Mathematics - 2020-07-01 (page 1 of 2)

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12:57 AM

Wrote ≈1.8k characters for apparantly no reason... Boredom gets me high sometimes

2 hours later…

2:42 AM

@user675453 what was it about?

1 hour later…

3:56 AM

@CalvinKhor @robjohn I want to prove that if $f$ is integrable on $[a,b]$ then if we change it value at some $x \in[a,b]$ then also it will be integrable. I have done a proof, I want to know if that’s correct.

hi, sure

Given that $f$ is integrable on $[a,b]$ we have a partition $P=\{x_0, x_1, \cdots x_n\}$ such that for a given very small $\epsilon \gt 0$
$$
U(f,P) - L(f,P) \lt \epsilon
$$

Now, let's say in the interval $[x_{i-1}, x_i]$ at some $x$ we change the value of $f$ from $f(x)$ to $\alpha$. To prove that $f$ is still integrable, make the partition $P'$ such that
$$
P'= \{x_0 , x_1 \cdots x_{i-1}, x , x_i, \cdots x_n\}
$$
(a new point $x$, where we changed the value of $f$, is introduced in the partition points)

Now, since $P'$ contains more number points therefore, we have
$$
U(f,P') \lt U(f,P) \\
L(f,P') \gt L(f,P) \\
U(f,P') - L(f,P') \lt U(f,P) - L(f,P) \lt \epsilon \\
U(f,P') - L(f,P') \lt \epsilon
$$
Hence, $f$ is still integrable

Is this proof correct? The only thing I'm suspicious of is that I never used the information that $f(x)$ has changed to $\alpha$.

maybe you should try to answer the following version of your question: let $f$ be integrable on $[a,b]$, and let $g(y) = \begin{cases} f(y) & y\neq x \\ \alpha & y=x \end{cases} $. Prove that $g$ is integrable

just introducing notation to talk about $f$ both before and "after changing the value at $x$"

4:12 AM

Can I just hop in for a second to ask: What's this "U(f,P)" and "L(f,P)" notation in this context?

upper lower riemann sum wrt partition P

or maybe they call it darboux sums idk

Upper and Lower Darboux sums of function $f$ and on partition $P$

of a bounded function $f$

@CalvinKhor How should I start Cal ?

@Drathora I know what he's talking about because this is continuing off some earlier stuff we talked about

4:13 AM

(Cal is short for Calvin, it's not Calculus)

@Knight lol

I don't know yet, I haven't really thought about it, except that this notation will make it very clear when you need to use the fact that $g(y)≠f(y)$

Ah I see, thanks.

Was my proof sloppy somewhere?

Hello Ted

I studied some weirdass integral (regulated integral) in my uni and never got round to getting Riemann / Darboux fully rigourous :P but lets see. I think the way to prove it will kind of depend on what you have proven already

Howdy.

4:16 AM

Yes, you used $U-L<\epsilon$, but this is only true for $f$, not for the modified $f$

Ted I read every comment of yours in your voice

You shouldn't know my voice :P

Was that text-to-speech on your youtube videos then :P

Aha.

@Knight: No, that isn't a proof. As Calvin said, you ignored the whole point.

@TedShifrin Where I ignored the things?

I mean which step was sloppy?

Where did you use $f(x)=\alpha$? By the way, I would suggest using a different letter, like $x_0$ or $c$ ...

I don't think using $c$ as a partition point is wise.

Have you drawn some pictures?

4:19 AM

you used the following result: If $U(f,P) - L(f,P) < \epsilon$ and I change $f$ and at one point, then $U(f,P) - L(f,P) < \epsilon$ also. But that's basically the result you wanted to prove

It's also false.

lol

@TedShifrin No pictures till now.

@CalvinKhor but I used the fact that $P'$ would make the difference even smaller.

let me be more careful, its not at all clear if that implication is true

@Knight write out the calculation, maybe

Take $f(x)=x$ on $[0,1]$ and change $f(1/2)$ to be $20$.

If you can write out a valid proof for this example, you should see how to do it in general. Rule of thumb: Examples and pictures are good.

Does this example (and some pictures) suggest an idea?

4:25 AM

user image

Well, yes, I know what the function looks like. What does this tell you?

The upper sum in interval $[x_{I-1}, 1/2]$ and $[1/2, x_i]$ has increased

So, if you draw inscribed and circumscribed rectangles for a partition, do you really want $1/2$ to be a partition point?

Well, what I meant was the total upper sum has increased.

Yes, right.

The sup increased on both those intervals.

4:29 AM

@TedShifrin I think it will be better if $1/2$ fall in one interval only, so I would prefer it to be not a partition point.

I concur.

So, the expression for upper sum gets changed only at the interval in which 1/2 occurs

Yes, but it may change by a huge amount, as this example illustrates.

that is , $$\sum_{I=1}^{n} M_i (x_i - x_{I-1})$$ gets changed

@TedShifrin Yes.

So what should you do to get $U-L$ back to small?

4:33 AM

Increase the number of partition points

Hey guys!

You want a very small interval around $1/2$. Now you need to make the $\epsilon$s work.

Heya Demonark.

How've you been Ted?

Body's falling apart, dental implant yesterday, but no Covid, so that's good :) How're you?

its also OK if 1/2 happens to be a partition point

@AminIdelhaj hi

4:34 AM

@TedShifrin Yes, but what to do with $\epsilon$ s ?

Yes, and your original good $\epsilon/2$ partition might include $1/2$. Then you just need to fix up two small intervals.

Hopefully the pain of that cools down soon. Things are going alright on my end

@TedShifrin good to see you're still in one piece then

Well, the major hint is that you should start with a partition $P'$ with $U-L<\epsilon/2$, @Knight. That's always the approach. Then you have some wiggle room to fix things.

Thanks, @Calvin.

You're getting to be an old man, Demonark.

Been going to the gym a bit to try and undo n years of soda, also been doing some reading with a tentative advisor!

Hahaha, yeah I'm 23 now. Bones aren't holding up like they used to

2

4:36 AM

Good. Oh puleeze. Just wait 'til you're 60+.

that previous proof that we did @Knight, which did the estimation with the gap of the partition going to 0, that might be something to revise

@CalvinKhor Really?

I don't think you need to do that.

All you need to do is modify the partition $P'$ I just named.

@TedShifrin Can you guide me to the formal proof?

ah sure yeah didnt see that.

4:37 AM

I want you to try to do it yourself first.

Assume that we have refined $P'$ and now it contains large number of partition points.

@Calvin: There are proofs like this (granted, in two variables mostly) in my lectures to which you alluded.

No, no, @Knight.

Just make a small change to $P'$ in accordance with what we talked about earlier happening at $1/2$.

Is all of this actually necessary by the way? Can't we just note that $f$ is bounded on $[a,b]$ and continuous almost-everywhere? Therefore so is $g$, and hence it's integrable on $[a,b]$? Or was the point of this to do this using the concepts being used here?

This is where pictures are good, by the way. You can draw the rectangles in there. Don't be cluttered with the picture.

I've only gone through the bits on forms

@Drathora yeah this is first-principles

4:39 AM

@Drathora: This is supposed to be a self-contained argument, I presume, not quoting a hard theorem.

Okay, let's say $1/2 \in [x_{I-1}, x_i]$ then make $x_i - x_{I-1} \to zero$, is that right?

Ah, ok, @Calvin. Some of my favorite lectures are at the end of that section.

No. Don't be vague like that. You want to refine the partition, so you will keep those points and then add one or two more.

Knight only a few pages back defined the Darboux integral, and I'm not sure if Knight is ready for continuity almost-everywhere

@CalvinKhor Yes, Calvin

Spivak I assume?

4:41 AM

Advanced Calculus by Patrick M. Fitzpatrick, I believe?

chapter 6ish

@TedShifrin Do I need to add more points to the interval $[x_{I-1}, x_i]$?

if memory serves me right

@AminIdelhaj Close, very clsoe

Ah

@CalvinKhor Yeah Calvin you know everything about me

4:42 AM

I appear to have skipped any formal mention of Darboux integrals on my journey.

it was easy to find because it uttered a rather unique phrase "Archimedes-Riemann Theorem"

@Drathora People say that it's trivial problem if we were to use Lebesgue integrals

@CalvinKhor LOL

I think Spivak just talks about "the integral" meaning Darboux but mentions Riemann sums in an appendix?

@Knight don't worry about all that rubbish about Lebesgue for now

Really my class just did the Darboux but called it Riemann

4:43 AM

Yes, it is extremely trivial in that context, which is why I was taken aback at first by the proof you were attempting, as I'm much more familiar with Lebesgue integration

@CalvinKhor Okay :)

Spivak actually added the Riemann appendix and applications because I made him, Demonark :P

4

lmao

what's up

Heya @Alekos

4:45 AM

how's it going in here?

I'm on a brief "vacation"

Finally hit 75 reputation. Might be time to burn it desperately on a bounty D:

You not traveling to Europe, I assume?

Ted, please before going back to sleep guide me with that formal proof

I guess not

...

I don't think any of us are for quite a while. :(

4:45 AM

hold up what time is it for you Ted

yeah, cornell issued some comical statement

I don't know if I read that bit, is it for applications to numerical integration?

about how their model predicts less cases if all of the students are on campus

or whatever that could possibly mean

So we had $P'$ with $U(f,P')-L(f,P')<\epsilon/2$. You said $1/2\in (x_{i-1},x_i)$. Then what are you going to do?

No, no, stuff on volumes and surface area wasn't in the early edition, Demonark.

And the exercises on numerical integration I wrote, yeah.

So far, the Regents in GA are not allowing masks to be required. The faculty are rioting, @Alekos. This country is so backward all of a sudden.

@Knight: Hint: You have to use the fact that $M_i = 20$ now.

@TedShifrin Add more points to the interval, or let the length of the interval go to zero but in decreasing the length we would increase the lengths of adjacent intervals and hence Upper sum may not be smaller than before.

4:48 AM

@Calviin: I'm in CA.

Get rid of the "length of the interval go to 0" crap.

maybe it was always backward, but it took Trump to bring the grime to the surface of the pool

That's sloppy engineering thinking.

Add more points to the interval $[x_{i-1}, x_i]$

Well, since Newt Gingrich, the congress has been unconstructive, @Alekos. It does go back to the 90's ... but not the hatred openly encouraged.

Choose a length for the interval that results in the inequality you desire holding Knight

4:49 AM

How many points, @Knight?

@TedShifrin I really don't know

Did you draw a picture?

yes

How much room do we have for $U-L$ on that one new interval if we want the whole $U-L<\epsilon$?

It's really crazy

I just hope people stay safe and healthy.

4:50 AM

Oh, there are hundreds of thousands who're going to kill a lot of us.

I'm pretty scared for me right now.

Even though I hardly do anything.

indeed. I hope I can avoid campus a lot in the fall

my father is somewhat immunocompromised and works in a hospital

so naturally I've been on edge

tbh even in MA and NY, people are not all wearing masks

@TedShifrin We have $ (20 -x_{i-1})(x_i - x_{i-1})$

It's the same in the UK, people are very indifferent. There's a lot of "if I get it I get it"

Hopefully California's doing a somewhat better job. In Texas it's getting a bit oof

It's not $x_i-x_{i-1}$ any more. You're putting in new points.

No, California is going really badly. I am convinced that the non-mask wearers are all Tromponistas.

4:53 AM

@Drathora this mentality really annoys me, you know? people should know that they have a social responsibility to not spread the disease even if they are themselves at minimal risk

Well, leadership in the UK is a carbon copy.

masks are a political statement these days (imagine that!)

@TedShifrin I cannot do more, please write it out sir.

@Drathora no book i've ever seen bothers to reprove results for Lebesgue in the case of e.g. nice smooth functions on [a,b], i think they normally prove something like Riemann integrable implies Lebesgue integrable and therefore the old results apply, so its not presented in a way that lets you forget that Riemann integrals exist

Add two points on either side of $1/2$. How far apart should we make them so that that rectangle gives $<\epsilon/2$ contribution to $U-L$?

My usual first exercise on this kind of stuff is something you should start with, @Knight. I don't think you're ready for this exercise. You have to practice easier ones.

4:55 AM

@TedShifrin btw did you ever learn grothendieck riemann roch?

Take the function $f(x) = \begin{cases} 1, & x=0 \\ 0, & x>0 \end{cases}$. Prove it's integrable on $[0,1]$.

Ah, I feel like the course I took on it didn't really refer to Riemann integrals at all

Yes, @Alekos. I even used it in a seminar talk or two.

Although a 16 lecture undergraduate course hardly spans the whole field

They passed a law in (at least my county in) Texas now where businesses have to require employees and visitors to wear masks

4:56 AM

@TedShifrin I will have $20 (x_i - x_{i-1})$ where $x_i$ and $x_{i-1}$are newly added points

Nowadays I'm in Computer Science so I'm just using everything with no regard for the foundations.

@TedShifrin oh great. I'll likely have some qns as I'm trying to learn it in the next weeks.

I heard on Rachel tonight that in Nebraska that governor has forbidden mayors and county heads to require masks. He said he would stop federal Covid funding if they do that. I just can't.

Jeez

Not that I remember, @Alekos. And I no longer have Hirzebruch's lovely book.

You can't use the same letters for the new points, @Knight. But you can fix that. So how should you choose them to get $<\epsilon/2$ for this?

4:57 AM

precisely answering "did you ever learn". love it

See, the UK doesn't have that problem. Everyone acknowledges that masks etc are a good idea. Just people are very blasé about it.

More of a "this is too much effort, I'm sure I'll be fine" than a "this mask is a breach of my freedom"

Let's call that length $\delta$ then we have
$$20 \delta \lt \epsilon /2 \\
\delta \lt \epsilon/40
$$

Right.

ahh I see. @TedShifrin

And the contribution of all the other intervals is still $<\epsilon/2$, @Knight. Why?

4:59 AM

@TedShifrin I cannot see that.

Think back to your original non-proof.

But easier.

I added more points to the partition but the difference of upper and lower sum was still less than $\epsilon/2$

We removed one piece of one rectangle, but everything else is still there.

Is it bad etiquette to post an unanswered question from the site of mine in here by the way? I've been wrestling with it for some time now and I'd love some suggestions for how to tackle it

sure, let er rip

5:02 AM

No, people do that a lot. You can link to it.

i have muted a number of users who post questions and then run away

usually people trying to post things from other sites

0

Q: If a function has a jacobian with linearly independent columns on a set B, is the following union equal to B?

Suppose we have a function $f : \mathbb{R}^m \rightarrow \mathbb{R}^n$ with the condition that the columns of the Jacobian of $f$ are linearly independent on $B \subseteq \mathbb{R}^m$. Let $B_p$ = $\{ t \in B : \forall s \in \mathbb{R}^m \; \; \lvert s - t \rvert < \frac{1}{p} \implies \frac{\l...

real-analysis measure-theory functions lebesgue-measure jacobian

but if its at least maths and you're gonna stick around, i think by all means

The idea I have for tackling this involves "hopping" in axis-aligned directions until we get from a point $t$ to a point $s$, within a small around around $t$, which should let us verify that $\frac{\lvert s - t \rvert}{p} \leq \lvert f(s)-f(t) \rvert$

So a differentiable function must be locally Lipschitz except on a set of measure 0, or something. I honestly work only with $C^1$, so I don't have good intuition. But this reminds of me of a result I knew in grad school.

5:06 AM

(the verification of that coming from the linearly independent columns of the jacobian at these points we "hop" to down the axis-aligned directions)

alright, i'm going to take off and watch some lord of the rings with the girlfriend :) talk soon

night all

Later!

@AlekosRobotis night!

Night, @Alekos.

I do know that in the one dimensional case we have equality between the union and B

5:08 AM

See you Alekos

But this equality doesn't hold in dimensions greater than 1

is there an easy example to see that?

Consider the function f(x,y) = (x,y) when xy = 0, f(x,y) = (0,0) otherwise

that function is differentiable?

Then B in this situation would be {(0,0)}

5:11 AM

I think you have rows and columns messed up. If we have $m$ linearly independent vectors in $\Bbb R^n$, then of course $m\le n$.

what ungodly computer science would need you to to consider non $C^1$ functions only differentiable on weird measurable sets

The Jacobian is defined and has linearly independent columns on {(0,0)}

Probabilistic Programming :P

actually, you mean the jacobian, as in the matrix of partial derivatives of components, not even that the function is differentiable anywhere

Yes, that's correct Calvin

Did I use the wrong term in my question? I don't really work with this stuff that much, I just happened to need it to show that a pushforward measure is absolutely continuous

Erm, one second Ted while I think about that

i guess its not wrong per sé, just that if I have a jacobian I tend to assume I also have a derivative, because I don't normally work with pathological examples

5:17 AM

The derivative map is injective.

Oh, I was assuming differentiability, just not $C^1$.

All we assume is existence of partials?

With regards to the rows and columns, the idea is that each column represents a dimension of the "output". Each row represents an "input variable"

@TedShifrin apparently

Ugh.

Yeah

No, @Drathora, you need to relearn linear algebra.

Each row represents a coordinate in the range.

5:19 AM

so your jacobian choice gives say $f(x+h) \approx f(x) + \nabla f(x)^Th$?

Unless you're multiplying by matrices on the right.

if $f$ was nice

It's important to get the shape of Jacobians right.

This is the main reason that in my book and my course I wrote vectors as columns always.

Oh right, I see what you're saying.

It doesn't affect my result, but I'll change it to be consistent with the literature

or tossing transposes all over the place

5:21 AM

For jacobians, the domain variables go across the rows, the range coordinates go down.

Well, it's important to understand injective versus surjective.

Anyhow, I'm not going to think about this further tonight.

Stuff along these lines I remember being a bit of a stumbling point

This is why I made a big point of being pedagogically logical, Demonark. My students rarely messed it up.

They messed lots of things up. Don't get me wrong. But not this. :)

haha

When I first saw differentiation in R^n I was just like wait hold on why are we taking a dot product out of nowhere

And then next quarter I was talking to some friends after we did like, Riesz rep in Hilbert spaces

Anyhow, night, folks. Talk soon!

5:24 AM

And we were like ohhhhh okay so Df is the linear functional and grad(f) is what you dot by thanks to RR okay

Later, thanks for your comments!

See you Ted!

But yeah, my idea is that since $B$ has positive measure, we can move around in a small ball centered at $t$, only moving to and from points in $B$ and only travelling in axis-aligned directions.

And eventually reach the other point $s$

Then something like the approximation you gave above Calvin will get me where I need to be

Except this "path" from $t$ to $s$ is a little hard to justify since $B$ can have some awkward shapes

night ted!

you dont get small balls yeah. If you prove this for open sets $B$ and $C^1$ functions can you then do some approximation in computer-science land

To be honest, $C^1$ functions would probably suffice for all the practical applications of this result

But as a stubborn half-mathematician half-computer scientist, I'm not overly fixated on the practical applications haha

There aren't small balls, that's true. But I think that since $B$ has positive measures, there should be maybe some "almost-everywhere ball" that we can bounce around in axis-aligned directions and eventually end up at any other point in $B$ that's within this "almost-everywhere ball"

But I could be way off on that

5:38 AM

lol

For full context, this result is what I need to show the main result I'm trying to prove, which is that a function $f$ that has a Jacobian with linearly independent rows almost-everywhere is non-nullifying (i.e. the preimage of null-sets is null)

(And base measures are Lebesgue)

Didn't expect when I entered the world of Computer Science research to spend so much time being beaten up by measure theory

maybe some geometric measure theory...? dont think the standard texts are very easy reads though

Quite possibly. Might just fix it up regarding Ted's comments and then stick a bounty on it and see what happens haha

:) gl

put a +1 on it

Thanks, that's very kind of you. I'm going to head to bed now because it's almost 7am, so I'll make the fixes etc tomorrow.

See you guys, I'll be around here now that I know it exists. Thanks again for being so helpful

5:54 AM

see ya, goodnight!

hello

anyone know where i can see a solution of mordell's technique for evaluating the gauss sum using a complex contour integral?

6:15 AM

never heard of it, sorry

6:28 AM

np!

6:51 AM

In a matrix can I add a number to a row and then add that to another row?

Example:

R2 + 1 + R1 > R1

7:26 AM

@TedShifrin I think I heard the world's smallest violin playing...

Hello Robbie sir

3 hours later…

10:13 AM

2

A: Is there any easy way to simplify the following term?

Well... This bizzare thing $$[p (p+1) (p+2) (p+3) (p+q)^3] - 4[ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ] +6[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)] -3[ p^4 (p+q+1) (p+q+2) (p+q+3)]$$ Can be split like : $T_1-T_2-3T_2+3T_3+3T_3-3T_4$ , Where $T_i$ denotes the $i^{\text{th}}$ term. $[p (p+1) (p+2) (p+3) (p+q)^...

@Beliod, you switched the sides, it should be R1ー> R1+R2+1 and no, you can't do that. That will be like adding a matrix having non zero entries only in that column (thus not the same matrix anymore)

2 hours later…

12:44 PM

What is meant by “1- periodic function” ?

I know what what periodic functions are, but what is one periodic function?

12:58 PM

a function with period 1, probably

Where did you read that?

1:26 PM

Hi ya'll
In General Topology, if U is a family of all neighborhoods of s then are all elements of U positioned within each other geometrically?

What do you mean by "positioned within each other geometrically"?

Are you asking whether they form a chain under inclusion?

Yes, sorry. Just learning.

Alessandro Codenotti

$\{(0,5), (1,6)\}$ is a family of nbhds of $\{3\}$

I see. Hence another Q: Is intersection of any members of U non-empty?

Alessandro Codenotti

1:43 PM

Surely any two member of U contain s

$\{(3,4), (2,3)\}$ is family of nbhds of $\{3\}$, but intersection is empty?

Just came up

neither of these contain $3$

so they're in particular not neighborhoods of $3$

Right. A neighborhood of a point contains an open set where the point belongs.

I am in the neighborhood of Rumpelstiltskin

Checkmate topologists

Haha
You mean you are reading the story?

1:50 PM

@BalarkaSen Qxd2!!!!!

https://duckduckgo.com/?q=qxd2&t=canonical&atb=v208-1&ia=web

??

@LeakyNun This is in reference to yesterday's game?

yeah

Well Magnus gave up the game

because Ding lost on time for bad internet earlier

2:22 PM

Why do these sound like YouTube channel

Channels*

2 hours later…

4:37 PM

@Hippalectryon believe me or not but your first name “Hippa” has taken thoughts with it down to a secret place. I cannot resist liking it.

4:49 PM

By the way Knight, I'm curious, what was the context of what you were trying to learn last night?

Are you self-teaching?

@Knight Wait what does that mean x)

lmao

Weird

What's this event?

Manolis Lyviakis

5:04 PM

is it obvious that the fundamental group of figure 8 is not abelian?

@Drathora Yes, I have no access to any teacher whatsoever.

@Hippalectryon I just named my 4 years old nice as “Hippa”

@Manolis No. You compute by passing to the universal cover.

She laughed called me Hippa

Manolis Lyviakis

ok thanks

@Knight What if I'm your niece ;)

5:06 PM

@Hippalectryon I have no problem, Hippa!

There was a anime character named Jippo

user image

ninjahattori.fandom.com/wiki/Jippo_finally_finds_Hattori

I see, but what is it you're trying to learn?

Just trying to get an undergraduate education in maths without a teacher? Or do you have a specific project in mind?

@Hippalectryon Lmao

Bringing back the low tech memes

Oh damn, that's impressive dedication

5:18 PM

Thanks. May I know something about you?

Of course

What would you like to know?

What's your current level of education?

Student, researcher, job man?

Oh oh. @Hippa is meme-ing again :)

5:33 PM

Morgen Ted

Michael Albanese

Hi @TedShifrin

Erm, PhD student

In Computer Science

My undergraduate masters was in Maths & Computer Science joint honours though

My research in computer science is very maths oriented though. I study probabilistic program via measure theory

Oh, that reminds me, I should make those edits so that Ted doesn't banish me for my scuffed linear algebra

@Drathora Thats an awesome educational background.

Wowee

I think that Maths & Computer Science is a very nice foundation for Computer Science research

I'd feel quite incapable of doing my current research if I'd done sole Computer Science, unless I did a lot of extra catching up

Although it has resulted in quite a few gaps on either side. I have no background whatsoever in any kind of geometry, calculus or really any kind of applied mathematics

@Drathora Do you like Real Analysis?

5:45 PM

Yes, I teach a bit of it to our undergraduates

Are you a teacher?

The university pays PhD students to do marking and teaching where required

For when professors are busy, or there aren't enough professors to do the required undergraduate teaching

That's pretty cool

So I've taught some real analysis, complex analysis and then lambda calculus & types

How much do they pay though?

5:47 PM

Hi, @Knight, @Michael

Michael Albanese

I don't suppose you're familiar with the book Compact Complex Surfaces by BHPV?

Good question user. I honestly don't know off the top of my head. I find teaching the undergraduates more fun than my research a lot of the time, so I didn't read what the deal was too hard

And they just throw a lump sum to me at the end of each term depending on how much teaching I've done, which I've never bothered to calculate

Hi @Ted, @Michael

Michael Albanese

Hi @BalarkaSen

@Drathora That means you can help me with analysis problems :) ?

5:49 PM

@Michael: Um, no. By whom?

Hi, a @Balarka!

Sure, to the best of my abilities. I may be a little rusty in places but I'm sure I can help most of the time

@Knight, xD

Michael Albanese

Barth, Hulek, Peters and Van de Ven

They use both 'smooth surface' and 'non-singular surface'. I don't know if these are supposed to be distinct concepts.

I know of the authors (most), of course, but no, I don't know the book. I think they're synonyms, unless they have some scheme-theoretic convention (perhaps $x^2=0$ is a smooth, but non-reduced, hence singular, subvariety). I doubt it.

Alright, the Jacobian in my question is now actually a Jacobian. We're making progress haha

5:56 PM

@Drathora: Presumably you can do it if you assume $C^1$? Hard to believe you'd have an applied setting where the function has discontinuous partials.

Michael Albanese

OK, thanks, that's what I figured. I don't think they use such a convention.

Does anyone know of an example of a manifold which is not a connected sum, but admits a cover which is a connected sum?

I mean, I certainly have done such things. It gets boring using identical phrases over and over and over. :)

Oh, that's a good question for @Balarka, @Michael.

You don't mean a branched cover.

Michael Albanese

Fair enough. Just seems like a confusing thing to do if the phrases are within the same page.

No, a genuine cover.

There are examples of connected sums which admit a cover which isn't, e.g. $\mathbb{RP}^3\#\mathbb{RP}^3$ admits $S^1\times S^2$ as a double cover.

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