutually inverse.

Benjamin Lim

so $M \cong \hom(M',N)$

t.b.

No, $\operatorname{Bil}{(M,M'; N)} = \Hom{(M,\Hom{(M',N)})}$.

Benjamin Lim

oh sorry yes

t.b.

23:37

and before we said that $\Hom{(M \otimes M', N)} = \operatorname{Bil}{(M,M';N)}$.

Benjamin Lim

So we have a one - one correspondence between bilinear maps $M \times M'$ to $N$

with elements in $\hom...$

yeah ok

so done

t.b.

1 hour ago, by t.b.

$$\DeclareMathOperator{\Hom}{Hom}\begin{align*}\Hom{\left(\left(\bigoplus M_i \right) \otimes M, N\right)} &=\Hom{\left(\bigoplus M_i, \Hom{(M,N)}\right)} = \prod \Hom{(M_i,\Hom{(M,N)})} \\ & = \prod \Hom{(M_i \otimes M, N))} = \Hom{\left(\bigoplus (M_i \otimes M),N\right)}\end{align*}$$

Benjamin Lim

@tb So now I only need to understand how you have the equality between two terms in the direct product

because each term individually, they are isomorphic

I know some stuff can mess up when you take the direct product

t.b.

@BenjaminLim yes, that's it.

Benjamin Lim

Oh but then

can we just say since $\hom(M_i,\hom(M,N)) \cong \hom (M_i \otimes M, N)$ for all $i$

t.b.

23:41

yes

Benjamin Lim

really???????????????????????

t.b.

well, I swept some naturality checks under the rug, but let's not enter that just yet.

Benjamin Lim

"naturality"?????

t.b.

whatever that means :)

Benjamin Lim

you make me laugh theo (not in a bad way)!!!!

t.b.

23:42

good :)

So, you see, each of those equalities is really simple.

Benjamin Lim

yes

but the direct product one I am a bit doubtful

because you see we know that tensor products distribute over direct sums and not direct products

so that tells me that somehow direct products don't behave so well

t.b.

@BenjaminLim and if you unravel the whole thing, you'll probably end up with the "long proof" you had before :)

@BenjaminLim I'll discuss that in a moment.

Benjamin Lim

@tb That proof was really ugly

@tb Can I ask you something?

t.b.

yes?

Benjamin Lim

Do you enjoy teaching me like this?

t.b.

23:44

No, that's why I never do it :)

Benjamin Lim

huhuhuhuhuhuhuhuhuh????????????????????????

(chuckles)

t.b.

see: as soon as we've done all the painful work of constructing the gadgets and checking the universal properties, we can now just manipulate the homs.

Benjamin Lim

ok

t.b.

It would never make sense to you to say $\Hom{(M \otimes M', N)} = \Hom{(M,\Hom{(M',N)})}$ without that painful work, but that's what it boils down to!

Benjamin Lim

yeah

that; s right

@tb So

t.b.

23:48

@BenjaminLim now let's see what happens when we try to replace $\bigoplus$ by $\prod$.

Benjamin Lim

replace it where?

t.b.

here:

1 hour ago, by t.b.

$$\DeclareMathOperator{\Hom}{Hom}\begin{align*}\Hom{\left(\left(\bigoplus M_i \right) \otimes M, N\right)} &=\Hom{\left(\bigoplus M_i, \Hom{(M,N)}\right)} = \prod \Hom{(M_i,\Hom{(M,N)})} \\ & = \prod \Hom{(M_i \otimes M, N))} = \Hom{\left(\bigoplus (M_i \otimes M),N\right)}\end{align*}$$

Benjamin Lim

ah ok

t.b.

No problem with the first equality, since that's just the tensor product doing its job.

Benjamin Lim

ok

that's what we just proved the first equality

t.b.

23:52

But then we're stuck: the universal property of the product tells us nothing about maps out of $\prod M_i$.

Only about maps into $\prod M_i$.

Benjamin Lim

so now

ffffffffffffffuuuuuuuuuuuuuuuuuuu

t.b.

so that argument breaks down terribly at the second step

and if we try to start at the rightmost end, we're immediately stuck.

Benjamin Lim

you mean the step where $\hom(\bigoplus M_i, \hom(M,N)) = \prod \hom(M_i, \hom(M,N))$??

t.b.

that's what I called second step.

Benjamin Lim

why are we stuck here?

t.b.

23:56

remember we replaced $\bigoplus$ by $\prod$.

Benjamin Lim

Don't we have the isomorphism $\hom(\bigoplus M_i, A) \cong \prod \hom(M_i,A)$?

for any module $A$?

t.b.

we do.

Benjamin Lim

$\hom(M,N)$ is module

so....?

t.b.

remember we replaced $\Hom (\bigoplus \dots)$ by $\Hom(\prod\ldots)$.

Benjamin Lim

okokok

ok clear now

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$Mathematics$ Mathematics