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00:12
@TedShifrin as it turns out, the situation was basically that they meant that M was isomorphic to NxR, R with the standard metric. So then it follows that the sectional curvature of M is basically blind to the R part.
00:25
Isometric? Isomorphic in what category?
Riemannian manifolds.
So say isometric :)
Oh sorry I didn't realize I used isomorphic.
So the $t$ coordinate is global, too.
Sounds sensical, now.
In this particular case it's actually a factor of R^n, but that doesn't change anything.
Isometry of Riemannian manifolds induces an isometry of metric spaces, but does the converse hold? I have never thought of this before.
Apparently this is one of the Myers–Steenrod theorems
00:43
yeah. i forget how technically difficult it is. is it clear that an isometry of the metric spaces sends geodesics to geodesics?
it ought to reduce to just working on a single ball, maybe with nicely chosen coordinates.
can any1 help me with this?
Distance-minimizing is preserved, no, @leslie?
You make efforts first, @Muzammilahmed.
yeah. it's subtler than i think, but the whole result ought to fall out of some fiddling with the exponential map.
it oughta only depend on limits of the lengths of geodesics.
I have tried but unable to get it
Let’s see what you’ve tried.
00:49
is it talking abt linear transformation when it talks abt linear
My guess is that you haven’t.
Yes, that’s the definition.
a) bit is ok coz 0 polynomial lies and linear combination of 2 functions also results in a real value again here
Huh? What is a subspace?
Where are you getting polynomials?
its a smaller region in vector space satisfying properties that of a space
sorry 0 function
So what must you check?
Nothing with $0$ function here.
00:54
why?
as i know we check 0 for non empty set
What is the definition to check $U_f$ is a subspace?
i think i don't know thatt
When you have an exercise, first thing you do is write down the definitions. Bye.
ok sir it, wud be more helpful if u can give me reference to this concept
I’m not here to be your nursemaid.
01:28
@copper.hat Irish
01:57
@robjohn :-)
 
2 hours later…
04:18
If D is a PID, then every proper ideal of D is contained in some maximal ideal of D.
Solution: Let $I\subset D$ be any proper ideal of D. Then there is some p in D such that I=<p>. Every PID is a unique factorization domain so p has an irreducible factor p'. So suppose that p=p'k for some k in D. Then p is contained in <p'>. That is, <p> is contained in <p'>. <p'> is a maximal ideal as p' is irreducible.
Is this correct? Thanks.
Hint to this exercise wants me to use ascending chain of ideals.
Justification for why S:=<p'> is a maximal ideal:
Suppose that J is any any ideal that contains S. J=<q> as D is PID. It follows that p'=qr for some r in D. Since p' is irreducible, either r or q is a unit. If q is a unit then J=D. If r is a unit, then J=S.
04:42
Hi all. where does this inequality arise? is there a name for it or something?
04:56
mm, it's a version of something sometimes called jensen's inequality, for concave functions.
see en.wikipedia.org/wiki/Jensen%27s_inequality subsection "finite form"
05:09
@leslietownes thanks sir
05:39
what does this symbol mean? C[x,y]
C is the field of complex numbers.
Is C[x,y]={rx+sy: r,s in C}?
koro: perhaps polynomials in two variables with complex coefficients.
is there more context?
0
Q: Maximal Ideals of $\mathbb C[x, y]$

caffeinemachineI recently learnt that the maximal ideals of $\mathbb C[x, y]$ are precisely the ones of the form $(x-a, y-b)$ for some $a, b\in \mathbb C$. I am unable to prove it. So I considered an easier version of the problem. Let $M=(p, q)$ be an ideal in $\mathbb C[x, y]$, where $p$ and $q$ are elem...

yes, polynomials in two variables with complex coefficients.
thanks. I saw it in a previous year question papers of an exam.
I have not yet seen this symbol.
similarly R[x_1, ..., x_n] for any commutative ring R and any number of variables. for noncommutative things i think different notation is used.
at least, the people who i knew used different notation for that.
 
5 hours later…
10:21
Suppose that I want to show that Z[$\sqrt 5$] is not a UFD.
4=2.2 and 4= $(\sqrt 5+1).(\sqrt 5-1)$.
I have shown that 2 and $\sqrt 5-1$ are irreducible. Should I in addition also show that 2 and $\sqrt 5-1$ are not associates?
x and y in integral domain D are said to be associates if there is a unit element u in D such that x=uy.
 
2 hours later…
12:05
@Koro yes I think you would need to
Thanks. I asked because sometimes the book skipped this step.
 
1 hour later…
13:35
can i decompose any matrix into a product of a symmetric and antisymmetric matrix?
i should say factor
13:49
I am reading Serre's book on Trees and he says that "there is an evident notion of morphism of graphs". But I don't find it so evident. Would someone help me spell this out?
@user193319 could he just mean a function from vertex set to vertex set that respects adjacency?
I think so, but I don't know how to express "adjacency" using the notion/terminology he uses.
I think I might have figured it out lol...Let me think about it for a second.
14:10
If F is a field then every prime ideal in F[x] is a maximal ideal in F[x].
Proof: Let I be any prime ideal in F[x]. Let J be any ideal that contains I. F[x] is PID so there are f(x) and g(x) such that I=(f(x)), J=(g(x)). f(x)=k(x) g(x) for some k(x) in F[x]. k(x)g(x) is in I , which is prime so either g(x) is in I or k(x) is in I. If g(x) is in I then J=I. If k(x) is in I then k(x)=r(x)f(x) for some r(x).
It follows that: f(x)=r(x)f(x)g(x) which gives: r(x)g(x)=1. It follows that g(x) is a unit. Therefore J=F[x].
So I is a maximal ideal.
Is this correct? Thanks.
@Koro yeah, seems fine
thanks. I had a little confusion over $f(x)=r(x)f(x)g(x)\implies 1=r(x)g(x)$.
This step is correct right? Because F[x] is an integral domain.
Oh yes, it is correct!!
basically, in a PID , irreducibles and prime elements coincide, prime ideals in a PID are (as the language suggests) generated by prime elements, and one can show maximal ideals need be generated by irreducibles, which coincide with prime elements in a PID
14:25
f(x), $\mathbb {f(x)}, \mathcal {f(x)}$
@porridgemathematics yeah, book solution also used this idea :).
14:38
why is the prime ideal (0) not a counterexample to this theorem?
hmm, the theorem should say 'every non zero prime ideal'.
14:54
If we have this Hyperboloid with two sheets than $x^2 + y^2 -z^2 <-1$ is inside of only second hyperbola?
@Slate: good morning.
or whatever
@robjohn Mornin'!
How's it going?
Given 1=dc+6k, how does this imply that c and 6 are relatively prime?
Converse to Bezout's identity is not true.
1=dc mod 6. so d=c=1 is the only possibility. 1=6k mod c
15:16
@Slate busy irl. Last week was my wife's first week of retirement, and a friend of the family had cancer surgery on Thursday, so we were caring for them and driving them from the hospital to their house 90 miles away. Hopefully, this week will be quieter.
@user193319 It's a function between vertices that respects the edge relation
@Koro if there were a common factor, it would divide $dc+6k$
what divides $1$?
Thanks @robjohn
So converse to Bezout's identity is true if gcd=1.
I mean given m, n and if there exist a and b such that 1=ma+nb then gcd (m,n)=1.
@Koro yes, and vice versa (iff)
yes :).
15:23
@robjohn Whew, that's a significant week. I do hope this week will be a bit quieter overall for you (and that your friend recovers well as they can). Congrats to your wife, hopefully retirement treats her well. :)
@Slate I'm sure it will. I plan to follow suit within the year.
Well, advance congrats to you, too!
Can someone help to imagine what is $x^2+y^2-z^2 < -1$? I know that $x^2+y^2-z^2 =1$ is two sheet Hyperboloid
Consider a function
\begin{equation}
Q^{(p)}(\alpha) := Q_p + \alpha (Q_{p-1} - Q_p), \quad \alpha \in [0,1]
\end{equation} where $Q_p, Q_{p-1}$ are real matrices in $\mathbb{R}^{n \times n}$.

The question is: at what point $\alpha$ the norm $Q^{(p)}(\alpha)$ achieves its maximum value?
15:54
@Slate Thanks!
@unit1991 $x^2+y^2-z^2 =1$ looks like a one sheet surface; the radius in the $x$-$y$ slices has a minimum of $1$
@robjohn thanks for reply my question is with $x^2+y^2-z^2 =-1$
that was my mistake sorry
$x^2+y^2-z^2 < -1$ is the two parts that don't include the origin of the three separated by the hyperboloid of two sheets.
Note that in the $x$-$y$ slices, $x^2+y^2\lt z^2-1$ is the inside of a circle of radius $\sqrt{z^2-1}$.
That means that $|z|\ge1$
16:18
So it's only inside of hyperboloids not including "border"?
16:33
Am i saying something wrong?
what do I call two matrices $A,B$ such that $Ax\cdot Bx=0$?
$A,B$ square matrices, $x$ any vector.
monty: without more context, i don't know that there's a word for that (note A0 \cdot B0 will always be 0)
Anyone know a reference for the result in this post? math.stackexchange.com/questions/439866/…
@leslietownes yes for any vector $x$
not just $0$
like when does that hold for all $x\in \mathbb{R}^n$ , $A,B\in \mathbb{R}^{n\times n}$
sorry, so Ax dot Bx = 0 for all vectors x? there is some ambiguity in your use of the word 'any' above. it sometimes means all and sometimes doesn't.
16:41
@leslietownes yes thats what I mean, you phrased it properly.
is there a name for such matrices?
My reason for asking is the following. For a matrix $C$ I write $C_a$ for its skew-symmetric part. Let $C$ be invertible and $D$ is symmetric and invertible (i.e pos definite). I want to know if $C_a x \cdot (e^{C^T}D^{-1}e^{C})x$ is zero.
Hence I was asking the more general question if there is a name for when matrices are such that $Ax\cdot Bx=0$.
@leslietownes where by any I mean all :)
B^T A ought to be the zero matrix. check out math.stackexchange.com/questions/2800484/…
16:56
greetingz
17:24
A bad sloppy proof of the uncountability of the reals: mathonline.wikidot.com/the-set-of-real-numbers-is-uncountable
The idea is correct but it implements wrongly.
Does anyone disagree with my assessment?
certainly the right idea. a little bit of static around the numbers that have two decimal expansions (one terminating and one non terminating).
Why
@leslietownes do you say it’s correct?
According to me it’s wrong, because the reasoning is horrible
for instance just because the decimal created by the diagonalization has different digits does not mean it’s not equal to another decimal
I think such proofs should not be in wiki @leslietownes
Well that's what Leslie mentioned. It glosses over where the number has another deimcal expansion.
A much better proof would map each number to a different number that is not 0 or 9.
Infact if one is really pedantic the that proof also uses the axiom of choice, since some numbers have two decimals expansions. So one needs choice to pick one, but of course that’s easily avoided.
The real question is why that proof is in wiki @anak
What's wrong with it?
17:36
What I mentioned above
if you're that bothered by the fact that a site on the internet has a suboptimal exposition of cantor's diagonalization argument, i have some very bad news for you
it tries to use the ‘fact’ that different decimal expansion digits implies different numbers. When proving the diagonalization number is not in the range.
@anak
decimal expansions just suck, proofs involving them suck. for this exact reason
In fact this is one of the reasons I present my proof using decimal expansion of binary equences instead no problem then!
It's a wiki, if you are bothered that it only gives a general (correct) idea of the proof, then you can practice your pedantry by creating an account and editing it, can't you?
17:40
and one does not even have to prove that every number has a decimal expansion
@leslietownes it’s not really a decimal problem, every base will have the same problem
i guess there's an interesting question about how much you need to know about the real numbers for any of the usual proofs to work. proofs tend not to go down to the level of constructing them from rationals or anything like that.
Nothing more than the fact that bounded monotone sequences converge, some knowledge of geometric series and some basic set theory.
the usual proofs involving sequence spaces show that sequence spaces are uncountable. that is uncontroversial. but you do need something to port this proof to a statement about real numbers. intervals in R do look like a kind of quotient of a sequence space but the usual proofs tend not to belabor this point or what makes it work.
this site doesn't seem to be a public wiki at all, although it invites people to 'sign in.' the main page suggests that it is a set of personal notes from a single person.
that is surprising. there are a lot of pages on this site.
oh, i've got a long day ahead of me of submitting typos and corrections.
i'll put this on my list of things to do the next time i'm sick and can't do anything but sit in bed on a laptop.
17:49
I think there are still better things to do in that state.
18:00
a connected subset of a locally connected space is trivially locally connected, right
no it isn't
I wouldn't say "trivially".
It seems to be wrong.
Oh, maybe not. shrug
Aren't connected sets open in a locally connected space, or am I thinking of the wrong thing?
No that's wrong what am I thinking
I only think about these things when I am teaching point set topology. And I haven't done that in 15 years.
When taking a subspace, the connectedness properties can break in the subspace, so it doesn't have to be locally connected
@anak no
18:12
You can just take any example of a connected + not locally connected space which sits naturally as a subspace in a locally connected space.
I was thinking comb.
yeah, intersections only really works ok when we are taking decreasing chain of continua
How comb?
Comb is connected but not locally path connected or locally connected if I recall correctly.
Correct. So I don't see what is next.
It's a connected subspace of R^2
18:18
Oh, duh.
tfw you pick up the last student midterm and it's markedly worse than the rest
I was trying to be more interesting than that.
@Ted I am sorry I am not interesting :((((((((
@Semiclassic There's a reason it was last.
well, it was (in essence) a makeup exam
so that's not quite relevant
18:36
@robjohn I posted a diagram showing how the rhombic dodecahedron is related to the cube & octahedron: chat.stackexchange.com/transcript/message/60100571#60100571 But it's probably easier to see in this interactive version:
@PM2Ring I changed the link I sent yesterday to an animation.
@robjohn Nice. My Sage version isn't very elegant, but I just wrote it on my phone, without looking at my ancient POV-Ray source code.
Sage uses three.js for its 3D stuff, but it only exposes very minimal features, so it can't do fancy surface textures, shadows, etc. Stuff that fully utilises three can look amazing... but I'm not keen on writing in JavaScript. ;)
Sage can do ray-tracing though, using Tachyon, but it's only for static images, not interactive stuff or anims. And you can export 3D in several popular formats, so you can import it into proper 3D renderers, or make 3D prints.
18:55
@PM2Ring I have been unable to do the ray tracing that I had hoped to do, but getting the view from different points of view provides a lot of information.
19:05
Rhombic dodecahedra make nice dice. They don't roll as much as platonic dodecahedra, which are too close to spherical. And if you have a bunch of them, they pack nicely. :)
19:53
I've just used Sierpiński theorem to prove a space is not path-connected
:D
@PM2Ring it doesn’t seem as if all sides are equiprobable
20:11
what dictates the probability? i'd have thought it's the relative surface areas of the faces, and if those are congruent then the sides are equiprobable
20:26
Happy pi day
I will repeat the same puzzle I gave last year
Circumscribe a regular $n$-gon around a circle (of radius $r$). Call its area $A_n$ and its perimeter $C_n$. Which is closer to $\pi$, $\dfrac{A_n}{r^2}$ or $\dfrac{C_n}{2r}$?
@Semiclassical There are more factors than area of the faces. There is the distance from the center of mass of each face. A major factor is the solid angle subtended by a face as seen from the center of mass. The faces of a rhombic dodecahedron subtend different solid angles, so I doubt it is a fair die.
@AkivaWeinberger they are both the same if my mental visualization is working right.
$n\tan\left(\frac\pi{n}\right)$
20:42
Yup! @robjohn
So, same question about inscribed polygons.
20:55
@robjohn I think the perimeter gives you a closer approximation
@robjohn Hey, can you help me with a question on Fourier coefficients?
I'm pretty sure there's a theorem about how, if $f$ if $C^1$, then $|\hat f(n)|\le(\sup f')/n$
or some such
How do you prove that
integration by parts
Depending on the normalization used, $2\pi in\hat f(n)=\widehat{f’}(n)$
i always forget the normalizations
@robjohn All isohedra are fair dice, as mentioned here: mathoverflow.net/q/46684 Ed Pegg Jr proved it for his master's thesis. Unfortunately, I can't find it online, but Ed mentions it briefly here: mathpuzzle.com/MAA/37-Fair%20Dice/mathgames_05_16_05.html There are more details hiding somewhere on the MAA.
21:11
i don't follow how a rhombic dodecahedron would have different solid angles depending on the face, based on the figure here: mathworld.wolfram.com/images/eps-svg/…
seems like each would subtend pi/3, as expected from 4pi/12
@robjohn The faces of a rhombic dodecahedron subtend different solid angles I don't think that's correct. Sure, there are two types of vertex, but all the faces are identical. But I'm too sleepy to calculate solid angles right now. ;)
you can also think of it as attaching right square pyramids to each face of a cube
@Semiclassical I posted an interactive 3d version earlier: chat.stackexchange.com/transcript/message/60645650#60645650
yeah, i saw
Yep. It's a stellated cube. It's also a stellated octahedron.
21:17
not sure what a good way to describe the particular stellation is. "put square pyramids on the cube in such a way that you end up with 12 faces instead of 24?"
tho said 24-sided dice should still be fair
the stellated cube would be a good name for pub. ideally one with a flat roof.
i do wonder, though. is having faces with the surface area really not equivalent to having faces subtend the same solid angle?
Well, having the same surface area isn't sufficient. You do need the solid angle. Ed's geometric method projects the edges of the die onto a sphere that's centred on the centre of mass of the die. If the spherical polygons have equal area, the dice is fair.
My guess was kind of correct. meta.stackexchange.com/a/377143/334566 We were hit by a DDoS attack. While the attack itself was mitigated by our systems, the conditions set off a series of errors that managed to uncover an edge case in one of our backend systems.
21:57
@PM2Ring You're right. They do subtend the same angle. I see that the polyhedron looks the same from each face (being isohedral). So it does make a fair die.
@AkivaWeinberger Yes. The perimeter gives $n\sin\left(\frac\pi{n}\right)$ where the area gives $\frac n2\sin\left(\frac{2\pi}{n}\right)$
The area is smaller by a factor of $\cos\left(\frac\pi{n}\right)$
@PM2Ring Yeah. I said that it acted and tasted like a DDoS attack.
I have changed my avatar from the St Patrick's Day avatar to a Pi Day avatar. I will switch back tomorrow.
@robjohn Right. But Catija said it wasn't. But I guess that's because technically the DDoS didn't succeed, but it rattled the system sufficiently to expose other problems.
22:13
Well, that is the purpose of a DDoS attack. Stress the attacked system.
So in a sense, it succeeded.
Yeah
My favourite rational approximation of 1/pi = 113/355. I'm normally not fond of "numerological" stuff that's base dependent, but that one's too cute & handy to resist. :) I also like (22/7) × (1 - 4/10000) = 3.1416
22:33
@PM2Ring Well, you hang out on that astrology site all the time ;-)
I hate when people mix up astronomy and astrology
me too. but i'm a scorpio, so of course i would hate that.
@robjohn You’re just over the moon!
22:49
On a now-defunct science site I used to frequent, one of the regulars was an astrophysics / cosmology professor from Sydney University. He was interviewed by one of the leading Sydney newspapers. In the article, they said he was an astrologer. He was not impressed. en.wikipedia.org/wiki/Geraint_F._Lewis
23:04
I made a bookmark to improve articles about Putin. javascript:(()=>{document.body.innerHTML=document.body.innerHTML.replaceAll(/\bputin\b/ig,'\u{1f4a9}\u{1f96b}')})()
Feel free to delete it if you don't think it's appropriate here.
i object to the implicit use of british english
or non-US english, i guess i should say
wrong english
Hell if I know what that says.
it's a poo character and a can character. like a toucan, but poocan.
as i understand it.
Australian English combines features of British & American English, and we have our own features too, although some of the old expressions are fading away.
i think sardines still come in tins in US english, but that is a can. i will die on this hill.
23:13
You guys also use "canning" for stuff preserved in glass jars. That amazed me when I first learned about it.
hah! yes. i'd never thought about that.
The footwear Americans & British call flip-flops are usually called thongs in Australia. This can lead to some hilarity with American tourists in Australia. en.wikipedia.org/wiki/Flip-flops#Etymology
One of my uncles calls them Japanese safety boots.
I'm a solipsistic descriptivist. Language is determined by my usage of it, and no one else's.
Down with communication!
In opposition to Grice's cooperative principle, which states that communication is inherently cooperative, I assert that all communication is based on lies and deceit
23:27
rubber
DogAteMy has been brainwashed by Tromp.
I suppose a better alternative to the cooperative principle would be, all utterances should be as confusing as possible
23:44
I have that album on CD. I still listen to it regularly.

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