1:43 AM
Is there general form of factorization of x^{p^n}-x over a finite field F_p?

it factorizes into exactly all of the irreducible monic polynomials with degree dividing n

Yes but I mean like x, I want to find a polynomial f(x) | x^{p^n}-x
Well it's awkward I will cancel the question

1:59 AM
yo
0

I came across an interesting integral: $$\int_{-\infty}^0 \left(\log \left(1-e^{x^3} \right)\right)^{1/3}~dx$$ And I used Mathematica to get this expression where $\beta(x,y)$ is the Beta function: \frac13\frac{\partial^{-1/3}\beta(x, y)}{\partial^{-2/3}x\partial^{1/3}y}\bigg|_{x \rightarrow 0,...

found a RARE integral
I would like to know the steps involved to calculate the result

2:18 AM
Kind of a silly question, but can anyone think of a well-known Theorem that has a double-barreled name in it? So something like Cauchy-Schwarz does not count because it is named after two people, but if a Mathematicians surname was Lorem-Ipsum it would count.

@JessicaK Levi-Civita connections count?

Mittag-Leffler

Cohn-Vossen's inequality

Sz.-Nagy's dilation theorem

2:33 AM
Yeah, it does. I think Mittag-Leffner is my favorite though thanks

Although not math and also not a theorem, but in CS we have these special kind of balanced binary trees called AVL-trees. 'AV' is one guy and 'L' is another.

It just came up when editing some math slides and using a hyphen vs an endash

I see.

2 hours later…
5:04 AM
@JessicaK de la Vallee Poussin is another mouthful
Choquet-Bruhat
Choquet-Bruhat is a funny double-surname because Choquet and Bruhat are both famous surnames in math. I think the famous Choquet is in fact Choquet-Bruhat's husband and the famous Bruhat is her brother, not to mention Choquet-Bruhat is pretty famous as well.

4 hours later…
9:05 AM
@robjohn You have just reached 300K reputation on Math.SE. I acknowledge and appreciate your invaluable contribution to Math.SE and your honest moderation on this site for more than 9 and 8 years, respectively. It is worth saying that the beauty of math chat rooms (and other rooms using TEX codes) is indebted to you for your MathJax rendering code.

2 hours later…
10:42 AM
Okay. I am working on the last part of the last problem and I cannot figure it out. I could use some help: Let $k$ be a field, and let $k[[x]]$ denote the ring of formal power series. Give an example of two non-isomorphic $k[[x]]$-modules annihilated by $x^3$, which are 7-dimensional $k$-vector spaces.

11:11 AM
@user193319 if it's annihilated by x^3, then you might as well say that it's a module over...?

$\Bbb{Z}_3$?

integers? Hmm...

how do you get a k[[x]]-module given a Z/3Z-module?

do you know any $k[[x]]$-modules

11:13 AM
No, I don't know any $k[[x]]$-modules.

given a ring R, do you know any R-module?

Lol, yes I know some modules. E.g., any vector space is a module.

but that doesn't work for my ring R

$R$ itself is an $R$-module

11:14 AM
right
anything else?

what other examples can you construct from $R$

I was thinking maybe a quotient of $k[[x]]$ would be one of them,
Quotients?

yep
exactly

yup, playing with quotients should get the job done

what's a quotient of k[[x]] that's annihilated by x^3?

11:15 AM
So, if you mod out by polynomials of degree $\ge 8$, you'll get a 7 dimensional vector space, right?

right, but it won't be annihilated by x^3

Yeah, you're right...hmm...Oh I'm an idiot. $k[[x]]/(x^3)$.
So that'll be one of the modules...now I need another one.

are they any other quotients of k[[x]] that's annihilated by x^3?

that's not a 7-dimensional vector space, tho

Oh yeah...hmm...

11:17 AM
but you're on the right track

hmm...I have to mod out by an ideal right?
I proved that they all take the form $(x^k)$ for $k \in \Bbb{N}_0$.

0

Consider a 2 submanifold M of $\mathbb{R}^3$. We define a vector field $pr(X)$ by $pr(X)_p=pr(X)_p=pr_p(X_p)\in T_pM$ where $pr_p: T_p\mathbb{R}^3=T_pM\oplus N_pM\rightarrow T_pM$ is orthogonal projection. if X is a smooth vector field along M in $\mathbb{R}^3$ then sufficiently near $p$, we can...

$k[[x]]/(x)$ will be a finite dimensional VS, but will any of the other quotients be finite dimensional?

what about $k[[x]]/(x^2)$?

Hmm...oh yeah...that's finite too. So both modules I am looking for quotients of $k[[x]]$?

11:30 AM
not quite, but almost
what's the dimension of $k[[x]]/(x^n)$ as $k$-vector space?

So one will be a quotient of $k[[x]]$, but what about the other? Where do I look to find the other module?
I would think $n+1$ dimensional.

why?
you can give a basis explicitly

Wouldn't $1,x,...,x^{n-1}$ form a basis...oh wait, that $n$ elements.
I can't count.

right
so what's the only quotient that's $7$-dimensional as a $k$-vector space and does that satisfy the hypothesis of being annihilated by $x^3$

$k[[x]]/(x^7)$, right?

11:33 AM
@user193319 what other methods are there to form modules?
perhaps given some other modules?

direct sums

bingo

Hmm...so what modules am I "direct summing" to get a $7$-dimensional vector space annihilated by $x^3$.
?

1. Classify the quotients of $k[[x]]$ that are annihilated by $x^3$
2. Convince yourself the direct sum of modules annihilated by $x^3$ is annihilated by $x^3$
3. Take direct sums to get the right dimensions

Okay. I'll give that a try. I have to take my dogs out now, so I'll try it a bit later.

11:47 AM
this is less trivial, but in fact all modules satisfying your conditions arise this way and there are up to isomorphism $8$ such modules unless I'm too stupid to count

12:33 PM
0

Consider a 2 submanifold M of $\mathbb{R}^3$. We define a vector field $pr(X)$ by $pr(X)_p=pr(X)_p=pr_p(X_p)\in T_pM$ where $pr_p: T_p\mathbb{R}^3=T_pM\oplus N_pM\rightarrow T_pM$ is orthogonal projection. if X is a smooth vector field along M in $\mathbb{R}^3$ then sufficiently near $p$, we can...

any ideas?

the equality is automatic when $N$ is a unit normal field by linear algebra
so you just need to prove local existence of a smooth unit normal field
which can be accomplished in a chart

1:10 PM
Sanity check: kernel and range are always disjoint, even in an infinite dimensional space, right?
I forget if I've seen a proof for the infinite dimensional case

they never are
so I assume you mean something different than what you're saying
never for endomorphisms*
ofc, if domain and codomain are disjoint, so are kernel and range

1:26 PM
Well, yeah okay, good point: Disjoint aside from the identity
So, $\text{Ker}\bigcap\text{Ran}=\{0\}$

consider $\mathbb{R}^2\rightarrow\mathbb{R}^2,(x,y)\mapsto(0,x)$

@Thorgott Okay. I finally got around to thinking about the problem again. I believe the two modules are $M_1 := k[[x]]/(x^2) \oplus k[[x]]/(x^2) \oplus k[[x]]/(x^3)$ and $M_2 := \bigoplus_{i=1}^{7} k[[x]]/(x)$.

yeah, those work
there are 6 more that work as well
they're in bijection with the ways of writing 7 as a sum of 1,2,3

I'm pretty certain $M_2 \cong k^7$, so $M_2$ is a free module of rank 7.
Hmm...$M_1$ probably isn't, but why not...hmm

@Thorgott Darn, not even in the finite dimensional case

1:34 PM
Wait...$k[[x]]/(x)$ is isomorphic to $k^7$ as a vector space, not a $k[[x]]$-module, right? So $M_2$ might not be a free module.

@Rithaniel let $T\colon V\rightarrow V$ be a vector space endomorphism, prove $\ker(T)\cap\operatorname{im}(T)=(0)$ iff $\ker(T)=\ker(T^2)$
@user193319 how is $k$ a $k[[x]]$-module

Hmm...you're right...so $M_2 \cong k^7$ is definitely an isomorphism of vector spaces.

yeah
they're also isomorphic as rings fwiw, but that isn't particularly interesting

Hmm...Okay. So I just need to argue that $M_1$ and $M_2$ are not isomorphic as $k[[x]]$-modules.

@Thorgott $X_q-<X_q,N_q>N_q \in T_qM$? what's the linear algebra argument
?

1:40 PM
neither of your modules is free as $k[[x]]$-module, by the way
since they have torsion
@orientable compute scalar product with $N$

you mean, <N,N>?

no, I mean the scalar product of your term and $N$

its 0

bingo

that means $X_q-<X_q,N_q>N_q$ is in $T_qM$, right?

1:48 PM
Correct, $T_q M$ is the orthocomplement of $N_q$
Another way to think about this is you throw away the component of $X$ in the direction of $N$... that's the term you subtract (think projection of a vectors)

AHA, $((T_qM)^{\perp})^{\perp}=T_qM$

@Thorgott Do you have any clues on how to show $M_1$ and $M_2$ are not isomorphic?

note that this alone doesn't yet show that this is actually the orthogonal projection onto that subspace
but it's not hard to see that it's the identity on TM, idempotent and self-adjoint
@user193319 look at which elements of $k[[x]]$ annihilate each module

@Thorgott the algebraists' definition of projection

what's yours lmao

1:57 PM
find the nearest point

ok fair enough

not hard to establish they're equivalent tho

@Thorgott yeah, but elements in direct sum are expressed uniquely so if $\mathbb{R}^3= T_pM \oplus N_pM$ then once establishing that $X_p-<N_p,X_p>N_p\in T_pM$ we automatically get the result?

it's more of a hassle to even verify that your definition is well-defined

1:59 PM
my definition is euclidean geometry. it's clear that it's well-defined
do some euclidean geometry
lol

as a sanity check, this can easily be extended to general reimannian manifolds, right?

@orientable well, once you verify that these summands are in fact image and kernel of the map, yes

@orientablesurface Hold on. What do you mean by extending to general Riemannian manifolds?

if you have a codimension 1 Riemannian submanifold, there locally exists a smooth unit normal vector field, yes

Let him state it man
Don't state it for him

2:02 PM
@BalarkaSen So if you have a reimannian manifold M and a regular submanifold $\overline{M}$ then we can get something similar to the projection connection on surfaces, right?
aha

Not from the whole of $M$ to $\overline{M}$.

what's a regular submanifold

embedded @Thorgott

ah, a sane persons submanifold

From a small neighborhood of $\overline{M}$ to $\overline{M}$, yes. Look up the tubular neighborhood theorem.

2:04 PM
@Thorgott I think $x$ annihilates $M_2$ but not $M_1$. Is this right?

Thanks @BalarkaSen
any good references on tubular neighborhood?

@user193319 y

Cool. Thanks bro!

whats a cool overlap between measure theory
and reimannian geometry
?

@orientablesurface For starters, read Guillemin-Pollack's section on tubular neighborhood theorem. They don't work in a Riemannian setup, so the orthogonality of the projection is not dealt with explicitly. But you can try to think about that

2:09 PM
Do they have a section about it? I thought it was a problem. The book is in my office so I can't check rn.

There's a section yeah but I don't really like it

hey chat
reading about manifolds. so a manifold is a triple $(M, U, \mathscr{A})$ s.t.:

@LucasHenrique everything is a manifold

Oh god

lmao

2:12 PM
$U$ is an open covering of $M$ and for every $U_\alpha \in U$, there's an $f: U \to \mathbb{R}^n$, $f \in \mathscr{A}$ homeomorphism, right?

@orientablesurface The point is as follows. If $N \subset M$ is a regular submanifold of a Riemannian manifold, there is always a projection map $TM \to TN$. This is given by orthogonal projection of $T_p M$ to $T_p N$ for any point $p \in N$ (you can write down a formula not unlike yours, but you need to locally choose a basis of normal vectors at $p$ - there's an issue here, why is this map "smooth" as you vary $p$? Food for thought).
Using an elaborate inverse function theorem, you can upgrade this to a map $f : U \to N$ where $U \subset M$ is an open neighborhood of $N$ in $M$, which is a smooth retraction, i.e., $f^2 = f$ or in sane person's language $f|_N = \text{id}$, and the derivative of $f$ along $N$ is exactly this projection $TM \to TN$ at the level of tangent spaces.
$f^2 = f$ is the map-level version of $P^2 = P$, the definition of projection operator that @Thorgott so likes.

I read that you can get $f: U \to B \subseteq \mathbb{R}^n$ instead of simply $\mathbb{R}^n$

But the self-adjointness of $P$ (wrt Riemannian metrics in domain/codomain) translates to something weirder for $f$... the right notion is a "submetry", which is like an isometry but surjective (so isometry upto some "kernel")
All of this is waffly and takes work to check though.

@LucasHenrique yeah , but manifolds that aren't second countable and hausdorff aren't manifolds, they're not even math, english probably
@BalarkaSen whats the elaborate inverse function theorem youre referring to?

That was a loose terminology invented by me. The philosophy is the same but it takes work to get $f$ out of the projection.

2:18 PM
what's the precise definition? Wikipedia is kinda vague and I can't really read a book about manifolds (differentiable ones are my interest in this case). I'm writing something that uses Lie groups but general Lie groups aren't my focus.

A topological manifold of dimension m is a top space that is second countable, hausdorff and for each point p there exists a neighborhood U of p and a homeoprhism $\phi: U\rightarrow \mathbb{R}^m$

The problem is that the precise definition is painful to state and takes time to grok. Beginners looking at the definition tend to lose the forest for the trees, because there are a shitload of trees and each tree seems very large.

equivalently you can get homemomorphisms to open balls

Are you working with matrix lie groups?

Thanks, is that all in the book you mentioned? @BalarkaSen

2:20 PM
Nope :) Unfortunately there are no good books on smooth manifolds
@MikeMiller Free groups should be called Tree groups

Lee's book is too wordy, G&P's book too basic for the advanced student, Bott&Tu too advanced for the intermediate student. I've never read Tu's solo book maybe it's fine.

I am petitioning to change names

Warner's book too tediously pedantic

Tu writes a new self-contained book everyday

@orientablesurface that's my problem with the definition. how so?

2:24 PM
A smooth manifold of dimension $n$ is a triple $(M,\tau,\mathfrak{A})$, where $M$ is a set, $\tau$ is a topology on $M$ that is Hausdorff and second-countable and $\mathfrak{A}$ is a maximal smooth atlas,
i.e. a collection of homeomorphisms $\varphi_i\colon U_i\rightarrow\varphi_i(U_i)$, indexed by some set $I$, where $U_i\subseteq M$ is open, $\varphi_i(U_i)\subseteq\mathbb{R}^n$ is open, such that $\varphi_i\circ\varphi_j^{-1}\colon\varphi_j(U_i\cap U_j)\rightarrow\varphi_i(U_i\cap U_j)$ is smooth for all $i,j\in I$, such that $\bigcup_{i\in I}U_i=M$ and such that this collection is maxi

using a function that tends to infinity?

Cuz open balls are diffeomorphic to R^n bro

Of course Thorgott had to post this

this was indeed painful to state

x mapsto x/(1-|x|^2)

2:24 PM
I'm just proving Mike right

I'm leaving

Hahaha

Tu is a very nice introductory text

@MikeMiller yup. I'm not sure if you were in the chat a few days ago, but I'm proving that $\mathrm{GL}_n(\mathbb{C})$ is path-connected

Nah, I wasn't here then. If you're using matrix groups I STRONGLY suggest you don't get into the general defn of smooth manifold.

2:26 PM
@Thorgott is $\mathfrak{A}$ maximal really necessary?

yesn't
every smooth atlas is contained in a unique maximal smooth atlas

A smooth manifold is a subset M of R^N for some integer N so that, near each point p in M, there is an r>0 and a diffeomorphism phi: B_r(p) -> B_r(0) which "straightens out M", aka, phi(M cap B_r(p)) is a linear subspace intersect B_r(0).
2

so to specify a smooth manifold, it suffices to exhibit a smooth atlas

I have never ever ever ever used a maximal atlas in my life. EVER.
It's a stupid, stupid, stupid definition.

nah, you want all the charts
for flexibility

2:28 PM
Give me a theorem I need to use maximal atlas to prove

Even Thorgott is staring at a single redwood in the forest and he's worked with this stuff

my point is: if you want to show a function is smooth, it suffices to check it's smooth on a collection of charts covering the manifold. the maximal atlas contains more charts than just any smooth atlas, so allows you to choose which charts to pick to test the smoothness of a function. this can be useful.

Give me the theorem I need to use maximal atlas to prove
Otherwise everything you say is continental philosophy

you know you're asking for the impossible
working with either is essentially equivalent

No I don't. When someone says X is useful I ask what can I prove using X. If there's nothing to prove using X it's not useful
By definition of "useful"

2:32 PM
you act as if proving things is the only thing one does in mathematics

@BalarkaSen Cmon continental philosophy is more interesting than maximal atlases

I thought you like examples

Give me an example where maximal atlases are useful to understand them

@LucasHenrique for one direction , the image of a point in the manifold is contained in an open set take , the preimage of the open ball it is contained in, for the other direction just note that every ball in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$

I don't consider an example of a maximal atlas to justify why maximal atlases are useful
lol

2:35 PM
either definition give you the same amount of structure, so it's not like either generates more insight
I mean, I hope we agree that a "chart" should be a diffeomorphism from an open subset of the manifold to the open subset of some Euclidean space, right
so the difference is just whether we choose a maximal atlas, which already contains all charts, to begin with or just pick a smooth atlas and then add the other charts by checking smoothness in the charts already given to us
it's the same thing, really

if it's the same thing why associate a name to it? it's a useless jargon

but considering the collection of all charts instead of just the ones in a given atlas can be crucial at times, so we may as well include them all in the atlas to begin with

i can break the definition of a group into 10 pieces by repeating the same thing over and over again - that doesn't help anyone

for example, if you want to exhibit a subspace as an embedded submanifold, you want to look for adapted charts among all possible charts, not necessarily among just the ones in one smooth atlas you exhibited to show the manifold is smooth at all as that might fail

if i want to check a subspace is an embedded manifold i pick a point and find an adapted chart
i dont know what among etc etc means

2:40 PM
yeah, my point is that this is a general chart, but not necessarily one of the charts contained in the smooth atlas you exhibited to show the manifold is smooth
like, I can find a circle on the sphere such that neither of the stereographic projection charts is adapted to it, but other charts will be
those other charts aren't part of our initial smooth atlas, but the maximal atlas it determines

yes, so? why carry around the smooth atlas with you in the first place?

cause we need a smooth structure?
that's what this is all about

give me an example, i am not following anything you are saying
because its all words and not math
lol

Let's come to a compromise here
You're both stupid assholes because you got into this argument to shout over a confused beginner
9

Lol
Look. Maximal atlases would be worth defining if they were algorithmic, it's not like I can write down a maximal atlas for a smooth structure. It's a name for the process we do, which is shrinking a chart or something
That's stupid

2:47 PM
ofc, to equip a top manifold with a smooth structure, we just exhibited one smooth atlas. that smooth atlas uniquely determines a maximal one. my point is that, once we have that smooth structure, we don't care about which particular charts we used to show that the manifold has a smooth structure anymore, but we start caring about all charts equally, so it makes sense pedagogically to define the manifold as coming equipped with the maximal atlas.

hi!
would anyone give me a helping hand here?

hmm, what's the context?

@LucasHenrique fourier series for the step function in the picture

clearly, we should just agree that the best definition is that a smooth manifold is a Hausdorff, second-countable differential space that is locally isomorphic to the differential space R^n

x = 1 , for 0<x<1 ; x = -1, for -1<x<0
I was trying a quick trick to find the sum of $1/n^2$ for odd n by the energy theorem
almost there
missing a factor of 2
damn

2:52 PM
@MikeMiller: when talking about Lie groups, what's the underlying geometrical structure of $\mathrm{GL}_n(\mathbb{R})$?

What's happening in the first equality

Looks natural to consider the induced topology by the induced norm $||\cdot||_2$

@Lucas it's an open subset of $\mathbb{R}^{n\times n}$

@Thorgott some Plancharel identity maybe

@Thorgott f(x) is the step function, x = 1 , for 0<x<1 ; x = -1, for -1<x<0
I used fourier series to expand it

2:54 PM
Yeah, for GL_n we can avoid all this nonsense. Since it's an open subset of Euclidean space, we still know what the topology/metric is, as you say (open balls should be open with respect to your favorite norm), as well as what a differentiable function is (well-approximated by its derivative at every point)

then I used the orthogonal property to calculate the value of 1/2 * int[ |f(x)|^2] (the energy theorem)

ah, energy theorem = Parselval's theorem

@Thorgott yes! I believe so

Does the statement that there exists an invertible, unitary intertwiner between two representations mean the two representations are "unitarily equivalent"? Just want to make sure I'm using the term correctly

lolol intertwiner

3:00 PM
>:(

@MikeMiller hmm... not sure if I know everything in this phrase

i have never seen that word

$\mathbb{R}^n$ can get a lot of topologies, so I don't know how the study of $\mathrm{GL}_n(\mathbb{C})$ is global with all the possible differentiable n-manifold structures

It is a funny word but I think it's standard :P

in physics maybe

3:01 PM
I mean, sure, but nobody would imagine you mean anything but the standard one, induced by a norm
Any two norms induce the same topology, which is a nice exercise in analysis / topology

What is the word mathematicians use?
There should be a dictionary for this stuff :P

@MikeMiller my metric spaces topology exam is today. I'll probably have this claim as an exercise, lol
there's an etymological dictionary too, iirc

@Shing I believe your Fourier coefficients are wrong

3:21 PM
Working really hard to prove that $\text{Ker}(T)^\perp\subseteq\text{Ran}(T^\ast)$ because I thought I had to show that $\text{Ker}(T)^\perp=\text{Ran}(T^\ast)$. But nope, I had to show that $\text{Ker}(T)^\perp=\overline{\text{Ran}(T^\ast)}$, which is infinitely easier

Infinite Dimensional Hilbert spaces
proceeds to hold up

And what is T

A bounded operator
Also, in this context the overline represents the closure of the space

So what you're saying is you're trying to prove that T* has closed image for free
But T** = T. So this is false in general.
What you actually proved is all that's true

3:27 PM
No no, I had misread the original question. What I needed to show was that $\text{Ker}(T)^\perp=\overline{\text{Ran}(T^\ast)}$

Gotcha. Yes, I agree with that statement

I should have probably caught the issue in what I had believed was the original, as I thought it was asking for $\text{Ker}(T)^\perp=\text{Ran}(T^\ast)$, but the LHS is closed and the RHS isn't necessarily

Right, exactly my point

(This is why we do problems, to weed out the mistakes lingering in our heads)

I think they are correct?
https://mathworld.wolfram.com/FourierSeriesSquareWave.html

3:44 PM
@MikeMiller I'm going to read some ergodic literature

lol
I'm also reading some ergodic literature, but of a different kind

4:10 PM
@Later Thanks! I've been busy offline for a while, but I hope to have more time to spend online again soon.

@BalarkaSen sheesh that is actual mumbo jumbo

no, it's just ergodic
you're not fulfilling the extranoematic responsibilities the article places on you

my fault, sorry

Stuff that is difficult to read. Like some math textbooks
Or perhaps a book that intentionally doesn't use punctuation of any sort
(and other avant garde ideas)

4:30 PM
A group law on $\log(x)\log(y)=1$ is:
$(e^{-a},e^{-1/a}) \oplus (e^{-b},e^{-1/b})= (e^{-ab},e^{-1/ab})$
and is isomorphic to $(\Bbb R^+,\times)$ via $e^{-a}\mapsto a.$
A group law on $\log(1-x)\log(y)=1$ is:
$(1-e^{-a},e^{-1/a}) \oplus (1-e^{-b},e^{-1/b})= (1-e^{-ab},e^{-1/ab})$
And is also isomorphic to $(\Bbb R^+,\times).$
How do you combine the two group laws with the direct product?
And the resultant group is isomorphic to $(\Bbb R^+ \times \Bbb R^+)?$
I'm having trouble with the direct product because multiplying the ordered pairs by components doesn't give the correct result

if $G\cong G^{\prime}$ and $H\cong H^{\prime}$, then $G\times H\cong G^{\prime}\times H^{\prime}$

Nevermind

I saw that

See? That's why people study amenable groups, because they wish non amenable ones didn't exist

!!
Almost all groups don't exist - Alessandro

4:36 PM
It's the same as commutative algebra, why do people study it? Because they wish noncommutative algebra didn't exist

@Thorgott but see how the two group laws are attached to functions? How can I find the function with a group law from the direct product of the two group laws?

idk what that means

By estimating the coefficients of the Laurent series, prove that if $z_0$ is an isolated singularity of $f$, and if $(z-z_0)f(z) \rightarrow 0$ as $z \rightarrow z_0$, then $z_0$ is removable. Give a second proof based on Morera's theorem.
Here is my attempt: Since $(z-z_0)f(z) \rightarrow 0$ as $z \rightarrow z_0$, and because $(z-z_0)f(z_0)$ is analytic near $z_0$, Riemann's theorem says that $z_0$ is a removable singularity of $(z-z_0)f(z)$. Hence, its Laurent series expansion is of the form $(z-z_0)f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n$ which implies $f(z) = \sum_{n=0}^{\infty} a_n(z-z_0)^{n-1}$....
I want to argue that $a_n = 0$....If $a_0 \neq 0$, then $z_0$ is a simple pole of $f(z)$, but I don't see what this contradicts.
I should have mentioned that $(z-z_0)f(z) \rightarrow 0$ as $z \rightarrow z_0$ means $(z-z_0)f(z)$ is bounded near $z_0$.

$(z-z_0)f(z)\rightarrow0$, substitute the Laurent series for $f$ and calculate the limit

Oh, I see.

4:49 PM
Stupid notation question:
Is $\varnothing \times \varnothing = \varnothing$, where $\times$ is the Cartesian product?

yes

Well, how many elements does that Cartesian product have?

Zero, and... oh. $\varnothing$ is the unique set with zero elements.
Thanks

There you go
Now, for my question, if you know that $0=\langle x-Tx,y\rangle+\langle y-Ty,x\rangle$ for all $x,y\in\mathcal{H}$, how can you conclude that $T$ is the identity operator?
(maybe take $x=y$?)
(Yeah, I think that works)

Ask me maybe in 2 months when I figure out what Hilbert and Banach spaces are.

4:53 PM
Clearly they're black magic

5:04 PM
@Rithaniel Let $A$ be any skew-Hermitian matrix. Then $\langle Ax, y \rangle + \langle x, Ay \rangle = 0$. Define $T = I - A$.
Check a statement for finite dimensional linear algebra before going to Hilbert spaces
2

Okay, then I have a slightly different statement to make
I'm trying to prove that $T$ is an isometry if and only if $T^\ast T=I$
And I got it to the form: $0=\langle x-T^\ast Tx,y\rangle+\langle y-T^\ast Ty,x\rangle$

So you have proved $I - T^* T$ is skew-adjoint. But it is self-adjoint, so you're done.

what about hilbert spaces in characteristic 2 man
3

lol

So, what's the detail that I lose between the statement I just made and the statement before, where I replaced $T^\ast T$ with $T$?

5:09 PM
brb gotta dinner

Alright

@Thorgott too funny
good joke

(I have never heard this term "skew-adjoint" before, so I don't think I can use it in an argument effectively)
(Thus, I'm still looking)

A just mathematician only wants to be good than those who donâ€™t study maths, an unjust mathematician always aim at getting better than other mathematicians.

Taking a momentary break to share a random thought: those moments are always funny in retrospect. When a person provides a proof to a statement that relies on stuff you just don't know. It's like "Why is this function unitary?" and the answer is "Oh, yeah, the moon aligned with Saturn the other night. I assume you can do the work from there."

5:20 PM
you can prove it directly from non-degenracy
non-degeneracy meaning that $x=x^{\prime}$ iff $\langle x,y\rangle=\langle x^{\prime},y\rangle$ for all $y$

@Rithaniel For matrix operators, this seems to mean that $I-T$ is skew-hermitian. For real matrices, that is skew-symmetric.

2 hours later…
7:39 PM
Thank you guys for the help, by the way. I managed to piece my way through to a proof I'm happy with
Terminology I'm not familiar with also helps me to find new topics, so there are always upsides

8:04 PM
yay, lockdown for 5 weeks

Oh no RIP

Meh I wasn't going out much anyway
University will be online though

We already had a curfew, and I'll still be able to jog outside
Not the end of the world (yet)

yet
but yeah should go out in the sun sometimes
dont be me

8:13 PM
Bah, who needs the sun, it's nothing but blinding, as Bane would say

lmao
"I was born in the darkness, you merely adopted it"

8:52 PM
@Astyx here too :(

In germany ?

I think most of Europe is going to

1 hour later…
10:11 PM
0

Let $V$ be a set of nodes, $c : V \times V \rightarrow \mathbb{R} \cup \{\infty\}$ be an edge cost function, and $h : V \times V \rightarrow \mathbb{R}$ be an admissible heuristic. Suppose we want to solve multiple independent least-cost path problems: from $a_1$ to $b_1$, from $a_2$ to $b_2$, an...

@LucasHenrique Don't forget $S^0$.