Mathematics - 2015-08-02 (page 1 of 3)

David Wheeler

12:06 AM

waiting for signs of life

Mike Miller

You're in the wrong place.

Kelechi Nze

Abandon hope, all he who enter here.

pjs36

These are the signs of life I saw asterisks for? 1 star.

David Wheeler

I left Hope years ago. She was too clingy.

Kelechi Nze

So I've started reading "A First Course in Abstract Algebra" during my holidays to try and get a jump on group theory, rings, monoids etc. algebra in general before I start my second year. I'm about a fifth of the way through and I just can't see the point of it. It all seems so obtuse. So I ask, what makes groups so useful?

Mike Miller

12:11 AM

what kind of math do you like?

Kelechi Nze

I loved calculus and differential equations in first year. I thought linear algebra was pretty neat too.

David Wheeler

@KelechiNze the set of all inveritble linear transformations of a vector space forms a group. The determinant map is a group homomorphism from that group to the multiplicative group of the underlying field. These are useful things to know.

Mike Miller

More important than the axioms (a product with inverses etc), the notion of 'group' might best be described as 'the symmetries of an object'. The symmetric group $S_n$ is the symmetries of a set; the dihedral groups $D_n$ are symmetries of $n$-gons; the group $GL_n(\mathbb R)$ of invertible $n \times n$ matrices are the symmetries of the vector space $\mathbb R^n$... groups capture the notion of symmetry, and if you like an object, it's only natural to think about the symmetries of that object.

pjs36

Indeed: Groups are the "movers and shakers" of the math world: If something is happening, chances are, a group is doing it; groups act on objects.

Mike Miller

Now, I tend to want to justify group theory not in terms of what it's good for (what's most of math good for?) but why it's sort of inherently interesting. And the above would be my pitch.

David Wheeler

12:17 AM

In organic chemistry, for example, the molecular symmetries of a compound, can determine its chemical properties. So it's good to know how these kinds of things can act.

In linear algebra, you've no doubt encountered the rank-nullity theorem. This is one of the versions of the First Isomorphism Theorem that cuts across many structures.

pjs36

I never considered that perspective on rank-nullity; good show, @DavidWheeler

Kelechi Nze

Okay I'll keep pushing at it. I totally understand the argument of beauty in mathematics. I also enjoyed studying number theory which was entirely thinly veiled cyclic groups. It's the lack of context putting me off. Number theory had the context of cryptography. I felt I could use what I had learnt to do something new as opposed to the theory I've seen so far which lets me do things I could already do but with more headache.

David Wheeler

Personally, I find it fascinating that just 3 axioms can cause so much trouble

2

Mike Miller

@KelechiNze I sympathize with this. One of the hard things about justifying mathematics is that everything needs a starting point, and in many senses, group theory is a starting point. Maybe the sort of justification that would please you is that Galois studied groups to study polynomials - his proof that you can't write down a general solution to a quintic equation with radicals follows, essentially, because the group $S_5$ isn't simple.

pjs36

And I think it's a similar story about Gauss's compass-and-straightedge construction of a regular 17-gon (or was it 7?), if I remember correctly

Mike Miller

12:26 AM

17.

David Wheeler

Arthur Cayley, one of the pioneers of group theory, discovered many of his results by playing around with matrices. Here is a small example: the group $$G = \left\{\begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}-1&0\\0&-1\end{bmatrix}, \begin{bmatrix}-1&0\\0&1\end{bmatrix}, \begin{bmatrix}1&0\\0&-1\end{bmatrix}\right\}$$

Mike Miller

You can't construct a 7-gon. That's part of this same theory.

Kelechi Nze

@MikeMiller That was actually one of the reasons that I picked up this particular book. But all the juicy Galois stuff is over in "Section 56" and I'm struggling my way past "Section 10".

Mike Miller

It's a lot of work to get that far.

If it makes you feel better, I don't get terribly excited about finite groups and the theory thereof (Sylow theorems, say) that shows up in these sorts of books. I tend to like how groups show up in other parts of mathematics, and I've never used the Sylow theorems except when I was taking a test.

But they're good toys to play with, especially when you're first learning.

David Wheeler

In general, when you have a set of "invertible" things, there's usually a group lurking in the background.

Sometimes, especially with things we add, it's easy to forget there's groups involved. Addition seems so natural.

pjs36

12:34 AM

And since you mentioned liking calc and DiffEQ, I remember reading that Cayley in particular treated differentiation in a sort of "operator" way, which seemed to motivate him to study trees. And in conjunction I believe he treated differentiation in an algebraic way... I never learned enough about his viewpoint, unfortunately.

Kelechi Nze

I suppose group theory will feel natural to me at some point and a handy tool as opposed to the too wide, too long, too heavy sledgehammer it feels like at the moment.

My father sings the praises of operator calculus.

David Wheeler

One of the beginning approaches to integration is through "step-functions" (what we use to form Riemann sums). This "abstracts" to the notion of a "vector lattice", a powerful tool in measure theory, which originates in algebra. The various structures are sort of like "tinker-toys", we can make intricate structures from the little bits.

Cyclic groups make a re-appearance in complex analysis as roots of unity. Permutations crop up all over the place in combinatorics.

pjs36

I don't know much about Fraleigh's book though, I (perhaps shamefully) learned from Gallian, then Isaacs, with a little Dummit and Foote for an exercise or two

David Wheeler

Homotopy groups show up in topology-an easy proof that the circle is not homeomorphic to the real line is to observe one has a trivial fundamental group, and the other does not.

Fraleigh isn't as good (in my opinion) as D&F, nor quite as well-written as Herstein. Better than Pinter, though.

Artin is good, but it's not an easy read.

All told, though, some people just don't have a taste for "pure" algebra. And that's OK.

David Wheeler

1:02 AM

and...chat dies. memorial services will be held next Wednesday. BYOB.

pjs36

It's a sad day indeed

PVAL

@David Wheeler That is not an easy proof the circle is not homeomorphic to the real line.

David Wheeler

well, i suppose people have impotant things to do. like getting drunk. or cherchez les femmes.

@PVAL which part are you objecting to?

pjs36

Getting drunk is definitely on my radar, as I do not want to figure out the final grades for this summer class (also, that involves grading the final).

PVAL

The easy part

David Wheeler

1:06 AM

lmfao

Paul Plummer

Yah not an easy proof

PVAL

Here's an easy proof: the real line is not compact.

Here's another easy proof $S^1$ without a point is connected.

David Wheeler

If you already know what the fundamental groups are, it's easy. Showing they have those fundamental groups is, of course, maybe a wee bit tricky. Pfft, details, gentlemen.

My comment was not about topology, it was about groups. Sure, sometimes there's complications in computing them. That's in another room, down the hall.

I think Wildberger uses the connected idea in one of his video lectures on algebraic topology.

David Wheeler

1:25 AM

ah well, I must leave now. Failure to entertain has exceeded pre-determined parameters.

pjs36

Fare thee well, your attempts were admirable

Alec Teal

2:18 AM

leo

Hi there

@AlecTeal Corn?

s

David Wheeler

2:51 AM

gee, someone doesn't know how to spell porn

pjs36

I had assumed they were in California and forgot an 'h'

Soham Chowdhury

3:12 AM

@Balarka, tell me about the uses of knowing exterior products. Also, how does one learn about diff. forms from there?

(I want to know about diff. forms (later!) because 1. they apparently allow one to define $\operatorname{d}x$ rigorously or something and 2. I once saw a cool sequence in an expository paper on the arXiv which talked about all this and I'd like to understand that sometime).

arxiv.org/abs/0910.0047

Is that sequence exact or something?

Mike Miller

Differential forms are things you integrate. The sequence is not usually exact. Its failure to be exact is measured by the de Rham Cohomology groups.

They're, pointwise, alternating bikinear functions from ($T_x M)^k$, aka elements of $\Lambda^k T_x^* M$.

Soham Chowdhury

3:28 AM

Okay.

Why would one learn about exterior products?

Seems like they allow you to generalize cross products.

Hmm, googling revealed some ultra-advanced book called "Differential Forms in Algebraic Topology". Wow.

Mike Miller

I don't understand the question. To me exterior products are something you do to vector spaces (the exterior product of $V$ and $W$). By the way; why should these be the things you integrate? Answer: they change right under change of coordinates. You'll see the details when you read about this stuff.

Soham Chowdhury

Right. No, most linear algebra books have nothing about exterior products, that's all.

Do people use that $\wedge^k$ formalism outside of lin. alg.? That's my question.

Mike Miller

I would learn about them precisely because they show up in differential geometry and homological algebra.

The thing I just wrote? Yes, that's precisely what a differential form is.

Soham Chowdhury

Oh, okay, cool.

Thanks.

Mike Miller

If one's being pedantic a differential form is a section of the bundle $\Lambda^k T* M$. But poinwise one just says it's an element of $\Lambda^k T^*_x M$.

Soham Chowdhury

3:35 AM

It makes no difference to me either way (yet), as you can guess. :P

Kaj Hansen

3:46 AM

http://math.stackexchange.com/questions/1381652/why-this-sigma-pi-sigma-1-keeps-apearing-in-my-group-theory-book-cycle

^This question deserves upvotes, IMO. It's a good question that, unfortunately, is somewhat "swept under the rug" for at least a few sections in some algebra texts.

Soham Chowdhury

6

Q: Why this $\sigma \pi \sigma^{-1}$ keeps apearing in my group theory book? (cycle decomposition)

I'm studying cycle decomposition in group theory. The exercises on my book keep saying things like: Find a permutation such that $\sigma (1 2) \sigma^{-1} = (123)$ Prove that there is no permutation $\sigma$ such that $\sigma (123) \sigma^{-1} = (456)$ Show that $\pi(x_1 x_2 \cdots x_n)\pi^{-1}...

Better.

Kaj Hansen

@SohamChowdhury, ugh, I saw wedge products in my multivariable course.

Soham Chowdhury

Didn't like them, I guess?

Kaj Hansen

Great course, no doubt. But I was not prepared whatsoever for the extent of abstraction I got.

Soham Chowdhury

Oh, that.

Kaj Hansen

3:49 AM

That was my first course after a 3 year hiatus I took from math, and up to that point the most advanced math I'd seen was plug-and-chug calculus.

Soham Chowdhury

Well, I'll have to learn multivariable properly sometime, so I figure I might as well get comfortable with the stuff.

@KajHansen ow

Kaj Hansen

You can actually watch my course for free online. They're Ted's videos advertised on his profile.

Soham Chowdhury

4:36 AM

Oh, cool.

Balarka Sen

5:50 AM

@Soham I really don't know about differential forms, so not the right person. (I'll learn those after exam). From what I understand, differential forms are multilinear maps (e.g., a 1-forms is a linear functional), and the wedge products are operations on the forms : if $df$ is a $p$-form and $dg$ is a $q-form then $df \wedge dg$ is a $(p+q)$-form.

Forms do show up in topology : you can define a cochain complex of $k$-forms on the tangent space of a manifold with boundary maps being the exterior derivative and compute it's cohomology. The resulting thing is called de Rham cohomology.

Soham Chowdhury

Yeah, that Bott-Tu book covers all those, I guess?

Balarka Sen

de Rham cohomology? Yes.

But you have to learn forms from elsewhere.

Soham Chowdhury

And Mike told me it's useful to be comfortable with exterior products because they show up in hom. alg.
I'm doing eigenstuff now (and all this seems stupidly easy for some reason!)

Balarka Sen

Classify vector spaces of dimension n over a field.

Soham Chowdhury

Uh, what do I need to know to do that?

Balarka Sen

5:54 AM

You know basis and vector spaces, not? Then you should be able to do that. If you can't, revise :)

Soham Chowdhury

Okay.

You also need linear maps, I think.

For isomorphisms.

Balarka Sen

And we'll see how easy eigenstuff really are when you do Artin's exercises.

Soham Chowdhury

Haha, okay.

But at least the ideas are easy to understand. Difficult exercises are another matter altogether.

@BalarkaSen You promised to tell me something once I managed to do the classification thing, right?

Balarka Sen

Any mathematical idea is easy. And linear algebra is basic undegrad stuff, so you need high-school background.

Just that the same thing can't be done with modules, but I have already told you about classification of modules.

(f.g. modules)

Soham Chowdhury

Uh, are all vses of dim n isomorphic?

Balarka Sen

6:00 AM

Reminder : you have a lot of exercises due. 1) every grp of idx 2 is normal 2) every f.g. abgrp can be broken up into it's torsion and nontorsion part 3) every ses with a free object at the end splits, (2 follows from 3)

@Soham prove it.

Soham Chowdhury

If there is a bijective linear map between the bases (which are of course of the same cardinality) then you can extend this to any vector by linearity.

How the hell am I supposed to do (2)?

Balarka Sen

by applying (3)

in AbGrp

Soham Chowdhury

okay.

Balarka Sen

and (1) is very easy. you're thinking too fancy.

what does being normal mean?

Soham Chowdhury

I'm not even thinking about groups right now.

Balarka Sen

6:04 AM

ok. let me know when you do, it has been months since i gave you that problem.

Soham Chowdhury

months?

okay, done.

the vs classification.

Balarka Sen

a month, yep.

ok, show me.

Soham Chowdhury

Any vs of dimension n over a field $k \cong k^n$.

proof: choose a basis.

then $v = \sum\lambda_i e_i$.

send this to $(\lambda_1, \lambda_2,\cdots,\lambda_n)$.

this is bijective and linear.

done?

I'm learning diagonalization now. Seems like an interesting idea.

BBL.

Balarka Sen

@Soham at the basis level, you're sending the unit coordinate axis basis of $F^n$ to the chosen basis of $V$, right?

that's the idea. but you have to prove it's bijective

leo

6:23 AM

And there's at least two ways of doing it

David Wheeler

Well, its injective because only ONE LC is the 0-vector. And its surjective because $(\lambda_1,\dots,\lambda_n)$ has the pre-image $\sum_i \lambda_i e_i$ (perhaps it would be better to call the basis vectors $b_i$ to indicate they are not necessarily the "standard basis").

Hello@DavidWheeler

Hello.

vice versa, in fact.

wait, what?

6:31 AM

Hey,soham

Soham Chowdhury

well, two vectors with the same expansion wrt the same basis are the same, so doesn't that imply bijectivity?

hello, @Rem.

David Wheeler

it certainly implies injectivity

Remember me

Oh you are trying to prove the classification theorem .. Just consider the map which takes basis

David Wheeler

...to basis, yes we are doing that

Remember me

That should do the job .. I guess

David Wheeler

6:35 AM

In finite-dimensional vector spaces, actually, a proof of injectivity will suffice. Rank-nullity.

Remember me

Yes .

@DavidW What kind of maths have you been doing?

David Wheeler

So, up to isomorphism, there is only one vector space over a given field, of dimension $n$.

I don't "do" math. I think about it, sometimes, but only because alcohol is expensive.

Remember me

Thats a weird excuse to do maths :p

@DavidW you have done group theory right?

David Wheeler

Hmm...I remember something vague about Norwegians and automorphisms.

Remember me

I have a very simple question though ,

How do groups and rings look like , Without using Cayley graphs and lattices , Is there a geometrical picture for them?

David Wheeler

6:42 AM

That question is rather like asking: what do letters sound like?

Remember me

So is there no geometrical image for them?

David Wheeler

Groups, strictly speaking, are not geometrical objects. Sometimes, you can visualize a group by how it acts on a geometric object, but that is not quite the same thing.

Remember me

I can think of a group as a topological space and then I can think of that geometrically , Cant I

David Wheeler

Well, let me give an example: sure, you can visualize $\Bbb Z$ (under vector addition, let's say) as a subgroup/subset of $\Bbb R$, and give it the relative (discrete) topology. But that won't tell you that its abelian, or cyclic.

Remember me

Ahh .. true

David Wheeler

6:50 AM

I suppose I tend to think of groups as quotients of free groups: that is-strings of letters with rules.

Remember me

Well I know construction of quotient spaces in topology .. but not in groups , though both construction might be similar

David Wheeler

Not even close.

They share one thing in common: they both have a theme of "identification".

Remember me

So we dont think of quotient spaces by identifying the space and then constructing a map

Oh identifying is there

David Wheeler

But in topology, you can identify stuff almost willy-nilly. Groups are constrained by their multiplication.

Remember me

Constrained ? What do you mean by that?

David Wheeler

6:53 AM

Put another way-functions can be a bit more "free" than homomorphisms, which have stricter rules.

Remember me

Oh.

David Wheeler

continuity constrains functions "somewhat", but if you relax the topology enough, you can make any function continuous.

you can't turn any function into a homomorphism.

Remember me

Oh, So instead of continuous functions we use homomorphisms in quotient spaces in groups ...

David Wheeler

For example, in any topology, a constant function is continuous. But the only constant functions in groups which are homomorphisms must map to the group identity of the target group.

Remember me

Yes I get that

David Wheeler

6:58 AM

There are similarities-in topology a quotient map is usually taken to be surjective, and in groups, quotient maps are the same as surjective homomorphisms.

Remember me

Well then whats the problem with isomorphisms ? I mean to say that why cannot we define quotient spaces in groups by isomorphisms . They are also homomorphisms with a little more strict constraints on it

David Wheeler

A group's multiplication introduces a certain "uniformity" to a group-in sets, or topological spaces, we can have subsets, or subspaces of varying "sizes".

With groups it's different, a subgroup always evenly divides the size of its parent group.

Isomorphisms are more restrictive than surjective maps, true. But the quotient maps so obtained are not very interesting.

Remember me

Oh.

David Wheeler

An analogy would be this: imagine the identity function on a topological space. This is a quotient map, as well, but the quotient topology so obtained isn't any different than our original one.

Soham Chowdhury

David Wheeler

7:05 AM

So it's a very uninteresting quotient map

Soham Chowdhury

Isn't $\operatorname{im} \hat A$ a subspace for any linear operator?

David Wheeler

@SohamChowdhury yes, that is a definition.

Remember me

That is why we dont think it in terms of an isomorphisms

Samuel Yusim

try \hat{A}

Soham Chowdhury

phew. thanks.

so why are projectors specifically useful for this?

Remember me

7:07 AM

@SohamChowdhury Hmm .. I didnt know that subspaces can be defined using special linear operators .. Nice!!

David Wheeler

@SohamChowdhury yes, but...for an arbitrary $\hat{A}$ there is no guarantee that $\text{im }\hat{A}$ is invariant under $\hat{A}$

Soham Chowdhury

is it because you also get some more information about the subspace (invariance under the projector)?

ah, okay.

@Rememberme the image of any linear operator is a subspace.

David Wheeler

The idea of a projection (which is what they are usually called, but terms can vary) is based on maps like this:

Remember me

Yes I got that .. Just didnt think it at the time I was doing Lin. operators @Soham Because then I did not know what subspaces were

David Wheeler

$P: \Bbb R^2 \to \Bbb R^2$ given by: $P(x,y) = (x,0)$

Remember me

7:10 AM

projection

David Wheeler

With projections, we have a nice decomposition: $V = \text{ker }P \oplus \text{im }P$

Soham Chowdhury

When @Rem said "projection" out of nowhere, I was reminded of this:

David Wheeler

@Rememberme it turns out that quotient maps that are not injective are more useful-they allow us to "filter out" informtion.

Soham Chowdhury

David Wheeler

The most powerful of think, lol

Soham Chowdhury

7:15 AM

@DavidWheeler sort of like how it's stupid to quotient by the trivial group?

David Wheeler

it's not "stupid", it's just "boring". Sometimes, it's one of the few choices available.

Soham Chowdhury

Well, it does nothing, that's all.

There's a whole subreddit dedicated to these.

:P

Link, if you're interested.

David Wheeler

It does "something"-we have to put curly brackets around group elements, then. Not very exciting, though.

Soham Chowdhury

Yep.

David Wheeler

The basic idea behind a quotient group is a congruence. It turns out that the equivalence class of the identity in such a congruence has a special structure.

Remember me

7:21 AM

@Soham you have done quotient spaces in topology?

David Wheeler

Fun fact about vector spaces-they are merely ring-homomorphisms from a field into the endomorphism ring of an abelian group.

Remember me

Many undefined and unknown terms for me there :p^^

David Wheeler

Most axiomatic presentations on vector spaces have 8-10 axioms, or so.

The first 4 (sometimes 5) or so basically state $(V,+)$ is an abelian group.

The rest typically deal with the scalar multiplication.

Now, most texts present this as a map: $F\times V \to V$

Remember me

@David Do we have unsolved problems in group theory? (just out of curiosity)

David Wheeler

But what you can say instead, is: given $a \in F$, that $a$ induces a map $V \to V$ given by $v \mapsto av$.

To say this map is an abelian group endomorphism means that $a(u+v) = au + av$.

To say that $a \mapsto a\cdot(-)$ is a ring homomorphism says that $(a+b)v = av + bv$ and $(ab)v = a(bv)$

And as a consequence $1v = v$.

@Rememberme There are many unsolved problems in group theory-even in finite groups.

Remember me

7:30 AM

So ring homomorphisms have nothing to do with rings ?? According to that map I cannot see a ring anywhere

David Wheeler

Of course, all the easy problems have been solved already.

@Rememberme Fields are rings.

Remember me

Sorry , I have not done ring theory yet so i dont have an idea

David Wheeler

The abelian group homomorphisms $A \to A$ (where the operation of $A$ is written as +) also form a ring, if we define $(h_1+h_2)(a) = h_1(a) + h_2(a)$, and define $h_1h_2 = h_1 \circ h_2$

These are called endomorphisms of $A$.

Remember me

endomorphisms .. I have heard those being used in altop

Though I might be wrong

David Wheeler

Typically "endo-something" means from a thing to itself.

Remember me

7:34 AM

Just like linear operators

David Wheeler

And "auto-something" is similar, but invertible.

Some structures use different words, mathematically terminology isn't always uniform.

For example, for a finite set, a set-automorphism is called a permutation.

whereas a set-endomorphism is usually just called a transformation.

Remember me

Oh. I have learnt about permutations in symmetric groups

David Wheeler

The "full symmetric group" on a set is its group of set-automorphisms.

anon

7:50 AM

@KajHansen, @SohamChowdhury I've written an answer to that question about conjugation, using three examples to illustrate my point: permutations, matrices and loops.

David Wheeler

which question?

you are Whacka, as well? confuzzled.

anon

yes

Jul 23 '14 at 11:09, by anon

I will farm an army of 10k+ers and take over as supreme leader.

David Wheeler

dammit, Jim, I'm a doctor, not a mind-reader!

Remember me

Well two elements of a group $G$ a,b are called conjugate if
$gag^{-1}=b$ , for some g in G
Am i right?

anon

yes

Remember me

7:59 AM

So what do we call b ? conjugate of a?

David Wheeler

yes

Remember me

Okay

David Wheeler

more correctly, conjugate of $a$ by $g$.

Remember me

So @anon You will give each account of yours a single topic in MSE? .. As in one will take care of abstract algebra , other topology, etc

anon

naw

Remember me

8:03 AM

Wont that be amazing .. As in every account will have a gold in the particular topic and you will get all golds for every topic :p

David Wheeler

Some texts write that as ${}^ga$ because of the nifty rules ${}^h({}^ga) = {}^{hg}a$ and $({}^ga)({}^gb) = {}^g(ab)$

anon

yes

Remember me

These are just exponent rules which I had learned in 6th grade

David Wheeler

Similar "form", different content

Remember me

Yes

Hello@r9m

David Wheeler

8:09 AM

the generic "form" of conjugation is something I use on a daily basis-like this: suppose I have a problem, but the arena it's in is difficult to work in. I then "translate" it to an area that's easier to work in. Then I solve my problem the "easy way", and translate back.

For example, I might have to calculate a measurement that is complicated. So I might make a faithful drawing of the measurement scenario I need in a CAD program, and then make a dimensional inquiry on the CAD program. Then I input the numerical measurement I needed in the original problem.

Remember me

Hello @Balarka

Balarka Sen

One way to visualize discrete groups is by looking at spaces on which it acts nicely.

Cayley graphs are just examples of those.

David Wheeler

hello balarkamanujan

Balarka Sen

hi David.

/Deveno

Soham Chowdhury

does anyone consider fields with a point at infinity added?

David Wheeler

8:14 AM

@SohamChowdhury in what context?

Balarka Sen

huh? what's your motivation for doing that?

Soham Chowdhury

like $\Bbb R \cup \{\infty\}$.

anon

sometimes fields are also topological spaces, and sometimes we compactify a topological space using a single point, and sometimes geometrically this point looks like it's "at infinity"

Soham Chowdhury

oh, ok.

Balarka Sen

Fields don't have a topology in general

David Wheeler

8:15 AM

I mean, sure, there is the notion of en.wikipedia.org/wiki/Extended_real_number_line

Soham Chowdhury

yeah, like that.

anon

extending R using two infinities is useful for describing limits and asymptotics, but diverges from the idea in complex analysis of adjoining just a single infinity (which handles mobius transformations better, and converts meromorphic functions X->plane into holomorphic functions X->Riemann sphere)

Balarka Sen

If you have a topology, you can compactify pretty much anything.

Soham Chowdhury

> diverges

heh

David Wheeler

you can also extend the real numbers by identifying $-\infty$ and $+\infty$

aka the real projective line

Balarka Sen

8:19 AM

@anon nice fact - the $\ell$-adic reps attached to two ell curves coming from action on the Tate modules are same iff the curves are isogenous.

Remember me

0

Q: The topology of $\mathbb{Z}_p$

I don't know much about topology, but anyway... Assuming $\displaystyle\prod{A_n} =\prod_{n\geq 1}{A_n}$, why is $\mathbb{Z}_p$ closed in a product of compact spaces? Googling I found Tychonoff's theorem on ProofWiki, so I see why the product is compact.

general-topology p-adic-number-theory proof-explanation

What is $\Bbb{Z}_p$ here

Balarka Sen

p adic numbers

Remember me

I know what p-adic metric is .. But what are p-adic numbers

Balarka Sen

Cmetric completion of Z under p adic metric

*metric

Remember me

WHat is metric completion .. Is it related to a complete metric space

Balarka Sen

8:23 AM

Yes. Google it.

I gotta go.

Remember me

Okay got it it is a pair of a complete metric space with an isometry $\phi$ on it such that $\phi(X)$ is dense in the complete metric space
Where X is the metric space

Okay I also gotta go

Soham Chowdhury

8:57 AM

Ey, Balarka. dvt.name/2015/…

skull patrol

Hi pal

Soham Chowdhury

You might understand it. I don't really have the time to look at it now. :P

Hey, @skull.

Remember me

10:38 AM

Ahh.. So that is what you mean when you say $\lambda$ calculus @Soham

Soham Chowdhury

Indeed, yes.

Remember me

11:08 AM

Done with the proof of the classification theorem?@SohamChowdhury

Soham Chowdhury

11:26 AM

I'd think so.

Okay, little lin. alg. question. Consider a basis $\{e_i\}_1^n$ of a space $V$.

We have functions $e_i^{\ast}$ which "extract" the coefficient of a vector corresponding to the basis element $e_i$.

Balarka Sen

@SohamChowdhury You mean those are the corresponding basis for $V^*$?

Soham Chowdhury

Yes.

Balarka Sen

OK, so what's the question?

I can't open that link of yours, btw.

Soham Chowdhury

Now the book says that changing any element in the basis of $V$ must also change every element in the basis of $V^{\ast}$.

Balarka Sen

What does "changing" mean?

Soham Chowdhury

11:30 AM

For example, consider $\{\hat i,2\hat j\}$.

This is a different basis from the standard one.

Right?

Balarka Sen

what is \hat{i}?

Soham Chowdhury

Basis elements of $\Bbb R^2$, sorry.

Balarka Sen

just write $e_1$ and $e_2$.

Soham Chowdhury

okay.

Balarka Sen

go on.

Soham Chowdhury

11:33 AM

But this doesn't change all the basis elements of the dual space: $e_1^\ast \mapsto e_1^\ast, e_2^\ast\mapsto \frac12 e_2^\ast$.

Huy

@SohamChowdhury: Can you copy the exact statement given in the book?

Balarka Sen

$e^1 : \Bbb R^2 \to \Bbb R$ sends $e_1$ to $1$ and $e_2$ to $0$. If $e_2$ is changed, surely $e^1$ hasn't changed.

Soham Chowdhury

Balarka Sen

So the book must be saying something different.

Soham Chowdhury

I think it should say "may change".

Balarka Sen

11:36 AM

yep.

I think so too.

Soham Chowdhury

for instance, changing the basis of $V$ to $\{e_1, e_1+e_2\}$ will change everything, right?

Balarka Sen

yes.

Soham Chowdhury

thank goodness. I was very confus.

Huy

@SohamChowdhury: Is $e_i^*(e_j) = \delta_{ij}$ supposed to hold?

Balarka Sen

that's how the dual basis are defined.

*dual basis for the standard coordinate basis

@Soham What is that book, btw?

Huy

11:39 AM

Maybe what's meant is that in order to find a single $e_i^*$ you have to consider the whole basis $(e_k)_k$ and it doesn't suffice to only look at $e_i$ itself. However rather badly formulated if that's what it means.

Balarka Sen

yeah, that makes sense. but I agree it's badly formulated.

Huy

This is my interpretation of "In other words, $e_1^*$ is not a result of some "star" operation applied only to $e_1$".

Danu

Yeah, I agree that this should be the interpretation.

Huy

@Danu: What are you doing here? ö.ö

Balarka Sen

The point is that linear functions $V \to \Bbb R$ are determined by what it does to a give basis $B$ of $V$. So just knowing the value of a single element in the basis won't suffice.

(this is meant for Soham)

Danu

11:42 AM

@Huy Math chat is always so hostile haha

Huy

@Danu: Just wondering, because you're usually here when you have a question. Same is true for me about physics chat. :P

@Danu: I actually find physics chat a lot more hostile. Last time I checked there were some guys debating about how ill-defined QM was. One of them basically kept saying "it isn't" and the other said "it is".

Danu

@Huy Haha, really? I meant on a personal level... The "hard core" of physics chat always treat each other very friendly

Here, I really feel I annoy people when I ask a question.

Huy

@Danu: Well there are some people in here who take things rather personally from time to time but I hope you do realize when I greet you with a "what are you doing here" it's nothing personal but more of a joke ;)

Balarka Sen

Exactly how I felt when I asked questions in the homotopy theory chat. I don't think the annoying-them-feeling is chat topic-dependent.

Huy

@Danu: If you were annoying people wouldn't talk to you, at least that's what I think.

Danu

11:47 AM

@Huy ...which is what happens :P most of the time, in any case.

Huy

@Danu: I'm talking to you!

Danu

@Huy Note my second sentence :)

Huy

@Danu: I asked a question in physics chat the other day and didn't receive a satisfying answer. I'm sure you would be able to give an elementary explanation?

Danu

@Huy maybe

Huy

@Danu: Why do we have $R_{ijkl} = g_{ik} g_{jl} - g_{il} g_{jk}$ in $S^n$? I have as an example the computation of sectional curvature of $S^n$ and it starts with this formula, but I've never seen it before. Is this a special case of the more general formula for $R_{ijk}^l$ expressed by Christoffel symbols which simplifies on $S^n$?

Danu

11:49 AM

Oh, I saw you asking that one

Lemme finish my chess game real quick (few minutes)

Huy

@Danu: Sure. I got an answer saying it's because $S^n$ is maximally symmetric and I can find out this formula using Killing vectors. Unfortunately I hardly know anything about those two things.

@SohamChowdhury: Which book are you using anyways?

Danu

@Huy Okay, well the answer you got is the correct one

Huy

@Danu: I'm sure it is but unfortunately it doesn't help me at all. :(

Danu

For a derivation see e.g. page 140-141 of Carroll

Huy

The book or his lecture notes?

Danu

11:56 AM

Book

Anyways, I can outline the argument for you if you want

It's really quite easy

Huy

What chapter would this be in? It doesn't appear to be on springerlink and I could try finding it in the lecture notes.

Sure.

Danu

So, first we should define what it means to maximally symmetric, I guess

We can just define it as a spacetime with $\frac{1}{2} n(n+1)$ Killing vectors, or "symmetries" (loosely speaking), which is the same number as flat space.

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