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00:12
Problem: Let $\alpha$ and $\beta$ be paths in a space $X$ from a point $x_0$ to a point $x_1$. If $\pi_1(X) = 0$, show that $\alpha$ and $\beta$ are path-homotopic...Is this as easy as I think it is? $[\alpha * \overline{\beta}] \in \pi_1(X) = 0$ so $[\alpha * \overline{\beta}] = 0 which implies $[\alpha] = [\beta]$...?
00:24
@user193319 it is as easy as you think
Sweet! Thanks!
Is someone from here familiar with the book Operational Research by Hillier? The thing is there is an example there that I don't understand (part of it) but I wouldn't like to ask the question because it's like 5 pages the complete example
 
1 hour later…
01:50
Hi guys, Consider the theorem: let H,K be normal subgroups of G so $K\leq H$. if $G/K$ is cyclic then $G/H$ and $H/K$ are cyclic.
Does this theorem has it own thread? I tried to prove it but without any luck. Wondering if the question was already asked.
 
6 hours later…
08:13
@vesii
https://math.stackexchange.com/questions/1290546/if-g-is-cyclic-then-g-h-is-cyclic
Is anyone here?
Just arrived. Good morning!
Hy can you help me with https://fe228cd5-a-62cb3a1a-s-sites.googlegroups.com/site/lovroshr/Home/zad_mat_ind.png?attachauth=ANoY7cozhV3BrLrU29G352ugNUxJCgnoeGrH8mNyV_5Gxe4fT6FiVHqEiNjZmrUQ64Ko82MJR2noXKLq1He5DsNDN_MpkdzaS9S3Ac4Iki42UYtBYYS4bCVnypNOuBCn6Mjnx-BPp-Viuhw2YResN_wdMbpfs7wqJumT3I57RH0zQjTQzmBnzj0t9WL0aVmb4zVknTzbLjCX-FHAa1xaaGxvypJouKGLbQ%3D%3D&attredirects=0 ?
My question is can this problem be solved by one mat. inudctoin ?
I solve it
and I need 3 mat. induction
to prove it
08:35
Folks, say I have a flat shape, and I know its area and a plane it lays on. There is also a second infinite plane, which I also know.
How do I find the projection of the shape onto the second plane?
Is it dot product of the vector normal to the shape's plane with magnitude equal to the shape's area and the unit vector normal to the second plane?
dot product returns a scalar, if you want to find the projection onto a plane you need to obtain a set of vectors don't you?
 
4 hours later…
13:05
Why is in above theorem, it is assumed that $D$ an open disk? is it to make sure that we can surely apply mean value Theorem, ie, walking parallel to x or y axis, we are not moving out of domain? Can a convex open set do that job? Also, this may sound silly, but why open disk? so that derivative is defined without fuss? @LeakyNun
13:37
@Silent The disc is irrelevant. All that matters is that the domain is connected. The given argument shows that under the given conditions, your function u + iv is locally constant, and locally constant means constant when you're connected.
13:57
Is it true, that the nullspace of a matrix and the image of its transpose are orthogonal and make up the complete vector space?
14:18
I mean, we usually want an open domain if we want to have derivatives
@FelixCrazzolara yes
Here's a challenge I have for myself. Consider the set of symmetric n-by-n matrices with unit diagonal. These are parametrized by the n(n-1)/2 matrix elements above the diagonal. I'm interested in the subset of this which are positive semidefinite.
If n=3, then there's 3 matrix elements; call them x,y,z. you can readily plot the set of such (x,y,z) for which the matrix is positive semidefinite.
If I go up to n=4, though, then there's six such matrix elements. I'm trying to figure out what some useful graphics would be in that case, since visualizing subsets of 6D space is not something I'm equipped to do.
In the context of vector spaces, if $B$ is some subset, why would $\emptyset + B = \emptyset$? Why not $\emptyset + B = B$?
@user193319 the latter is right
Hmm...But Rudin says $\emptyset + B = \emptyset$...
typo?
wait
is it $+$ as in $A+B := \{x+y \mid x \in A, y \in B\}$
if so then $\varnothing + B = \varnothing$
I got confused because in the context of vector spaces we should only add subspaces
14:35
Ah, of course. Thanks!
hello,sorry friends i have a question...its been a long time that i didn't ask any question on this website...now i forgot how i write my questions in somewhere of them which need math language(a form of Latex).could you give me the address of guidance of writing math stuffs in this website?
@MikeMiller Thank you very much
@pershinaolad See this
@Silent thank you a lot :X:X:X
What is the largest possible $c$ in $||a_1x_1+a_2x_2+...+a_nx_n||\ge c(|a_1|+|a_2|+...+|a_n|)$. if $X = \mathbb R^2 $and $x_1 = (1, 0)$, $x_2 = (0, I)$? If $X = \mathbb R^3$ and $x_1 = (1, 0, 0), x_2 = (0, 1,0), x_3 = (0, 0, I)$?
we have $||a_1x_1+a_2x_2+...+a_nx_n||\ge c(|a_1|+|a_2|+...+|a_n|)$
what we want to find is $c=\inf\{||a_1x_1+a_2x_2||:|a_1|+|a_2|=1,a_1,a_2 \in \mathbb R\}$
So this value will be the largest value for $c$.
14:55
Is there an easy way to see that the Hausdorff 2-dimensional measure of R is 0?
15:09
Hi, could someone help me work through Miller & Freund?

 Probability and Statistics

Any discussion on Probability and Statistics.For rendering LaT...
^I'll be here. Thanks.
15:58
we have $||a_1x_1+a_2x_2+...+a_nx_n||\ge c(|a_1|+|a_2|+...+|a_n|)$
what we want to find is $c=\inf\{||a_1x_1+a_2x_2||:|a_1|+|a_2|=1,a_1,a_2 \in \mathbb R\}$
So this value will be the largest value for $c$.
16:09
@LorenzoQuarisa I want to find the scalar area of the projection of the flat shape onto the second plane.
> Think about a time you did a major project in collaboration with others. Describe the process, and then give two after-the-fact performance assessments, one of yourself and one of the team.
^This is from a job application for the summer
and the problem is I always did school projects alone so I have no idea what I should write
I know none of you can help me with this but I just wanna write this somewhere
16:31
notation problem
AxB mean set of all order pairs (a,b)
R^n mean list of (x1,x2,....)
But what R^
R^s
(RxR)xR is the set of all ordered pairs ((a,b),c), which is technically different from R^3, but there's a natural bijection between them so we don't care
Similarly, (AxB)xC is the set of ((a,b),c) and Ax(BxC) is the set of (a,(b,c)), so they're also technically different, but we don't care because there's a natural bijection
Hello! Do you think I have written this question properly? Have I chosen the right tags? math.stackexchange.com/questions/3066386/…
in linear algebra done write he says "F^s or R^s denotes the set of functions from S to F" i don't understand what this mean and how to relate this with my understanding of cross product of sets
16:37
@AbdelrhmanFawzy Think of 3 as a set with three elements in it
{0,1,2} for example
What does a function from {0,1,2} to R look like?
f(0) is some real number, f(1) is another real number, and f(2) is again some real number
Basically, it's a choice of three reals, right?
It'll look like: f(0)=a, f(1)=b, f(2)=c
On the other hand, an element of R^3 looks like (a,b,c)
It's also a choice of three reals
So there's a natural bijection between R^3, and the set of functions from {0,1,2} to R
That means they're essentially the same thing
aha thanks
natural in both components?
I'm not using "natural" in the category theory sense
At least, I don't think I am
I just mean that it's an obvious bijection
16:47
Hi @LorenzoQuarisa
@AbdelrhmanFawzy Oh also: Think about (A^C)x(B^C) and (AxB)^C
3
(Remember that A^B is functions from B to A, not the other way around)
Also think about (A^B)x(A^C)
and (A^B)^C and A^(BxC)
17:03
Hey @AlessandroCodenotti , how are you doing in Bonn? I applied there for the phd
There's a "bonne année" joke here and I have no idea what it is
17:22
Pretty well thanks! That's nice, maybe we'll meet next year then, I don't remember if we ever met in Trento! How's Turin? I was going to study there too if I didn't get into Bonn
I guess you can say "@AlessandroCodenotti bonne chance with your studies there"
hi demonic @Alessandro, @Leaky.
oh, and DogAteMy
I haven't seen you since your question about the characteristic polynomial of $A^2$, @Leaky. I assume my silly answer was enough?
yes
I've heard that $TM$ and $T^\ast M$ are isomorphic
for Riemannian manifolds?
17:25
obvious statement: visualizing 4D polytopes is hard, and visualizing 4D convex sets even more so
Hi @Ted @Leaky
@Leaky: Sure. When you have an inner product, you get an isomorphism $V\cong V^*$.
@Semiclassical polygons are ok, through shadow study.
heya @Semiclassic
yeah, or slices
hey ted
17:26
@TedShifrin so for example $M=\Bbb R^3$?
@Semiclassical yeah, that's more analytically standard.
Any manifold admits a Riemannian metric, @Leaky. No big deal. But the isomorphism is not natural, in the categorical sense.
sure
I'm just interested in the induced correspondence between vector fields and differential forms
Right. That uses a metric.
great!
 
1 hour later…
18:33
Hi guys, I asked it yesterday but no answer. Consider the theorem: let H,K be normal subgroups of G so $K\leq H$. if $G/K$ is cyclic then $G/H$ and $H/K$ are cyclic.
Does this theorem has it own thread? I tried to prove it but without any luck. Wondering if the question was already asked.
anyone here know some r?
@vesil subgroups and quotients of cyclic groups are cyclic. Do you see how this applies here?
Howdy, @Mathein.
19:01
@Ted I feel like I should learn Lie theory at some point, but I'm not sure how to approach it
Well, you know a lot of representation theory, so that's already a lot of both Lie groups and Lie algebras, from one aspect. I don't have a great book to recommend, sadly.
how much differential geometry does one need? I don't feel comfortable with that, although I technically did a course on it
Why is mathematics so spectacularly effective at describing the physical universe?

The answer is that the universe itself is simply a timeless mathematical construct. Of what type, we’re not exactly sure yet; it’s looking like it’ll be some kind of mashup of a 3+1-dimensional pseudo-Riemannian manifold with tensor fields obeying certain partial differential equations, plus operator-valued fields on ℝ4 with certain commutation relations acting on an abstract Hilbert space. But when we do have our theory of everything, plus the initial conditions of the universe, it’s clear to me that within
There's some beautiful geometry tying in with homogeneous spaces and symmetric spaces, but for the beginnings you just need to know manifolds (which you do).
Could anyone take a look at my question about showing that the vector of independent Markov processes is a Markov process? This seems to be super simple - and is claimed in many books - but I'm absolutely not able to figure out how we can prove that.
19:07
I want to learn about homogenous spaces, too. I've read multiple times that Bruhat-Tits buildings are p-adic analogs of symmetric spaces and they show up in the research of my advisor, so I hope it's useful for motivation
@0xbadf00d: Your question is way too abstract for me. Sorry.
@Mathein: To me it's important and beautiful how the basic Lie algebraic structure determines the geometry of a homogeneous space, and then there's a bit more for symmetric spaces (e.g., all the invariant forms are closed and hence cohomology is given by the complex of invariant forms, at least in the compact case).
Heya DogAteMy
Suppose $a$ and $n$ are relatively prime. Under what conditions is $(x+a)^n\equiv x^n+a\pmod n$, as polynomials?
I'm gonna go take a shower
and that's the problem I'm gonna think about during it
Well if $n$ is prime then it's automatic isn't it
Does it ever happen if $n$ is not prime?
@AkivaWeinberger if you want to spoil yourself, look at the theory behind the AKS-primality test
That's the "PRIMES is in P" one, right?
Yeah, it's automatic when $n$ is prime. Or a power of a prime.
19:13
I should get around to it at some point
@AkivaWeinberger yes
@TedShifrin $a^n\equiv a\pmod n$ when $n$ is a power of a prime?
(and $a$ is coprime to it)
I think that's part of the generalized high school students' binomial theorem :P
No, there are counterexamples
I think you misread it as $(x+a)^n\equiv x^n+a^n\pmod n$
@TedShifrin a cool thing about Lie groups: each connected locally compact Hausdorff group is the inverse limit of Lie groups
19:15
It's just $x^n+a$
Oh, no, just mod p.
Ah, sure
$a^{p^k}\equiv a\pmod p$
I shouldn't weigh in on anything number theory, anyhow.
I'mma shower, bye
so when you have a locally compact Hausdorff group $G$, you can look at the connected component of the identity $G^o$, then $G/G^o$ is totally disconnected locally compact Hausdorff which is also known as locally profinite, one can show that $G/G^o$ has compact open subgroup $K$ which is then automatically profinite and the quotient $(G/G^o)/K$ will be discrete
19:19
Well, @Mathein, you know this sort of stuff doesn't appeal to me ... nor do I know anything about it.
I think it's cool that something as general as "locally compact Hausdorff group" leads to something as concrete as profinite groups or Lie groups
is the upper-half plane a homogenous space for $\mathrm{SL}_2(\Bbb R)$ (or $\mathrm{PSL}_2(\Bbb R)$?)
Exhaustive list of people who call profinite groups concrete: Mathei.
For me it's an exhausted list.
19:35
Rehi Ted
I've been wondering if the differential geometry of the (hyperbolic) upper-half plane can be applied to give something interesting about modular forms
the answer is probably yes, but I guess knowing some hyperbolic geometry is necessary to understand why
The answer is indeed probably yes
in our modular forms course, we've been integrating a lot against the $\mathrm{SL}_2(\Bbb R)$-invariant modular form $\frac{dx dy}{y^2}$ which looks a lot like the standard metric on $\Bbb H$
20:11
Not the metric, but the area 2-form.
 
1 hour later…
21:21
@MatheinBoulomenos, thanks for the reply. K and H are subgroups of G so G/H and H/K are subgroups of G/K?
21:45
@vesii H/K is a subgroup of G/K, G/H is quotient of G/K (think 3. isomorphism theorem)
 
1 hour later…
22:55
3
Q: Projective-invariant differential operator

user76284Suppose we want a differential operator $T : (\mathbb{R}^n \rightarrow \mathbb{R}^n) \rightarrow (\mathbb{R}^n \rightarrow \mathbb{R})$ such that \begin{align*} &T(g) = 0 \Longleftrightarrow g \in G \\ &g \in G \Longrightarrow T(g \circ f) = T(f) \end{align*} where $G = \text{Aff}(n, \mathbb{R}...

Should I migrate this question to MO?
Quite likely. You want someone like Robert Bryant to see it.
 
1 hour later…
23:58
Hi chat.
If $X$ is path connected, does it follow that any two constant map into $X$ with the same domain are homotopic?

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