1:03 AM
Hello

11 hours later…
12:25 PM
Hello.
Only 2.7k messages per week, on average :O.
I think there used to be upwards of 15k per week, on average, back in 2015.
Surely there are more math students on the internet though? Maybe they have went elsewhere.

12:42 PM
regression to the mean

how to determine is a curve in R^2
is a closed loop
if*

check the images of the endpoints

1:42 PM
@JackDon It's an inside joke understood only by fans of Brian Lumley and the Necroscope himself. But I would say Harry used his higher maths to trap the math students in the void.
@JackDon Ignore that I couldn't resist. I'm just here cause I am terrible at math and I seek a numbers wizard.

1:56 PM
Is anti-social on your page meant to be asocial? @GWarner
I think most people that use anti-social actually mean asocial, that's what I ask.

Numbers are not difficult. It is all about counting the numbers, connecting the dots, and coloring in.

That's drawing numbers Mats :P.

What?
It is?

Connecting the dots, and colouring in.

@JackDon True, but it was necessary to use a more widely understood meaning.
I like connecting the dots and coloring in...

2:05 PM
180

@JackDon Harry is the protagonist in a series of books about a guy that can speak to the dead, hence learning math from past geniuses (Mobius especially) to travel through space and time to hunt vampires.

if we know the highlighted angle to be 90-\theta, why how the rest two angles are established and why in this order?

Anyway, I was looking for help to figure out the progression of degrees to turn and how often based on speed to cover a grid of squares of a set size. I realize I could go in a linear fashion, turn 90 degrees for one grid and then 90 again. . im looking to spiral outward instead

3 hours later…
4:45 PM
Suppose $A_{i}$, $B_{i}$, $C_{i}$ are chain complexes such that the sequence $0 \rightarrow A_i \rightarrow^{f} B_i \rightarrow^{g} C_i \rightarrow 0$ is exact for all i. Does this imply that $f,g$ are chain maps?
I would expect it to be false in general, but because of my lack of examples of chain complexes I couldnt find one

That property has nothing to do with the differential, so certainly not
As a quick example, take $B_0 = B_1 = \Bbb Z$, with differential $d(b_1) = b_0$, with $A_1 = \Bbb Z$ and $C_0 = \Bbb Z$. Let $f$ be the obvious inclusion and $g$ be the obvious projection
But $d f(a_1) = db_1 = b_0 \neq 0 = f(0) = f(da_1)$, and $g(db_1) = g(b_0) = c_0 \neq 0 = d(0) = dg(b_1)$
So neither $f$ nor $g$ are chain maps in this example
This doesn't seem like what you wanted to ask, is it?

5:04 PM
@Mike This is actually what I wanted to ask. Actually I am proving that if such a short exact sequence exists then it induces a long exact sequence of the homology groups. But you have the map $H_p(A_i) \rightarrow^{f*} H_p(B_i)$ in the long exact sequence which can only happen if $f$ is a chain map, but my notes don't mention f being a chain map.

Your notes are just being sloppy
$f$ is a chain map

Guessed so, thanks!

5:35 PM
I absolutely hate tex-ing commutative diagrams
Another thing, in the long exact sequence $\cdots \rightarrow H_p(A) \rightarrow^{f}* H_p(B) \rightarrow^{g}* H_p(C) \rightarrow{\partial}* H_{p-1}(A) \rightarrow^{f}* \cdots$ is the map $\partial *$ unique?

what do you mean by unique

In the sense that (atleast from what I have been given) you define $\partial*[c] = [f^{-1} \circ \partial \circ g^{-1}(c)]$, but looking at the commutative diagram there are many other ways you can define this map

@BalarkaSen do you want time odds
I'm definitely not strong enough to give piece odds

@Sayan the snake lemma produces long exact sequence of homology groups out of a short exact sequence of chain complexes in a natural way. if you have a map of exact sequence of chain complexes (0->A->B->C->0) -> (0->A'->B'->C'->0) then it gives a map of long exact sequences (H(A)->H(B)->H(C)) -> (H(A')->H(B')->H(C'))
if you have two recipes of producing a connecting homomorphism such that naturality is satisfied, you can use it on the identity map (0->A->B->C->0)->(0->A->B->C->0) to get that they are the same
@LeakyNun naw man really cant play right now

a'ight

5:46 PM
What do you mean by naturality?

I explained it

Oh, okay, I kind of get it.

Hi @Sayan, @Leaky, a @Bskarka, @MikeM

Hi @TedShifrin

5:50 PM
You butchered my name, @Ted!!
Hi

that's an interesting consonant cluster

I'll just call you Malarkey.

That's an accurate description of me tbh

A synonym of malarkey is bullshit, aka BS aka my initials

5:54 PM
Perfect :)
Maybe typing BS would be easiest.

Haha, uh-oh

@TedShifrin I think I'm dropping maths

@TedShifrin Suppose I have a Riemannian surface $(M, g)$ with a triangulation of $M$. Can you give me a homotopy $g_t$, $t \in [0, 1)$ of the metric such that the curvature function $K_t$ has a well defined limit $K_1 = \lim_{t \to 1} K_t$ to a function which vanishes everywhere on $M$ except on the vertices of the triangulation $M$, on which it agrees with the angle defect?
That is, are simplicial surfaces just limits of Riemannian manifolds?
It's probably easy to construct

6:06 PM
@Leaky For?

@TedShifrin for a degree in April Fools' jokes

Lmao

low quality bit

Leaky, it didn't work. You have no other skill.

LMAO
rekt

6:07 PM
cries
@Thorgott you play chess?

Leaky stop chess is sucking you in

you'll turn into a rook

it's irreversible

@BS If we think of this as embedded, you can deform a polyhedron to the manifold. But you have singularities on edges of faces, not just at vertices, no?

6:09 PM
one day gregor samsa woke up and found himself turned into a rook

wow, never heard of that book

nah

I have a quick question

I don't have a quick answer

How many parameters does the student t-distribution have?

6:11 PM
@TedShifrin I thought of that, but not sure how to encode that into the curvature

BS loves his Kafka.
Use the induced metrics from Euclidean space.

Is it just one parameter?
which is the degrees of freedom?

Or I suppose you start with the flat metric on the faces of the (embedded) triangulation and straight-line homotop to the metric on $M$.
Not sure what happens on edges of the triangulation, though.

In probability and statistics, Student's t-distribution (or simply the t-distribution) is any member of a family of continuous probability distributions that arises when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown. It was developed by William Sealy Gosset under the pseudonym Student. The t-distribution plays a role in a number of widely used statistical analyses, including Student's t-test for assessing the statistical significance of the difference between two sample means, the construction...
@Mathphile according to wiki, it has one parameter, namely $\nu \in (0,\infty)$

@Mathphile It's one parameter; always centered at 0.

6:14 PM
I am kind of confused with that

The other parameter control the standard deviation

Aren't the mean and variance also parameters?

@TedShifrin I like the idea.

according to wiki, the mean is $0$ for $\nu > 1$ and otherwise undefined
and the variance is $\frac{\nu}{\nu-2}$ for $\nu > 2$ and etc
@BalarkaSen why does it say that the variance is $\infty$ for $\nu \in (1,2]$ and undefined for $\nu \in (0,1]$?
what's the difference?

ah so the variance depends on the degrees of freedom?

6:17 PM
according to wiki, yes

@LeakyNun I really don't think about the $t$-distribution with non-integral degrees of freedom. To me it's just $t_{n-1}$ where $n$ is a natural number.
Your $\nu$ is my $n-1$

so undefined for $n=2$ and $\infty$ for $n=3$

I have couple stats doubts so I hope you guys won't mind clearing my concepts

Balarka is secretely a statistician

No way

6:20 PM
when is it a secret

@BalarkaSen Let $C \subset M$ be a Morse-Bott critical locus, i.e. $\nabla f = 0$ on $C$ whereas $\text{Hess}^N f$, the normal Hessian, is everywhere nondegenerate on $C$. Is there anything special about the map $\nabla f: M \to TM$ along $C$ that we can see just at the level of manifolds, not using special structure of $M$ or $TM$?

Self learning stats is hard

So $f$ is a Morse function along the fibers of the normal bundle of $C$ in $M$
Hm

The map $\text{Hess} f: TM/TC \to TM$ is recoverable just using manifold-level constructions (if $M = X$ and $TM = Y$ with the $0$-section being $Z \subset Y$, this is the composite $TM \to TY \to TY/TZ$, factored as $TM/TC \to TY/TZ$) but the normal Hessian requires that you kill off $TC \subset TM$, and knowing that subspace of $TM$ requires that you know the codomain is a tangent bundle (so we can take $TC$ for submanifolds $C$ of the domain)

Hi demonic @Alessandro

6:26 PM
Hi Ted

$E[2X-Y]=2E[X]-E[Y]$ right?

It's not totally obvious to me whether that $TC$ is very important. Here, let's try this. Maybe the abstract notion we're getting at (Morse : transversality :: Morse-Bott : ?) is a map $F: X \to Y$ and a submanifold $S \subset Y$ so that a) $f^{-1}(S)$ is a submanifold $C$, and b) $DF\big|_C: TX/TC \to TY/TS$ is injective.

Oh you're trying to say some transversality condition for $\nabla f$, I see

yeah

Hi everyone

6:30 PM
i dont want to use the normal hessian since that requires special structure
i do not think there is a good set of assumptions that gets around assumption (a) --- I think you always need to assume the inverse image is a submanifold and then state the transversality property

@Mathphile yes. that's called "expectation is linear"

Hrm
That makes sense

Now choose a projection $TY\big|_S \to TS$, aka, choose a normal bundle. I guess we already have a metric

@loch welcome back

lol
hey

6:36 PM
@BalarkaSen So I'm trying to encode the condition $g: M \to \Bbb R$ such that $g\big|_C$ is a Morse function

If $V[X]=1$ and $V[Y]=4$, I am trying to find $V[2X-Y]$

Hi @Captain
Hi @loch

$V[2X-Y]=2V[X]+4V[Y]+2ab Cov(X,Y)$ right?

hi @TedShifrin

@BalarkaSen

6:40 PM
Hola, Ted

@MikeM, forgive the interruption. Can't you just say the gradient is transverse to the zero section along $C$?

@TedShifrin I'm trying to take the following situation: $f$ is Morse-Bott, and $g$ is Morse along the critical submanifolds of $f$, and I consider $F_t: M \to TM$ given by $\nabla(f+tg)$. I would like to phrase this setup abstractly, not using that the codomain is the tangent bundle of the domain. That's where I run into trouble --- I already don't know quite how to encode "$f$ is Morse-Bott".

@TedShifrin I'm gonna do like another week or so of logic/proof practice and then move on. I feel like I have a better understanding of the techniques and stuff now.

Write $\psi = DF|_C: TX/TC \to TY/TS$. Then the condition I decided was the right encoding of "$f$ is Morse-Bott" above was just that $\psi$ is fiberwise injective. But $g$ itself is not apparent in this abstract setting, since we can't add elements of $Y$.
Only $F'_0$ is, and that's a map $X \to TY$, not a map $X \to Y$. It's tough.
I'm probably making my life harder than it has to be

is the answer for this question "it cannot be determined"?

6:56 PM
Hi
i got this problem : A set S contains 0 and 1 and the mean of each finite set. Prove
that S contains all rational numbers (in [0,1]).

@BalarkaSen OK, "a pseudotransverse map" is all the data of: $(X,C)$, $(Y,S)$, a map $f: X \to Y$ so that $F^{-1}(S) = C$, and such that the map $\psi = DF|_C: TX/TC \to TY/TS$ is injective.

i can't see how to get all the numbers of the form $m/n$ in $[0,1]$ , but i can see we can get many of them..

@BalarkaSen a little help with my question above please?

random pings don't really encourage people, misspellings even less so

It's not random I was asking some of my stat doubts from Balarka before too

7:00 PM
fair enough

@MikeM I confess I don't think about the Morse-Bott. I guess I shouldn't butt in, as I never saw who $g$ is.

@MikeMiller OK, seems a fair definition
@Mathphile I am busy with some other things at the moment; maybe others can help

@Captain: I am not here to approve or disapprove :)

@TedShifrin I'm just struggling with some calculations so trying to abstract a bit.

@TedShifrin I see ;)

7:02 PM
@BalarkaSen alright
Can someone help me with some of my probability and stats doubts

@BalarkaSen This is hellish

Rip

@infinity what do they mean by “each finite set”?
You need to allow repetitions.

I think there's a way to write it but I'm not going to bother
I've forgotten how to prove the Morse lemma
Oh is this like a Moser trick
If $Q(v)$ is your function with $Q(0) = 0$ and $D_0 Q = 0$ but $H(v) = \langle (D^2 Q) v, v\rangle$ nondegenerate, then set $Q_t(v) = Q(v) + t(Q(v)-H(v))$; then $Q_t$ has its first two derivatives constant in time

7:18 PM
I dunno why it's like a Moser trick. I think of it as first order Taylor expansion with a second order reminder term
I think that's literally the proof
If I am not missing any subtlety

what? how do you cook up the diffeomorphism you precompose with to get a quadratic function

The Hessian is nondegenerate, so find coordinates where it diagonalizes. You immediately have the desired expression of $f$ in those coordinates, given by writing $f$ in Taylor expansion w/ reminder, no?

I don't at all understand what you're talking about. I didn't say $f(v,w) = |v|^2 - |w|^2$ plus remainder term

You misunderstand me
I am using $f(x) = f(a) + Df(a)(x - a) + \frac{1}{2} (\xi - a)^T Hf(a) (\xi - a)$
Second order term is the reminder
$\xi = \xi(x)$

I don't get what you're saying but whatever

7:22 PM
OK, I am not being helpful so I won't comment.

I don't see what $\xi(x)$ is, presumably it is the reparameterization. I don't understand how you've gotten rid of the higher order terms.

I do think this is how the proof goes, though

I mean, do 1D for me
$f(x) = x^2 + x^3$
How do you find the diffeomorphism $g$ so that $g(f(x)) = x^2$?

@MikeMiller OK, I got the subtlety.
You can find a matrix of functions $G$ around $a$ such that $G(a) = Hf(a)$ and $f(x) = f(a) + Df(a)(x - a) + 1/2 (x - a)^T G(x) (x - a)$
My expression is also true (this is Taylor's theorem with reminder) but is not helpful

Oh, I see

7:29 PM
Here's the key lemma. Take a smooth function $f : \Bbb R^n \to \Bbb R$, $f(0) = 0$. I claim you can find functions $g_i : \Bbb R^n \to \Bbb R$ such that $g_i(0) = \partial f(0)/\partial x_i$ such that $f(x) = \sum x_i g_i(x)$
The proof is to integrate: $f(x) = \int_0^1 \nabla f(tx) \cdot x dt$ like you were writing earlier, I think.
Last expression is coming from $f(x) = \int_0^1 \partial_t f(tx) dt$
I always forget this variant of the reminder term
Oh alright

Nonsense

I wrote the Taylor's theorem with remainder term wrong, it is $f(x) = f(a) + Df(a) (x - a) + 1/2 (x - a)^T Hf(\xi) (x - a)$. $\xi$ is a function of $x - a$ which vanishes at $0$.
I forget how to diagoalize it as a function of $x - a$. I can diagonalize it as a function of $\xi$, but that's not what we want.

8:00 PM
@BalarkaSen I just have two questions
On basic stats concepts
1 hour ago, by Mathphile
is the answer for this question "it cannot be determined"?

8:26 PM
Try coming up with different random variables for which each numerical answer is right
So find an r.v. for which Pr[X=c]=1, another for 0, another for 0.5
(The last one is the hardest)

9:01 PM
Hey chat
How do I prove that the set of functions that solve a $n$-th order homogeneous linear ODE is a $n$-dimensional subspace of a set of (all?) functions?

9:15 PM
Understood that question

9:33 PM
How would you describe the geometry of $\Bbb H^n$ at infinity? Would you say it has a "conformal structure", in the sense that angles still makes sense at infinity.
I suppose that's why if you have a finite group $G$ acting by isometries on $\Bbb H^n$ then the quotient, if noncompact, can always be "cusp compactified"; the only way it can be noncompact is if the fundamental hyperbolic polyhedron has vertices at infinity, in which case you can just put that point in
I suppose those manifest as orbifold points downstairs because of this "conformal structure at infinity"
Let's do $n = 2$ because higher dimension is too hard for me

Nah just random questions
What I said is meaningful, right? Angle at infinity makes sense for two geodesics converging to a point at the boundary in any negatively curved space

9:54 PM
I don't really know anything about curvature and hyperbolic spaces

Me neither

I'm not even sure what the definition of $\Bbb H^n$ is

Haha I did not know either until like yesterday
My knowledge of H^2 comes from playing the hyperbolic maze

"the maximally symmetric, simply connected, n-dimensional Riemannian manifold with a constant negative sectional curvature"
quoting wiki

Lmao

9:57 PM
No idea what maximally symmetric means
Extremely vague idea what sectional curvature means
So this is probably the wrong question to ask, but if I forget about the metric, what is $\Bbb H^n$ topologically?

Ah just $\Bbb R^n$

Oh ok
So can the metric be written down explicitely as a matrix or?

Yes, take the open unit disk $\Bbb D^n$. The metric is uh
$ds^2 = \sum_{i = 1}^n dx_i^2/(1 - \|x\|^2)^2$
Note that $\sum_{i = 1}^n dx_i^2$ is the Euclidean metric, so it's a scaling of the Euclidean metric, pointwise
That means angles between vectors are the same as Euclidean angles (angle doesn't care about scale in the metric)

Right
Hmm I probably shouldn't try to understand this at midnight

Lmao
I am writing a blogpost on this (for $n = 2$, but everything generalizes). You might be interested

10:01 PM
I'll listen to you tomorrow if you want to try and explain me what $\Bbb H^n$ is though
Sure, send me a link here or on dc when it's posted!

Yup for sure

Yeah it's basically the reference point to study negatively curved spaces. Alexandrov curvature is defined by comparing a space with $\Bbb H^n$ and see if triangles in the former are more deflated than triangles in the latter
That's being CAT(-1)

CAT spaces are another thing people in ggt talk about a lot, but I don't understand it
ggt looks like a niche thing, but it's actually a huge field

i like the terminology
do u know what CAT stands for
@AlessandroCodenotti yeah lol

10:04 PM
Nope

CAT = Cartan, Alexandrov, Toponogov
but also
CAT = Compare with Alexandrov Triangle
godlike terminology coined by Misha God Gromov

There's people who do ggt that are pretty much topologists, but some are pretty much group theorists, some are actually analysts in disguise, and there's even some that are model theorists
@BalarkaSen lol
@BalarkaSen Who else

They say Grothendieck was an eloquent coiner of terminologies
>etale
try and beat CAT
such a meaningful terminology, whereas etale means nothing
@AlessandroCodenotti Oof
Model theorists are the automatic group guys?

@BalarkaSen is there a fast way to prove triangle inequality for R^n using the standard metric?

I know lots of decidability problems are studied in context of ggt, eg hyp groups have solvable word problems

10:08 PM
one that does not involve Cauchy ?
@BalarkaSen is there a fast way to prove triangle inequality for R^n using the standard metric?

@BalarkaSen not only
For example a big problem in this flavour was whether any two f.g. free groups are elementary equivalent
Look at Tent's pubblications for example, the whole model theory group in Muenster is big on the intersection of model theory and algebra

Huh cool
@Alessandro check dc
@JackOhara you have to use something, i mean. you can "prove" triangle inequality by knowing triangles in R^n lie on a 2d subspace, and on 2d you "know" it by euclidean geometry

@BalarkaSen Ok thanks, how do you prove it in R^2 ?

euclidean geometry

trying to find pure algebraic way

10:18 PM
but you're also trying to avoid cauchy??
that doesnt make sense

but seems that it is hard to work with square roots in that formula

10:29 PM
Having difficulty with probability and expected value
Anyone good at this?

Hi
I have been told that if the Jacobin is non-zero over an interval than a system of non-linear equations will have a unique solution. Is that right?
@Hopper I maybe able to help with your probability question

A language question: If someone says an element is 4-torsion, do they specifically mean that the element has order $4$, or just that the order divides $4$?

10:45 PM

@Thorgott So what I was told is correct?

It's not a precise statement, the precise, correct and more general statement is the one I sent

I find the wiki article to be hard to read
@thorgott Thanks
and bye

11:31 PM
1

Someone has asked similiar question before, but the answer to that question omits the proof. And I couldn't find the proof on Rudin's Real and Complex Analysis or Stein's Real Analysis. Prove that :$\int_a^b |f'(x)|\,dx \leq V(f,[a,b])$, where f is a function of bounded variation. I tried to ...

Hi, anyone would like to help?