Mathematics - 2018-02-18 (page 4 of 7)

Sam

15:05

Hello

Balarka Sen

@Akiva u learning riemannian fam

Akiva Weinberger

Haven't actually started yet, had a small vacation

Balarka Sen

ah

Akiva Weinberger

I continued a tiny bit into a measure theory thing I had though

Sam

Can someone explain this to me I think Im being a retard.
I have two vectors: a,b. I can rewrite ||a-b||^2 as (a-b) . (a-b) which is fine. But, on the example I'm looking at, they expand it out further:
a . a -* a . b - b . a +* b . b . I've put an astrix next to the minus and plus signs because I'm unsure why those operators are used. Should it be a . a + a . -b - b . a - b . -b Or am I being stupid?

Assume the period is dot product.

Balarka Sen

15:12

my knowledge of measure theory is infinitisimal

Akiva Weinberger

Hmm where did I put my glasses

Balarka Sen

@Sam the expression you wrote and in the example are equal

rearrange some minus signs

a . (-b) = - (a . b)

- (b . (-b)) = + (b . b) 'cuz (-1)*(-1) = 1

Akiva Weinberger

Found my glasses

Balarka Sen

@AkivaWeinberger Do you want to hear a story

Akiva Weinberger

Sure

^ That's the book, by the way. Real Analysis for Graduate Students

Balarka Sen

15:16

So once upon a time there was a manifold

Akiva Weinberger

The "affordability" bit means a) it's self-published and b) there is the occasional typo in the math

Balarka Sen

Oh yeah I've heard of Bass

you're in treble

Akiva Weinberger

Not like misspelled words, like a $\cap$ where there should be a $\cup$, stuff like that

Sam

Ok kinda get that. But not the 'cuz (-1)*(-1) part

Akiva Weinberger

@BalarkaSen Why?

Balarka Sen

15:17

just making dumbjokes.jpeg

Akiva Weinberger

Arright

It's pretty good so far, to be honest

Balarka Sen

Daminark and Eric knows that book I think

@Sam Well - (b . (-b)) = -(- (b . b)) = (-1)(-1)(b . b) = b . b

Akiva Weinberger

@Daminark @EricSilva I a book have

Balarka Sen

Oh speak of the devil @Daminark

Akiva Weinberger

Whoa hey whatup

Sam

15:19

Ah got you. Thanks a lot for clearing that up

MatheinBoulomenos

Sup @Daminark

Balarka Sen

Okay so my story was this

$(M, g)$ be a Riemannian manifold

Akiva Weinberger

Mrmhrm

Balarka Sen

That is to say, for each $x \in M$, $g_x$ is an inner product on $T_x M$ and they vary smoothly (i.e., for any two smooth vector fields $X, Y$ on $M$, $f(x) = g_x(X, Y)$ is a smooth function on $M$)

Maybe I should have explained how to define smooth vector fields?

Akiva Weinberger

You can if you want

Balarka Sen

15:25

Here's an indirect definition like the smoothness of the metric. Let $X$ be a vector field on $M$. If for any smooth function $f : M \to \Bbb R$, the directional derivative $g(x) = D_{X(x)} f(x)$ defines a smooth function $g : M \to \Bbb R$ too, $X$ is said to be smooth

There are other ways to define it but hopefully that suffices

0celo7

Are you corrupting @AkivaWeinberger

Akiva Weinberger

"Smooth" means $C^\infty$ here? @BalarkaSen

(I'm back)

Eric Silva

oh bass is dec

Akiva Weinberger

Decent?

Eric Silva

ya

user193319

15:37

Suppose that $n \in \Bbb{N}$ is such that $(n-1)$ divides $n$. I am having trouble showing that $n$ must be $2$. Clearly it cannot be $n=1$, for this would imply $1=0$, a contradiction. How do I rule out the $n \ge 3$ case?

ÍgjøgnumMeg

$\gcd(n, n-1) = 1$

Akiva Weinberger

What numbers does $n-1$ divide?

mercio

I'm bored

user193319

@AkivaWeinberger Only multiples of $n-1$.

Akiva Weinberger

$0$, $n-1$, $2(n-1)$, $3(n-1)$, etc., right?

If $n>2$, then $2(n-1)>n$.

So none of the things on that list can equal $n$; the first two are too small, and the rest are too big.

That's one way to do it, anyway; @ÍgjøgnumMeg's thing is another way

Daminark

15:41

Hey

Akiva Weinberger

To spell his thing out: If $n-1$ divides $n$, it also divides $n-(n-1)$

which means it divides $1$

The only thing that divides $1$ is $1$, so $n-1=1$, so $n=2$

That's a better way, actually

ÍgjøgnumMeg

xoxo

Daminark

@Balarka, @Mathein, @Akiva strangely, I was asleep when you said hello

And yeah Bass is excellent

user193319

@AkivaWeinberger Ah! Yes! Very good. Thanks!

Akiva Weinberger

Arright @Daminark

You agree with me on the occasional typo, right?

I'm not very far in, I just finished the chapter in which he defines the Lebeg integral

Daminark

15:43

It has happened, yeah

Akiva Weinberger

Hey, you know the exercise that says,

> Suppose $A$ is a Lebesgue measurable subset of $\Bbb R$ and$$B=\cup_{x\in A}[x-1,x+1].$$Prove that $B$ is Lebesgue measurable.

We don't need $A$ to be Lebesgue measurable, do we?

MatheinBoulomenos

Are you sure? That union doesn't have to countable

Eric Silva

uncountable unions are craaazy

Akiva Weinberger

Of constant length intervals, though?

Eric Silva

upon reflection you do not need A to be measurable no your intuition was right

Akiva Weinberger

15:50

Wait I forget my reasoning though

Eric Silva

$B \cap [n, n + 1]$ is fairly tame

and measurability of that suffices

Akiva Weinberger

Ah, yeah, that makes sense. It's either two intervals or the whole thing

MatheinBoulomenos

Is it wrong if I write "Lemma 1 [bla] Proof: this is a routine verification [...] Lemma 3 [bla] Proof: the proof is similar to Lemma 1"

0celo7

@XanderHenderson

see above

Akiva Weinberger

Like, $[n,\alpha)\cup(\beta,n+1]$, where the round parentheses might be square

Eric Silva

15:53

i guess the question wanted you to make the easy argument only available when its measurable (decompose into 3 things that are obviously measurable)

Akiva Weinberger

Right, I see it now

Incidentally:

I'm pretty sure that $\mu^*(A_n)$ doesn't${}\uparrow\mu^*(A)$ in general

ÍgjøgnumMeg

$\notuparrow$ looks disturbingly similar to a really tall stickman

or whatever the command was

Akiva Weinberger

\not\uparrow

ÍgjøgnumMeg

$\not\uparrow$

maybe from a distance idk

Akiva Weinberger

\not{\uparrow}?

Consider the outer measure on $\Bbb Z$ with $\mu^*(\emptyset)=0$, $\mu^*(\Bbb Z)=2$, and $\mu^*($anything else$){}=1$. That's an outer measure, right?

ÍgjøgnumMeg

15:56

$\not\not\uparrow$

Akiva Weinberger

For reference

And then if $A_n$ is any collection of sets that $\uparrow$s to $\Bbb Z$, then we have a counterexample

Eric Silva

did he mean to prove it for measures

it's true there

Akiva Weinberger

But then the second part of the exercise wouldn't make sense

Right?

Eric Silva

right yeah it wouldnt lol

Akiva Weinberger

Are there errata for this book somewhere

Eric Silva

16:00

bass.math.uconn.edu/real-errata31.pdf

Akiva Weinberger

@EricSilva Actually, I'm looking at the "Version 3.1" pdf that's online, and it looks like he fixed it there

Eric Silva

yeah

Akiva Weinberger

0celo7

"You must restart your system before using Adobe Reader again"

God dammit I didn't ask for this update

Akiva Weinberger

Right, OK, I'm holding the "Second Edition"

Eric Silva

16:06

right he has errata for that one on his page

damn bass is on top of it

Akiva Weinberger

Kinda annoying that I'm not holding the "right" version

I'mma have to climb back through all the errata once I'm done reading this

Or skim through the most current version online and see what's different

Xander Henderson

@0celo7 The simple solution is not to use Adobe Reader

0celo7

@XanderHenderson what do I use

Xander Henderson

What OS are you using?

0celo7

win10

Xander Henderson

16:19

List of PDF software

This is a list of links to articles on software used to manage Portable Document Format (PDF) documents. The distinction between the various functions is not entirely clear-cut; for example, some viewers allow adding of annotations, signatures, etc. Some software allows redaction, removing content irreversibly for security. Extracting embedded text is a common feature, but other applications perform optical character recognition (OCR) to convert imaged text to machine-readable form, sometimes by using an external OCR module. == Multi-platform == === Converters === These allow users to convert PDF...

I don't know windows, but there are lots of free ones out there

0celo7

adobe is free

Xander Henderson

free as in speech, not free as in beer

I've used the Linux version of Evince, and it seems fairly useful

Akiva Weinberger

@EricSilva If a set function sends every nonempty set to infinity, it's a measure, right?

And it sends the empty set to zero

Eric Silva

lol

MatheinBoulomenos

very useful measure

Akiva Weinberger

16:24

No I'm trying to understand a change Bass made

Xander Henderson

@AkivaWeinberger Sure... just check the axioms: it is a set function, sends the empty set to zero, and is otherwise additve

Eric Silva

what'd he change

Xander Henderson

some authors do require that there be at least one set of finite positive measure, however

Akiva Weinberger

He added "and $\mu(B)$ is finite for some non-empty $B\in\mathcal A$."

That didn't use to be there.

Eric Silva

maybe it was just to throw out the trivial case

Xander Henderson

16:26

well, it is (wait for it, @0celo7) "trivial" in the other case

0celo7

It's trivial, obvious, well-known, and routine to check.

Xander Henderson

(though I'm using trivial in both the vernacular and technical sense, here)

the zero measure and the infinite measure are both trivial measures that can be defined for any space! Yay!

Akiva Weinberger

Speaking of trivial things, do there exist topological spaces $X$ such that the only continuous maps $f:X\to X$ are the trivial ones (identity function and constant functions)?

With more than two elements.

Xander Henderson

would you take a set with THREE elements?

(even there, I think that the answer is "no") :\

Akiva Weinberger

With more than one element is what I meant

Hm, wait, Sierpinski?

Oh wow

Hm OK with more than two elements then

Eric Silva

16:33

finite spaces are for nerds cough @Daminark cough

Akiva Weinberger

Or maybe I should just specify "Hausdorff"

Xander Henderson

What about Sierpinski?

Akiva Weinberger

Right, Sierpinsky would work, since swapping the two elements isn't continuous

Xander Henderson

that was the two element example I had in mind

but I'm not sure that I see how to generalize beyond two elements

but I'm kind of slow, so that doesn't mean anything :)

actually... what if you just nest the open sets? $\{ \emptyset, \{0\}, \{0, 1\}, \{0,1,2\} \}$? doesn't that work?

swapping any two elements is discontinuous, no?

Akiva Weinberger

Can't you map $2\mapsto1$ and keep the other two the same?

$f(2)=f(1)=1$, $f(0)=0$

Xander Henderson

16:39

oh, shit

yeah

I wasn't thinking of locally constant functions :\

THE POWERSET IS TOO BIG!

Akiva Weinberger

$f(2)=2$, $f(1)=f(0)=0$ would work as well

Xander Henderson

yeah, I was only thinking about permutations---I wasn't tracking the locally constant guys

Akiva Weinberger

Wait a moment

@Slereah was talking about something like this a while ago

except it was no nonidentity automorphism, rather than no nonidentity, nonconstant map

bolbteppa

What's better than adobe, they are all horrible and blockey

Daminark

@EricSilva 0_0

Eric Silva

16:54

get rekt punk

Akiva Weinberger

Who needs infinitely many points in your $\rm\Bbb RP^2$ when you can have a perfectly good finite approximation with just 13

Daminark

^^

Eric Silva

i guess youre right

i am now a finite topologist and renounce analysis

0celo7

is @Daminark into combinatorial topology or what

Daminark

Norman Wildberger approves of algebraic topology now

Xander Henderson

16:58

@bolbteppa I don't know for Windows; for Mac, the already-installed Preview works well, and I rather like Skim

Eric Silva

@Daminark what an icon

Daminark

@0celo7 I'm not really into anything at the moment, I've had a small taste of a bunch of things, but I did have a reading course which started off talking about how you can jump back and forth between finite spaces/posets, small categories, simplicial sets, etc

Akiva Weinberger

@Daminark He has a course on algebraic topology on YouTube

Xander Henderson

but a PDF VIEWER which requires a system reboot for an update is crap...

Transcript for

$Mathematics$ Mathematics