hey question

I have this problem

@TedShifrin Points here are lines, right?

FInd the volume using the shell method of f(x) = x^-4 from -3 to -1 around x =4

No, no. We're talking natural things in $\Bbb R^3$. Points are points and lines are lines.

@Link: Have you drawn a sketch and labeled radius, height, etc.?

I set this up as the 2pi*integral of (x*(4-x^-4)) from -3 to -1

yes

but, the book shows this as 2pi*integral of ((4-x)*(x^-4)) from -3 to -1

why is this?

What in the formula you just wrote down is the radius and what is the height, and does it match your picture?

LOL, because the book is right :P

I know that

but I don't get why thats true

Isn't the radius the distance from your $x$ to the line $x=4$?

yes..

oh

loves the "oh" :)

hmmmm

Shouldn't we all, @Jasper? :D

I threatened to disappear. I might still, but not until @Peter has finished this investigation :P

@TedShifrin Our line is $t(1,0,0)$ so our line must have a solution for $(t,0,0)=sv+u$?

@Jasper: Don't worry. I've spent over a decade being "addicted" to chatrooms.

@Peter: Yes, we want to see that this imposes one condition on our set of lines $\ell$.

@TedShifrin OK.

@TedShifrin What is the best way to kick the addiction?

LOL@cyber. If I were kicking you, would I be here? :D

@TedShifrin I don't mean you kicking me, but just in general.

@Peter: Here's a hint. Don't do this algebraically. Go back to thinking about things the way we did with the original problem. Look at the plane spanned by the point $u$ and the $x$-axis.

@cyber: Do things with friends in real time, read books, work on learning. :)

@TedShifrin Thanks pal :)

@TedShifrin A hint for what? I am lost.

For answering the question for why it imposes one condition on $u,v$ for $(t,0,0) = sv+u$ for some $s,t$.

@Peter: Maybe I didn't tell you the best way. Look at the plane spanned by $(1,0,0)$ and $v$.

@TedShifrin But, the lines that cross $(t,0,0)$ one a fixed point "fill up" all the space, don't they?

For each point you give me, I can find a line that crosses $(t,0,0)$ and goes through that point.

But the direction of the line is supposed to be $v$.

If you fix $u$ and $v$ both ahead of time, there won't necessarily be a solution.

@TedShifrin Ah, sorry. I easily forget our set up.

I know, this is tough.

If we look at the plane spanned by $e_1$ and $v$, it's one condition for $u$ to lie in that plane. (That is, $u$ must satisfy the linear equation of the plane.)

@TedShifrin Aha.

So you agree that $\sigma_m$ is given by one condition, hence codimension $1$?

Yes.

So, now we have four lines in general position ... That gives us four different $\sigma_m$'s ... and we want to figure out in how many points all four intersect in our space of all lines.

As you expect from linear algebra, four "linearly independent" equations have a single point of intersection. So you expect a $0$-dimensional, finite intersection here.

Motivation: Go take a course in differential topology. (The ubiquitous MSE questions on Guillemin & Pollack notwithstanding, you should learn that book when you're done with multivariable analysis.) :P

@TedShifrin OK. I guess if I understood all this better I'd be more thrilled, =D

I told you this was a tough road.

@TedShifrin Right. But this was all standing in thin air. I'd be better off reading a book, building things bit by bit.

We're not done. So what we're doing is trying to understand the algebraic topology of this $4$-dimensional manifold. And we're going to answer the question by knowing a "basis" for the $k$-dimensional cycles. The $\sigma_m$ (for any $m$) generates the codimension $1$, and we need to know how to intersect two of 'em ... then figure out how to intersect $4$.

This is a classical subject called Schubert calculus.

I know, @Peter... I told you it was fancy and hard. Eventually, you'll want to look at something like Griffiths and Harris to get all the foundations, but you'll need a bit of algebraic/differential topology to lay the foundations.

@TedShifrin Ah, yes. Mariano told me about this, he mentioned Enumerative Geometry when I mentioned this problem to him.

Yes, exactly, that's what this is.

If Mariano and I gang up on you, you'll be a super-geometer :P

"... by knowing a "basis" for the k-dimensional cycles."?

But it should make you appreciate the fact that you can do this from our $z=xy$ viewpoint plus a smidgeon of classical projective geometry (in the chapter I sent you, there's a complete proof, in fact).

Yes, @Peter, those are called the Schubert cycles, and they give a basis for the homology of $G(2,4)$ ...

I don't expect you to believe all this ... but you have math to learn :P

@TedShifrin Wait. You never said what those were.

LOL @ explicitized :P

@TedShifrin I spanglished.

Well, in each dimension/codimension there are sets like our $\sigma_m$. For example in codimension $2$, we have $\sigma_{p} = \{\ell: p\in \ell\}$ for points $p$ and $\sigma_H = \{\ell: \ell\subset H\}$ for planes $H$.

@Peter: Is Mariano offering comfort since I'm scaring you so much? :D

@TedShifrin Heh, we haven't talked much lately, but he explains awesome stuff to me every now and then.

And he tells me about events like the one in Rosario last week.

Yes, I'm sure. I'd much rather do some of this stuff at an actual blackboard.

You have lots of opportunities opening for you. I hope you'll keep taking advantage :P

@TedShifrin Yeah, I'm sure it'd be great. =)

I think you've done enough for today, @Peter. You're dismissed and I'll go cook dinner :P

Maybe I can swap some math lessons for those tennis lessons soon :P

@TedShifrin I will be going to the USA mid year next year, I hope.

yes, you told me ... we'll make a plan. :P

@TedShifrin I'll go back to linear algebra then!

then? you mean now? :)

@TedShifrin Aha.

English is a horrid language :)

@Peter ... I should never have tempted you with the "fancy" approach, but it gives you perspective and, I hope, motivation to learn more :)

@TedShifrin ¡Tendrías que aprender castellano!

@TedShifrin Indeed. =)

¿ Is that your way at getting back at me for French ?

@TedShifrin Nah, just playing. I am not a vengeful person.

Well, we evil profs never know.

@TedShifrin It would be fun to attend on of your classes and see the faces of the students.

@TedShifrin I have a question.

could anyone answer a quick question about ideals?

@JackM I can try, I guarantee nothing.

if someone tells me an ideal is generated by a single element, f

does that means every element of the ideal is a sum/product of f?

or am I allowed to multiply by units aswell

@JackM Elements of the ideal are of the form $\sum x_{k_j}fy_{i_j}$

wait, what are x and y?

@JackM Arbitrary elements of your ring.

oh I see

I should have checked wiki first

@JackM Why?

so the concept of the ideal generated by a set

is actually specifically the smallest ideal that includes that set?

@JackM Yes. $$(S)=\bigcap_{S\subseteq I,I\text{ an ideal }}I$$

As with every definition of "generated by..." =)

I assumed it was like in linear algebra, where you just close the set under addition and multiplication

@JackM Well, here you close it under "what an ideal is."

in which case the set generated by a single element would just be sums of powers of that element, right?

@JackM Come again?

if you had a set A, and you took the smallest set containing A which is closed under adding and multiplying

that would just be sums of powers of elements of A

no wait, sums of products of powers of elements of A

@JackM An ideal is not closed under "adding and multiplying". It is an additive subgroup of an underlying ring $R$, with the additional property that if $a\in I$ and $b\in R$, then $ab,ba\in I$.

I know

So it is "closed under multiplication on the left and right" by elements of $R$.

I was just double checking that, had my mistaken definition of a generated ideal been correct, then I would have been right

or rather of a generated set

@JackM What was your mistaken definition?

that you just closed the set under adding and multiplying

@JackM And what do you mean by that?

intersection of all closed sets that include the set in question

@JackM "closed set"?

closed under adding and multiplying

So you're thinking about subrings.

yes

You'd get the subring generated by $S$.

Oh, well. There you have it then.

well anyway

thanks

trying to plow into field and galois theory

with only a little prior experience with rings and groups

@JackM Well, I guess that is not really wise.

I'm finding it painless

I'm learning quite a bit of algebra from it

the subject provides a good motivation for the general theory

anyway, plenty of time to tighten my foundations once we get to group theory next semester

@JackM That is important.

would anyone have any idea why there's a physics class who has to take the group theory class with us?

seems like an odd requirement for physics

group theory becomes pretty important once you get into what physicists call field theory

@Anthony classical or quantum?

@anon Yao, yao. I have a question for you.

yo

go ahead

@anon I have proven that if $f:V\to W$; then ${\rm im} \;f^\ast=(\ker f)^\circ$ and $\ker f^\ast =({\rm im} \;f)^\circ$. I used finite dimensionality for the first, but not for the second. Wonder if this is true for inf dim, guessing yes?

The inclusion ${\rm im} \;f^\ast \subseteq (\ker f)^\ast$ is trivial. Didn't think about the other really, used the fact dimensions coincide.

so $f^*$ is the transpose (dual map $W^*\to V^*$) and $(\ker f)^\circ$ is the set of all maps $\in V^*$ that vanish on $\ker f$ correct?

@JackM: both, quantum more so I think

a lot of the models seem to have a compact lie group involved, they always talk about "SU(2) theories" and such

@PedroTamaroff both are true for inf dim I believe

would anyone happen to have an example of a field of finite dimension over the rationals, but that isn't obviously algebraic?

not possible

I think the word "obviously" is key here.

It's also quite subjective.

it'll be an algebraic field, I just mean a construction where you can't immediately see that every element is a root

okay, write down a root $z$ to a polynomial equation with insoluble galois group using hypergeometric series (so it can't be expressed via radicals), then form ${\Bbb Q}(z)$.

then again, I suppose even with $Q(\sqrt2)$, it's not immediately obvious that every element is a root of a polynomial

your question is pretty subjective. to me at least any expression formed using nested radicals "obviously" sits in a tower of cyclic extensions

I know it's subjective

that's why I just brought it up in chat, not on the main site

are subjective questions discouraged in here too?

I don't come here often

@anon OK, take $\varphi \in (\ker f)^\circ$.

@Peter: Be careful. In infinite dimensions, linear maps needn't be continuous and subspaces needn't be closed. But the annihilator of a subspace is always closed, right?

LOL, derp.

Other containment.

@TedShifrin Closed, topologically?

Yes, assuming a topological vector space.

@TedShifrin What are you referring to now?

@KarlKronenfeld Right, so $\ker f\subseteq \ker \varphi$.

Babbling as usual :) A vector space with a topology so that vector space operations are continuous.

@TedShifrin But why should I be careful?

@PedroTamaroff mkay, so $\varphi$ determines uniquely a linear map $V/(\ker f)\to K$.

Because things that seem obvious in finite dimensions may go awry. Images may not be closed, but annihilators/orthogonal complements are.

Extend this new map however you want to the rest of $W$.

@TedShifrin Right, I kinda know inf dim makes thing go awry.

Don't know many examples though.

@KarlKronenfeld Ponders.

I'm sorry. I just don't get it. When I'm integrating the volume of x^-4 from -3 to -1 with the shell method. Why is the radius 4-x??

@PedroTamaroff I forgot the name of the theorem in group theory which says basically that.

My favorite example is that $(V^\perp)^\perp$ may be bigger than $V$ in a Hilbert space.

@KarlKronenfeld I think I never heard about it.

Going Dubuque all over the place =D @anon

@KarlKronenfeld Let's try and prove it.

@PedroTamaroff So you have a map $f:G\to H$ of groups and $K$ is a normal subgroup contained in the kernel of $f$.

@Link: That's why a picture is essential. The distance from $x$ to $4$ is $4-x$ when $x<4$.

@KarlKronenfeld (normal)

Find a map $\varphi:G/K\to H$ such that $\varphi\circ\alpha = f$ where $\alpha:G\to G/K$ is the canonical map.

@PedroTamaroff Thanks.

@KarlKronenfeld Okey.

@KarlKronenfeld (I know you know, just for the record.)

So if we have $g+K$, what would be a natural place to put it in $H$?

@KarlKronenfeld $f(g)$.

@TedShifrin hmm. I made a picture but it doesn't seem to help

Yipi.

@PedroTamaroff The real work is in showing this map is a well-defined group homomorphism.

@KarlKronenfeld OK. Will do that.

they're called the isomorphism theorems

@Link: Say in words what you think the radius of the shell should mean.

@anon It's a combination of the first iso theorem and the lattice theorem right?

why do you include lattice?

The distance from the centre (what it's being rotated around) to the edge.

$K$ is contained in the kernel, not necessarily the whole thing.

@Link, ok. So the center is at 4 and the edge is at $x$.

So we have to have some sort of map $G/K\to (G/K)/(\ker/K)\to H$.

The first is given by the lattice theorem.

Why is it at x??

Because the shell you're looking at has its edge at $x$ ... We put the shells together as $x$ goes from $-3$ to. $-1$, or whatever.

Wait what do you mean?

Oh okay. So x is the point that is being rotated around?

So If I have x as negative one

The vertical slice of the graph at $x$, yes.

My radius is five?

Got it that makes sense then.

@anon

Thanks!!

It varies from $5$ to $7$ ....

Yup that makes sense

Thank you

Ok, great. Now go explain it to your friends :)

Gotta love when you can condense a proof down to a single sentence.

Let $A\in K^{n\times n}$. Then $$\{P\in K[X]:P(A)=0\}=(m_A)$$ where $m_A$ is the minimal polynomial of $A$. @anon

XD it was just me that didn't get it

Time to go finish my hw then

Have fun @Link

@PedroTamaroff ?

@Daniel: I can't do that very often.