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00:00 - 22:0022:00 - 23:00

12:34 AM
@MatsGranvik I have attempted to generalize your argument here.
Maybe it's easy to prove/disprove with a counterexample.
 
1:00 AM
@KarlKronenfeld Thanks. Sylow theorems seem to be what I was looking for.
 
1:43 AM
Asaf and Aloisio are the new mods, everybody.
8
 
Thanks for the update pal :-)
 
*Aloizio
@user2646 more than welcome. next year i hope a mod wins without needing a godzilla level rep like the two of them.
:P
 
:P
 
Well, high rep is better no?
The more the merrier kind of stuff?
 
1:52 AM
Is it to me? If it is then I don't want to
 
nope, it was a joke
:-D
 
:P
 
 
1 hour later…
3:11 AM
0
Q: On the difference set of all primes is $\Bbb{Z}$, elementary observation.

EnjoysMathLet $\Bbb{P}$ be the set of prime numbers in $\Bbb{N}$. Well known open problem: $$ \Delta\Bbb{P} := \{ p - q : p,q \in \Bbb{P}\} = \Bbb{Z} $$ If $2n$ doesn't occur, then adjacent gaps $\{ n\}, \dots , \{2, 2n-2\}$ do not occur. Let $(g_1, \dots, g_k), g_i \in \Bbb{Z}$ represent an adjacent g...

 
 
3 hours later…
6:11 AM
How not to get stuck at degenerate saddles
 
Are there any good ways to show a Galois group is cyclic aside from just exhibiting an automorphism of the right order?
In particular, right now I have a Galois group that I'm 99% sure is $\mathbb{Z}/8\mathbb{Z}$, but so far I haven't eliminated $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$
Namely $\mathbb{Q}(i,\sqrt[4]{12 + 8\sqrt{3}})$ over $\mathbb{Q}(i)$. I know the automorphism sending $\sqrt[4]{12 + 8\sqrt{3}}\mapsto \sqrt[4]{12 - 8\sqrt{3}}$ is at least order 4 since it sends $\sqrt[4]{-3}$ to $\pm i\sqrt[4]{-3}$
But I'm not really sure how to get farther
 
@Daminark Did you compute the fourth power of it?
 
Right now I'm trying to compute its square and find out what it does to $\sqrt[4]{12 - 8\sqrt{3}}$ but I'm not really seeing it yet
 
@Daminark What is the minimal polynomial of that field?
 
6:26 AM
$x^8 - 24x^4 - 48$
 
Is that irreducible?
 
Yeah, 3 is prime in $\mathbb{Z}[i]$ so you use Eisenstein
 
great, so there is your answer
(assuming you already know that the group has order $8$)
Since the Galois group of an irreducible polynomial is transitive
 
Ah transitivity
 
6:45 AM
@Daminark how euclidean of them :P
 
$$\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$$
 
I wish the ICM would hurry up and get those lectures uploaded
 
Actually wait maybe it's just really late at night and I'm not thinking straight but would this argument not just imply that every Galois group is transitive? Take a Galois extension, find a primitive element and write your field as $K(\alpha)$, then our group is the Galois group of its minimal polynomial, which should have degree equal to the order of it
 
@Daminark it implies that if K(alpha) is Galois then it's transitive
 
@Daminark What is the requirement for the primitive element theorem?
 
6:52 AM
Finite separable extensions have primitive elements
 
@Daminark also I don't know what you mean by a Galois group being transitive. Take $\Bbb Q(\sqrt2,\sqrt3)$ for example
the orbit of $\sqrt2 + \sqrt3$ has 4 elements
but the orbit of $\sqrt2$ only has 2
but then of course $\Bbb Q(\sqrt2,\sqrt3) = \Bbb Q(\sqrt2+\sqrt3)$
 
Transitive subgroup of $S_n$
 
is there a canonical identification with S_n?
the Cayley embedding always gives you a transitive subgroup
 
@LeakyNun using the action on the roots of the polynomial
 
It's acting on the roots of the minimal polynomial of the primitive element
 
6:57 AM
then I think the galois group of K(a)/K is transitive on a
 
@Daminark Ahh, what happens is that when we change from one polynomial to another, we also change how we embed into the symmetric group
So if we take the polynomial $(x^2 - 2)(x^2 - 3)$ we get the Klein group in the "obvious" way.
But if we do it using the minimal polynomial of the primitive element, we get it via double transpositions, which do form a transitive subgroup
 
Oh oh oh okay
 
So we do need to be a bit careful when using this sort of argument
And indeed, it is now no longer clear to me that the second possible group cannot be transitive on $8$ elements (it probably can actually)
So instead, you should show that there is some prime such that the polynomial remains irreducible mod that prime.
If there is one, you should be about 50% to find one by chance as long as you avoid divisors in the discriminant
 
@TobiasKildetoft since the icm2022 is going to be in St Petersburg, and it's closer to you, are you considering going?
 
@user2646 I am not sure yet. For one thing, I have no idea if I will be able to still be doing math research in 4 years
 
7:08 AM
why not?
 
@LeakyNun My current employment ends in January, and there is currently no real sight of an extension or of a permanent position.
 
Don't they let independent researchers attend?
 
@user2646 Sure, but then I would need to arrange everything on my own and pay all costs
 
:(
 
And if I am no longer in academia, I would find it a bit hard to justify the cost
(now, if it had been in Paris, it might have been easier to turn it into a vacation with my wife, possibly leaving the kids at the grandparents)
 
7:21 AM
@LeakyNun Are you there?
 
@Abcd no idea
 
@LeakyNun ??
 
35 mins ago, by Abcd
$$\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$$
 
@LeakyNun Not this one. Another one.
 
ok then what
 
7:23 AM
$$(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 - 8)y dy$$
 
no idea
 
I just realized that I may need to rename the file I use for this review, since I am not sure if the authors might be able to see the file name of the uploaded file, and I just submitted a review of a paper written by the editor this review will go to (and I use a fairly recognizable naming system for the reviews).
 
8:10 AM
@Abcd could you please check the coefficients in your problem (just a hunch but if one of them was different, its straightforward)
 
@Shobhit ^^^
I hope this question is not wrong
@LeakyNun WA is not able to solve it. Does it mean the question is wrong??
 
maybe
 
what a time waste for me then!!
 
9:01 AM
hey how do I message another user I cant see the button when I visit the profile
please may someone assist me I would be very grateful
 
9:17 AM
What do you mean by message them? @Adam
If you go to one of their questions or answers and leave a comment they will get notified of it.
If they reply, in the comments, you will get a notification.
 
Like I am studying a specific proof of a specific postulate and ive gathered up a batch of posts made about 5 years ago on the same subject and so because my questions are even more basic that the ones he asked in terms of the what I don't understand I just prefer if I can ask for his help directly instead of posting dumb questions about it
they where all asked by the same user, so yeah it would just be ideal to get help from a person that has focused their attention on this specific problem in the past
but sure that's fine I guess I can just leave a comment that's true I was just shocked that you cant pm people having said that I guess it would get very annoying for the really good users in terms of people spamming them
 
Ask them, in the comment section, if they would like to start a chat room
 
@Tobias it turns out element counting did the trick
 
9:47 AM
You have a nice explicit formula for the roots of that polynomial, namely $i^k \sqrt[4]{12 \pm 8\sqrt{3}}$. Let $\sigma_k:\sqrt[4]{12 + 8\sqrt{3}} \mapsto i^k\sqrt[4]{12 - 8\sqrt{3}}$
Well, those guys all send $\sqrt{3}\mapsto -\sqrt{3}$, so in particular you need that $\sqrt[4]{-3} \mapsto \pm i \sqrt[4]{-3}$
So now those 4 maps all have to have order at least 4, and then you have $\sqrt[4]{12 + 8\sqrt{3}} \mapsto \pm i \sqrt[4]{12 + 8\sqrt{3}}$ which has order exactly 4
But you can't have 6 elements of order 4 in an abelian group of order 8 so someone's gotta be order 8
 
ngn
anyone into knot theory? this answer looks wrong to me and I'd like a second opinion on it
 
Is the total space of a holomorphic line bundle a Riemann surface?
I believe the trivialisation transition functions being holomorphic, and the trivialisation neighbourhoods being isomorphic to $\mathbb{C}^1$ makes it a complex one-dimensional manifold?
@Adam What is your postulate?
 
10:46 AM
Oh sorry it's just Bertrand's postulate
 
 
2 hours later…
12:24 PM
0
Q: A Question on certain Hilbert space of continuous functions, and a characteristic of convergence in it

Rajesh DachirajuDefine $T^k(\Omega)$, $\Omega$ an open subset of $\mathbb{R}^m$ (with a smooth boundary), as a space of function equivalance classes, with the norm defined as $$ \|f\|_{T^k(\Omega)} = \|f\|_{L^2(\Omega)} + \|(\sum\limits_{i=1}^m(\frac{\partial^{k}f}{\partial x_i^{k}})^2)^{\frac{1}{2}}\|_{L^2(...

 
12:36 PM
I am profoundly unready for college, and I have no reason to believe this will change between now and next year (when I start college)
(Don't ping me, I just feel like I need a place to put down my thoughts)
 
Why? What do you lack?
 
sounds like you have a profound lack of confidence
watch this^
 
12:54 PM
Replacement?
 
yup, the original was stolen
 
What? That's crazy!
 
check it out
 
Where?
 
Rio
Brazil
 
12:56 PM
Yeah, but is there a tape of when the medal got stolen?
 
Yes
 
Really? And they didn't catch the guy?
 
Nope
The whole story is on the link.
I just fast forward it to his speech.
 
Okay. Sad that people can be so small minded.
 
well, this is a starving country
 
1:03 PM
I guess.
 
melted roads, more to come this summer
When Global Wet Bulb Temperature reached 35 C, that will be the end of humanity as we knew it (unless you happened to have a whole summer supply of air conditioner)
for the general picture
 
$$\int_{0}^{-1} x\cdot{(1+x)}^n dx$$
How should I solve this integral?
 
do a substitution u = 1+x
that swap the troublesome term away
 
Tq
 
1:18 PM
Unrelated
$$\int_{2018}^{2100} \text{We are all going to die by the end of this century} d\text{Human}$$
 
@Secret you scared of death?
 
nope
I have experience the death of my grandparents, I am pretty neutral about it
I want bigot hypocrites to die however
especially those who pressed the ignore button exactly 3 weeks after falling out with me
but again, they are nothing compared to the real problem: the politicians
sorry, got a bit overheated due to recent politics in australia
pardon my strong language...
 
I feel this home as dead as a ruin.
 
1:34 PM
hmm
 
what recent politics
 
I have never had an experience of having a grandfather, so don't know how a grandfather should be like.
 
and yes you are afraid of death you just have come close enough to know it
anyway I need hints for this one $$\forall x,y,q,z\in \mathbb N \quad z^y=x-w \cdot \bigl\lfloor \frac{x}{w} \bigr\rfloor \Rightarrow y \leq w$$
q should be a w there sorry $$\forall x,y,w,z\in \mathbb N \quad z^y=x-w \cdot \bigl\lfloor \frac{x}{w} \bigr\rfloor \Rightarrow y \leq w$$
 
Politics: Someone in the paliament is going to resurrect the White Australian Policy, which is known to kill many people in the past
Death: Maybe, definitely not want to be killed by having the body boiled slowly inside out
 
um white Australian policy?
 
1:39 PM
The term White Australia policy comprises various historical policies that effectively barred people of non-European descent from immigrating to Australia. There was never any specific policy titled as such, but the term was invented later to encapsulate a collection of policies that were designed to exclude people from Asia (particularly China) and the Pacific Islands (particularly Melanesia) from immigrating to Australia. These policies were progressively dismantled between 1949 and 1973.Competition in the gold fields between British and Chinese miners, and labour union opposition to the ...
how they conclude it involves genocide I don't know
 
so how did this kill people and what makes you think they are going to enact this
 
it's on the news just yesterday
that MP get roasted or something
Anyway back to maths, what is the question you need help from that relation?
 
ok you need to relax just because one old retard says something bigoted in parliament house does not mean that his policies are ever going to become a reality
just how to prove it
 
1:54 PM
@Adam You should not use such words in the chat.
 
ok sorry is idiot acceptable?
 
bigot hypocrites sounds mild in comparison, and assuming the data is valid, my suspicion is confirmed
I don't think anyone deserved the descriptor "retard" or "idiot", because no sane person is insane enough for that
In fact, the 1960s of calling mentally ill people with these words is an insult when looked under present date lens
Retard when used as a noun is a pejorative word used to refer to people with mental disabilities. The word has gained notoriety for causing a growing number of mentally disabled people to feel unfairly stereotyped. == Etymology == The word retard dates as far back as 1426. It stems from the Latin verb retardare, meaning to hinder or make slow. The English adopted the word and used it as similar meaning, slow and delayed. The first time the word "retard" was printed in American newspapers was in 1704. At this time, it was used in a way to describe the slowing down or the diminishing of something...
how much society have changed over the years...
 
2:10 PM
What is happening here?
 
They replaced r with r-3
 
@Nebulae why I need to replace?
I am a simple man and If I write r=0,1,2,3... I am getting wrong answer
@Secret ?
 
Today I wrote $1912-1900 = 6$ in exam! :((
 
I don't know why $C_3$ appeared, but the reindexing allows the symmetry of the binomial coefficients to be exploited. The sum vanishes because of the binomial coefficients get paired up
 
Would you give any marks if you asked a student to calculate $(1912)^{218}\mod{19}$ and they started with saying it's equivalent with $6^{2018}\mod{19}$? When it should be $12^{2018}\mod{19}$.
And they calculated $6^{2018}\mod{19}$ correctly?
 
2:22 PM
They will need to use some theorem to justify that otherwise it is not obvious to the marker
 
@Secret It's wrong (i.e. no theorem). They are not equivalent. It's basically saying $1912-1900 = 6$ :(
But everything else from that point on was done correctly, albeit obviously getting the answer to a question that wasn't asked. xD
 
well, depending on how the exam is designed, yo might get points for showing workings despite the hiccups in the logic in the middle
 
I had to use Fermat's Little Theorem like three times. But I didn't realise that I got the easiest step wrong right at the start!
 
@Secret well the word really isn't as significant here as made out to be, no matter what I use, I was still implying the individual has diminished mental capacity
 
I wrote $(1912)^{218}$ earlier when I meant $(1912)^{2018}$.
 
2:30 PM
just bear in mind that the modern society have near zero tolerance on the use of those words, so be careful
That's what happens when being in a society, it has its own rules
 
well I'm hardly in a situation that requires maintenance of notoriety or respect in said society
 
2:50 PM
true
 
 
1 hour later…
4:14 PM
Hello!!

Let $f:A\rightarrow B$ be an injective map. Do we have then that $f(A)=A$ from the definition of injectivity?
 
you mean $f(A)\sim A$, where $\sim$ means "there is a bijection" ?
if so, yes: this bijection is precisely $f$
 
it is \sim
 
thanks lol
 
:P
 
@MaryStar consider $f(x)=x+1$ and $A=\{0\}$
 
4:21 PM
Or any function with $A\cup B=\emptyset$
 
Oh I meant |f(A)| = |A|. Is this correct then? @robjohn @fonini @Holo
 
yes!
since $f$ is injective, if you consider its codomain to be f(A), then it is also surjective, and then bijective, and then |A|=|f(A)|
of course, I'm supposing that your definition of "$|A|=|f(A)|$" is exactly "there is a bijection between A and f(A)"
 
I am an idiot... I start explaining how, assuming injective, we have $g:\mbox{Range}(f)\to A$ injective and then using Schröder–Bernstein theorem...
 
I don't see why you need Schroder-Bernstein
what exactly do you want?
 
I don't
 
4:26 PM
oh :p
 
I was stupid and didnt thought about $g=f\restriction f^{-1}\mbox{Range}(f)$
 
So, you mean that in general it holds that $f: A\rightarrow f(A)$ is surjective, and since we also have that f is injective, it is a bijective map? @fonini
 
Well, you can look at $f: A\rightarrow f(A)$ as the definition of surjective, so yes
 
@MaryStar yes; am I wrong?
 
I just realized that I won't have any course on set theory for at least a 2 years... That is bummer
 
4:34 PM
What is the radius of the circle traced out by rotating a square about its centroid?
 
If I understand you correctly diagonal/2
 
well there are two circles generated. yes the outer circle has a radius of $sqrt(2)/2$
 
@fonini And at a bijective map the domain and the range have the same number of elements, right?
 
The inner has side/2
 
@Holo you know what i'm talking about in terms of the "inner circle" traced out by the envelope?
 
4:39 PM
@MaryStar the definition of equal cardinality is $|A|=|B|\iff$ there exists a bijective map from $A$ to $B$
 
how did you get 1/2?
 
@MaryStar that's usually the definition of "the same number of elements"
 
@geocalc33 I assume you take the square and "bold" every place it pass through and so there are 2 "special" circles, one with the biggest radius and one with smallest, for the smallest take the dot that is the closest to centroid: the middle of the sides, so side/2 is the radius
 
Ok! Thank you very much!! @Holo @fonini
 
You are welcome
 
4:44 PM
@MaryStar if the set is finite, then there is another definition of "same number of elements". You can prove that if there's a bijection between A and $\{1,2,...,n\}$ for some natural $n$, then that integer $n$ is unique (called "the number of elements of A"), and then the existence of another bijection between $A$ and $B$ will imply the existence of a bijection between $B$ and $\{1,2,...,n\}$
 
okay i understand now cool
 
@fonini I think you are confusing with the definition of finite set no? A set $A$ is finite if there is bijective between $A$ and $[n]=\{0,1,...,n\}$
 
I was mentioning that, if the set is finite, there's a more intuitive way of defining the concept of "same number of elements", and that this concept agrees with the more general concept given by the relation "there is a bijection"
that is, when the set is finite, you can define the number of elements of the set without going through a lot of set theory
 
Oh, that what you meant
 
what is a good resource to learn about optimal transport theory
 
5:08 PM
I don't see why you cannot have $|A|=|f(A)|$ and $f(A) \subset A$ if $|A|$ is infinite, e.g. robjohn example which is missing {0} in f(A)
proper subsets of the same cardinality are commonplace if axiom of choice holds
 
no one said that $|A|=|f(A)|$ implies $A=f(A)$
 
ah nvm I misread
 
I also don't see what this has to do with robjohn's example (it is a finite example, motivated by a typo in the question, not by the question itself)
(:
 
consider A=$\Bbb{N}$ and $f(x) = x +1$
 
I think Secret is didn't saw the original question(without $|A|$ and such)
 
5:18 PM
@MaryStar yes, as I see that fonini has said.
@Secret my counterexample, using $A=\{0\}$ is the same idea. However, the question was modified.
 
I see
 
@fonini when the question is stated as $f(A)=A$, it is not clear whether it is a typo or not. If this had not been chat, I would have suggested that the question intended might have been $|f(A)|=|A|$ and that that would be true.
But chat has a quick turn-around
 
of course, I think you did well to provide an answer assuming it was not a typo (it was really not clear)
I was only explaining to Secret that your example did not relate to the actual question we had been discussing
 
5:48 PM
I made this today:
$$\boxed{N(t)=\frac{1}{n^c}\left(\vartheta (t)-\Re\left(\sum _{n=1}^{\text{nn}} \left(\underset{d \mid n}{\sum\limits_{d=1}^{n}} \left(f(\frac{1}{2}+it, d)-f(\frac{1}{2}+i0, d) \right)\right)\right)\right)}$$


$$f(s,d)=-i \mu (d) \left(\sum _{n=0}^{q-1} \left(\sum _{i=0}^{2 n+1} -\frac{d B_{2 (n+1)} k^{-2 n-1} \left|S_{2 n+1}^{(i)}\right| \log ^{-i-1}(d k) \Gamma (i+1,s \log (k d))}{(2 (n+1))!}\right)+\sum _{n=2}^k -\frac{d^{1-s} n^{-s}}{\log (d)+\log (n)}+\text{If}\left[d=1 \text{ then } 0\text{ else }-\frac{1^{-s} d^{1-s}}{\log (d)+\log (1)}\right]+\frac{d^{1-s} k^{-s}
The function $N(t)$ has a Fourier series like jump of size one at every zeta zero on the critical line.
But I wonder why Andrew Guinand did not consider this formula. I mean is it because the truncated Euler Maclaurin formula does not converge all the way to infinity?
And his (Andrew Guinands) formula in contrast does converge all the way to infinity?
I now notice that I have put the fraction $1/n^c$ outside the parentheses. How embarrasing. The Mathematica program is correct though.
 
6:19 PM
And division by $\pi$ is missing too.
 
6:40 PM
math.stackexchange.com/questions/162/… - I'm pretty tempted to post an answer to this question which reads, in its entirety: "It's not."
(But I won't.)
 
Sam
Can someone show me how $(\sqrt(x)-1)\frac{1}{\sqrt(x)}$ is simplified?
 
@Sam I think the traditional way is to multiply the numerator and denominator by $\sqrt{x}$, distributing the $\sqrt{x}$ over the $\sqrt{x} - 1$, and calling it good.
And by "the numerator", I mean $(\sqrt{x}-1)$.
 
Sam
6:55 PM
You able to write out the Math? I'm not that great
 
Well, $\frac{\sqrt{x}-1}{\sqrt{x}}$ becomes $\frac{(\sqrt{x}-1)\sqrt{x}}{\sqrt{x}\sqrt{x}}$...
Do you see a step that can be done there?
 
Sam
7:10 PM
Ah okay
 
 
2 hours later…
9:03 PM
@Sam: Let's look at $\dfrac{a-1}{a}$ first. This is $\dfrac{a}{a} - \dfrac{1}{a} = 1-\dfrac{1}{a}$. That's the simplification. Whether you want to "rationalize" $\dfrac{1}{\sqrt x}$ and write it as $\dfrac{\sqrt x}{x}$ is another question. I personally don't want. How is this showing up for you?
hi @Eric
 
hlo!
 
Any news to report?
 
not atm
currently prepping my lecture
for souganidis
 
more stochasticicicicity?
 
If I have a topological space $Y$ and a set $X$ with a bijection $h: X \rightarrow Y$ and I endow $X$ with the inital topology with respect to $(h,Y)$, and with the final topology with respect to $(h^{-1}, Y)$, do these coincide? I don't think so but I'm not sure.
 
9:17 PM
yup stochasticiciciciciciciciciicicicic
 
I don't understand the question, @Jannik. I'm not familiar with the terminology.
 
They should coincide I think
 
Probably demonic @Alessandro can help, too.
 
Initial topology with respect to $h$ is generated by sets of the form $h^{-1}(U)$ where $U\subset Y$ is open
Or actually it's just those sets since there's only one map here
 
Final and initial topology are just fancy names for "finest topology on the codomain making a (family of) function continuous" and "the coarsest one on the domain blablabla"
 
9:21 PM
And final is the same
Also hey Ted, Eric, and Alessandro!
 
@Alessandro: But Jannik's question referred to two topologies on $X$?
 
Hi everyone
 
Hi, one.
 
@TedShifrin One is final because it has a function going into X, one is initial because it has a function going out of $X$
 
So that wouldn't make them equal?
 
9:28 PM
I believe they are equal by some annoying argument with universal properties
 
any one know how to find equal points along a bezier curve? i thought i had it but i evidently flopped it.
 
Put on $X$ the initial topology wrt $h:X\to Y$, by the universal property of the initial topology we have $h^{-1}:Y\to X$ is continuous iff $h\circ h^{-1}:Y\to Y$ is continuous, which it certainly is
So the initial topology is weaker than the final one
Using the universal property of the final topology we get the other direction
I'm starting to talk like an algebraist...
 
no comment
 
imgur.com/a/b1jSoA3 this is what i managed with my attempt . but they start to stretch
kinda a nuisance
 
@TedShifrin There's surely an honest proof by dealing with open sets, but that's likely to be too long to be typed on my phone!
 
9:39 PM
I don't think it's so horrid to use UMP arguments :)
I was just "no comment"ing on your talking like an algebraist.
 
Apparently the initial topology has a characteristic property while the final one has an universal property, this made me realize I don't actually know the formal definition of "universal property"!
 
Yeah @alessandro
 
@AlessandroCodenotti thanks, using the universal property it really becomes obvious
 
oh @Ted a student pointed out a little weirdness in typical presentations of the proof of gauss bonnet in undergrad geo texts that i hadnt thought too hard about before
 
Can a infinite dimensional banach space be complete?
 
9:48 PM
Yes, @Eric?
 
What's your definition of Banach space?
 
@quallenjäger: It has to be, by definition :P
 
I see:D stupid one
 
well typically one states the hopf umlautsatz for planar curves and then applies it to polygonal curves on a surface region thats orthogonally parametrized without mentioning that one needs to fill in the little gap
 
I mean I am interested in what kind of norm do I have to choose on $V \otimes V$, such that then tensor product is complete.
 
9:50 PM
But, the vector space of differentiable functions, with the $C^0$ norm on it, is not complete, for example.
 
of course its not too hard to fill that gap
 
Huh? You're mixing apples and oranges, I think.
 
If I am only given that $V$ is a infinite dimensional banach space.
 
but it is curious that we dont mention it
 
I mentioned it, @EricS, I'm pretty sure. Let me look.
 
9:51 PM
oh i didnt check your notes
 
@TedShifrin what do you mean? what is apple and what is orange?
 
do carmo old editions didnt seem to
idk about the 2016 edition bc i havent looked at it
i have an old portuguese copy
 
Oh, I misread your question, @quallenjäger. So how're you putting a norm on $V\otimes V$ in the first place?
I haven't seen anything of doCarmo's since the 70s, Eric.
 
@TedShifrin There are projective tensor norms
Or I can define a norm like for $\Bbb R^n$
 
Actually, I just said it was a consequence, @EricS. It just comes from the chart.
I don't know what projective tensor norms are.
 
9:54 PM
well if you use an orthogonal parametrization a priori the rotated angle function could be different
 
It has to be a multiple of $2\pi$, Eric, and so by contractibility it can't be anything but $\pm2\pi$.
 
ya ik it has to be a multiple of 2pi and so you just interpolate but it wasn’t obvious to one of my students
 
@quallenjäger: It's not obvious how to define a norm on the tensor product other than something like $\|v\otimes w\| = \|v\|\|w\|$.
Well, you need winding number to be a continuous function of the curve, of course. @Eric
 
This is actually the idea of projective tensor norm
 
yes i agree
 
9:57 PM
Would $V \otimes W$ be complete with the norm given by you?
 
I had this discussion with Jack Lee several times when he taught out of my notes, @EricS. He wanted them to be as rigorous as a typical analysis course, and I said I wasn't aiming to be that pedantic.
@quallenjäger: You mean assuming $V$ ($W$ too, I suppose) is Banach, i.e., complete?
 
well i think at least its important to be able to articulate how you go from one to the other
 
@EricS: Well, I wrote a paragraph about it :P
 
maybe its not so important on a first run through though idk
 
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