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04:00 - 15:0015:00 - 00:00

04:01
A ring is just a group with certain homs present?
1
Q: Is this an equivalent formulation of a ring?

EnjoysMathLet $G$ be an abelian group such that there exist two families $\{\psi_g, \phi_g\}_{g \in G}$, not neccessarily disjoint, of homomorphisms $\psi_g, \phi_g : G \to \langle g \rangle$, each mapping onto the corresponding cyclic subgroup; such that $\phi_g(h) = \psi_h(g)$ for all $g,h \in G$. Then ...

 
2 hours later…
06:27
Hey chat
I have already posted the following question, but there hasn't been any response. I think it might be better to discuss about it in chat since it is rather a soft question.
Do any of you know examples of smooth, single-sheeted (i.e. not intersecting itself) curves (preferably closed), which do not have a "simple" shape?
Here "simple" is meant in the geometric sense, that is: A circle, ellipse or many other famous curves have many symmetries and other geometric properties (constant or at least too simple curvature).
For my thesis I need to test a matlab code on some exemplary curves, but a circle for instance doesn't really represent a "random" smooth curve due to these strong geometric properties.
If you could think of a method (that can be implemented) to generate such random curves, that would be even better of course.
28
Q: How can I generate "random" curves?

Herman TullekenIn game programming (my profession) it is often necessary to generate all kinds of random things, including random curves (for example, to make a procedural island or for an agent to follow some path). For one dimensional things, we usually use some random generator that generates say floats (w...

06:43
$\land$ stack things
$A,B$
$A \land B$
$(A \land B) \land B$
etc.
$\land$ commutes
$A \land B$
$B \land A$
one can undo $\land$
$A \land B$
B
A
Therefore:
$(A \land B) \land C \implies (B \land A) \land C \implies B \land (A \land C)$
06:56
ah interesting:
commutative + associative -> cyclic permutation for right associative expressions
Because:
$A \land (B \land C)$
$(B \land C) \land A$
$B \land (C \land A)$
$(C \land A) \land B$
$C \land (A \land B)$
$(A \land B) \land C$
well it kinda make sense:
Every left association translate the bracket to the right
and every commutation flip things from right to left
Thus a cyclic orbit is set up
Meanwhile, commutativity itself is ideompotent
$A \land B$
$B \land A$
$A \land B$
So to generalise, let P be a program
Then we have the following:
P(A,BC), P(BC,A), P(B,CA), P(CA,B), P(C,AB), P(AB,C), P(A,BC)
Hey @Secret
Thank you for the link.
no problem
I forgot to mention that I need implicit curves
But maybe in that post I can find something helpful
yeah you will probably need to generate some Brownian motion somehow, they are some of the most random curves that are still smooth
Okay
Actually while thinking about this, I was wondering: Is it not possible to obtain ANY such curve approximatively through polynomial expressions?
In other words, is there not some equivalent statement of the Stone Weierstrass theorem for implicit curves?
07:15
I thought stone weierstrass said that for any continuous curve on a closed interval you can always approximate it with polynomials?
This just ended up on the front page of MO again. It is a really neat proof of something that is so intuitive mathoverflow.net/questions/60375/…
(also, the original version of the answer about $S^2$ is pretty funny).
hmm: $\to$ is right associative: $A, A\to B, A \to (B \to C) ,...$
Meanwhile $\implies$ is useful to de-nest something i.e. $(A \to A), A \implies A$
@Secret Yes, that's what stone weierstrass says. Also, there are constructive proofs of this (the one I know is due to Bernstein using his Bernstein polynomials). So what I meant was: Maybe there is some analogue statement for smooth (and perhaps single-sheeted) curves. Maybe any such curve can be approximated with polynomial implicit curves, i.e. (for planar curves) with some zero set $P(X,Y)=0$ where $P$ is a polynomial in two variables.
07:40
That I am not sure. I don't remember if there is a multi variable version
 
2 hours later…
10:01
Hi all there!
@Rudi_Birnbaum Hi
Just read about "denialism" in the Guardian and thought about if there are cases of that as well in mathematics. Then one mathematicians talk I once saw on youtube came to my mind. But I forgot his name. I remember he is elderly, from the US (possibly immigrated), and says "infinity" does not make sense. He is shouting almost for the whole talk and in gives in general a quite surreal impression. The talk is in some University (maybe it was NY area). Anyone an idea what the name of the guy is?
@Rudi_Birnbaum Sounds like Wildberger from the content (not sure if he usually shouts though)
Also, I recall Wildberger being at a German university
@TobiasKildetoft This talk I saw was surely the most bizarre maths talkI ever saw. He really was shouting very loud and often. But it was often that he spoke calm and just all of a sudden started to shout. I just checked a Wildberger talk, and I think its not him. The guy also had some noticeable accent (when I remember correctly).
Anyway, Wildberger is certainly also an example of denialism. There is nothing in itself wrong with being a finitist, but Wildberger is so in a completely absurd and confrontational way
10:15
@TobiasKildetoft OK maybe then it was even him on a very strange performance then.
I have not actually seen any of his videos, so I have no idea
hi @TobiasKildetoft
@LeakyNun Hi
German: written "dich", pronounced /diç/
Danish: written "dig", pronounced /dai/
????
@LeakyNun What's so weird about that?
10:29
lol
how does that "i" become /a/ and how does that "g" become /i/
ohh, there is no good reason. It is just that way we do things
Danish is just a very fascinating language for me
Same with "sig" (except that also has a different possible meaning where it is pronounced completely differently)
brilliant
@leaky: The "i" ist just pronounced as i should be ... and the "g" is weakend to some other stuff which is typical in germanic languages. Like the Englisch "I" which was an "Ig" or "Ich" ...
You hear that in some German that the ending "...ig" like in "farbig" is pronounced "farbich".
10:33
more like and "ik" or "ich"
g and k are closely related.
I think the "g" all underwent lenition in the Scandanavian languages
@Rudi_Birnbaum I don't think it was "ig" in English
especially with the tendency of Germanic languages to devoice the final consonant
no that was scandinavian. in englisch I think it was "ich"
well it was written "ic" and "c" could be k or ch
Wiktionary includes both
I found also "Ich" somewhere ..., but yeah.
10:36
@Rudi_Birnbaum that's a different "ch" in "farbich"
"g" did not undergo lenition in the beginings or the middle like "gamla" old
@Rudi_Birnbaum I see
@Rudi_Birnbaum Except before certain vowel sounds in Swedish where g is pronounced more like a soft j
Well its the same as the German ch in farbich as in "ich"/I
@TobiasKildetoft : Yes but it also depends on the "variant" of the Swedish. My reference I have in my ears is Finnlandswedish ...
@Rudi_Birnbaum Right, there are some pretty big regional differences. Especially when it comes to "sj"
10:39
@Rudi_Birnbaum then it's different from the Middle English "ich"
but "egentlig" is an exampl
which would be "itsch" under German orthography
@LeakyNun Oh yes, that was what I saw. How comes that?
The Italians here quite like the way we pronounce "selvfølgelig"
@TobiasKildetoft I can never in my life guess which letters are silent
10:40
"sæføli"
ok you win
@TobiasKildetoft Now I'm curious how you pronounce it
I would understand "Igsch" or something ...
(this is not really how it would be pronounced correctly, but it is how it is pronounced usually)
Or you possibliy mean "Isch"?
10:43
no, I mean, exactly as "itch" in English
"itch" the real English word
because "k" after "i" becomes /t͡ʃ/ in Old English
and "k" becomes /x/ in German
@LeakyNun OK :-)
11:01
@ group theory: Is the dimension of the sign-representation always 1?
@Rudi_Birnbaum Yes, whenever it makes sense
@TobiasKildetoft When it doesn't? I think for example about the symmetry group of the (chiral) tetrahedron (order $12$)?
Well, for the sign to make sense, we need to embed into a symmetric group
Explanation of the trammel of Archimedes:
(though for certain groups, we have a larger collection of "sign" representations)
all of them are still $1$-dimensional though
11:10
(Prove that if the orange segment has length $a$, and the blue segment has length $b$, then the ellipse is the circle stretched vertically by a factor of $\dfrac{a+b}b$.)
o..o
@TobiasKildetoft "for the sign to make sense, we need to embed into a symmetric group" Why?
Because group elements do not in themselves have a sign
because we actually have a definition for the sign of a permutation
Isn't every group a subgroup of a symmetric group?
11:13
sure, but it can be so in many ways
All groups are also subgroups of alternating groups, which would make the sign representation trivial
I suppose the canonical choice would be permutation on the group elements themselves
that's not what canonical means
"canonical choice" is an oxymoron
11:14
:)
you can also get one dimensional representations by embedding the group into a GLn(R) and taking determinants
canonical case?
maybe every real 1-dimensional representation occurs from some embedding into a symmetric group
@mercio you mean one dimensional sign representations?
I forgot to say "real"
11:18
But is that true? if you have rotations of order $>2$?
well cyclic groups don't have many representations
also I wish I could delete some of my dumb posts sometimes o..o
who cares about representation of random groups
we should only care about representation of the Weil group
chemists apparently
what is a random group?
group of random permutations?
11:23
it's a random variable having groups as values
What are the representations of the Weil group?
first you should ask what the weil group is
Thats in wikipedia
well then they are representations of that group
And I had the random hope that it might be easier to state what the representations are, then what the group is...
11:29
x)
well that's probably the case
which is the whole point of studying them
OK, then go on ...
idk anything about the weil group sadly
1
A: Definition of the Weil group: Question about exact sequence with Inertia Group and absolute Galois group over a local field

Kenny LauLet $K$ be your non-archimedean local field, $\mathcal O_K$ be the ring of integers, $\mathfrak p$ be the maximal ideal of the ring, and $k := \mathcal O_K / \mathfrak p$ be the residue field. If $k$ is a finite field and $|k| = q$, then we assemble $\overline k$ this way: by the classificatio...

... oh dear, maybe you have a simple, enlightening example ?
Weil group is an integral part of local class field theory...
 
1 hour later…
12:38
does a (Hurewicz) fibration, as in Peter May's Concise Algebraic Topology, have to be surjective?
Peter May says that if $A \to X$ is a cofibration, then $B^X \to B^A$ is a fibration. But nowhere does he prove the latter map is surjective. That would be another extension theorem (like Tietze).
@LeakyNun Need help in spotting mistake
$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$
x-2 = t^2
$\implies dx = 2t dt$
$$2 \int \sqrt{\dfrac{1}{1-t^2}} t^2dt$$
$t= \sin \theta $
$dt = \cos\theta d\theta$
$$\int (1- \cos 2\theta )d\theta $$
$=\theta - \dfrac{\sin 2\theta}{2}$
$= \arcsin(t)- t\sqrt{1-t^2}$
@Abcd is this meant to be a definite integral?
@LeakyNun no
12:53
lemme guess: the answer is off by a constant
$$= \arcsin (x-2)- \sqrt{(x-2)(3-x)}+C$$
where C is constant of integration
It may be worth noting that $\sqrt{c^2}=|c|$
@LeakyNun Do you see any mistake.
@SimplyBeautifulArt it's alright
@Abcd what is the given answer?
Answer given is:
12:54
drum roll
$-\arcsin(2x-5)+ \sqrt{(2-x)(3-x)}$
@SimplyBeautifulArt my teacher told us dont worry about all this while doing indefinite integration
hi @loch
@SimplyBeautifulArt @LeakyNun Do you see any error in my attempt please let me know or should I ask on main
@Abcd the given answer is wrong
your answer is correct
12:56
@LeakyNun How can you claim that?
the given answer isn't even defined anywhere
and I plotted it on desmos
??
??
3 mins ago, by Abcd
$-\arcsin(2x-5)+ \sqrt{(2-x)(3-x)}$
this expression is only defined at $x=2$ and $x=3$
this is calculator's answer^
13:00
arcsin(t) is only defined for |t|≤1, and √(t) is only defined for t≥0.
6 mins ago, by Abcd
$$= \arcsin (x-2)- \sqrt{(x-2)(3-x)}+C$$
btw you mean $\arcsin(\sqrt{x-2})$ in your answer, I automatically corrected it in my head
@LeakyNun ya of course
$$\arctan\left(\sqrt{\frac{x-3}{2-x}}\right) = -\arcsin\left(\sqrt{x-2}\right)+\frac{\pi}{2}$$
so it's alright
Note however the calculator's answer involves square rooting negative numbers
13:02
@LeakyNun Mine has positive sign with arcsin
The sign difference may be accounted by the fact that $\sqrt{x^2}=|x|$ was ignored
but the one with the positive sign is correct
@SimplyBeautifulArt :\
at least inside $(2,3)$
idk what you guys want if you guys are doing it some "imaginary" domain
antiderivative needs a function defined somewhere
and don't tell me complex numbers without choosing the branch cut for the square root
If you simplify the radicals in the calculator's answer so that you don't get imaginary numbers, the result is exactly your answer up to a sign difference and a constant.
@LeakyNun :P
13:08
Now I am confused
We were supposed to not worry about all this like sign of square roots etc
Like I always took $\arctan(\tan x)= x$ (its true only for limited range)
during indefinite integration
let's just live in an algebraic vacuum
3
@Abcd It doesn't mean it can't show up, especially when you get different answers than something else. All it means is that you're not wrong, over a limited range.
could someone point me in the direction of where I shoudl look/ what I should try to understand to get at this question:


if I have a mean and variance from a gaussian at time t, that dependent on the outcome of a draw from a bernoulli at time t, how do I write/compute the conditional probability of the mean given the bernoulli outcome?
Take $t = \sin x$, then $x = \arcsin(t)$, $\mathrm dx = \dfrac{1}{\sqrt{1-t^2}} \mathrm dt$: $$\int_{-3\pi/2}^{\pi/2} f(x) \ \mathrm dx = \int_1^1 \frac{f(\arcsin(t))}{\sqrt{1-t^2}} \ \mathrm dt = 0$$
so every definite integral is $0$
@Abcd @SimplyBeautifulArt ^
13:19
@LeakyNun i dont see how thats relevant
is that correct?
13:43
Hi everyone
I'm studying linear algebra
Hi!
Is there a name for the roots of the unit matrix?
Let V be a finite dimensional vector space, B a basis of V and W a subspace of V. Is it correct to say that there exist a subset of B that is a base of W?
Why?
pick your favorite basis B of your favourite vector space V
for example (1,0) and (0,1) in R²
then if you pick W to be the line directed by (1,1)
well W doesn't have a basis made from elements of B
also your sentence would imply that there are only 2^n subspaces of V if V has dimension n
13:55
So dumb
Thaks
at most
*Thanks
And how are they distributed among the orthogonal matrices?
there are roots of the identity matrix that are not orthogonal
Yes, but I like not to regard these for the moment.
14:10
the orthogonal ones are probably dense in On(R)
I just thought about the "random group". Like: pick a random matrix $A$ (say from the Gaussian Orthogonal Ensemble, thats why) and check what order $n$ one can expect (for the matrix interpreted as group element): $A^n=\Bbb 1$. Then the question is can you give sense to "expect".
In the case of $1\times1$ matrices, you'd have either $n=2$ or $n=1$, letting you expect order of $1.5$.
o..o
my dad is a physics crank
and because of him i have to learn all of the physics
I know that terrible feeling.
A crank is a person who wants to know science without making the effort to learn science.
he gave me a 3 pages paper by some chinese person that he said "gave an alternative to special relativity" while it was just someone finding the very same Lorentz transforms but with different assumptions and apparently they didn't know matrices
For $2\times 2$ matrices you get lots of infinite order ones, so I asked myself if there would be any "symmetry" in the distribution which would still let one expect a finite expectation value of $n$.
14:19
also my dad says that the doppler shift doesn't exist and instead the light changes color because it goes through some weird matter
I tried explaining Occam's razor to him and that it was not cool to posit strange weird matter just because he didn't like the big bang
but he won't listen to anything
@mercio I have heard that the Doppler shift is in terms of maths a quite hard one to properly describe.
The elements of finite order are dense in any compact Lie group
really ? isn't it just that a wave emitted by a moving source appears to have a different frequency when you hear/see it ?
@mercio: Yes it is, but a clean mathematical way to describe that is not apparent.
it looked pretty clean to me
14:23
@mercio That (at times) doesn't mean anything for the proper description.
@MikeMiller OK, so what does it mean for this question, I have no clear conception of the complact Lie group.
@mercio you have the trouble with the "steady" medium and stuff, google it.
@mercio Stuff like that might give you an idea researchgate.net/publication/…
The orthogonal group is compact.
The proof in the special orthogonal case is as follows. Every element in SO(2n) or SO(2n+1) is conjugate to rotation in n planes. The space of rotations in each separate R^2 of (R^2)^n is the n-torus R^n/Z^n, parameterizing one angle for each plane.
in special relativity we assume the medium is steady, right ?
Q^n/Z^n is the set of elements of finite order and is dense.
These are the rotations through rational angles on each plane.
@mercio You know in the first instance I think on sound waves ...
ah yes
well if I happen to be in a tornado or a typhoon I will try to listen to all the weird Doppler shifts
14:36
@MikeMiller OK. but integrating all these, I suppose would still yield divergence to $+\infty$, wouldn't it?
what do you mean with integrating ?
I thought you were trying to make random groups
Yes, the expectation for the random group.
Which is an integral, usually.
expectation of what ?
The order of the group element
are you using the Haar measure ?
14:39
Hence the group (when I only pull one).
draw
hence the size of the group ?
the elements of finite order probably have measure 0
so yeah, it's infinite cyclic almost always
except for $1\times 1$ matrices
with |det| = 1.
I should try to find a better word than "probably" when dealing with probability topics
that is not a very interesting case
14:41
but its non-trivial
The elements of fixed finite order form a finite set.
Yeah that's false
I think that's true in connected groups
But clearly the non-identity component of O(2) all squares to 1
My b
oooh that's right
so they have positive measure in O(2)
I still think they have measure 0 in O(3)
It probably depends on the parity of n.
Unfortunately powers are hard to control in the non-identity component. I prefer SO. :)
14:48
well the issue with O(2) came from how the part of determinant -1 are pretty rigid since their eigenspaces have dimension 1
. . .
and they square to the identity
Im not sure Im making any sense
I get what you mean though.
Hey everyone
So let $p \in \mathbb{R}^n$, then the tangent space $T_p(\mathbb{R}^n)$ depends on the basis of $\mathbb{R}^n$ and the cotangent space in turn depends on the basis for the tangent space. Do I need a diffeomorphism of $\mathbb{R}^n \to \mathbb{R}^n$ to induce an isomorphism of the tangent spaces and thus perform a basis change on the tangent space?
I don't really understand the question. The tangent space is the tangent space, it doesn't depend on a basis.
what's important is how the tangent space and cotangent space transform when you change the charts
A diffeomorphism does indeed induce an isomorphism on the tangent space and provides a different canonical basis.
14:52
Ohh I meant the way we express the basis for the tangent space depends on the way we express the basis for R^n (I think, call me out if this is wrong)
I don't know if your questoin is asking if you need to find a diffeomorphism to get an isomorphism from tangent space to some mysterious thing, or if you are asking if there is a requirement in the definitions of diffeomorphisms that they induce isomorphisms
My chief feelings about linear algebra and diff geo questions regarding basis are that they are always trivial but take forever to explain, since people just write and think differently
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