Mathematics - 2015-08-04 (page 2 of 3)

Huy

08:20

@anon: Is there a notion of bounded sets in non-metrizable topologies?

anon

being bounded is not a topological property. for instance (0,1) is a bounded metric space while R isn't, even though they are topologically indistinguishable (i.e. homeomorphic)

Tobias Kildetoft

@Huy Any metric space can be made bounded by defining a new "bounded" metric, which is topologically equivalent.

anon

there is an analogue of closed and bounded - compactness

Huy

That I know.

anon

(I say analogue, not generalization.)

Tobias Kildetoft

08:23

@anon Even that is not quite the correct analogue, as it is not generally the same in any metric space

anon

indeed

Tobias Kildetoft

though it is probably the closest one can get

Huy

I'm just going over different diffgeo notes and one reminded me what "bounded" actually means, I somehow completely forgot because it was geometrically clear.

anon

any discrete metric space is closed and bounded, but compact + discrete iff finite. so closed+bounded needn't be equivalent to compact for metric spaces. but heine-borel applies for manifolds though, which is usually our intuition for spaces

I always confuse heine-borel and bolzano-weierstrass, can never remember which is which

Huy

I always remember Heine-Borel but I forget what Bolzano-Weierstrass actually says.

Ionică Bizău

08:52

Hello there!

I posted a question on SO that requires some math I guess.. Maybe there are some experts around:

6

Q: Create svg arcs between two points

I want to connect two SVG points (e.g. the centers of two circles) using arcs. If there is only one connection, the line (<path>) will be straight. If there are two connections, both will be rounded and will be symmetrical, this way: So, in fact, there are few rules: Everything should be sym...

Help is always appreciated. :)

Chris's sis the artist

09:30

I think I have great news! I just developed a new proof to the Euler's sine product that I consider to be far easier than anything I saw so far (and it's rigorous, not a guess).

Ionică Bizău

@Chris'ssistheartist Hey, I'm Romanian too! :)

3

Chris's sis the artist

@IonicăBizău Hi. Glad to see Romanians around. :-)

Ionică Bizău

Me too! :)

Huy

12:15

@DanielFischer: Is it correct that only left-translation is a smooth action from a Lie group on itself and right-translation isn't?

Tobias Kildetoft

@Huy Well, the thing that fails is being an (left) action, not smoothness

Huy

@TobiasKildetoft: I didn't say smoothness fails? :P

Tobias Kildetoft

@Huy Right, but you did say smooth, while this has nothing to do with the Lie structure at all

just a general fact about groups

well, non-abelian ones

Huy

@TobiasKildetoft: I've only come across the structure of Lie groups in the setting of smooth manifolds, that's why.

Tobias Kildetoft

@Huy Sure, Lie groups are smooth manifolds, but the failure here is not related to that

Huy

12:19

@TobiasKildetoft: I know, but that $R_a \circ R_b \neq R_{ab}$.

Tobias Kildetoft

@Huy Right, that is true in general for non-commuting $a$ and $b$ (since it equals $R_{ba}$).

Huy

@TobiasKildetoft: Should it be obvious that the two translations are orientation preserving?

Tobias Kildetoft

@Huy I am not really familiar with oriented manifolds, so no idea

Balarka Sen

hi @iwriteonbananas

iwriteonbananas

oh hi

what's happening?

Balarka Sen

12:29

got some exam tomorrow for which i am pretty well-prepared, thus chatting

iwriteonbananas

exact same here lol

dunno what else i could do to prepare

Balarka Sen

wanna do a new kind of problem?

iwriteonbananas

on the other hand, i also have an exam on thursday for which i could prepare some more (but it's not math so i dont care about it)

sure

Huy

@BalarkaSen: Do you see how to quickly prove that $\operatorname{Ad}: S^3 \to SO(3)$ is surjective?

Balarka Sen

prove or disprove : every self-homotopy equivalence of a space $X$ is homotopic to a self-homeomorphism of $X$.

iwriteonbananas

12:33

hmm ok

im guessing it's wrong

Balarka Sen

@Huy what is that map?

@iwriteonbananas correct guess. prove it.

Huy

@BalarkaSen: The adjoint map, $u \mapsto (v \mapsto uv\bar{u})$.

Balarka Sen

eh, no, not off the top of my head.

@PVAL interesting. i don't understand the last bit about why $f^{-1}([a, \infty])$ is a handlebody, though. (dunno anything about morse theory other than the statement of the main theorems)

iwriteonbananas

@BalarkaSen hmm im unable to construct a counter example

Balarka Sen

[whistles] aspherical spaces [/whistles]

Mike Miller

12:38

@Huy: Here's the easiest way: show that it's a homomorphism, and then show that it maps to all rotations along a plane.

Huy

@MikeMiller: I've shown that it's a homomorphism. Is it obvious that left/right-translation are orientation-preserving? I don't see it immediately for the adjoint.

Mike Miller

@iwriteonbananas: That's fine, neither was Balarka when he first asked about this.

iwriteonbananas

ok i've not heard of aspherical spaces :P

Balarka Sen

i mean eilenberg maclane spaces, @iwriteonbananas. K(G, 1)s.

i came up with the idea that aspherical spaces are gonna work later on though, Mike. :) but yeah, it's a hard problem, i agree.

Huy

@MikeMiller: Even if I see that it's orientation preserving, I don't see how it maps to all rotations.

Mike Miller

12:41

@Huy: Write down an explicit example.

Huy

How trivial is it allowed to be to still demonstrate what you're trying to tell me?

Wait, I think I've got it, @MikeMiller.

Mike Miller

I can't parse that sentence. You can write it out explicitly for rotations along some plane - or more helfpully, perpendicular to some axis.

@Huy: Here's an easy way to show that it's always orientation-preserving: the adjoint map is continuous!

iwriteonbananas

@BalarkaSen ok, how about $\Bbb{R}P^{\infty}$ then? (im just guessing :P)

Balarka Sen

you have to come up with a self-htpy equiv that isn't htpic to a self-hmeo, not just a space.

Mike Miller

I don't think he's asked you something fair, because you don't have the tools you want.

Huy

12:49

@MikeMiller: I realized that you said "show it's a homomorphism" for a reason. I think the idea is to get a "basis" for all of $SO(3)$, the rotations around the three axes and then because it's a homomorphism we can get any other rotation by composition.

iwriteonbananas

yes, this seems beyond what im capable right now

Balarka Sen

? @iwriteonbananas doesn't know that any self-homeo of K(G, 1) comes from an aut of G?

it's in hatcher 1.A

iwriteonbananas

i had to look up what a K(G,1) is :/

Mike Miller

He doesn't know homotopy classes of maps K(G,1) -> K(G',1) correspond to homomorphisms G -> G'. You also didn't.

iwriteonbananas

@BalarkaSen 1.B by the way

Balarka Sen

12:51

ok, i didn't realize @iwriteonbananas didn't know that.

sorry.

iwriteonbananas

looks at balarka in disgust

@BalarkaSen no worries lol

Balarka Sen

@MikeMiller i did when i worked the problem out. i didn't, say, about a week ago. you were the one who told me, yes.

well, didn't really work it out - only suggested something with which you worked it out.

it was a fun little problem.

@iwriteonbananas better problem -- assuming the fact that every closed odd dimensional manifold has euler charecteristic 0 (which i don't know how to prove), prove or disprove : there is a 5-manifold s.t. $\Bbb CP^2$ appears as a boundary of that manifold. (this is tricky)

iwriteonbananas

T__T

where did you get that problem from?

Balarka Sen

came up by myself (with the help of Mike and prof) while fiddling with cobordisms.

iwriteonbananas

nice

Balarka Sen

13:00

you know the relevant fact which can be used to solve this exercise this time. but as i said, it's tricky to figure out how it can be used.

Huy

@MikeMiller: I have a statement that for compactly contained $U \subset \subset M$ there exists a $\delta > 0$ such that the local flow is defined on $U \times (-\delta,\delta)$. I assume this is just applying the open cover definition or is there more to this statement?

iwriteonbananas

@BalarkaSen ok, i dont see right now what to do (i'm trying to disprove it)

Balarka Sen

you're on the right track, as it is indeed false. try a proof by contradiction.

iwriteonbananas

i also dont have time right now, but i'd be interested in talking about the problem later

Balarka Sen

and don't forget about the fact, it serves as a hint

@iwriteonbananas let me know when you disprove it.

iwriteonbananas

13:07

you never said what fact you are referring to

Balarka Sen

"assuming the fact that every closed odd dimensional manifold has euler charecteristic 0"

iwriteonbananas

oh right

by the way, you meant to write "closed 5-manifold" right?

Mike Miller

@Huy: Yeah, this is just: you know that the function $f$ (such that $f(x)$ is how long it's defined) is continuous and nonzero, so because $U$ is compact, it has a minimum, $\delta$.

Balarka Sen

@iwriteonbananas not just 5. it works for any odd dimensional manifold.

but in particular, it works for 5.

Huy

Ah.

iwriteonbananas

13:08

@BalarkaSen i mean at the bottom of this message

Mike Miller

@iwriteonbananas: He meant to say show it appears as the boundary of a compact 5-manifold. Certainly not closed.

Balarka Sen

closed 5-manifolds don't have boundaries... but yeah, i meant compact.

iwriteonbananas

right, for some reason i was automatically assuming that he meant compact

i gotta go now, talk to you later

Balarka Sen

bye.

i have to run too.

PVAL

14:43

@MikeMiller There's non-connected examples which are trivial and a simple connected example which just requires the basics of the fundamental group (nothing about Eilenberg-Maclane spaces). Take the space $S^1$ wedge an annulus.

Simple is quantifier for how complicated the example is (i dont mean simply connected).

Szabolcs

14:54

Network motifs are (usually directed) induced subgraphs. In principle it should be possible to express the number of each 3-vertex subgraph as linear functions of A, A^2 and A^3 where A is the adjacency matrix.

Does anyone know of a reference for this to spare me the calculations?

Mike Miller

@PVAL: Fair enough.

Soham Chowdhury

15:39

@TedShifrin computations are useful, but multiplying pairs of complex numbers gets old soon. :P

Remember me

True @Soham

Hello, anyways @Soham

Can someone here explain me why do we care about "Monster Groups"

Huy

15:56

@Rememberme: It's one of the 26 sporadic simple groups and thus is part of the complete classification of all finite simple groups?

Remember me

sporadic? .. Can you describe that a bit ?@Huy

Huy

@Rememberme: You can classify all finite simple groups: There's cyclic groups with prime order, alternating groups of degree $\geq 5$, groups of Lie type and then 26 other groups which are called sporadic groups. One of those 26 is the Monster group.

Remember me

Oh .. So these groups can be thought as building blocks of other groups?

Huy

??

Remember me

nvm

Huy

16:01

I think you're reading wiki sentences out of context.

Remember me

No I don't have a wiki page open...

Huy

Ok.

Remember me

I remember of monster groups from this page @vzn once linked.. It said monster groups and j functions @Huy

Huy

Sorry I don't see and check every link ever posted in chat.

Remember me

Not in this chat .. I guess in the number theory one.. Wait let me try finding it

Huy

16:03

We actually have a seminar coming up about this this fall: physics.rutgers.edu/pages/keller/HS15-MonstrousMoonshine.pdf

Remember me

Yes that page was also titled Monstrous Moonshine

@Huy Seems fascinating

Huy

Not really my cup of tea.

Remember me

There we go @Huy :
https://www.quantamagazine.org/20150312-mathematicians-chase-moonshines-shadow/

Huy

@Rememberme: Cool!

Remember me

It got me hooked into algebra badly :p

Huy

16:11

For me it was mostly Abel-Ruffini.

Remember me

@Huy You understand the proof of
The classification theorem of finite simple groups?

Huy

@Rememberme: We didn't do it in the abstract algebra course and I never read up about it myself.

Remember me

Oh.

PVAL

@Rememberme No one understands the proof of the classification of finite simple groups.

apnorton

17:00

I'm back... ish. Anyway, I was really tempted to cast the final vote to close this question as "too broad" (people have literally written books on the subject), but the user just put a bounty on it. I need someone else to tell me I'm not being a monster by closing it... :P

PVAL

You can't close questions with a bounty (unless you are a moderator or super special) for some expletive deleted reason.

apnorton

... oh. Just confirmed. I guess I'll wait 6 days.

Mike Miller

That's a clever trick: wait til you have 4 close votes, add a bounty, close votes expire...

apnorton

I think I'm going to write a "kill this question" script... give the script a question, then it makes sure it dies.

(If your vote expires, it votes again. If it gets reopened, it votes to close again. When it is no longer "on-hold," it votes to delete. :P )

Andrew Thompson

Just to be sure, contraction of a tensor of type $\binom{k}{l}$ to one of type $\binom{k-1}{l-1}$ is simply sending $(\omega^1 \otimes \cdots \otimes \omega^l \otimes X_1 \otimes \cdots \otimes X_k)$ to $\omega^l(X_k)(\omega ^1 \otimes \cdots \otimes \omega^{l-1} \otimes X_1 \otimes \cdots \otimes X_{k-1})$, right?

Pedro Tamaroff

17:15

@AndrewThompson Never heard of that.

Andrew Thompson

Tensor contraction

In multilinear algebra, a tensor contraction is an operation on one or more tensors that arises from the natural pairing of a finite-dimensional vector space and its dual. In components, it is expressed as a sum of products of scalar components of the tensor(s) caused by applying the summation convention to a pair of dummy indices which are bound to each other in an expression. The contraction of a single mixed tensor occurs when a pair of literal indices (one a subscript, the other a superscript) of the tensor are set equal to each other and summed over. In the Einstein notation this summation...

Mike Miller

You should be able to contract by any choice of $\omega^i$ and $X_j$.

Andrew Thompson

Ah, yes, the book did mention that, in particular for weird choiced of indicing.

Mike Miller

I say "should" instead of "can" only because I haven't thought about this in a while and don't have Kobayashi next to me to check. But I think it's "can".

Andrew Thompson

I see, thanks again!

Mike Miller

17:18

What are you reading?

Andrew Thompson

Lee's Riemannian manifolds.

Read through chapter 2. Skimming a bit further it seems to get hard quickly.

Mike Miller

I see.

PVAL

If you post a question on this site about that book(or his other books), Lee will likely answer it.

Andrew Thompson

Ohhh, that's cool. I'll do so once I have a bigger question.

Owatch

17:39

Hello.

skull patrol

Hi pal

Soham Chowdhury

@Huy Same here! internet high five

skull patrol

Being ignored is like a internet low five

Vibhav Pant

17:59

amazon.com/Million-Random-Digits-Normal-Deviates/…

Huy

@SohamChowdhury: Same here about what?

Balarka Sen

hi @SohamChowdhury

skull patrol

2 hours ago, by Huy

For me it was mostly Abel-Ruffini.

Huy

Ah, right.

Chris's sis the artist

The production of problems and solutions is simply amazing here these days. Hope things will go on like that!

Huy

18:05

I'll never get used to this response system.

skull patrol

:-)

Huy

@PVAL: Is he on MSE? What's his nick?

PVAL

Jack Lee

Huy

Ah, I see it now. I was searching for John.

Balarka Sen

@Rememberme I am not sure from where you get the motivation for studying about finite simple groups. I found them truly boring when I started algebra. The point is that to each group, there is an associated thing called composition series. These are unique, in a sense (Jordan-Hölder). The factors in a composition series, by defn, are simple groups. So once you know what the finite simple groups are, you can build your finite group as you like.

moonnoon

18:16

hi, is it ok to ask mathematical questions here on chat? (My question is a pretty easy one , but I cannot figure out one thing about integrals. I dont want to make it a question since it is really really basic.)

Pedro Tamaroff

@moonnoon Yes.

moonnoon

My question is: why do I have to have a < 1 in this example? wolframalpha.com/input/?i=integral+of+1%2Fx%5Ea+from+0+to+1

Pedro Tamaroff

Because the integrals fails to exist elsewhen.

Soham Chowdhury

@BalarkaSen well, if people can like chemistry, they can like finite simple groups . . . /s

Huy

How is chemistry not likable?

Pedro Tamaroff

18:20

@Huy People losing eyebrows and stuff.

You don't want to walk around with drawn eyebrows.

Huy

I wouldn't do experiments if I was doing chemistry.

Balarka Sen

lol

Soham Chowdhury

@Huy It is! I was just joking. :)

Balarka Sen

hi @Pedro.

Soham Chowdhury

Is that Hölder the same guy from the inequality? And the Jordan the same guy from the theorems?

Balarka Sen

18:21

yes, I think

Owatch

$\int{\frac{-3}{x+1}}dx$, is $-3ln| x+1| + C$ or $-3log|x+1| + C$ ?

Soham Chowdhury

It scares me how versatile people used to be.

Huy

Look at Gauss and Euler.

They show up everywhere.

Balarka Sen

Or Bernoulli.

Owatch

ln, or log?

skull patrol

18:25

What base?

Huy

Always log.

Owatch

@Huy The examples I am looking at have their answers in terms of ln

dawsoncollege.qc.ca/public/72b18975-8251-444e-8af8-224b7df11fb7/…

Check it out.

Huy

It's also written in Word.

Mike Miller

@Owatch Log is how you write Ln when you realize you don't care much about other bases.

Balarka Sen

@Huy except for analytic number theorists, to whom its loglog

Huy

18:27

@MikeMiller: Because you know I'm all about the base 'bout the base

Owatch

I am aware that Ln is a special case of Log with base e. But I wasn't sure what to include when integrating the above example.

Sometimes I would see log, and ln

Depending on where I got my problems

Mike Miller

Unless you're talking to a computer scientist, those are synonymous.

skull patrol

Or chemistry

Owatch

Okay then.

moonnoon

@PedroTamaroff, thank you. However I still dont understand it. If you could give me another hint I would be grateful :)

Mike Miller

18:29

Computer scientists sometimes write Log for the base-10 log and lg for the base-2 log. But they're not to be trusted.

Pedro Tamaroff

@moonnoon Wolfram is interpreting that as a line integral through the real segment $[0,1]$.

Owatch

I will keep an eye out for such deviants then.

Pedro Tamaroff

So it's just evaluating, say if $a=s+iy$, the integral $$\int_0^1 x^{-s}dx$$

This converges if and only if $s<1$.

moonnoon

@PedroTamaroff, I think I get it - this integral does not converge for $a>1$ because the range is from 0 to 1.

@PedroTamaroff, thank you :)

Pedro Tamaroff

@moonnoon By that reasoning, does the integral from 1/2 to 1 converge?

Gato

19:06

@Chris'ssistheartist did you see that : math.stackexchange.com/questions/364452/… =D

JC574

aayy

Owatch

19:20

lmao

JC574

19:31

ayy lmao

Owatch

oaml

Huy

Back to reddit you two.

Daniel Fischer

@Huy Cruel and unusual punishment.

PVAL

@DanielFischer Did you read my solution here math.stackexchange.com/questions/1384379/… (you commented on every post on that thread besides my answer)?

Moses

hi

Daniel Fischer

19:42

@PVAL Not yet.

Moses

Hi @DanielFischer

Daniel Fischer

@Moses Hi.

Pedro Tamaroff

19:57

Daniel.

I know that for every $|c|\leqslant \infty$, $P(\limsup S_n=c)$ has either probability $0$ or $1$. It is clear no two events can have probability one simultaneously. How can I show that at least one has probability one?

Here $S_n=X_1+\cdots+X_n$ where the $X_i$ are i.i.d. and take values on $\Bbb R^d$, say.

I have to go, drats.

Chris's sis the artist

@Gato I never worked on that integral.

Pedro Tamaroff

If you have an answer, let me know.

I was meaning to ask this to @saz.

I know the claim follows from the Hewitt-Savage 0,1 law.

Moses

@DanielFischer How would you interpret the notation $\lim\limits_{n,m \to \infty} \sum\limits_{k = M + 1}^{N} \| b \|^{k}$? Is this equivalent to $\lim\limits_{N \to \infty}(\lim\limits_{M \to \infty} \sum\limits_{k = M + 1}^{N} \| b \|^{k}) = \lim\limits_{N \to \infty} \sum_{k = 0}^{N}\| b \|^{k} - \lim\limits_{M \to \infty}\sum^{M + 1}_{k=0}\| b \|^{k}$?

Chris's sis the artist

@RandomVariable Why did you delete your answer to one of my questions? You shoud let your approach there I have no problem you used complex analysis despite the fact that I asked for real analysis ways. For me it's OK.

Daniel Fischer

@Moses For all $\varepsilon > 0$ there is a $K$ such that for all $K \leqslant M < N$ $$\Biggl\lvert\sum_{k = M+1}^N \lVert b\rVert^k - L\Biggr\rvert < \varepsilon.$$

Where $L$ is the limit.

Random Variable

20:04

@Chris'ssistheartist Because I convinced myself it was Parseval's identity in disguise. And there was already an answer that used Parseval's identity.

Moses

@DanielFischer So does what I wrote not follow?

Daniel Fischer

@Moses Well, it does, but not directly. If the limit exists, it is $0$. Then the things you wrote also exist and are equal to $0$. But that's not what the $\lim\limits_{N,M\to\infty}$ is defined as [note: the restriction $M < N$ in my formulation is derived from the meaning of $M$ and $N$ as bounds on the summation index; we can omit that restriction if we have an interpretation of $\sum_{k=u}^v \lVert b\rVert^k $ for $v < u$], it's a consequence of the behaviour.

Chris's sis the artist

20:21

@Gato besides, at the moment I'm only interested in the brilliance of Ramanujan, that level, not in simple applications of well-known stuff. :-)

Moses

@DanielFischer Oh okay I see. In order to see that this sum goes to zero would it suffice for me to split it $|\sum\limits_{k = 0}^{N} \| b \|^{k} -\sum\limits_{k = 0}^{M+1}\|b \|^{k}|$ and then evaluate each finite sum as a geometric series and show that this tends to zero as $N$ and $M$ tend to infinity?

Daniel Fischer

@Moses Yes. You're basically showing that $\sum \lVert b\rVert^k$ converges [for $\lVert b\rVert < 1$], so the sequence of partial sums is a Cauchy sequence, i.e. "$\lvert s_N - s_M\rvert \to 0$ for $N,M\to \infty$".

evinda

Hello!!! :)

Moses

20:51

@evinda what is the latex symbol for direct sum

Mike Miller

\oplus

Moses

Oh okay thanks.

evinda

@MikeMiller You were faster than me :)

@Moses Do you study maths?

Karl Kronenfeld

Oh, I was thinking evinda transformed into mike

Moses

@MikeMiller Do you know if you have a banach algebra $X$ which does not have a unit you can follow the process of unitiization of $X$?

evinda

20:55

@KarlKronenfeld Don't worry... We are not the same person :D

Moses

@evinda I try to from time to time.

evinda

Where do you study? @Moses

Moses

@evinda Just online mostly. Yourself?

evinda

@Moses In Greece

Moses

@DanielFischer Do you know if you have a banach algebra $X$ which does not have a unit you can follow the process of unitiization of $X$?

@evinda Kewl. How's it going there...you guys okay?

Daniel Fischer

20:57

@Moses What do you mean by "follow the process of unitization"?

Moses

@evinda What area of maths you studying?

evinda

Yes, it's ok. @Moses

@Moses Applied mathematics

Moses

@DanielFischer You define $X^{+} = \{ (a, \lambda): a \in X, \lambda \in \mathbb{C}\ }$

@DanielFischer and $(a, \lambda) \cdot (b,v) := (ab + av + b \lambda, \lambda v)$.

Daniel Fischer

@Moses Yes, we can adjoin a unit to make it a unital Banach algebra. But what does "follow the process" mean?

Moses

@DanielFischer I just mean that you define a unital Banach algebra from one without a unit in this way.

Daniel Fischer

21:03

@Moses Yes, that's a standard procedure. You seem to know that it exists and how it is done. So I have trouble understanding what you're asking.

Moses

@DanielFischer Why is the symbol $\oplus$ used in when defining this unital Banach algebra. It is given as $X^{+} = X \oplus \mathcal{C}$.

Daniel Fischer

@Moses Because as topological vector spaces (or Banachable spaces), it is the direct sum of $X$ and $\mathbb{C}$.

Moses

@DanielFischer The notion of direct sum I am familiar with is from linear algebra where $W = U \oplus V$ if $W = U + V$ and $U \cap V = \{ 0 \}$. Is this a similar type definition?

Daniel Fischer

@Moses Yes. In addition, it says something about the topology. Have you already heard of categories?

Moses

@DanielFischer Yes, just the basic idea.

Daniel Fischer

21:07

@Moses A rough idea about products and coproducts?

Moses

@DanielFischer Cartesian products and product topology yes, not any coproducts.

@DanielFischer Can I report a crime on MSE someone is going crazy on one of my posts. In there defense they are spewing maths and nothing else.

Daniel Fischer

@Moses Well. The definition of products by their universal property? If so, in coproducts you reverse the arrows, and in some nice categories, coproducts are called direct sums.

@Moses I'm afraid I know who. Try to ignore, maybe that helps.

Moses

@DanielFischer I will have to read up on that, thanks.

Daniel Fischer

@Moses Well. But for vector spaces (pure, topological, normed ...), a direct sum of finitely many (nontrivial) spaces is isomorphic to the direct product. Only for infinitely many nontrivial spaces is there a true difference. So for the moment, understanding products suffices.

Moses

@DanielFischer Okay kewl so for the unitization I can just think of it as the usual cartesian product?

@DanielFischer They should have never allowed the word 'normal' to be used as a technical term. Now we can't describe anything as normal...unless it is normal in some mathematical sense.

Daniel Fischer

21:19

@Moses Yes. Take $X \times \mathbb{C}$ as the vector space, define the multiplication as above to have a unital $\mathbb{C}$-algebra, and endow the whole shemozzle with a norm inducing the product topology such that the inequality $\lVert xy\rVert \leqslant \lVert x\rVert\cdot\lVert y\rVert$ holds.

@Moses Well, how about "perfectly normal"? Oops. "Completely normal"? Oy, gevalt.

Morning @Mike.

Mike Miller

Morning.

Moses

:) Exactly

Mike Miller

Oh, that guy...

Chris's sis the artist

@robjohn you're pretty quiet today.

Daniel Fischer

@MikeMiller Who?

Mike Miller

21:23

The guy you're afraid you know, above.

robjohn

@Chris'ssistheartist Just working and trying to compute an integral in the off time.

Daniel Fischer

@MikeMiller I didn't think you had much overlap with him. Didn't expect you to be familiar.

@robjohn Compute a simpler integral, that's quicker.

Chris's sis the artist

@robjohn Some interesting integral?

robjohn

@Chris'ssistheartist Don't know if it's interesting or just complicated. I'll know when I finish.

Chris's sis the artist

@robjohn OK

Karl Kronenfeld

21:26

@robjohn Beats writing turing machines when you have access to high level programming languages

Mike Miller

I read operator algebra questions sometimes, but his titles are rather singular.

Daniel Fischer

@MikeMiller As is his penchant for non-standard unclear notation.

Mike Miller

I don't usually engage in the actual questions. The reason I keep coming back is that I never get bored of "show 22 more comments".

Chris's sis the artist

Ups, there is a mistake

The closed form looks amazing and short.

robjohn

@Chris'ssistheartist I imagine the product can be simplified or the integral telescopes

Chris's sis the artist

21:37

@robjohn That product can be worked, indeed. There might be more ways than what I presently have in mind.

robjohn

@Chris'ssistheartist Yeah, one or the other, or the thing looks intractable.

Chris's sis the artist

@robjohn Yeap.

Mike Miller

@DanielF: I hadn't realized how trivial our intersection is. I guess it's the complex-analysis and operator-algebras tags, and nothing else, since you don't like derivatives.

Daniel Fischer

@MikeMiller That is an exaggeration. I don't like PDEs, I have nothing against derivatives.

Karl Kronenfeld

lol I found the "show 22 more comments" one

Mike Miller

21:44

So you claim.

Karl Kronenfeld

well I had a good time reading them regardless

Mike Miller

Oh, I'm not disagreeing with your claim, @Karl...

Karl Kronenfeld

there may be a second operator-theory guy who has those same tendencies

Mike Miller

Not one I've met, at least

Karl Kronenfeld

i can change that

Mike Miller

21:56

Sure.

Karl Kronenfeld

Hot_T

Mike Miller

I was referring to him. I don't know who your 22 commentor is, then.

Karl Kronenfeld

ah, but does it make a difference?

unless, of course, Hot_T decides to kill(melt) that guy or something

Mike Miller

heh

Stan Shunpike

22:22

@TedShifrin You want me to do Chapter 6 as opposed to Chapter 5?

I assume u meant what u said but am just clarifying before I start

PVAL

22:47

I found the "show 22 more comments". Do I win a prize?

robjohn

22:58

Doggummit! The integral that I was working on, and have finished, was closed as off-topic 9 hours ago. I guess it was a while since I refreshed that page. >8(

It had 2 close votes last I looked.

@Chris'ssistheartist: I guess it was not interesting.

alan2here

what types of values can be represented by, or are equal to sureal numbers?

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$Mathematics$ Mathematics