@Remember: It's perfectly standard, but a lot of matrix algebra courses hide all the good stuff :P

Yes they do.....

Can you explain this: "0 is the only constant in a maximal ideal of K[x]"

@TedShifrin my econ course is kinda boring on the math side. I'm in the "honors" class and its just not very interesting math. The econ itself is cool though.

@TedShifrin

@zed: What other constant do you propose?

@TedShifrin Now since we have defined Matrix multiplication with the help of linear transformation how can i show that complex number multiplication is the same as we do it....for example explaining or defining complex number multiplication with the help of something like linear transformations?

Is it possible

Any k \in K? @TedShifrin

@Stan: I'd expect Chicago to have one of the most sophisticated econ programs but, at the undergraduate level, no one seems to be all that mathematical. They save it for grad work.

Yes, @Remember, sure it is.

Well, if $k\ne 0$ is in the ideal, then so is $1$, right, @zed?

How do you do it

Well, @Remember, write out $(\alpha + i\beta)(x+iy)$ as a linear map on $(x,y)$.

this is, I hear, a problem for people applying to econ grad school... the good grad schools don't want to accept anyone who doesn't have a lot of math ability (usually they demand real analysis - not because of real analysis itself, but as proof that one can do math with rigor), and people who aspire to be in econ but don't know about this screw themselves on the math they need

What's the $2\times 2$ matrix?

@TedShifrin correct

The only ideal with $1$ in it is the whole ring, @zed.

Yes

@Mike: We have econ majors who try to talk their way into real analysis and just can't handle it. The kids who come through my multivariable math course seem to do just fine.

right, that's the problem... they should be doing math when they start undergrad to keep up

Many of them are not good enough at math to pull that off, it turns out :)

@TedShifrin : From this how do we deduce that: the canonical projection \pi K[x] \to K[x]/(f) = F, when restricted to K, is a monomorphism. (f is a maximal ideal of K[x])

But @Ted Isnt linear mappings between only vector spaces....?? And complex plane is what i think i have to show that it is somehow related to the $\mathbb{R^2}$ then only i can tackle this problem right....so how should i do that?

Yes, I'm saying you should identify $x+iy$ with the vector $(x,y)\in\Bbb R^2$, @Remember.

So, we just proved it, @zed. If $k\ne 0$ were in the maximal ideal, the ideal would be all the ring, and that is not a maximal ideal.

So $T:\Bbb{C} \to \Bbb{R^2}$ right?

I wanted you to give the matrix representing the map $\Bbb R^2\to\Bbb R^2$, @Remember.

Identify $x+iy\in\Bbb C$ with $(x,y)\in\Bbb R^2$ in both cases.

@TedShifrin It is a monomorphism because: k \to k + (f) and k \neq 0 \in K[x] does not exist in f?

in (f)

To say that $k$ maps to $0$ in the quotient is to say that $k$ belongs to the ideal you mod out by.

@TedShifrin Can you explain the second statement a bit i have written the matrix But what do you mean by the second statement

I'm going to take real analysis next fall. I'm very excited about it.

What second statement, @Remember? What's your matrix?

What book do they use, @Stan?

@TedShifrin this one

To write the $2\times 2$ matrix you need to be thinking of both inputs and outputs as vectors instead of as complex numbers, @Remember. What's your matrix?

I hope Rudin. I haven't looked yet.

OK, I have to go to my last geometry seminar :) Let's see what I can learn. :D

Well, that'll kick your butt a bit, @Stan, but good :)

When will you come back

Heh, I don't mind. I get bored without a bit of that

Probably after I get home, @Remember. A few hours.

See you guys ...

Adios amigo

Peace!!!!!

@Josué You can read an inequality both ways

@MikeMiller Not sure what you mean by "what came before". Was that about the $\pi_1 SP^\infty \cong H_1$ isomorphism?

Thanks for those infos about symmetric products, btw. Provides a lot of food for thought :P

@BalarkaSen a doubt....

ok

that seems quite elementary. what have you tried?

think about what being linearly dependent means.

some constant C1 and C2 times the respective vectors when added gives me 0

write it down explicitly, please.

let there be two vectors V and K such that $c_1(V)+c_2(K)=0$

and now you have to prove that $V = \lambda K$ for some $\lambda$.

...

yes

don't you see what $\lambda$ is?

V/K?

huh? how do you propose to multiply and invert vectors?

it looks nonsense.

Yes, @BalarkaSen, you said something about wanting to find something analagous to that argument for $\pi_*(SP^\infty)$, or something.

@Balarka got it.... $V=\frac{-c_2{K}}{c_1}$

right.

see, that wasn't too hard.

You know whats my problem i need to see few proofs before i start going

@MikeMiller My argument for $\pi_1(SP^\infty)$ is okay, right?

@Rememberme you need to see a few proofs to do basic manipulations?

No i mean the hard ones

Yes. You'd have to show that it's actually an H-space, but that'e asy.

yeah, it's $\bigcup_{n \geq 1} SP^n$

@Rememberme but this was not hard

Yes it was not........

@Balarka to show that a few vectors are linearly independent i have to show that there is a point after where i cant further reduced them in a matrix right?

"i can't further reduce them" is vague.

try to see how much you can reduce the block matrix formed from some linearly dependent vectors first, and formalize it.

but you're "kind of" right, yes.

Wait @Balarka I just thought of something we can use determinants right in the following question?

nah, no need to

Okay

@Balarka I am getting that two vectors are linearly dependent and the other two are not and the question asks that are the vectors linearly independent or not so what should i write?

do it yourself.

use your brain.

I think so they are not linearly independent

Because in order to be linearly independent all scalars should be equal to 0 and in this case some are both other are not so they are linearly dependent @Balarka Am i right?

prove what you're saying. i am not going to do each and every exercise for you.

convince yourself whether you are right or wrong.

So it should contain the trivial solution

But in my case yes there are some rows with trivial solution are 0 but there are some of them which give a non zero solution and even if the linear combination has a one non zero solution it leads to it being linearly dependent

So am i right with my argument?@BalarkaSen

$v_1, v_2, v_3, v_4$ be your vectors, $v_1, v_2$ linearly dependent and $v_3, v_4$ linearly indendent. You say that $v_1, v_2, v_3, v_4$ are linearly dependent. What are the coefficients $c_1, c_2, c_3, c_4$ you have in mind such that $c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4 = 0$?

Yes i have it

What are they?

Write them down.

You want the vectors?

I want the coefficients.

Oh a matrix.....so tiring

$c_1 = ?$, $c_2 = ?$, $c_3 = ?$ and $c_4 = ?$. If the vectors are linearly dependent then they must exist, right?

What matrix? I just want you to write down some darn scalars, @Remember

Instead of rambling nonsense, just write them down and be done with it.

OK wait then

there we go

i don't get it.

what is c_1?

what is c_2?

Dosent this show that these are linearly dependent

i want to know what your $c_i$s are...

if you claim these are linearly independent, surely you have a set of coefficients in mind?

what are the coefficients?

Hello. I am looking for some help. I have calculated an equation for a line but don't know how to calculate if a given point is on the line or not =/

@Balarka i say that these are linearly dependent because the matrix shows that there will be non zero solutions for the c's and hence linearly dependent

i don't want to know the matrix. i want you to do it from first-principles, definitions.

it's super-easy that way.

never mind i just worked it out :)

Even from definitions it shows that there will be a non zero c for which the following combination will be equal to 0 and even if there is one the definition says that it is linearly dependent

Because of the matrices@BalarkaSen

i give up. consider $c_1 = a, c_2 = b, c_3 = 0, c_4 = 0$ where $av_1 + bv_2 = 0$ (which happens for some $a, b$ by the statement of the question). no nonsensical matrices needed.

i'm going. bye.

Hi @Ted

Good night, @Mike.

Sadly, that often happens

How did the student who wanted more rigor do?

nevermind, I don't think that's appropriate for this chat.

well, ok, but it's all anonymous

Nonetheless, I'm uncomfortable with it :)

I suppose one never knows who might be anonymously lurking ...

Since several of my students show up here unanonymously, I like to insult them personally :D

it's the nonymous ones you don't have to worry about

In my last 35 minutes of my multivariable class, I did a whirlwind complex variables class today ... from $\partial f$ and $\bar\partial f$ to the Cauchy integral formula to the power series expansion of a holomorphic function. :)

Why hello there guys

hi @Karim

We just did Green's theorem.

You're supposed to be on vacation, @Karim

hi @TedShifrin :D

well, all my students are experts on forms and Stokes's Thm., of course, @Mike

yeah but I want to study as well I am bored doing nothing @TedShifrin

lol

you need a vacation for a while, @Karim

hides from Balarka who's supposed to be asleep

@Ted: I hate these interims between things. My fridays have 1hr breaks between things I do, so I never have time to sit down and get anything done.

yeah, @Mike, it's better to have everything back-to-back ...

yeah I will take vacation starting from tomorrow xD

class 10-11, lunch at 12, office hours at 1, seminar at 3...

We always had diff geo seminar at 4:00 on Fridays at Berkeley, @Mike ... and coffee hour which was wine and cheese hour once in a while at 3 PM first

well, you could make office hours 1-3 and not have an extra hour.

i'm awake, @Ted. just like every kid should be at the crack of a dawn.

my girlfriend works in the morning usually I just do math and gym in the morning and then hang out with her during evening @TedShifrin though if she knew I was studying she would kill me xD

@Ted: I stay with students that want to stay. I always schedule my office hours so that I can stay over.

see, @Karim, more reason not to study

yeah :D

how is your last few days in diff geo class @TedShifrin?

"the crack of dawn," @Balarka, no [a]

haha

He's up at the crack of this specific dawn

@robjohn I cracked $$\int_0^{\infty} \frac{\cos(x)}{x} \left(\int_0^x \frac{\sin(t)}{t} \ dt \right)^2 \ dx$$ without CAS, and then the same for the cubic version. The squared one is famous since it looked like no one did it without CAS. Of course, the cubic one is far more advanced.

We finished, @Karim. I talked about hyperbolic geometry the last few days ... including some classical stuff and how you could make a many-holed torus hyperbolically

so that is it for the class only finals and they will be done ?

yeah, @Karim ... one more class (review) for the young'uns and then finals.

cool :D

are you expecting some people to ace the class?

@evinda bist du da?

I rarely have students "ace" finals.

I already like grammar, @Balarka.

Guten Abend, @Alessandro.

haha

Guten Abend @TedShifrin

ich bin nicht da @Alessandro

@KarimMansour wo bist du denn? :P

Alessandro muß mir italienisch lehren.

ich bin hier und da !!! @Alessandro

@Karim: Are you exemplifying Heisenberg uncertainty?

hahaha

yeah xD

I've read a very bad joke earlier on Heisenberg playing hide and seek and just going around shouting his exact velocity instead of hiding

LOL, yup, bad joke :P

lool

I stopped drinking caffe now I am getting withdrawl effect headaces :S

yup, that happens

I prefer my addiction

When I started having serious heart/blood pressure problems, I cut way back. Now I drink about one mug of cappuccino a day.

yeah I used to drink like 2-3 cups a day of caffe

that's not that much, actually, amongst mathematicians

you won't be able to produce enough co-theorems that way, @Karim

@Mike: Are you still off the cigs?

for him they'd need to be ca-theorems, @Balarka

hahaha

I think I drink too much coffee too

Italians are spoiled ... very good espresso

Of course, @Ted

Good to hear, @Mike.

A friend of mine has a very nice espresso machine... so it's a treat to go over there

I will be bringing mine, @Mike ...

I have a spare one for when one breaks down :P

Here in Germany it isn't as good as in Italy sadly :( I should have brought my moka like I did in China

aha

But the next time one breaks down, I won't be getting it repaired.

Yeah, @Alessandro, not as good in Germany or in France.

There's a place with good (drip, not espresso) beans in Santa Monica I should force you to go to

force, huh?

well, you can come willingly if you want

Ah.

@Mike @Balarka: Here's a question one of our grad students posed this afternoon. If two subsets of $\Bbb R$, say, are homeomorphic and one has measure 0, must the other have measure 0?

No

cantor sets are homeomorphic no matter the lengths you delete in the construction; doing so as the lengths go to 0 sufficiently fast gives you something of positive measure

So called fat cantor sets

Although I think it holds if you instead look at strong measure zero sets

Yuppers. They didn't believe that fat Cantor sets were homeormorphic to thin ones.

i mean, the same way you write down a homeomorphism to $\{0,1\}^{\Bbb N}$ works perfectly well

just a bifurcating tree etc

well, the homeomorphism is not too hard.

You pass, @Mike :)

I gave my answer with a suggestion, and left them to spend the weekend on it.

@Discipleof Wilma: What is a strong measure zero set?

looks suitably unpleasant

apparently borel conjectured something about them, which leads to conflicts with much more interesting borel conjectures

I'm not sure I've ever heard of these.

i haven't fiddled much with measures.

@TedShifrin Given any sequence of $\epsilon_n$ you can cover the set with intervals $I_n$ with length less than $\epsilon_n$

No, @Balarka, that would entail learning some analysis. :D

right

i'd prefer staying out of there until i have good reasons to study those

Its independent of ZFC whether or not only countable sets satisfy the strong measure zero property @TedShifrin

but that it's false follows from the continuum hypothesis, which is true

@MikeMiller Haha

I presume you mean that $I_n$ is a union of intervals of total length $<\epsilon_n$?

lol

No, now I'm confused.

@TedShifrin $I_n$ is just a single interval less of less than $\epsilon_n$ (although I am sure they are equivalent

@TedShifrin: as in $(I_n)$ forms a cover of your set, and the length of $I_n$ is less than $\varepsilon_n$

Why is that any different from the usual definition, where $\sum |I_n|<\epsilon$ ?

If $\sum \epsilon_n$ diverges, not much harm.

you can do this for every sequence $\varepsilon_n$.

is the hypothesis

Right, so make $\sum \epsilon_n = \epsilon$.

Oh, I see, you're controlling the individual intervals more harshly.

in any case, i'm not willing to think about this more

LOL

Well, I'll just leave.

i've got 15 minutes til the seminar...

today's on the vafa-witten equations

really

i guess i should just forget about symmetric products for a while

@TedShifrin here are proofs the cantor set is not strong measure zero

what university are you in @MikeMiller ?

i'm a grad student at UCLA.

cool

@MikeMiller sounds oddly physics-ish.

so does all of low-dimensional topology

flees

where do you think anybody gets ideas, other than from ed witten? :)

in any case, the best invariants for 4-manifolds we have come from studying the moduli space of solutions to a PDE on them. the vafa-witten equations form one such system of PDEs, and i suppose i'll learn if they're any better than the ones we've already got.

if low-dimensional topology intersects with physics, i guess i shouldn't think about being a geometric topologist.

@MikeMiller weird

rolling my eyes so hard they're falling out of the sockets

lol

@MikeMiller Do you think there are any cardinals between $2$ and $2^2$? :P

of course I believe in the three hypothesis

I think that is you think the four hypothesis is false

i've heard of the guy Vafa somewhere, i think.

probably from some visiting guy in the uni rambling about conformal field theories and monodromies. blergh.

@Mike: Interestingly, Cumrun Vafa was a student in my Guillemin & Pollack class at MIT in fall 1980. So were Rafe Mazzeo and Lorenzo Sadun. Some amazing mathematicians ...

@Ted: The invariants look interesting. It's hard to say what they'll do, if anything - their study is fairly fresh. They're somehow supposed to capture data related to the Euler characteristic of Donaldson's ASD moduli spaces - which depends on the metric - even when it's not 0-dimensional

But I think nobody knows quite what that means yet.

OK, people know what it means - you need to make some curvature restrictions on the metric, and it makes the Euler chadacteristic well-defined. But it's not clear if the VW-invariants are actually defined yet. :P I think there's serious progress.

Can someone tell me what the heck Tobias was probably referring to in regards to inversing $I+N$ where N is a strictly upper triangular(and thus nilpotent) matrix?

HI @Gates Wang