Mathematics - 2020-11-22 (page 1 of 2)

Unknown x

00:00

0

Q: Consider the function: $f(x+iy)=\sqrt{|xy|}$. Then

Consider the function: $f(x+iy)=\sqrt{|xy|}$. Then $f$ is continuous at $0.$ $f$ satisfy Cauchy-Riemann equation. $f$ is differentiable at $0.$ $f $ continuous everywhere. My Attempt:- Let $x=r\cos \theta, y=r \sin\theta $ $$f(x+iy)=f(r,\theta)=\sqrt{|xy|}=\sqrt{|r\cos \theta \cdot r \sin\...

complex-analysis

I understood that C-R equation satisfy only at z=0. Can I say f is differentiable at z=0. then?

Ted Shifrin

I just commented on your question, @Unknown. What does it mean to say $f$ is differentiable at $z=a$?

Unknown x

question is asking general. whether it satisfy C-R equation or not?

Ted Shifrin

Is that answering the question I just asked you?

You can prove for yourself that a real-valued function satisfies the C-R. equations on an open set if and only if it is constant on that open set.

Unknown x

@TedShifrin Since it has only real part. the C-R equation doen't satisfied

only at zerp C-R equation satisfied

Ted Shifrin

Right. But I have asked the same question several times. What does differentiable at $z=a$ mean?

Unknown x

00:07

$\lim_{h\to 0}(f(a+h)-f(a))/h$ exists.

Ted Shifrin

OK, so that's the definition, not differentiability as a function $\Bbb R^2\to\Bbb R^2$.

So, because the partial derivatives vanish at the origin, the complex derivative you just wrote down is $0$ at $z=0$.

Oh, wait. Is this right? What if you take $h = (1+i)t$. What happens then?

Unknown x

Let me check sir

Limit goes to zero

Ted Shifrin

How did that become $0$?

Unknown x

sorry sir

limit doesnot exists. I forgot the squareroot.

Ted Shifrin

Yup. That's correct :)

Unknown x

00:18

But C-R equation satisfied at z=0. there. there must be extra condition there. right?

Ted Shifrin

Read your theorem carefully. You need C-R. equations in an open set, not just at a point.

Unknown x

Okay sir. Thank you. I will do.

Ted Shifrin

Good question, though.

Unknown x

:)

robjohn

00:32

@TedShifrin It might not advance math very much, but it would sure be something to see.

Ted Shifrin

Indeed.

AttractorNotStrangeAtAll

Hello!

I was given the following challenge to be solved only with tools from intro analysis (continuity, compactness etc) and would like some hint on where to start.

Let $\pi_1,\pi_2\colon \mathbb{R}^2\to \mathbb{R}$ with

$$\pi_1(x,y)=x\ \text{and}\ \pi_2(x,y)=y$$.

Let $I$ be a nondegenerate interval and $\Gamma\colon I\times [0,1]\to X\subseteq \mathbb{R}^2$ a continuous function such that $\forall t\in I,\ \exists!\ u(t)\in [0,1]$ with $\pi_2\left(\Gamma(t,u(t)\right)=0$. This uniqueness of $u(t)$ for each $t\in [0,1]$ defines a function $u\colon I\to [0,1]$. Show that

Also, first it said that $t\in I$, then it says that $t\in [0,1]$, so this kept me confused

Mike Miller

@user2103480 This is really totally obvious though

Your category is the category of $\Bbb N$-graded abelian groups, whose objects are: an abelian group $A_n$ for all $n$, and morphisms $A \to B$ are a choice of homomorphism $A_n \to B_n$ for all $n$

Ted Shifrin

You're right, @AttractorNotStrangeAtAll. Should be in $I$.

Mike Miller

Quite literally the Cartesian product $\mathsf{Ab}^{\Bbb N}$

If you have a bunch of functors $H_n: \mathsf{Top} \to \mathsf{Ab}$, they package together (trivially!) into a functor $H = (H_n)_{n \in \Bbb N}: \mathsf{Top} \to \mathsf{Ab}^{\Bbb N}$

There is no content to this statement. It's phrasing the same thing a different way.

If you want to think of $\bigoplus_n H_n$ as a functor $\mathsf{Top} \to \mathsf{Ab}$, you're just composing with the direct-sum functor $\bigoplus: \mathsf{Ab}^{\Bbb N} \to \mathsf{Ab}$

This statement has only slightly more content

AttractorNotStrangeAtAll

00:49

@TedShifrin This question was posed by a colleague in our google group, and I'm really curious because he arrived at it while thinking about continuity of roots of continuous functions. He also said that this is related to homotopy. These things got my curiosity.

user2103480

@MikeMiller not if you must google most group theory terms again

Mike Miller

Do you get my point though? A graded group is just a bunch of groups. If you have a bunch of functors landing in groups you rather tautologically have a functor landing in graded groups

user2103480

yeah yeah, but we didn't already check that a chain map induces maps on the homology groups, which is the equivalent statement, and the trivial equivalence is what you're getting at

Mike Miller

Yeah that's the nontrivial part.

Ted Shifrin

@AttractorNotStrangeAtAll Some things are more difficult than they look. Like the statement that the roots of a polynomial vary continuously as you vary the coefficients. E.g. The eigenvalues of a matrix vary continuously as you vary the matrix.

user2103480

00:55

also, that we didn't again have a chain map again made me realize the obvious fact that all boundary maps on homology groups are 0, and totally by definition so

Mike Miller

I don't see the relationship to homotopy. Seems like a somewhat difficult question, the notation might be the hardest part

user193319

0

Q: Banach Bimodules

If $A$ and $U$ are Banach algebras, what does exactly does it mean for $U$ to be a Banach $A$-bimodule? I'm guessing it means than that it is a left and right module over $A$. Do the left and right module structures have to agree in some sense? Does the adjective "Banach" mean the module actions ...

modules banach-spaces operator-algebras banach-algebras bimodules

user2103480

Ugh. It's best to calculate through the proof that says short exact sequences of chain complexes induce long exact sequences on homology groups oneself, right?

how proudly all lecturers say that this is an example of diagram chasing

Mike Miller

@user2103480 Yeah man cmon you can o it

You pick an element

you see what the assumptions give you

and you see what that buys you

You can do it out loud if you want

user2103480

ugh. alright I'll do it myself tomorrow

Mike Miller

01:03

Just do it now it's 5 minutes

Cmon just think it out

user2103480

it's 2am here, but thanks, I feel like in a sports class now, which is oddly motivating

"CMON TEN MORE PUSH UPS"

gotta work through the excision theorem for singular homology tomorrow

Mike Miller

@AttractorNotStrangeAtAll Real question: let $X$ be a space and $Y$ be a compact space. Suppose $f: X \times Y \to \Bbb R$ is a continuous function so that for each $x \in X$, there is a unique $u(x) \in Y$ so that $f(x, u(x)) = 0$. Show that $u: X \to Y$ is continuous.

The other stuff is auxiliary

user2103480

but I have good lecture notes available, which mirror my course closely

Mike Miller

This uses a lemma about compact spaces I suspect you don't know, that he's trying to get you to prove

AttractorNotStrangeAtAll

@MikeMiller Oh wow, much clearer proposition. Thanks

Mike Miller

01:10

I am unsure how much of a challenge you want this to be. I can point you to the lemma you need to prove or let you try to figure that out yourself.

Up to you.

AttractorNotStrangeAtAll

@MikeMiller As I also have other things to care about (i.e. the problem set that is actually graded), I would gladly accept the lemma

Mike Miller

Prove that if $u: X \to Y$ is a map (not assumed continuous) from a topological space $X$ to a compact space $Y$, then if the graph $\Gamma(u) = \{(x, u(x)) \mid x \in X\} \subset X \times Y$ is closed, then $u$ is continuous.

2

If you know metric spaces but not topological spaces, assume these are metric spaces

AttractorNotStrangeAtAll

Yes, I've seen metric spaces (professor introduced it concomitantly with intro analysis).

Thanks.

What makes me sad is how this colleague knew this if he's till in intro analysis. I don't know him personally, but makes me think I'm far behind

Mike Miller

As a concrete response, this fact is sometimes proven in analysis classes

And sometimes an exercise

But the better response is that it is usually unwise to spend time comparing yourself to friends or colleagues

AttractorNotStrangeAtAll

Maybe the fact that he posted it as a challenge and the notation itself scared me

Sure

Mike Miller

01:16

It is easy to say but a wildly important lesson

Took me a decade to learn

user2103480

the best approach is to learn from then

Ted Shifrin

And MikeM is only two decades old :)

user2103480

AND THEN STEAL THE TT FROM RIGHT UNDER THEIR NOSE

jk

Mike Miller

@TedShifrin I think understating my age might help contribute to our friend's imposter syndrome

AttractorNotStrangeAtAll

The fact is that I'll learn everything I'll have to learn in my own time. My Bachelor's is not in Math, so it should be obvious that I don't know everything.

I stopped worrying that maybe I'll finish grad school some years after the average age, that's ok. I really enjoy all of this and I can't see myself studying other things. And I want to arrive at grad school as sharp as possible, I won't rush just because I'm 3 years (or more) older than the average

Mike Miller

01:23

It really is crucial to have a good time

If you're not, you're really wasting your time, since most people with PhDs do not end up with TT jobs (assuming that's what you want)

user2103480

@MikeMiller I don't get this whole imposter syndrome terminology. it's just a collective inferiority complex, why not say that out loud

Mike Miller

My PoV is that if I am at least enjoying what you learn then no matter how the job market looks like on the other end, at least I had something of a good time learning this stuff and able to think about it

Ted Shifrin

If my sarcasm wasn't evident, I apologize.

Mike Miller

@user2103480 Sure I just use whatever terminology is clear to the largest audience

you're fine ted

Ted Shifrin

Relevant or not, I think @AttractorNotStrangeAtAll is not in the US.

AttractorNotStrangeAtAll

01:26

I actually admire people who got their PhD with ~22 y.o., but I don't really compare myself to them

Ted Shifrin

I only know a handful of such.

AttractorNotStrangeAtAll

Yep, I'm Brazil. It's not uncommon to know people in their 15~18 y.o. studying at IMPA for Master's and getting their PhD at 22 yo.o as well

But of course, the majority of people follow the traditional path of Bachelors-Masters-PhD here

And they are also good mathematicians, so...

user2103480

I don't know. I think it's quite nice to have a few more years of studying

As long as there's no urgent money trouble

AttractorNotStrangeAtAll

No, not really. To be sincere, I was an intern at a M&A boutique, which is a career that pays really good but you also work like hell. It sucked. I felt like I was doing non-interesting things with horrible people that only cared about money. I was not challenged, and worked +10h a day doing investor presentations with PowerPoint or doing bullshitting in Excel.

That's totally my opinion, tho.

user2103480

huh

I think you misunderstood. I just meant that as long as you don't constantly worry whether you can pay your next bills, it's cool not to rush through uni

and that job that you mention is valuable life experience I'd guess

AttractorNotStrangeAtAll

01:43

Ohhhh lol, true. Sorry about the rant, it's somewhat recent

epic_math

Hi there

700 years of secrets of the Sum of Sums (paradoxical harmonic series)

►

AttractorNotStrangeAtAll

@user2103480 And yes, it was a valuable life experience.

geocalc33

02:13

$$\sum_{n=0}^.5 f(n) $$

can you sum with upper bound and lower bound as such?

actually lower bound is $n>0$

epic_math

@geocalc33 how can a sum have non integral limits?

I would rather sum it as $f(0)+f(0.5)$ if you really want to do it.

geocalc33

@epic_math I'm trying to evaluate this $$ \lim_{n \to 0}\sum_{n}^{1/2} \exp\bigg(-\pi\bigg(\frac{1}{n}\bigg)\bigg) $$

where $\pi(n)$ is the prime counting function

epic_math

it doesn't make much sense to me

geocalc33

$n \in (0,1)$ so I don't know how to sum it up

epic_math

what is the summing variable?

geocalc33

02:26

$n$

epic_math

(the variable which runs through values)

if you rather sum it as f(0)+f(0.5), then the sum is nonsense

where does it come from?

the sum seems undefined to me in all ways

geocalc33

here's the plot of the function

epic_math

integrals are coming in my mind

geocalc33

yeah the integration yields.... approximately .3

epic_math

you said the summing variable is n

geocalc33

02:31

yep

epic_math

so the final result would not depend on n

geocalc33

it doesn't

epic_math

yes

geocalc33

.3 doesn't involve n

epic_math

so how can you take the limit wrt n?

and what is the starting value of the sum?

geocalc33

02:32

$$ \int_0^.5 \exp(-\pi(1/n)) \approx .29$$

epic_math

that sum is nonsense

@geocalc33 yes that is right

summing is wrong

do you want the exact value of the integral?

geocalc33

no

it's just for fun

epic_math

you sometimes come up with wild ideas

that are fun

@geocalc33 you like NT?

if yes, would you listen one of my ideas?

geocalc33

I'll listen to your idea

epic_math

okay

Let's take an example

can you factorize $$2+3$$

lol I am not joking

it's a prime

what about $$2^2+3^3$$

it's still a prime

now what about this: $$2^{2^2}+3^{3^3}$$

let me calculate

is it a prime or a composite?

I dunno

well it is PRIME

Now, what about $$2^{2^{2^2}}+3^{3^{3^3}}$$

sorry, I calculated wrong, $2^{2^2}+3^{3^3}$ is a prime

are you listening?

geocalc33

06:31

There are two metrics given: $ds_1^2=\frac{dudv}{v-uv}$ and $ds_2^2=\frac{dudv}{uv}.$ How do you form the product of the two metrics?

$ds_1^2 \otimes ds_2^2=\bigg(\frac{dudv}{v-uv}\bigg)\bigg( \frac{dudv}{uv} \bigg)$

is that right so far?

robjohn

06:52

@geocalc33 can you evaluate $H_{1/2}$? (harmonic number)

geocalc33

$H_{1/2}=2-\log(4)?$ @robjohn

robjohn

@geocalc33 and how did you evaluate it?

geocalc33

using Wolfram alpha

robjohn

so much for mathematical knowledge.

How did you do that problem? I plugged it into WA.

so the answer is: no, you can't evaluate $H_{1/2}$, but WA can.

geocalc33

Yeah I can't evaluate $H_{1/2}$

I'm not at that level yet

robjohn

07:05

can WA do $H_{2/7}$?

geocalc33

yeah but it's really complicated. There's also a representation using the Euler-Mascheroni constant and the digamma function

robjohn

@geocalc33 what's the really complicated version? I think you can use //TeXForm to get latex

$\frac{7}{2}+\frac{3}{7} \pi \sin \left(\frac{\pi }{7}\right)-\frac{5}{7} \pi \cos\left(\frac{\pi }{14}\right)+\frac{1}{7} \pi \cos \left(\frac{3 \pi }{14}\right)-2\left(1+\sin \left(\frac{\pi }{14}\right)\right) \log \left(2 \sin \left(\frac{\pi}{7}\right)\right)-2 \left(1+\cos \left(\frac{\pi }{7}\right)\right) \log \left(2 \cos\left(\frac{3 \pi }{14}\right)\right)+2 \left(\sin \left(\frac{3 \pi}{14}\right)-1\right) \log \left(2 \cos \left(\frac{\pi }{14}\right)\right)$

geocalc33

$7/2 - \log(14) - 1/2 π \tan((3 π)/14) - 2 \sin(π/14) \log(\sin(π/7)) - 2 \cos(π/7) \log(\cos((3 π)/14)) + 2 \sin((3 π)/14) \log(\cos(π/14))$

robjohn

A tangent in there? odd

geocalc33

$H_{2/79}$ is even more complicated

robjohn

07:21

@geocalc33 take a look at this answer

Equation $(7)$ in particular

I used the Mathematica implementation at the end of that article.

geocalc33

I looked at it. How does this relate to the question I asked before about $\sum_{n=0}^.5 f(n)$

robjohn

07:36

It tells how to evaluate $H_{p/q}$. It only relates in the sense that this is one way to evaluate a sum with non-integer limits.

Since $H_n=\sum\limits_{k=1}^n\frac1k$

geocalc33

oh okay

@robjohn I'm interested in extending $G_n=\sum_{k=1}^n e^{-k}$ to non-integer limits

this is of course a geometric series

robjohn

07:58

@geocalc33 That would be $\frac{1-e^{-n}}{e-1}$

That can be extended in the same way as the Harmonic Numbers: $$\sum_{k=1}^\infty\left(e^{-k}-e^{-k-n}\right)$$ This gives a convergent series for all $n\in\mathbb{C}$ and agrees for all $n\in\mathbb{N}$.

This is even simpler than the Harmonic Numbers because each part is summable by itself.

$$H_x=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)$$

gives a convergent series for all $x\in\mathbb{C}$ and agrees for all $x\in\mathbb{N}$.

love_sodam

If $f$ is a bounded variation on [a,b], then $\int_a^b |f'(x)|dx\leq sup\sum_{i\in I} |f(t_i)-f(t_{i-1})|$ where $I$ is a partition of $[a,b]$

I know that bounded variation is almost everywhere differentiable

To prove the above, can I assume $f$ is differentiable everywhere on $[a,b]$

??

robjohn

@love_sodam bounded variation is the difference of two monontonic functions, and monotonic functions are almost differentiable.

love_sodam

08:13

Yes I know but form that fact, can I assume that bounded variation is everywhere differentiable?

epic_math

Howdy

robjohn

@love_sodam If you use Riemann-Stieltjes integrals and use Dirac functions you get equality and the inequality comes from leaving out the DIrac functions.

love_sodam

I can't understand. Could you write down equations?

robjohn

Try looking at monotonic functions. It is easier and just as general

The integral of the derivative is the integral over the partitions where there are no jump discontinuities. The difference between the integral and the sum of the partitions are the sum of the jump discontinuities.

love_sodam

08:31

Yea that's intuitive understanding.

So the integral

Just assume $f$ is differentiable then everything is easier

epic_math

Do you people know

love_sodam

But I think @robjohn you think I shouldn't

epic_math

Some people in simpliFire's chatroom are doing research on an NT project.

Which is primes of various forms

Interesting

My research before this was partially related to this

My research was on prime numbers that were in the sequence of partitions

And related functions

love_sodam

08:46

We don't know $f'$ is integrable or not

right?

robjohn

@love_sodam I wouldn't just assume it's differentiable. You need to account for the discontinuities

love_sodam

09:06

@robjohn So write f as a difference of two monotone increasing $f_1-f_2$ and write each $f_i$ as a sum of continuous increasing and jump function you mean?

love_sodam

09:41

I think

for $x\in [a,b]$, $F(x) = P_F(a,x)-N_F(a,x)+F(a)$. Then as $F$ is differentiable almost everywhere, $F'(x) = P_F(a,x)'-N_F(a,x)'$ a.e.. Hence, $|F'|\leq P_F(a,x)'+N_F(a,x)' = T_F(x)'$ a.e.. Hence, $\int_{a}^b |F'|\leq \int_a^b T_F(x)'\leq T_F(a,b)$.

This proof is valid

Tobias Kildetoft

@BalarkaSen @BalarkaSen Yeah, especially their latest album, Alter Echo. I also like their first two albums, though those have a slightly different sound. The first two are from the late 90's after which they broke up and got reunited in 2016 (but I don't really like their album form that year). The latest is from this year and I really like that one.

love_sodam

09:55

No it's seems not clear

love_sodam

10:12

I think only the last inequality is unclear

epic_math

10:26

what's going on with the problem on existence of od perfect numbers?

What are some recent discoveries on it?

robjohn

@love_sodam something like that

love_sodam

@robjohn only need to verify the last inequality but how can I?

geocalc33

14:19

given a manifold furnished with geodesics, how can you decompose the manifold into two transversal manifolds and their respective geodesics?

I am aware that any manifold can be constructed using several transversal manifolds

love_sodam

14:59

0

Q: $\int_a^b T_F' \leq T_F(a,b)$ where $T_F$ is a bounded variation

If $F$ is a bounded variation on [a,b], then $\int_a^b |F'(x)|dx\leq T_F(a,x)$ where $T_F(a,x)$ is a total variation of $f$ on $[a,x]$. I know that $F(x) = P_F(a,x)-N_F(a,x)+F(a)$ and $T_F(a,x) = P_F(a,x)+N_F(a,x)$ where $P_F$ is a positive variation and $N_F$ is a negative variation of $F$ bot...

real-analysis bounded-variation

Anyone could answer this?

Thorgott

what's $T_F^{\prime}$

love_sodam

derivative of T_F

Thorgott

with respect to?

love_sodam

x

T_F(a,x)

x is in [a,b]

Thorgott

ok, then $\int_a^bT_F^{\prime}=T_F(a,b)$ is literally just the FTC, no?

epic_math

15:09

I am being very ignored here.

Lol

love_sodam

we don't know $T_F$ is absolutely continuous

Thorgott

15:25

what hypotheses are you making to ensure that T_F' exists

love_sodam

T_F is a sum of two bounded variation

love_sodam

15:51

I think I can handle that problem never mind

user2103480

17:32

@MikeMiller I calculated through the proof of existence of the LES and I don't feel smarter than before

Thorgott

"An easy Zorn's lemma argument will show that every complex
atlas is contained in a unique maximal complex atlas"

why would anyone do that...

user2103480

@Thorgott it is easy

Don't get the hate on zorn's lemma. Perfectly intuitive statement

Thorgott

it is easy, but the statement is just as easily proven in ZF

user2103480

take the union, eh?

Thorgott

ye

just one routine calculation required and you're done

I like Zorn's lemma, but invoking it to prove a statement that's just as easily proven in ZF is bizarre to me

user2103480

17:43

ok constructivist

fckn algebraists, always talking about this and that category and how it's intuitionistic logic

gimme that maximal atlas via zorn's, I don't want to prove compatibility

I VOTE AOC

my proofs, my choice

Thorgott

brb, using Zorn's to prove continuous functions form a sheaf by gluing through maximal extensions

user2103480

that's the spirit

Mike Miller

@user2103480 You shouldn't, a child could do it

But you have to prove you're a child or else you're a baby

user2103480

@MikeMiller fair, I definitely need the practice

Thorgott

the existence of the LES is just abstract nonsense

it can be fruitful to look at some concrete examples of boundary maps, though

user2103480

17:53

@Thorgott yeah, I'll probably see that for the LES of a pair

Thorgott

do you know M-V?

user2103480

nope

given the current pace, I will know it next week

Mike Miller

@user2103480 Calculate H_*(T^2) by comparing to H_*(T^2, W) where W = S^1 x {1} cup {1} x S^1 < T^2

I assume you know H*(X, A) = reduced homology H*(X/A)

nbro

2

Q: In GradCAM, why is activation strength considered an indicator of relevant regions?

EDIT 2 I've changed this question from the general version below to a more specific one. In the GradCAM paper section 3 they implicitly propose that two things are needed to understand which areas of an input image contribute most to the output class (in a multi-label classification problem). Tha...

neural-networks deep-neural-networks explainable-ai

Mike Miller

For any god-fearing pair of spaces (X, A)

user2103480

17:57

@MikeMiller we did start with reduced homology, but I haven't come to that yet. Although I do know that identity H*(X/A) = H(X,A)

@MikeMiller the asterisks are a bit confusing to me here. Should one be the usual homology?

Mike Miller

Yes I'm being lazy

Hn

Hn(X, A) = Hn(X/A) for n>0

user2103480

And only H_0 is trouble because of the triviality of the boundary map

So that's the reason for reduced homology I'd guess?

Mike Miller

I mean think about it explicitly.

user2103480

will do

Mike Miller

C0(X,A) is the C_0(X)/C_0(A)

Meh I don't want to do this on my phone

user2103480

18:02

chill chill

Rest your thumbs

Mike Miller

The point is that "points rel A" kills off extra junk

If X is path-connected and A is nonempty every point in X can be path-connected to a point in A

But in C_0(X, A) we kill off the points in A

user2103480

So everything's 0 I'd guess? I'm too lazy to calculate myself

And we don't want everything to be 0 since X/A is path-connected

Mike Miller

Right, H_0(X/A) is Z

The general analysis is some 5 lemma crap.

user2103480

* since X/A is nonempty I should've said but I know you get what I mean

Mike Miller

H_0 is junk

You should carefully understand what's going on in H_0 but it's junk

It's good to use reduced homology as much as you can to make it slightly less junk

robjohn

19:20

@love_sodam cover the set where $f$ is not differentiable with an open cover of measure $\epsilon\gt0$. The integral over the remainder is less than $f(b)-f(a)$ while the integral over the set removed is less than $\epsilon f(b)$. Since this is true for any $\epsilon\gt0$...

Huy

anyone know a nice geometric proof that the cardinality of C is the same as R? I know the construction with alternating digits but was hoping for something visual

Mike Miller

I think visual is a bit much to ask, since those are distinct as topological spaces

Huy

sounds reasonable, but a bit disappointing

Alessandro Codenotti

A space filling curve?

Tobias Kildetoft

19:36

They are isomorphic as abelian groups, hence have the same cardinality?

Astyx

Are they?

user2103480

@Huy cardinalities are not very geometric in nature. as seen when one removes choice and all hell breaks loose

Huy

-4

A: Do the real numbers and the complex numbers have the same cardinality?

One particularly nice class of bijections from $\Bbb R$ to $\Bbb C = \Bbb R^2$, which is in my opinion a little bit similar to the spiral around the grid, is given by the space-filling curves.

Leaky Nun

|R| = |C| doesn't need choice

Astyx

What matters is it's a surjection

user2103480

19:42

@LeakyNun you mean the existence of bijections. Defining cardinality without choice is the more difficult task

Leaky Nun

right

skullpatrol

70

Q: Books that teach other subjects, written for a mathematician

Say I am a mathematician doesn't know any chemistry, but would like to learn it. What books should I read? Or say I want to learn about Einstein's theory of relativity, but I don't even know much basic physics. What sources should I read? I am looking for texts that teach subjects that are not ...

reference-request soft-question big-list textbook-recommendation

Astyx

"Categories for the Working Mathematician" lol

skullpatrol

:D

"Working"

user2103480

20:04

5-lemma is also diagram chasing?

i.e. reading the proof takes as long as proving it oneself

Astyx

IIRC yes

skullpatrol

proof by time management

Thorgott

what's the least painful way of showing that every 2-manifold is triangulable

@user2103480 yes, diy

user2103480

alrighty thanks

Balarka Sen

Hey I am getting a little confusing in a calculation. If $X, Y \subset \Bbb C^n$ are complex submanifolds, what's the easiest way to see exactly what the critical points of $d(x, y)^2 : X \times Y \to \Bbb R$ are?

If I write it all out I seem to get that they are secants $\ell$ between points $x \in X$ and $y \in Y$ such that Hermitian inner product of $\ell$ with vectors in $T_x X$ and $T_y Y$ are equal

Which is a little strange for me to interpret. Eg, what does the dimension of the kernel of the Hessian (assuming it is nondegenerate) count at such secants?

OK, to be super precise what I'm getting is $T_y Y \subseteq T_x X + \ell^\perp$, I think.

OK just even in real coordinates $d(x, y)^2 = \sum_i \|x_i - y_i\|^2$ and if you differentiate that you get $(x_i - y_i) \cdot (\partial x_i/\partial u_i - \partial y_i/\partial v_j) = 0$ where $(u_1, \cdots, u_{\dim X})$ are the coordinates on $X$, $(v_1, \cdots, v_{\dim Y})$ are the coordinates on $Y$ and $x = x(u)$, $y = y(u)$ are the charts, right?

Which is the statement that $T_y Y \subseteq T_x X + \ell^\perp$. The Hessian looks annoying

I meant $(x - y) \cdot (\partial x/\partial u_i - \partial y/\partial v_i) = 0$

Bizarre, geometrically.

Mike Miller

20:47

@Thorgott Show every 2-manifold is smoothable and then show that smooth manifolds are triangulable

Not kidding

Thorgott

I don't mind just assuming smooth, scratch topological manifolds

but why is it true for smooth ones

Mike Miller

Look up proof by Cairns very clear 2pp

There's a beautiful 10-15pp document by Hatcher where he explains why every topological surfaces is uniquely smoothable

Balarka Sen

@Thorgott Morse theory

Mike Miller

If you want a proof that never talks abt smooth structures whatsoever (you didn't say this but idc) the point is massive repeated application of Schoenflies to extend "partial triangulations" along one more chart. Eventually this comes down to observing that if you have two arcs intersecting in a nasty way in Euclidean space you can manage those intersections and still triangulate the result. Rather tedious and unenlightening

@BalarkaSen Don't agree, this is too hard

Balarka Sen

I dont particularly care to find out an optimal proof

this is a boring question

Thorgott

20:51

projecteuclid.org/download/pdf_1/euclid.bams/1183524257

this?

Mike Miller

Yes

@BalarkaSen True, but Cairns proof is very clean and pedagogically useful

Balarka Sen

OK

Mike Miller

The Morse theory argument is conceptually clean too though I admit.

But the details are harder than we'd like to admit

Cuz you need the attaching maps to be simplicial and it's tedious

Balarka Sen

OK dude

Mike Miller

Don't be a dick I'm just talking

That's what I do

Balarka Sen

20:53

Haha sorry

No you're right

I agree there's details in the Morse proof which I haven't checked

Basically handlebodies should be triangularizable but dunno why.... it's because space of concordance classes of triangulations on a disk is contractible maybe

where concordance between two triangulations on $D^k$ is a triangulation on $D^k \times I$ connecting em

Mike Miller

Nah it's just that handles are clearly triangulable right? And simplicial approximation says that we can wiggle the attaching map slightly to make it a simplicial smooth embedding so long as we subdivide both sides

Thorgott

is the morse theory proof in Milnor

Mike Miller

It's just a tedious thing where you need to subdivide a shit load

Balarka Sen

@MikeMiller ah ok yeah

i was trying to go bottom up, no need

Mike Miller

@Thorgott He constructs handlebodies not triangulations since you're a smart boy and can do the triangulation part

Yeah just do it one cell at a time

Sha Vuklia

20:57

Guys, say $\mathcal F$ is a sheaf of $\mathcal O_X$ modules (with $X$ a topological space), is it then true that for each open $U\subset X$, the elements $s\vert_{U}$ with $s\in\mathcal F(X)$ generate $\mathcal F(U)$?

Mike Miller

The Cairns idea is pretty good. You take a fine mesh of points on your guy and subdivide M into open cells of points closer to one point than any other

To check that this is a regular CW complex you just project to R^n and do the analysis there

@ShaVuklia Doesn't quite parse, X is a ringed space to say O_X-modules

And no. Global sections could be trivial!

Sha Vuklia

@MikeMiller So... it could be true?

Thorgott

I don't wanna be a smart boy, I'll check out the Cairns thing

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