Mathematics - 2023-12-14

mick

00:03

This confused me

0

Q: About the Lacunary $f(z)$ such that $f(z) = b^{-1}(a(z)), a(z) = \sum_{0<n} \frac{x^{n^2}}{(4n)!} , b(z) = \sum_{0<n} \frac{x^{n^2}}{(4n+3)!}$?

Let $$a(z) = \sum_{0<n} \frac{x^{n^2}}{(4n)!}$$ $$b(z) = \sum_{0<n} \frac{x^{n^2}}{(4n+3)!}$$ Both have a natural boundary at $|z| = 1$. Let $c(z)$ be the series reversion of $b(z)$. So $c(z)$ is the functional inverse of $b(z)$. See : https://mathworld.wolfram.com/SeriesReversion.html Does $f(z)...

Jakobian

00:21

@Joe it sounds good though. I wish I had a book like this to learn from

monoidaltransform

is every holomorphic complex bundle over $\mathbb{C}$ trivial?

Bob

Hi

is it fair to post a problem and a solution and ask if the solution is correct?

it seems to me that it is, but assuming it is, how should the person answering the question respond?

leslie townes

i dunno about 'fair,' i think such questions are discouraged. one reason sometimes given is that they are really many questions in one - roughly, "for each step of the argument, is this step correct?" when the site policies call for one question per post.

if there is focus on a particular part of an argument, with background for why you think it might or might not hold up (e.g. a more general situation where the same type of argument does not work), such a question would be clearly OK.

Bob

the reason I do it is that I want somebody to check my work

leslie townes

i'm not a mod or even a high rep user, but my impression that "please check my work" is more or less exactly what various site policies are designed to discourage. i don't know if there is chapter and verse on this issue.

but if it's really "please check this step in my work, for reasons XYZ i am unsure of it, i only know how to justify it in situation W", that's fine. and the answer would engage with that.

Bob

00:37

I have had good responses when I post a question, my answer and the book's answer where my answer is wrong

personally, of all the stack exchanges, I think math is the best

leslie townes

separately from what the site would or would not allow or encourage, "please check my work" questions tend to get less engagement than more focused questions.

Bob

okay

Ted Shifrin

01:20

@Jakobian How does a book force you to do exercises? That’s silly.

Read Dieudonné for succinct, clear exposition in total generality.

4

Xander Henderson

01:36

@Bob Math SE is not a work-checking site. "Check my work" questions are not generally on topic.

peek-a-boo

@TedShifrin probably for the first volume. The later ones have random references to previous volumes :(

fortunately though I have access to all his volumes

Ted Shifrin

I actually bought the first 4 (all there was) in French in Paris many years ago.

@XanderHenderson Those and no-effort homework posts probably account for 80% of the posts these days.

peek-a-boo

01:56

it’s weird how all but one volume are available in English

oh that reminds me, is there a way/some standard reference for an abstract definition of ‘manifold with ___ boundary’ where ___ is some type of regularity? We all know the notion of smooth manifolds with boundaries modelled on $H^n$, and also manifolds with corners. Also, if we have a smooth manifold $M$ sitting in $R^n$, then we can characterize smoothness of the boundary in terms of graphs of functions of corresponding regularity ((piecewise-) Lipschitz/Holder/smooth etc)

monoidaltransform

anyone know if my last question is true? certainly true for vector bundles and fiber bundles but can trivialization be taken to be biholomorphic?

John Zimmerman

Xander have you ever wondered about functions defined by integrals of the form

$$\zeta(s) := \int_{A_\delta} d(x,A)^s \,\mathrm{d}\mu(x),$$

where $d$ satisfies a linear pde, and what effect the analytic continuation has on the pde?

Ted Shifrin

I do not know an answer, @peek-a-boo. I used to know about Whitney conditions for stratified (analytic ) spaces, but not this.

@monoidaltransform Why not? Local inverses of holomorphic bijections are holomorphic.

monoidaltransform

@TedShifrin Yes I agree with that statement but I don't see how the result follows. Could u please perhaps elaborate more on the intuition?

peek-a-boo

:( that’s sad. It’s not even that I need the theory in full generality, because in any given concrete situation I can usually work things out and check that things behave the way I want them to (same thing with ‘improper’ Stokes/divergence theorems), but it would be nice to avoid having to take this detour every so often

Ted Shifrin

02:11

Work with integral currents, peek-a-boo.

What is your concern, monoidal?

monoidaltransform

My concern is that I don't know of any other way of showing result without going over complete proof in the reference im using and make sure everything can be holomorphic. Ur answer suggests that there is more of bigger picture thing involved which i dont quite see

I think for real vector bundles the standard argument is using partitions of unity

in case of complex manifolds no tool exists

Ted Shifrin

The usual smooth stuff is based on partitions of unity. The holomorphic category is different. But stuff on Stein manifolds is well-known. Gunning & Rossi is a standard sirce.

monoidaltransform

On general complex maniflds its true?

Ted Shifrin

General? Of course not.

This is the opposite of projective submanifolds.

monoidaltransform

Sorry, I mean complex holomorphic vector bundles over contractible complex manifolds

Ted Shifrin

02:20

Stein is the key word.

monoidaltransform

aha I see. Thnx!!

Ted Shifrin

Happy learning!

Xander Henderson

02:53

@JohnZimmerman $d$ is the distance function, i.e. $d(x,A) = \inf\{d(x,y) : y \in A\}$, where the latter $d$ a metric on some metric space...

So... uh... no?

But the big theorem in the field is (more or less) that the Mellin transform of that function is meromorphic on some domain, and its poles correspond to the spectrum of the Dirichlet operator.

@TedShifrin Yeah, it is a frustration.

Ted Shifrin

03:34

@XanderHenderson I know nothing about Mellin transforms, but this sounds neat.

Xander Henderson

@TedShifrin It is neat. I am not nearly smart enough to figure it out. :/

Ted Shifrin

I’m not smart anymore.

Maybe I’m dumb enough to serve in Congress.

Xander Henderson

Heh.

Right... it is quite dark out. Bedtime. G'night.

leslie townes

04:03

dieudonne is the one good bourbakist.

one potato two potato

For $f\in H^2(\Bbb R^3)$, there exists a fixed constant $C>0$ such that $\Vert f\Vert_\infty\leq C(\Vert(-\Delta f)\Vert_2+\Vert f\Vert_2)$. Is this a known inequality?

oscarmetal break

04:59

Can I conclude that there are 8 sylow-7 subgroups in $GL_3(F_2)$, without explicitly writing two matrices of order 7 that are not conjugated to each other?

Ted Shifrin

Who knows. You could conceivably assume normality and arrive at a contradiction.

Are you trying to prove the group is simple?

oscarmetal break

No, it is part of the exercise of classifying all Sylow subgroups of $G$ but proving that G is simple does give what I want though

Ted Shifrin

05:17

If it were normal, you might get something by acting on the quotient group, but this may be harder.

oscarmetal break

I am not sure if the question in my review was designed to ask us to directly prove this, since just in the next part, it gives two explicit matrices and asks to prove they are order 7 and not conjugate to each other.

And as you may expect these two matrices are indeed order 7 and not conjugate to each other but it is next part

Ted Shifrin

You can see they have different Jordan form, I suspect.

oscarmetal break

Seeing they have different Jordan forms without knowing some explicit form of the matrices is possible?

Ted Shifrin

You said they were explicit?

oscarmetal break

Yes but it is second part of the question like the question is split in (a) (b), then (a) is to prove how many sylow-7 subgroups in $G$, and (b) is to prove the two explicit matrices are order 7 and not conjugate with each other..

It is awkward to take the second part of the question into the assumption of the first part although it is possible

I have the same problem in the latter part of the question. Where I have proved there is a subgroup of index 7 that is isomorphic to $S_4$ in $G$, then I need to use this fact to deduce the number of sylow-2 and sylow-3 subgroups and so on

I knew this must come out from the fact that there are 3 sylow-2 subgroups and 4 sylow-3 subgroups in $S_4$, but I don't know why it is 3 and 4 instead of 3 and 1 or something...

Ted Shifrin

05:40

Oh, have you tried the usual element-counting game?

oscarmetal break

Like summing the numbers the elements of different order and show the sum is the order of $G$?

But how exactly to start with?

Ted Shifrin

If there’s only one subgroup of order $7$, can you fill up the group?

oscarmetal break

What do you mean by fill up?

Ted Shifrin

Account for $168$ elements.

oscarmetal break

not sure how to fill up

ChoMedit

07:29

When you need to read huge amount of papers, like a thousands-page-book, to study or begin your career, what would you do? just read all things?

Mad

10:25

I am having a hard time to understand, why we use the derived lie Algebra [L,L] to classify Lie algebras, whatever "classify mean"
https://mathweb.ucsd.edu/~abowers/downloads/survey/3d_Lie_alg_classify.pdf
"Using this result, we can classify a given Lie algebra by finding a suitable
basis so that its Lie bracket multiplication table will be in a “canonical”
form. These canonical Lie algebras will be shown to be non-isomorphic by
considering certain invariants of isomorphic mappings. One such invariant

Mad

10:43

https://math.stackexchange.com/a/2103485/695930
can i get a tip here on how to prove that for dimension 2 and 1 that [L,L] is an ideal?

Thorgott

it's always an ideal

Mad

Oh you are right, i got confused

I meant to say, how to prove that for dimension [L,L]= 2 and 1, that L is solvable.

I get it for 0 it needs to be abelian, for 3 its simple, apparently for 2 and 1 it is solvable.

"because any non-trivial ideal I
would be of dimension ≤2
, hence solvable with quotient of dimension ≤2
. Then g
would be solvable,"
what does he mean with the qoutient?

oh i think i understand

One dimensional and two dimensional Lie algebras are abelian and thus solvable

In Two dimensions theres one more case, okay i got it.

Jakobian

11:58

@ChoMedit are we talking about mathematics?

there are things you just need to pull through and study

but sometimes sacrifices are needed, that you hopefully fill in later

when the right way fails, you need the helpful hand of the devil to allow you to progress faster

@TedShifrin I've read the introduction, and I like the way that the book is trying to present itself. And I also like the way that derivatives aren't introduced right away, instead the concept of a function being tangent to another function is introduced. It's such a little thing but very insightful

mick

0

Q: About the Lacunary $f(z)$ such that $f(z) = b^{-1}(a(z)), a(z) = \sum_{0<n} \frac{x^{n^2}}{(4n)!} , b(z) = \sum_{0<n} \frac{x^{n^2}}{(4n+3)!}$?

Let $$a(z) = \sum_{0<n} \frac{x^{n^2}}{(4n)!}$$ $$b(z) = \sum_{0<n} \frac{x^{n^2}}{(4n+3)!}$$ Both have a natural boundary at $|z| = 1$. Let $c(z)$ be the series reversion of $b(z)$. So $c(z)$ is the functional inverse of $b(z)$. See : https://mathworld.wolfram.com/SeriesReversion.html Does $f(z)...

complex-analysis taylor-expansion inverse-function analytic-continuation lacunary-series

I edited this one

PM 2Ring

14:26

@mick You can see recently deleted posts using the so-called moderator tools. math.stackexchange.com/help/privileges/moderator-tools You can see other deleted questions if you've saved a link, otherwise you're out of luck, because normal users cannot search for deleted questions.

I guess they might also be visible using the Wayback Machine, but I've never tried doing that. (I don't have enough points to see deleted posts on Math.SE, but I do on 4 other sites).

one potato two potato

15:17

$H^2(\Bbb R^3)\hookrightarrow L^\infty(\Bbb R^3)$?

11 hours ago, by one potato two potato

For $f\in H^2(\Bbb R^3)$, there exists a fixed constant $C>0$ such that $\Vert f\Vert_\infty\leq C(\Vert(-\Delta f)\Vert_2+\Vert f\Vert_2)$. Is this a known inequality?

o.w. this is false

Jakobian

16:21

Interesting how for Banach spaces differentiability doesn't imply continuity

because of how linear maps can be discontinuous

If $f:U\to Y$ is differentiable at $x\in U$, then $f$ is continuous at $x$ iff $f'(x)$ is a continuous linear map

Jakobian

16:32

I mean, Dieudonne only talks about derivatives of continuous maps anyway, but I don't know why require that

Ted Shifrin

I forget Dueudonné’s definition, but usually it’s assumed that the derivative is a bounded linear map.

Jakobian

Its not that $f'(x)$ is assumed to be continuous, but $f$ itself is

I don't need that much assumptions though

Ted Shifrin

In fact, Dieudonné proves that if $f$ is continuous and $f'(x)$ exists, then $f'(x)$ is a continuous linear map.

It's immediate from continuity of $f$ at $x$.

Jakobian

16:48

I know, I'm just saying that assuming continuity of $f$ is a bit excessive

its fine if its at one point

Ted Shifrin

As I said, it's customary just to assume the derivative is a bounded linear map. Lang does this, for example. I no longer have hundreds of books to check.

I don't find this an interesting question, anyhow.

Jakobian

Assuming continuity of $f$ at $x$ is fine. Assuming it on the whole domain?

Dieudonne isn't assuming continuity of $f$ at $x$ alone, but on the whole domain

Ted Shifrin

He's not writing for a beginning calculus student. Anyhow, enough.

Jakobian

I don't see how that's a point to make

Ted Shifrin

I wrote my book for students first learning multivariable calculus, and, since it was a proof-oriented course, I wanted them to play with examples where partial derivatives exist and the function is not continuous ... and more involved versions of that. I don't know what Dieudonné's rationale was, but he clearly has a different pedagogical goal.

Xander Henderson

16:54

@user85795 Your soul.

Ted Shifrin

It would appear @Xander had a good night of sleep and is restored to his usual essence.

Xander Henderson

Yay!

Though I discovered this morning that there is no more coffee in the freezer. Which is a problem. I have enough beans for one more cup. :(

Jakobian

@XanderHenderson I have 4 different kinds of ground coffee if you need a drink

Tchibo, Jacobs, Cafe d'Or and Woseba

Xander Henderson

@Jakobian Given that you are not in the same physical space I am in, that won't help.

Also, I buy whole beans.

Jakobian

I'll e-mail you

Xander Henderson

17:01

I do have a small stash of good instant coffee for emergencies. I also have tea, which will work in a pinch.

Ted Shifrin

@Xander I do, too, but I'm shocked you let the supply dwindle so far! Where do you buy your beans?

I cannot abide instant coffee. Epic fail.

Xander Henderson

@TedShifrin I get it from a roaster up the mountain from me. I usually buy 5 lbs at a time, divide it into 8 oz baggies, and freeze it (shipping is expensive, so I'd rather get more, and accept that the quality will decline a bit).

Jakobian

I have to agree, instant coffee is borderline disgusting

Xander Henderson

pinetopcoffeehouse.com

one potato two potato

The stated inequality is true

Xander Henderson

17:03

And the instant coffee I have is a stupidly expensive product which is... okay.

There does exist acceptable instant coffee. It is just hard to find.

Jakobian

I'd just use ground coffee if I had an alternative

I don't drink coffee from beans unless I go to the store and buy one

I've used to drink instant coffee until I swapped and its better. It could probably be even better if I were to drink one from bought beans. shrug

Xander Henderson

@Jakobian I mean, I make coffee with ground beans. But I buy the beans whole, and grind them myself.

Ted Shifrin

I get whole beans 3 or 4 pounds at a time and freeze (which I know is a bad bad boy thing to do). Years ago I used to get coffee shipped from a roaster in SF, then for many years I got it from Anderson's in Austin, TX. Now I use Peet's, which I like a bit less but ...

Jakobian

Oh, weird. The sound effect restarted and it notified me

Xander Henderson

@Jakobian !

@Jakobian !!

@Jakobian !!!

:P

Jakobian

17:09

It notified me twice actually

Ted Shifrin

smacks @Xander @Xander

Xander Henderson

@TedShifrin After moving out to Arizona, I continued to order coffee from the place where I spent many hours writing my thesis. But they didn't survive COVID (and, it turns out, probably had some shitty labor practices, too).

@TedShifrin AGAIN!

oscarmetal break

@TedShifrin Hi ted, I still don't get that how to fill up

Jakobian

@TedShifrin I'll keep on doing as little assumptions as possible, to see if this breaks somewhere along the way. That sounds more educational to me

Ted Shifrin

For someone with your interests and background, that is an excellent approach, @Jakobian.

@oscar This is a bad example to learn on because the group is so large. Basically, you want to consider elements of different orders and figure the maximum number of each you can have. Are there enough elements to account for all the elements of the group?

I last taught algebra 11 years ago, so I don't have immediate recall of examples I used to use. I'll see if I can give you one.

oscarmetal break

17:25

Thanks ted.

Joe

@Jakobian: I think if you drop the assumption that the derivative has to be a bounded linear map between normed spaces, then the derivative operator $D:C^{\infty}(\mathbb R)\to C^{\infty}(\mathbb R)$ is itself differentiable!

Ted Shifrin

17:40

@oscar Let's try this. Take a group $G$ of order $30=2\cdot 3\cdot 5$. Could the subgroups of order $3$ and $5$ both fail to be normal? Then we would have $6$ subgroups of order $5$. This accounts for $1+24$ elements. But we also have $10$ subgroups of order $3$, so this accounts for $20$ more elements. Way too many elements. So one of those must be normal. .... This is the sort of argument I had in mind.

Probably it doesn't work easily to argue that you cannot have a normal subgroup of order $7$, as you don't have simple control over the orders of all the elements.

@Joe That's an interesting point. But $C^\infty$ isn't a Banach space. So I'm confused.

Joe

17:57

@TedShifrin: From what I've read, I don't think you need a Banach space structure to define the Frechét derivative. See Wikipedia's definition for instance. I don't know the details, but I would assume that assuming that the normed spaces are complete just gives you nicer theorems

Jakobian

I think you need it on the domain

Ted Shifrin

Oh, I've never even thought about it in a non-Banach setting.

Jakobian

the image doesn't need it, but how are you going to divide by $\|x-x_0\|$ if the domain isn't Banach?

Ted Shifrin

This is all because Jakobian is a trouble-maker :D

oscarmetal break

Thanks, ted. I will think about this later. My hands and brain are frozen now because of the cold and the upcoming final in the next hour. :)

Ted Shifrin

18:00

@Jakobian But when you write down the definition, the limit is occurring in the range, not the domain, no?

Jakobian

You have $\lim \frac{\|f(x)-(f(x_0)+u(x-x_0))\|}{\|x-x_0\|}$

so the norm of both the domain and range is present

it won't be an issue to replace the norm in the range by a semi-norm and go over all the semi-norms

but in the denominator you divide by $\|x-x_0\|$ so its problematic

Joe

@Jakobian: I don't really see the problem with dividing by $\|x-x_0\|$

Jakobian

@Joe why not

I think I've explained myself well

Ted Shifrin

I don't see your point, either.

Jakobian

You can't replace $\|x-x_0\|$ by some kind of distance from $0$ either. You want the properties of a norm

Division by $0$?

Ted Shifrin

18:04

Oh, Joe wants a norm. He just doesn't want completeness.

But good point. There is still no norm on $C^\infty$, which was my point.

Jakobian

Oh. I was thinking he was trying to define Frechet derivative on a Frechet space

Ted Shifrin

No, he was removing completeness, but his example still fails.

Joe

Whoops.

I think if you replace $C^{\infty}(\mathbb R)$ with $C^{\infty}([a,b])$, then you can use the supremum norm.

Ted Shifrin

@Jakobian By the way, when you get to the Inverse Function Theorem, make sure you google and read Terry Tao's improvement on it. If I had known about it when I wrote my book, I would have included at least a comment on this, if not a (hard) exercise.

No, @Joe, you need norms on all the derivatives.

The domain isn't the issue.

Joe

But if you are thinking of the derivative operator as map from $C^{\infty}([a,b])\to C^{\infty}([a,b])$, then you have norms on both sets, and you can talk about derivatives, surely?

Ted Shifrin

18:11

We don't have norms on both. If you do $C^k$, sure.

Jakobian

@TedShifrin for Frechet derivatives?

Ted Shifrin

@Jakobian I forget whether he did it just in $\Bbb R^n$ or more generally. But he replaces the hypothesis that $f\in C^1$ with invertible derivative at $a$ with the hypothesis that the derivative is invertible at every point in a neighborhood of $a$.

leslie townes

terrytao.wordpress.com/2011/09/12/…

Joe

Surely the supremum norm is a norm on $C^{\infty}([a,b])$ though?

I just mean, on the space $C^{\infty}([a,b])$, the norm of $f$ is defined as the maximum value of $|f|$ on $[a,b]$, which exists due to the continuity of $f$. Then, the derivative operator is defined as the map $C^{\infty}([a,b])\to C^{\infty}([a,b]), f\mapsto f'$ (at the endpoints we take the continuous extension of $f'$ on $(a,b)$ to $[a,b]$).

Then, we can talk about the Fréchet derivative of $f$. This doesn't exist in the usual sense, since $f$ isn't even continuous.

Ted Shifrin

Your norm is the $C^0$ norm, nothing to do with derivatives. Totally wrong notion.

Jakobian

18:23

@TedShifrin That sounds about right. The condition that $f'(x)\neq 0$ in the one-dimensional case corresponds to $\text{det}(f'(x))\neq 0$ in higher dimensions

Ted Shifrin

You're wanting to include $C^\infty$ as a dense subset of $C^0$ and use the $C^0$ norm? The whole point of $C^k$ is to govern the behavior of the derivatives.

@Jakobian Sure, but the usual proofs all use continuity of $f'$ to get started.

Sine of the Time

18:47

Hi, I'm having trouble understanding the following problem. Consider the Hilbert space $H=L^2([-1,1])$ with the usual inner product, $f(x)=\sqrt{|x|}$ and $V=\text{span}\{1,x,x^2\}$. I have to find the projection of $f$ on $V$ and $V^{\perp}$. Now, solving a linear system, I found the projection on $V$ and since $V$ is a subspace and it's closed I can use the orthogonal decomposition

Is it correct to say that the projection on $V^{\perp}$ of $f$ is $f-g$ where $g$ is the projection of $f$ on $V$ or I have to find a basis for $V^{\perp}$ and express the projection in terms of the vector of the basis?

Ted Shifrin

You'll never find a basis for $V^\perp$. You can try, but you'll take infinitely much time.

Personally, I would use Gram-Schmidt to find an orthogonal basis for $V$.

Sine of the Time

Yes it's an option, but I used the matrix of inner products and found the projection on $V$

Ted Shifrin

Ah, that's what you meant by solving a linear system. OK.

Sine of the Time

Yes. So is it correct to write that the projection on $V^{\perp}$ is $h=f-g$ where $g$ is the projection on $V$ ?

Ted Shifrin

Absolutely.

Sine of the Time

18:58

thank you :)

Ted Shifrin

That's the definition, essentially, of a projection.

$\text{proj}_V x$ is the unique vector in $V$ so that $x-\text{proj}_V x \in V^\perp$.

Jakobian

Hilbert projection theorem

In mathematics, the Hilbert projection theorem is a famous result of convex analysis that says that for every vector x {\displaystyle x} in a Hilbert space H {\displaystyle H} and every nonempty closed convex C ⊆ H , {\displaystyle C\subseteq H,} there exists a unique vector m ∈ C {\displaystyle m\in C} for which ‖ c − x ‖...

depends on how you define it, this theorem says both definitions are equivalent

Ted Shifrin

What we're doing here is linear algebra. It has nothing to do with convex analysis.

Sine of the Time

yes, my doubt was if I have to compute it by hand with a basis or just write it using the decomposition theorem. As Ted said, it'd take an infinite amount of time to find a basis for the orthogonal complement

Jakobian

I wouldn't call it linear algebra, rather I'd call it functional analysis

Ted Shifrin

19:02

Oh geez. Do you have to quibble over everything?

Sine of the Time

-_-

Ted Shifrin

Not you, Sine.

Jakobian

@TedShifrin Since $(V^\perp)^\perp = V$ because $V$ is closed, and using either definition (helping ourselves with Hilbert projection theorem if necessary), from what Ted said here it follows that $h$ is projection of $f$ onto $V^\perp$

(finite-dimensional subspace of normed space is closed)

mick

19:22

I have the privalidge of 10000 and even 15000 since i have 15.4k now.
But I do not know how to find it.

oh wait i found it !

yay lol

finally starting to understand the site after 15.4 k :)

hmm i see deleted posts , but not a distinction between questions and answers ?

Jakobian

19:40

The reason why we want to define $f:U\to F$ as differentiable at $x_0$ iff $f$ is continuous at $x_0$ and linear map such that ... exists, is that without it the chain rule doesn't seem to hold.

Ted Shifrin

The proof I know of the chain rule needs continuity of the inner function, but it also needs boundedness of the derivatives.

Joe

19:57

@TedShifrin: I think that if $f:U\to V$ is a map between normed spaces, and $f$ is continuous at $x\in U$, then this already implies that the linear approximation to $f$ at $x$, if it exists, is bounded.

Koro

@mick one can usually see deleted posts even if their rep is <10k. There are ways to do so.

Ted Shifrin

@Joe Dieudonné makes that argument, albeit assuming Banach. But the completeness is irrelevant here.

Jakobian

If the approximation $f(x)\approx f(x_0)+u(x-x_0)$ holds where $u$ is linear, then $f$ is continuous at $x_0$ iff $u$ is continuous

Ted Shifrin

Right.

Joe

Nice. I have a habit of assuming that the normed vector spaces are finite-dimensional anyway. Then, we don't need to worry about discontinuous linear maps, and all norms are equivalent, which is useful for practical computations. I'd be interested in applications of the infinite-dimensional theory, though

Ted Shifrin

20:06

Oh, there are plenty of applications. The theory of ODE relies a lot on calculus in various function spaces (including $C^1$ and $C^2$). One often wants the implicit function theorem or fixed points of contraction mappings, so the completeness is germane.

Thorgott

all vector spaces are finite-dimensional until they aren't

2

Charlie

20:44

Could someone give me a hand with this, I don't really feel like I understand how to prove surjectivity properly. The specific case I'm considering is proving that for a topological space $X$ and topological group $G$ that for fixed $g \in G$, the right action on $X$, namely $p\mapsto p\cdot g$ with the standard axioms is a homeomorphism. I don't really understand how you show that every point in $X$ is mapped to by at least other point in $X$?

I can start with assuming that no such point exists, but I'm not really sure where to go from there

Jakobian

@Charlie Prove that $p\mapsto p\cdot g$ is invertible by showing that a certain obvious map is its inverse

Charlie

Oh I hadn't considered that that was an equivalent statement

leslie townes

about "I don't really understand how you show that every point in $X$ is mapped to by at least other point in $X$," do you mean that you aren't sure why p mapsto p.g is surjective?

Charlie

I get that $g^{-1}$ is the map

Jakobian

Yeah

I think you want $G$ to act continuously on $X$?

Charlie

20:48

Yes

Jakobian

Yeah then all is fine

Charlie

@leslietownes Yes and also how to show it

Is it not possible for me to construct an inverse of a map without the inital map being surjective?

Ted Shifrin

Nope.

Charlie

Couldn't it only be defined on a subset of the codomain of the initial map?

Ted Shifrin

Not if you pay attention to domains/codomains of functions.

Jakobian

20:49

Yes, but you have composition being identity from both sides

Ted Shifrin

By definition, if $f\colon X\to Y$ has an inverse, that inverse is defined on all of $Y$.

Jakobian

If $A_h:p\mapsto p\cdot h$, then $A_g\circ A_{g^{-1}}$ and $A_{g^{-1}}\circ A_g$ are both identity maps

Ted Shifrin

For example, if you consider $e^x$ as a mapping from $\Bbb R$ to $\Bbb R$, it does not have an inverse.

Jakobian

not just one of those compositions

This forces $A_{g^{-1}}$ to be inverse of $A_g$

Charlie

But a map that is just injective has an inverse

Or maybe I'm semantically wrong and it "is invertible"

Jakobian

20:51

@Charlie We usually don't call that an inverse, unless you specifically restrict the codomain

Charlie

Hmm ok

leslie townes

you seem to be fixing g throughout, so maybe it helps to simplify the notation, and write the map from X to X given by p mapsto p.g by something something like f. given x in X, can you fill in the blank with something in X that makes f(_) = x true?

Charlie

@Jakobian I think this answers my question

@leslietownes $x\cdot g^{-1}$?

I think I might just need to think about it for a minute, I think Jakobians comment above makes it clearer

leslie townes

yeah. so that's a proof that f is surjective. if you wanted to belabor everything you might write out what you're assuming about cdot that makes that work.

Jakobian

When talking about inverses of maps, the codomain and domain are very important

Ted Shifrin

20:55

In your case, you don't have to be so fancy. Just use the fact that $gg^{-1}=e$ and $p = p\cdot (gg^{-1}) = (p\cdot g)\cdot g^{-1}$.

Jakobian

You don't want to think of maps as just binary relations, you want to think of them as having both domain and codomain

Ted Shifrin

All we're using is the definition of group action (which involves this "associativity").

Too many people talking to Charlie at once. I'll go make my lunch.

Charlie

Am I right so say that if $f:X\rightarrow Y$ and I can construct a map $g:Y\rightarrow X$ such that $g\circ f=\mathrm{id}_X$ then $f$ is necessarily surjective?

Jakobian

@Charlie No, thats for injectivity

Charlie

Damn

Jakobian

20:58

Surjectivity would be $f\circ g = \text{id}_Y$

Charlie

Ohhh

Jakobian

Here we have both of these relations, and it implies that our maps are bijective

in fact, the maps involved are continuous, so they are homeomorphisms

Charlie

If we only had $f\circ g=\mathrm{id}_Y$ alone that would only show surjectivity and not injectivity right?

Jakobian

Yes

Charlie

Surjectivity of $f$ that is

Ahhhhh

Jakobian

21:00

Surjectivity of $f$ and injectivity of $g$

Charlie

That's cool that the requirements are kind of dual to each other like that

I do see what you mean though

Jakobian

One almost wants to mention term category theory. Almost

Charlie

So the fact that I can construct a map $g(p):=p\circ g^{-1}$ and that $f(g(p))=(p\circ g^{-1})\circ g=p$ proves that the map $f:p\mapsto p\circ g$ is surjective

Thorgott

use \cdot not \circ

Jakobian

Yes

Charlie

21:03

Oh yeah cdot is better I mistyped it

Awesome

Sha Vuklia

I come here to confess that only recently I learned how to count (I started tutoring a student in elementary combinatorics recently, which prompted me to finally actually learn combinatorics)

Ted Shifrin

22:14

@ShaVuklia Counting is hard!

Sha Vuklia

It is :) It never appealed to me during my undergrad years, and after a short (failed) attempt at a combinatorics grad course, I thought counting is just not for me - but the field of discrete mathematics seems to be booming right now, and many mathematicians around me are into it, so I've finally picked up a bunch of books, and I'll be taking my first course in combinatorics next semester (the course is a mixed bag with emphasis on the probabilistic method)

Ted Shifrin

I never took or taught combinatorics, but I was certainly proud of myself when I figured out stars and bars for myself in grad school in trying to derive the formula for $h^0(\Bbb P^n, \mathscr O(k))$ — in other words, for the dimension of the space of homogeneous polynomials of degree $k$ in $n+1$ variables. Only years later did I discover that the method was called stars and bars.

Sha Vuklia

Nice :D

Rithaniel

Hola math chat

Thorgott

22:31

@TedShifrin I always have to rederive that on the spot, never remember it

@Rithaniel hey

Rithaniel

How's it goin', Thorgott?

Thorgott

22:47

pretty stressed, but fine. currently preparing a seminar talk for next week. what about you?

Rithaniel

Finishing up the semester. Got through most of the stress earlier in the week. Had a poster session for a non-math topic on Tuesday, and, in the process of preparing for that, I learned that there are academic skills I haven't practiced that much in my math career (such as arguing something without absolute proof)

Currently finishing up a last couple of assignments in the factorization course and stuck trying to prove that Prüfer is equivalent to all overrings being integrally closed

oscarmetal break

23:13

I just wrote the exam. A complete misery.

Ted Shifrin

@Thorgott It's indelibly imprinted upon my brain. Probably due to my having figured it out for myself when I needed it. I think I must have known the answer from my first complex manifolds course in college and then algebraic geometry with Hartshorne, but I had to figure it out for myself to learn it.

@Rithaniel You mean like giving intuition or heuristics?

Oh, and hi.

Rithaniel

Heya Ted

Ted Shifrin

@oscarmetalbreak Sorry to hear.

oscarmetal break

Well the questions came from review, but not from review :(

Ted Shifrin

I don't understand.

Rithaniel

23:17

But yeah, like the poster was talking about a potential study to identify personality traits common to successful math students, and I had to explain why knowing those personality traits is helpful when teaching math, and why everything connects together in the ways that my work claims it did

oscarmetal break

For example, $S_6$ contains two subgroups isomorphic to $S_5$ and not conjugate to each other is a question in my review and I remembered I have done it. But what is different from the exam to the review was that in the review this question is separated into serval parts and I was guided from step to step, while in the exam this is the question itself and apparently I fail to work it out.

I also failed to prove in a commutative ring, a prime ideal is a maximal ideal... frustrated since I was reminded this yesterday, although I did prove the converse I guess.

Ted Shifrin

@Rithaniel Not to mention that people think about math (and learn) in different ways ... some formal, some algebraic, some geometric ...

No, that's wrong. What we discussed yesterday is that in a PID every prime ideal is maximal. It's certainly false in general.

Rithaniel

23:32

Yes, that's the bigger topic that the study was meant to be leading into. However, advice I kept getting was that I was biting off way more than I could I chew in a one-semester course, if I were trying to tackle "what are the different ways that people learn?"

Ted Shifrin

I got flack from some of my colleagues for approaching abstract algebra in a more geometric way when I teach (like fiber bundle pictures for $G \to G/H$ and color-coding according to the point in the quotient), but it clearly resonated with a number of the students.

oscarmetal break

@TedShifrin emmm, unless I missed something. What if $R$ Is a finite commutative ring?

I spotted a similar question in my review that says, let $R$ be finite commutative ring, then $R$ is an integral domain iff $R$ is a field

Ted Shifrin

A finite integral domain is always a field. It is true in a quotient of a PID.

Yeah, that question you just quoted is correct ... :)

oscarmetal break

23:50

I also did this on my review with the hint that consider the map $x\to ax$ for $a\in R-{0}$. I should have remembered this so that I can prove if it is a integral domain it is a field...

I am speechless to myself :)

mick

0

Q: Continuation beyond the natural boundary with a limit? $f_+(z)=\lim_{x \to 0+} f(z,x)=\lim_{x \to 0+}\sum_{n=1}^{\infty}\frac{z^{n^2}}{n^n}n^{-n^2 x}$

Consider $z$ is complex and $$f(z) = \sum_{n=1}^{\infty} \frac{z^{n^2}}{n^n}$$ This function has a natural boundary at $|z| = 1$. Now I was thinking about summability methods or continuations beyond the natural boundary. Define $$f(z,x) = \sum_{n=1}^{\infty} \frac{z^{n^2}}{n^n} n^{-n^2 x}$$ Notic...

complex-analysis limits taylor-expansion summability-theory lacunary-series

This confuses me

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$Mathematics$ Mathematics