@Semiclassic: First of all, after I said that I told you how the Poincaré residue occurs in this setting. But you only care about what Griffiths will prove directly — and that's how you give a basis for the holomorphic 1-forms for a curve in $\Bbb P^2$. You don't need to know the more general Poincaré residue construction to do this.

ahh

@TedShifrin where did you say the bit about 'Poincare residue in this setting'? I seem to have missed that

though if that was about 'curve as a divisor on P^2' I didn't really follow it

hey @TedShifrin

@arctictern If I have that for every convergent sequence in taken in ||.||_1 also converges in ||.||_2 and vice versa I would like to show that the norms are equivalent

what would be the best way to prove it ?

I was thinking in terms of contradiction ?

@Adeek Are two norms equivalent iff they induce the same topology?

I guess that is an equivalent way of defining it but my analysis prof defined in in a different way.

@Adeek Because if you know that (I think it's reasonable to say it's true), then I think you can show the closed sets are the same.

Two norms are equivalent iff there exists $c_1,c_2$ such that $c_1||x||_1 \leq ||x||_2 \leq c_2||x||_1$ for all x.

Using the characterization of closed in metric spaces via limits of sequences.

I know that.

But a norm induces a topology on your vector space.

I think that if two norms are equivalent, they induce the same topology, and vice versa.

yeah that is correct

but I would have to prove that which isn't bad.

Ok, so in the topologies induced by your two norms, the closed sets are the same.

Because any sequence that converges in one, converges in the other.

yes

But the closed sets being equivalent is the same as the open sets being equivalent.

So your norms induce the same topology.

And you agreed this is equivalent to them being equivalent.

That's an indirect proof.

yeah I guess you could translate this to topological proof.

(overuse of "equivalent")

yeah sounds good the topological proof isn't that bad

but wait

So using the characterization of closed in the topology of metric space. We have A is closed in a metric space iff if the sequence in A converges to an element x then x must be in A.

right ?

suppose A is closed in the topology generated by norm $|| . ||_1$

then we have that for every sequence in A using the first norm it also converge in the second norm.

but how do you know that A is closed in the second topology ?

@0celo7

@Adeek Yes.

Crap

Ok it's true but I just deleted by proof

@Adeek You know the sequence converges in $A$ because $A$ is closed in the 1 norm, right?

yeah

So $(v_n)\subset A$ convergent wrt. norm 1 implies $\lim v_n=v\in A$.

Ah, I see the problem.

We don't know that $v$ is the limit for both sequences.

Is that what you're saying?

yes

Let me think...

@Adeek Yeah a direct proof might be better.

yeah I think so

This can probably be fixed, but with work.

maybe a contradiction

Suppose that two norms aren't equivalent then what this means is as follows for every $c > 0$ we have either $||x|_2 > c||x_1||$

or the other way around

Yeah.

i.e $||.||_2$ is unbounded with respect to $||.||_1$

Hmm, I don't like that already.

It's wrong for $x=0$.

c > 0

oh

In Euclid's elements where is the proof that a parallelogram's diagonals bisect each other?

For any c > 0 there exists an $x \in X$ such that $||x||_2 > c||x_1||$

@Adeek yeah I'm very confused by how you actually negate that statement

Ah yes

the negation of the forall

I think that is how we negate it

Agreed

@Adeek So pick a convergent sequence wrt. one norm, show it diverges wrt. the other

yeah that is what I was thinking. @0celo7

@Adeek so what sequence do we want?

the problem is the negation gives you an $\exists x\in X$

so you're basically limited to those guys

oh, but you can pick $c_n=n$ or $c_n=\frac{1}{n}$ or something like that

what is $n$ ?

well you're trying to build a sequence here

we are working in arbitrarily normed vector space.

$n$ is just your sequence label

oh oke

$c$ is a scalar

Since this works for all x in X. So we can find a sequence $x_n$ such that $||x_n|| \leq 1$ but we have $||x_n||_2 \geq n$. agree?

Since what works for all $x\in X$?

The negation? No, you have $\exists x\in X$ above.

@Adeek Also note that you need more for (sub-)convergence than boundedness.

@0celo7 so we can construct the sequence like that

I have an idea how to do it now

let me make what I am saying more precise

just a sec

Suppose that $||.||_1$ isn't equivalent to $||.||_2$, then we either have there exists for c > 0 there exists $x \in X$ such that $||x||_2 > c||x||_1$. Suppose we pick a sequence $\{x_i\}$ in $|| . ||_1$ such that $||x_i|| \leq 1$ for each i. Construct another sequence $\{b_j\}$ as follows for each x_i we know that there exists $x \in X$ such that $||x||_2 > n||x_i||_1$, set $b_j = x$ for each j. Then, the sequence $z_i = x_i / \sqrt(n)$ goes to zero in first norm but not second norm.

@0celo7 :D

what do you think

Hmm.

How did you get $||x||_2>n||x_i||_1$?

The vectors in the norms are different, no?

yeah I am picking a different x for each of the $x_i$

we know they exist by the negation

and I am picking n = c.

The negation gives $||x||_2>c||x||_1$.

The same $x$.

yeah so set c = n

Unless you want to revise the negation in the first line.

The negation is for any c > 0 there exists $x \in X$ such that $||x||_2 > c||y||_1$ for any y in X.

yeah it should be revised

This is the actual negation.

I think

I'm fairly sure that's trivially true.

You can make a norm arbitrarily large.

The negation should be the same vector on both sides.

oh

That statement has to be wrong.

$||x||_2>c||y||_1$ for ANY $y$?

yeah maybe I said the negation wrong.

I think it is same vector.

sorry I didn't get enough sleep.

@0celo7 I think my argument still works but we just pick the same sequence

No, because you can't guarantee $||x_i||\le 1$.

why ? I am just picking a sequence in the first norm where it is less than 1

Yes, but you don't know that this sequence satisfies the negation inequality.

oh

hm I have an idea. We know that for any c > 0 there exists $x \in X$ such that $||x||_2 > c||x||_1$ this means that $\frac{||x||_2}{||x||_1} > c$ for any c > 0. Pick sequence $x_i$ such that this property holds.

I.e pick sequence $x_i$ such that $\frac{||x||_2}{||x||_1} > i^2$ for each i.

we know that this is guaranteed.

yes.

Then the sequence $y_i = \frac{x_i}{i*|x_i|_1}$ would be convergent to zero in first norm but in the second one it wouldn't converge right ?

ew is that a * for multiplication

yeah

yeah it converges to 0 (?) in the first norm

but what happens in the second norm ?

Its norm diverges like $i^3$ I think.

Since convergent sequences are bounded, I think that's good.

yeah

good

small detail

can you guarantee $||x_i||_1\ne 0$

you're dividing, always have to check such things

yes because by the construction of the sequence.

yup, that would imply $||x||_2=0$ so the inequality does not hold

looks good

Yeah analysis can be fun at a times

I still prefer algebra and topology

thanks @0celo7 for checking my work

np

hey @0celo7 do you functional analysis ?

not yet

oh ok I am having a really tough question have no idea how to solve it

what is it?

just curious

Oh that doesn't seem that bad.

I don't know how to do it, but I could see that being on my (non functional) analysis homework

unless there's some theorems you need

hm

I'm pretty tired, so I can't offer help

oh ok

I've been working on "If $(X,d)$ is a compact metric space, any family $\mathcal F\subset C_E^b(X)$ which is equicontinuous at each $x\in X$, is uniformly equicontinuous on $X$."

oh cool

in your analysis class ?

yes

I will ask this question on main.

cool question @0celo7

I'm 99% sure I need the Lebesgue number lemma

But we haven't talked about that

And when I went back to look at our proofs on uniform continuity, I think they're all wrong

oh

the prof said "this part is trivial" and I'm quite sure it's highly nontrivial

because we did some uniform continuity proofs in topology and they used the lebesgue number lemma

yeah I hate it when prof do that

a continuous function on a compact metric space is uniformly continuous

yeah

you can get it to work without the lebesgue number lemma but it's not obvious

the proof he gave in class is certainly not complete, probably wrong

also on the homework solutions he said something was "trivial algebra" but it took me a page to do it

so I need to ask him about that

I see

and something about principal bundles

which analysis course is that ?

that's not for the course

he's my advisor

(the prof is)

what is his field ?

I think I have an idea how to solve it.

@Adeek geometric analysis

Let X be a metric space, let A be a non-countable subset of X s.t there exists E > 0 for all $a,b \in A$ one has p(a,b) >= E. Then X is non-seperable.

cool @0celo7

@Adeek p?

metric

what's wrong with d :o

I dunno prof likes his p's

haha

you sure it's a p not a rho

aha

I have found the uniform continuity proof

I don't think I need Lebesgue for the homework, but I should point out he told us a wrong proof

oh

yeah

I think you should prof like it as well when you do that

Is (a,b) open in all (R,d)?

arbitrary d such that (R,d) is met space

R being the reals

use a bijection on R that sends (a,b) to a set that is not open wrt the Euclidean metric. transport said metric using the bijection.

(if f:R->R is such a bijection, define d as d(x,y)=|f(x)-f(y)|)

roger that

hey @arctictern would you like to discuss the problem above with me ?

nah. not in an analysis mood, youtubing off to sleep.

oke

I guess this is quite a common question so that are probably several posts about his on the main site.

good thing that your here @MartinSleziak I am struggling.

By a quick search I found this one: Let $(X,d)$ be a metric space and $M\subset X$ is uncountable. There is $\alpha>0$ such that $d(x,y)=\alpha$. Prove that $X$ is not separable..

I simply put uncountable separable metric "d(x,y)" site:math.stackexchange.com into Google. Perhaps if you try searching for some different - but related - phrases, you will find other questions about the same problem.

oh I see

assuming that result you posted above. if I consider for example $l_p(\gamma)$ how can I find f,g that satisfy this condition ? I mean the domain that we are talking about is arbitrarily uncountable set ?

BTW there is also general topology room. Your question would fit nicely in the topic of the room. However, for quite a long time there is almost no activity in that room, so it might go unnoticed if you post there. But let's hope that the general topology room will become active later again.

oh awesome I will go there

This is my question in case you didn't see it.

So we are talking about norm $\|x\|=\left(\sum_{i\in I} x^p\right)^{1/p}$ where $I$ is a set of uncountable cardinality.

yes

Ok, in your notation you have $\Gamma$ instead of $I$.

yeah

yes

@Adeek Wouldn't such functions which have 1 exactly at one position and zero at others work?

Sorry, Adeek, that's basically all what I can do now. I will have to leave soon - time to go to work. :-(

hm I will think about it thank you though

I dunno why people

didn't like my question in main

BTW I consider the space $\ell_p(\Gamma)$ a nice thing.

It is interesting to know that we are able to work quite nicely with uncountable sums.

yeah

I like that as well I generally like infinite uncountable spaces.

it is very nice thing to stretch imagination.

yes

@MartinSleziak defining it this way we get actually what we want

because if we lets say have f as you defined it

And also I like the result that each Hilbert space is isomorphic to $\ell_2(\Gamma)$ for some $\Gamma$. (Although I do not know about some reasonable application of this results. But still, knowing that they are completely characterized and that that there is one simple cardinal invariant which determines whether two such spaces are isomorphic is kind of nice.)

and we define another g in same way where $\gamma$ is chosen in a different than the other one that you did, then we get $d(x,y) = \alpha$

or I guess $d(f,y) = \alpha$ under that norm.

Here $\alpha=2^{1/p}$, right?

in particular $\alpha = 2^{1/p}$.

yeah

yes so that works.

also for $l_{\infty}(\Gamma)$ we can do same argument actually

@MartinSleziak I really like functional analysis

it is nice

@MartinSleziak before you go do you have an idea about which functions to use in $c_0(\Gamma)$?

@Adeek My guess would be that the same functions should work.

ok, I'll have to go, see you later

alright cya l8er

thanks a lot for your help @MartinSleziak

yes your righttttt

same function work

thanks a lot @MartinSleziak

@TedShifrin But what were you thinking? :P

Also @TedShifrin I've got another question about complex geometry. Huybrechts tells me that on a Kaehler manifold the Lefschetz operator (wedging with the fundamental form) $L$ induces isomorphisms $L^{n-k}:\mathcal H^{p,k-p}\to \mathcal H^{p+n-k,n-p}$ where $0\leq p\leq k\leq n$, $\operatorname{dim}_{\Bbb R}(X)=2n$ and $\mathcal H$ is the space of harmonic forms.

I understand injectivity, and about surjectivity Huybrechts says "we note that the dual $\Lambda$ commutes with $\Delta$". I'm not completely sure how to use that. I mean, I see that that fact gives me an "obvious candidate", namely $\Lambda^{n-k}\alpha$, which has the right bidegree to be mapped onto $\alpha$ by $L^{n-k}$. It's also harmonic by the vanishing commutator. But I'm not sure if it actually works. After all, $\Lambda$ is the dual, not some kind of inverse!

I tried using the expression for $\Lambda$ in terms of $L$ ($\Lambda=*^{-1}L*$), and get the condition that $L^{n-k}\circ *^{-1}\circ L^{n-k} \circ *=\operatorname{id}$. Note that this doesn't have to hold on all bidegrees (only on anything of the form $\mathcal H^{p+n-k,n-p}$ with above conditions on $p,k$), otherwise $1$ is an easy counterexample, as $L*1=0$.

It seems kind of likely-ish, at least when you try some obvious things like $\alpha=*1$ then it seems to work. I guess what I'm asking is if there is some clear way to prove this---since Huybrechts doesn't spend any time on it I'm getting the feeling that I'm missing something simple. Any help would be much appreciated @TedShifrin ;)

Oh, actually... Let's see... I can at least write any such $\alpha$ as $*\beta$ where $\beta\in\mathcal H^{p,k-p}$. So that reduces it to $L^{n-k}*^{-1} L^{n-k}\beta=(-1)^k*\beta$

(that doesn't help much :P)

Perhaps a (disgusting!) computation with explicit basis is the only way to go?

Hihi

wish I could help but I'm trying to finish my abstract algebra

ugh so mad!!!! I need to do surjection T_T!

is the set A = {(1,1), (2,2), (3,3)} symmetric or antisymmetric

Im struggling to understand how I can tell if any set is either reflexive anti or symmetric or transitive.

Could anyone provide simple examples of sets that are each one of the four?

reflexive is like $(a,a) \in A $

ugh late

tired T_T

but usually reflexive is just a single element.. or you can think if it as a single color red

so reflexive would be {(1,1),(2,2), (3,3)}?

yup

ugh I need sleep

thats if the set of A = {1,2,3} ?

I should.. I was just winging the last parts of abstract algebra homework

there's one tiny part left but zzz time

@usukidoll Btw thanks for help sleep well

yw

nighty night

what would symmetric and anti symmetric look like|?

How to convert a matrix into ket/bra notation: would that be a physics or mathematics question? (I mean which site)

can you show an example such as {(1,1),(2,2), (3,3)}

but make it symmetric?

so that example is symmetric?

{(1,1),(2,2), (3,3)}

^

how can it have 1_a and 1_b and 1_b and 1_a

(1,1) is only in that set one time

so {(1,1),(2,2), (3,3)} is in fact reflexive and anti symmetric

symmetric*

what would change in it to make it anti symmetric

{(1,1),(1,2), (3,3)} this would be anti symmetric correct?

@MAFIA36790 Thank you very much for the help off to learn programming in pascal -_- hopefully ill fully grasp the concept eventually

I think it would be anti because there is no (2,1)

but there is a (1,2)

@MAFIA36790, okay, thanks!

im back

Get to be on computers here haha

So I think it would be anti because there is no matching (b, a) for the (a, b)

Hey how do I insert a table at math.se?

thanks

homework isn't an allowed tag?

@MAFIA36790 cause it didn't find the tag

Well not allowed isn't the right word: not existent is a better word

Also not sure if it's truly homework: I just posted a question about a "problem" the textbook just skipped over without explaining how it is done; something that felt utterly silly but not trivial to me nonetheless

Well it's just for a maths course (probstat)

seasons change