6:33 PM
Is the concatenation between paths always defined?
I don't see why the set of Paths should not form a monoid

@AkivaWeinberger @quallenjäger how many times do you concatenate paths before it becomes a marathon

"If we have a Galois group, then we can look at its representations. Galois Representations are really important, but they are never done in isolation. Wiles did a whole bunch of work to show that for ever Galois Representation of a certain type, you get a modular form that "matches" it. This means that you can use what you know about modular forms to conclude something about these kinds of Galois representations (eg, that there is no Galois representation of our type with a certain signature needed by a counterexample to FLT).
What is class field theory

A lot

@quallenjäger loops form a group as long as you pick a common point of origin for all loops

6:35 PM
If I consider only paths?
Do they have a monoid structure?

but for paths to be concatenated you have you deal with the possibility that the end of the first path might not be the beginning of the next path
so you'd need a way to define that

I don't require my paths to be continuous
I mean, more is the problem with the associativity. Concatenation is not really associative because of different parametrization
But beside that, I don't see why the concatenation should not be defined for some paths
6

Given a graph G, and the set of paths in G called PathG. Is there a monoid structure on PathG? Will concatenation be the multiplication formula? even if it's not defined for some paths? What about the identity?

My question is related to this. I don't understand the answer.

@quallenjäger how can you concatenate paths if the destination of one is not the beginning of the other?

I thought it might have a jump at this point.

that question is talking about paths as a sequence of edges, not a function from [0,1] into a topological space
(and in any case, the second usage of the term "path" would have to be continuous anyway, as semi notes)

6:40 PM
Ah I see.
I am dealing with a set of general paths, where I only require it to be bounded variation
Does it have a monoid structure under equivalence of reparametrization?

if you're allowing your "paths" to have jumps, then you're not actually talking about paths.
do you think it has an identity element?

I don't know actually. I thought I could define a stationary point as an identity element.
But I need the reparametrization equivalence to achieve the associativity right?
This cannot be relaxed?

it's certainly not associative without reparametrization. are you sure a stationary point is an identity?

(As an aside, iirc one can do stuff involving homotopy of unbased paths. But then the space of paths forms a groupoid not a group.)

yes, fundamental groupoid

6:48 PM
I am not sure, I thought I can define the endpoint of the paths to be the unit element. Then under reparametrization, it will give me the original paths.

"the endpoint of the paths" - wait, so they all have the same endpoint?

Does it need to be unique? the identity element?
Ah ok I see.

@anon kk. A pity I only know that at the level of a slogan

you can't have a nonunique two-sided identity
(what do you think would happen if you were to combine distinct identity elements together?)

Ok then It will only have a semi group structure I think.

6:52 PM
yes

@anon are you familiar with the iterated integral as a homomorphism to the semi group of the paths, which I have mentioned before.

@quallenjäger If you want to stay in the realm of associative algebras (rather than move up to categories), then you just define the composition of paths to be zero if the start and end points do not match
usually you do this with quivers and get path algebras

I think Tobias is talking about quiver algebras,

What are quivers?

basically graphs

6:54 PM
just directed graphs

we already found out quallen is talking about functions from [0,1] to a topological space, rather than graphs in discrete math

@anon ahh, woops
but the linked question is about graphs in the sense I meant

yes

I am currently working on a paper by Chen, where he has developed a representation on the group of continuous paths (this can be constructed by introducing equivalence classes) into the tensor algebra. I was wondering, if the same applies to paths with bounded variation only. That's why I am asking. I need somehow a certain group structure on the set of paths with bounded variation.
Do you have an idea why it is useful to represent the group of paths into some algebraic object?
Is it because we can deal with concatenation of the paths more easily?

What does it mean for a path to have bounded variation?

7:01 PM
I consider a paths as a function from $[0,1]$
the function has bounded variation.
i.e. finite length

Where does it go? I mean, this is not a term that applies to just arbitrary paths

into $\Bbb R$
sorry

whats a groupoid missing to be a group
?

@GFauxPas You mean what it has that is too much

idk I've never heard of it before

7:03 PM
unless you mean groupoid in the older sense which is now usually called a magma
the old term is used for any set with a binary operation
the newer use is a category where all morphisms are isomorphisms

If you think of a group as the set of functions on one set, a groupoid could be a set of functions between many sets, which includes all the functions' inverses and all the sets' identity functions.

hm
thanks

The 15-puzzle is easier to think of as a groupoid than a group, honestly

@anon why? I find it more natural to consider it as a group action

there are more states than just "lower right corner missing a square"

7:06 PM
ahh, true

each individual shift is a transition between the different states

Is a polynomial function on the ring of polynomials always linear?

one usually finishes the phrase "polynomial function on []" with a vector space or a variety
the ring of polynomials is itself an infinite-dimensional vector space. are you thinking of it that way?

Yes
i.e. polynomials of polynomials.

then replace the phrase "ring of polynomials" with "infinite-dimensional vector space." is a polynomial function on an infinite-dimensional vector space always linear?

7:11 PM
That's sounds more correct :D Thanks.

by "correct" do you mean accurate portrayal of your question, or do you mean more likely to have an answer of "yes"?

I mean thats the formulation of the question sounds more correct.

well, is a polynomial function on a finite-dimensional vector space always linear?

yes

Do you think f(x,y) = x^2+y^2 is a linear map from R^2 to R?

7:18 PM
Hmm, I see.
So it is wrong.

very

How to show $d(a,b) - d(b,c) = d(a,c)$?

What's $d$?

some metric

It's not always true

7:23 PM
huh, interesting

It's true if $c$ is on the line segment $ab$ I guess
But if you take $a=0,~b=1,~c=2$ then you'll see $d(a,b)-d(b,c)=1-1\ne d(a,c)=2$
($d$ here being the usual metric on $\Bbb R$, the absolute value of the difference)

Hmm. I have to get $sup_{x\in X} |d(x,p)-d(x,q)| = d(p,q)$
Where, $p$ and $q$ are fixed

So that's attained when $x=p$ or $x=q$, so you just need to show that it can't ever be any higher than that

Oh, I see what you're saying
Thanks

In fact, in $\Bbb R^n$, we get $|d(x,p)-d(x,q)|=d(p,q)$ iff $x$ is on the line $pq$ but not on the interior of the line segment $pq$
(i.e. colinear with but not between)

7:32 PM
@TobiasKildetoft should it be obvious that $\mathfrak{gl}_n$-representations are not necessarily isomorphic to their duals?
Actually, I should ammend that.

@anakhronizein Sure, just consider multiples of the trace

should it be obvious that the standard ones on $\mathbb C^n$ are not necessarily isomorphic to their duals.

$X^{-T}=CXC^{-1}$, now take $\det$
that's for $GL$ anyway

Some days ago, someone posted a question regarding a possible circularity in the definition of <.

@anakhronizein Well, even for $\mathfrak{sl}_n$ it is only with $n=2$ that it is self-dual (until you modify the duality)

7:36 PM
If I remember correctly, the person pointed out that the definition that: a > b iff a - b >0 used > to define >.

So which duality are we taking?
@BillyRubina Rather, it reduces to defining what it means to be positive

Yes, but the person pointed it this way, I guess.

I just mean the dual representation as defined by $(\rho^*(a)f)v := -f(\rho(a)v)$, @TobiasKildetoft

@anakhronizein Are you familiar with highest weights?

Yes, a bit.
Does it relate to that?

7:41 PM
calculate the highest weight of these representations

Anyone know the Extreme Value Theorem? trying to prove it

@user525966 can you give it ?
because there many theorem, like the Hence theorem et other
the image by a continuous fonction of a compact is a compact

I will try that, @TobiasKildetoft thanks!

If $f:[a,b]\to\Bbb R$ is continuous, then $f$ is bounded and attains its maximum and minimum.
is how I usually see it phrased.

For some continuous function over interval $[a, b]$ there exists a $c,d \in [a,b]$ such that $f(c) \leq f(x) \leq f(d)$ for all $x \in [a,b]$

7:46 PM
yes it's this result

So this is from an analysis class?

the hence result

Self learning

Do you know the compacts ?

7:47 PM
"Compactness" is an idea from topology

do you know what's an open set ?

How do you define the reals? Is it in terms of completeness, or the least upper bound property?

@Dattier No but I assume it's a set with infinitely many elements with some kind of pattern to them that approach something not actually in the set
like an open interval (a, b) I imagine is an open set

How are you self-learning?
What resources are you using?

7:50 PM
A set $A$ is open if, for every $x\in A$, there exists an $\epsilon$ such that $(x-\epsilon,x+\epsilon)\subseteq A$

do you know this result : a sequence bounded increasing converged ?

*A bounded, increasing sequence is convergent.

Thanks

I'd recommend getting an analysis book, like Rudin.

@user525966 : do you this result ?
A bounded, increasing sequence is convergent.

Note that this is only true over the reals. There are counterexamples when it's over something smaller, like the rationals
For example
Consider $f$ whose domain is the set of rationals between $0$ and $2$ inclusive, and define it as $f(x)=\dfrac1{x^2-2}$
If you graph it, you'll see that it is definitely not bounded. However, it is defined everywhere in that domain, and it's continuous everywhere.

@user525966 : sorry, but theses questions are necessary, and if you don't answer to them, we can't help you

The one place where there "should" be a discontinuity, $\sqrt2$, is not in the domain, since I did this over the rationals.

Unfortunately my answer to most of these are "no, not yet familiar"
Still trying to understand the open set thing

Thus: The Extreme Value Theorem uses the properties of the reals in an essential way.

7:55 PM
familiar with epsilon-delta stuff already but not quite sure what this open set looks like

An open set is something like $(0,1)$ or $(0,1)\cup(2,3)$

for every element in A there are arbitrarily close elements in A on either side of it?

A union of open intervals, even an infinite union of them, is called an open set
@user525966 Well, that's true of $\Bbb Q$ as well, but it's not open.
We need an open interval surrounding every point such that everything in that interval is in the set
This is equivalent to the "union of open intervals" thing that I just gave

like the definition you mentioned earlier almost seems like it could be recursive

Open interval means $(a,b)$ for some $a$ and $b$; open set means an arbitrary union of open intervals
I mentioned that the Extreme Value Theorem uses, in an essential way, the properties of the reals. One such property is the least upper bound property.
Aka the supremum property.
An upper bound of a set is something $\ge$ everything in the set. For example, $2$ and $3$ are both upper bounds of $[0,2]$, but neither $0$ nor $1$ are upper bounds. @user525966
Similarly, $2$ and $3$ are also upper bounds of $(0,2)$ (and $0$ and $1$ are not).
Now, $2$ is the smallest number that is an upper bound of $(0,2)$. It is called its least upper bound, or supremum.
If a set has a maximum element, then the supremum equals that maximum. For example, the maximum element of $[0,2]$ is $2$, and that's the supremum.

8:05 PM
yes

However, some sets don't have maximums. No element of $(0,2)$ is a maximum, for example (note that $2$ is not in the set because the round parentheses mean that I don't include $0$ or $2$).
But $(0,2)$ still has a supremum, $2$. So supremums don't need to be in the set, and if they are in the set, they're the maximum element.
The least upper bound property, or the supremum property, says that every bounded set has a supremum.
(Except for the empty set but we can ignore that)
Again, this is true in the reals, but it's not true in the rationals. Take, for example, $X=\{3,3.1,3.14,3.141,\dots\}$.
Every element of $X$ is rational, but its supremum, $\pi$, is irrational.

Hey everyone
What do you guys use to denote your closed balls in a metric space?

One way to show the Extreme Value Theorem is to consider the set $B$ of points $x$ in $[a,b]$ such that $f(x)$ is bounded on $[a,x]$, and think about the supremum of $B$.

@Perturbative $B^c(a,r)$

Well, actually, once you fill in the details, that only shows that $f$ is bounded on $[a,b]$. It doesn't show that the maximum is attained.

8:10 PM
$B^o(a,r)$ for the opens balls

In other words, it shows that there exists an $M$ such that $f(x)\le M$ for $x\in[a,b]$, but it doesn't find a $c\in[a,b]$ such that $f(c)=M$.

o for open and c for closed

But most of the theorem is just showing boundedness.

@Perturbative : ok ?

@Dattier Yep, thanks :)

8:12 PM
The annoying thing is that the closure of $B^o(a,r)$ need not equal $B^c(a,r)$ in that notation @Perturbative
Consider $X=\Bbb Z$, $~a=0$, $~r=1$.

@AkivaWeinberger But correct me if I'm wrong the closure of $B^o(a, r)$ need not equal $B^c(a, r)$ in general...
So it isn't that bad a thing to have I'd guess

That's literally what I just said

It'd be really annoying in Euclidean spaces
@AkivaWeinberger Sorry I was confused why you were saying that it's annoying

Not an annoying thing about the notation, an annoying thing about the balls
For what it's worth, here's how Wikipedia writes it:

In Euclidean spaces I'm just gonna say denote an arbitrary closed ball as $\overline{B(x, r)}$
@AkivaWeinberger The wiki article is just .... ewww....

8:18 PM
$B_r(p)$ or $B(p;r)$ versus $B_r[p]$ or $B[p;r]$
Can't say I've ever seen that

Is "usually denoted" lmao

Oh weird it wrote $\overline B_{\overline r}\overline{(p)}$. I assume that's a typo
Yeah it's using its native formatting rather than LaTeX there

these questions may seem secondary, but in fact they may be the source of ease, or difficulty.

@AkivaWeinberger I still haven’t answered my $U=US$ question. It feels like it should have an obvious answer.

I have an exemple about that

8:24 PM
Like rk(U) < # of rows of U

$F$ function from $\mathbb R^n$ to $\mathbb R$, $A=\sum \limits_{\sigma \in S_n} s(\sigma)\times f(a_{\sigma(1)},...,a_{\sigma(n)}$.

@AkivaWeinberger How do we know if there is no max element of the set or not?

$i\neq j$ with $a_i=a_j$. Is it true that A=0 ?

grumble grumble

$F$ function from $\mathbb R^n$ to $\mathbb R$, $A=\sum \limits_{\sigma \in S_n} s(\sigma)\times f(a_{\sigma(1)},...,a_{\sigma(n)}$.
$i\neq j$ with $a_i=a_j$. Is it true that $A=0$ ?

8:25 PM
@user525966 Depends on the set

If you choose the good notation, one half lines, is enough.

@Semi what's the question in full?

You know how $\begin{bmatrix}1&0&0 \\ 0&1&0 \end{bmatrix} \begin{bmatrix}1&0&0 \\ 0&1&0 \\ d&e&f\end{bmatrix} = \begin{bmatrix}1&0&0 \\ 0&1&0 \end{bmatrix}$?

@Perturbative do you want try it ?

He wants examples of $US=U$ where $U$ is fat and $S$ is orthogonal

8:28 PM
@AkivaWeinberger that’s only orthogonal if d=e=0 and f=1, isn’t it?

Yeah I guess it is

Hmmm

And I don’t want S to be the identity, that’s important

@TobiasKildetoft Is it right that the highest weights are the 1st standard basis elements for C^n and its dual?

What do you mean by $i\ne j$ with $a_i=a_j$? As in, there exists such an $i$ and $j$? @Dattier

8:30 PM
I know at the least that rk(U) <= the number of rows of U

$\exists i\neq j \in \{1,..,n\}^2,a_i=a_j$ @AkivaWeinberger

What I think should further be true is that the inequality is strict

Then $f(a_1,\dots,a_i,\dots,a_j,\dots,a_n)$ and $f(a_1,\dots,a_j,\dots,a_i,\dots,a_n)$ will have opposite signs
One will be $+$ and the other will be $-$
However, since $a_i=a_j$, they'll be equal
so they'll cancel each other out

why ?
the signs are opposite

$f(\dots,a_i,\dots,a_j,\dots)$ and $f(\dots,a_j,\dots,a_i,\dots)$ will be equal I mean
So $s(\sigma)f$ will be opposite

8:33 PM
Well that is heighest weight vectors.

It's a problem to choose the good notation
@AkivaWeinberger

One hope I had was that the Sylvester or Frobenius rank inequalities would help

I guess "highest weights" then would correspond to then $-\varepsilon_1$....
Or negative that for the dual.

@Dattier Write it as $\sum s(\sigma)f(\sigma)$
Then define $\sigma'$ and $\sigma\circ(i~j)$
$s(\sigma')=-s(\sigma)$

But then I thought the coefficients for the weight vectors should decrease

8:37 PM
However, $a_i=a_j$ implies $f(\sigma)=f(\sigma')$
Finally, note that $\sigma\mapsto\sigma'$ is a bijection on the set of permutations

well, I give a good notation, let $h\in S_n, A_{h}=\sum \limits_{\sigma \in S_n} s(\sigma\circ h)f(a_{\sigma(h(1))},...,a_{\sigma(h(n))})$

Thus, $A=\sum s(\sigma)f(\sigma)=\sum s(\sigma')f(\sigma')=-\sum s(\sigma)f(\sigma)=-A$
Thus $A=-A$
and $A=0$, QED.
@Dattier You want $s$ in there somewhere
@Dattier And then $A_{(i~j)}$ equals both $A$ and $-A$, therefore $A=0$.

bravo

Well, $A_h=A$ for all $h$

so, when you choose the good notation, the proof is quick

8:46 PM
I think $A=\sum s(\sigma)f(\sigma)=\sum s\big(\sigma\circ(i~j)\big) f\big(\sigma\circ(i~j)\big)=-\sum s(\sigma)f(\sigma)=-A$ is quick

ok

I defined $f(\sigma)$ to mean $f(a_{\sigma(1)},\dots,a_{\sigma(n)})$

it's an other path
so the choice of a good notation is important

9:10 PM
@Semi wait $U$ being fat means here that it has more columns than rows?

Helo

hi
@Tuki

If you sit in centrifuge the force created by normal acceleration faces inwards or outwards ? i think this is outwards but not sure

@Daminark yeah

@AkivaWeinberger : Do you think that the Wiles-Fermat theorem can be proved in less than 10 lines?

9:17 PM
@Daminark fat transpose can make you thin

@AkivaWeinberger What is the subject that covers all that stuff re: supremums, infimums, sets, etc
or a (good) resource for learning it

@user525966 Analysis

@Dattier I saw that video but he just describes what it is, not how to prove or derive it

@Dattier Definitely not. We'd be lucky to prove it in less than 50 pages.

9:22 PM
@AkivaWeinberger why ?
why I think is possible because
when I proposed my problems in many case I know a solution of few lines (less of 3)
but when an other propose a solution is more long (ten lines or more)
@AkivaWeinberger and you, why you think it's not possible ?

Because it's been open for >300 years

I think It's possible to build a problem, like that (and with a solution of few lines)

Sorry can you please explain open set again? A set $A$ is open if $\forall x \in A, \exists \epsilon : (x-\epsilon,x+\epsilon)\subseteq A$, that "C" symbol means is a subset and possibly IS the set A itself?
And $(x-\epsilon,x+\epsilon)$ is an interval?

Yes
Try to find such an epsilon with $A=(0,1)$ and $x=0.75$

9:37 PM
I guess I don't quite get it since we're saying these "elements" x are in A, but we also have these intervals in A, like it's a mix of data types

Well that's why $\in$ and $\subseteq$ use different symbols
Try to find such an epsilon with $A=(0,1)$ and $x=0.75$

when you say that are you saying $A = \{0, 1\}$ or $A = \{(0, 1)\}$ for example?

Neither. $A$ is the set of real numbers between $0$ and $1$, not including $0$ or $1$.

that doesn't use the set brackets $\{ \}$?

9:41 PM
it's in french but he write on board

Maybe I am missing something but I don't see a solution to epsilon if .75-epsilon=0 and .75+epsilon=1, since its not epsilon=.75 or epsilon=.25
unless by the C symbol it means within those bounds or something,
so epsilon = .25 probably, which corresponds to (.75-.25, .75+.25) = (.5, 1)?

Yes
(.5,1) is the set of things between .5 and 1.
If you're between .5 and 1, you're also between 0 and 1.
That is, if you're in (.5,1), you're in (0,1).
In other words, (.5,1) is a subset of (0,1).
"Subset" means every element of the first set is an element of the second set.
Incidentally: Examples of things that are not open: (0,1], [0,1), and [0,1] are not open
(0,1] means the set of things between 0 and 1, not including 0 but yes including 1
Similarly for the others
The problem is that if $x$ is $1\in(0,1]$, you can't find any such epsilon

Howdy, DogAteMy.

Hey Ted!

Hi Ted

9:56 PM
hi Demonark, Tuki

@user525966 : $\exists (x_n) \in [a,b]^{\mathbb N}\sup(f[a,b])=\lim\limits_{n\rightarrow \infty} f(x_n)$ ok ?
case 1 : $\forall n\in \mathbb N, \sup\{x_k;n\geq k\}=\max\{x_k;n\geq k\}=x_{\phi(n)}$, $\phi(n)\geq n$
so $x_{\phi(n)}$ descreasing and bounded so converged to $z$ and by continuity

Here’s something I thought I’d never say: The simplest proof I have for something is to appeal to Hölder’s inequality

What's wrong with Hölder?

Nothing, I just didn’t expect to ever run into it in a physics problem :)

$f(z)=\lim f(x_{\phi(n)})=\lim f(x_n)= \sup(f([a,b]))$

10:02 PM
Well, certainly the $L^2$ case shows up in physics all over.

Isn’t that just Cauchy-Schwartz?

@user525966 ok ?

Sure, @Semiclassic. :)

But Hölder includes that.

10:03 PM
Of course
Strictly speaking I need to use Hölder twice so that in effect I have 1/2+1/2+1/infinity =1

I have no idea what that means.
Hölder for $p=1$ and $q=\infty$ is just trivial, so no one ever calls that Hölder.

hmm
I’m omitting context, mind

All the more reason for it to make no sense.

B/c my norms in this case are the Schatten p-norms

I don't know what that means, either.

10:07 PM
They satisfy a version of Hölder though

So let's skip it.

Fair. Back in a but

A pretty proof of Bolzano-Weirestrass theorem
$(x_n) \in [a,b]^{\mathbb N}$ then $\exists \phi, x_{\phi(n)}$ converged
case 1 : $\forall n\in\mathbb N, \sup\{x_k;k\geq n\}=\max\{x_k;k\geq n\}=x_{s(n)}$ $s(n)\geq n$.

Back now

so $x_{s(n)}$ discreasing and bounded, then converged
case 2 : is trivial
the negation of the case 1
What do you think about ?

10:22 PM
@Dattier I don't follow your earlier msg -- lots of new notation

10:33 PM
\o Alessandro

Hi chat
Sup @Dami
I pinged you earlier because you seemed online but I guess you actually weren't
So did you ever take that model theory course you mentioned a lot of time ago?

Is it true that $\lim \inf x_n \le \lim_{n \to \infty} x_n \le \lim \sup x_n$?

yes
If the limit exists

But if the limit exists, then aren't they all equal? I was thinking the inequality holds for any sequence.

Yeah the are all equal
If it exists
If it doesn't exist, it doesn't make sense to say liminf < lim
Does that make sense?