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12:01 AM
@robjohn hi sir
 
In my lecture notes, "locally uniform convergence in probability" is mentioned for discrete time stochastic processes that converge towards an ito integral. I'm not sure if this is just convergence in L^2 with respect to the product of the given probability measure with the lebesgue measure
A quick google search didn't bring up useful definitions
So, does anyone have a clue what this refers to?
 
12:52 AM
@user2103480 I think that's exactly right
 
@geocalc33 it actually isn't! I just found the relevant definition. It's defined as "X^n converges locally uniformly in probability if for all t>0, sup_{s in [0,t]} |(X^n)_s - X_s| converges to 0 in probability". The "popular" term for it seems to be "uniformly on compact sets in probability"
thank you still!
 
@user2103480 great :)
 
 
3 hours later…
4:19 AM
Gud mornin
 
4:41 AM
meh it's OK I guess
 
 
2 hours later…
6:34 AM
What up my shitters
:D
 
7:02 AM
see I would get a ban for that
 
I meant shedders
 
lmfao
 
that's fine I'll go back to facebook and take my frustration out on high functioning differently abled people
 
lmfao
 
7:49 AM
@AdamL didn't you get banned for consistently making sexist and casually racist comments on this chat?
Or was that a different Adam
 
 
2 hours later…
9:29 AM
Did he proof that $\pi_1(\Sigma_i)$ is hyperbolic for $i > 2$ in the previous session @BalarkaSen ?
I don't recall seeing it
 
No.
It follows from Svarc-Milnor lemma and quasi-isometry invariance of hyperbolicity
 
So we get the cayley graph is quasi-isometric to the universal cover of $\Sigma_i$
that's the plane, isn't it ?
 
That's the hyperbolic plane
$\pi_1(\Sigma)$ acts isometrically on the hyperbolic plane
 
oh ok. And that's the normal plane for $\Sigma_1$. I see
 
Genus >= 2 is crucial
 
9:34 AM
Maybe this is complete unrelated - but is there any way to prove it (that $\pi_1(\Sigma)$ acts isometrically on the hyperbolic plane) with using uniformization theorem, and the fact that $\Sigma_i$ is a Riemann surface ?
ok seems he is back
 
@Lelouch this is killing a fly with a nuclear bomb
$\pi_1(\Sigma)$ acts isometrically on $\Bbb H^2$ by drawing a picture
 
@BalarkaSen ok sorry, yes, this is obvious by drawing a suitable $2n$-gon in $\mathbb{H}^2$
my bad
 
9:55 AM
I thought general topology could get ugly, but topological dynamics is way worse, it's all nonsense
You can have a space $X$ and an automorphism $f$ such that every point has a dense orbit in the associated Z action, but the space is not homogeneous
And $X$ is like compact metrizable, as good as spaces can be
 
 
1 hour later…
11:00 AM
guys what's the procedure to start a new stackexchange website?
 
11:14 AM
 
11:48 AM
https://math.stackexchange.com/a/2684915/586524
My notes emphasise that the partials must be continuous. I know that if a real function is differentiable, it is continuous. That being said, is: "If continuous partial derivatives exist at some point a" less redundantly stated as "If partial derivatives exist ..."?
 
what does it mean for continuous partial derivatives to exist at a point
if you want to talk about continuity, you better have the partials exist in an entire neighborhood
 
That's how I understand it too...
Wait, let me quote the notes directly:
"Assume that u and v have continuous first-order partial derivatives in D and that they satisfy the Cauchy-Riemann equations at some point a in D. Then f is (complex) differentiable at a"
Is continuous here redundant? If they're (u and v) partially differentiable they must be continuous, or does this not follow for partial derivatives?
 
https://math.stackexchange.com/questions/3781829/calculate-int-0-infty-frac-sin-xx3x-mathrmdx-find-my-mistake
can someone tell what I do wrong here?
 
It's not saying $u$ and $v$ are continuous, it's saying $\frac{\partial u}{\partial t}$ and $\frac{\partial v}{\partial t}$ are continuous
 
I mean I take a pacman path
and everything zeroes out
and I can't understand why
 
12:00 PM
I'm pretty sure it's not redundant
 
@Thorgott are you speaking to me?
 
@Threnody For a more elementary example from real analysis, it's possible to have a function f : R^2 -> R whose partial derivatives exist at a point, but aren't continuous at that point, and this can in fact cause f to fail to be differentiable
 
@hash no
 
:(
 
The condition of the partial derivatives themselves being continuous is stronger than them simply existing
 
12:04 PM
@Fargle I see, so... for partial derivatives, it does not follow that (they exist => they're continuous)
 
Indeed
 
But I'm right in saying that it does follow for one variable function derivatives
 
no, that isn't true
 
Derivatives don't have to be continuous even in single variable
x^2 sin (1/x)
 
Oh..
No no... uhh
 
12:05 PM
good ol' sin(1/x)
 
The original function has to be continuous if it's differentiable, but the derivative doesn't
 
Yes! Ok that's what I was trying to say
 
here's a nice thing to read for a reality check math.stackexchange.com/questions/112067/…
makes you realize just how bad derivatives can be
 
Multivariable is a weird case even in real analysis, because even if partial derivatives exist, you don't have to have differentiability---in fact, it's possible for both partial derivatives of f to exist at a point, and yet f is discontinuous at that point
 
I see...
 
12:08 PM
yeah, however the existence of bounded partials (in a neighborhood) implies continuity
 
classic example is f(x,y) = xy/(x^2+y^2) for (x,y) =/= (0,0), f(x,y) = 0 for (x,y) = (0,0)
Partial derivatives exist at (0,0) but the function isn't even continuous there
 
Yeah I see it!
@Thorgott lol that's way above my paygrade, i don't even know what a measure is yet
 
This is why you have notions like the class of C^n functions, which means "(partial) derivatives of nth order and below all exist and are continuous"
That "and are continuous" does a lot of work
(C^0 just means continuous)
 
Oh, so there's interest in talking about a class of functions?
 
the point is just that derivatives can be discontinuous on large sets
 
12:13 PM
Ignore my choice of words there.
 
Ok :D
 
C^n(R^k, R^m) (C^n functions from R^k to R^m) is just a set
 
From wikipedia: "In mathematics, the term large set is sometimes used to refer to any set that is "large" in some sense" lol
 
But, for example, x^2 sin (1/x) is differentiable (hence continuous aka C^0), but not C^1
 
yeah, the ways in which large can be made precise are what's listed in the answer :P
 
12:17 PM
@Fargle I see.. we haven't done much about this, probably because I chose combinatorics and graphs instead of more analysis and set theory ._.
 
Can't blame you, I was never very strong at analysis myself
 
I like it, I just don't like being an automatic theorem prover
 
But it's helpful to carry around a bucket of counterexamples, just to have something to sharpen concepts against
 
Thank you all, have a good one
 
@Balarka @Alessandro open.spotify.com/album/…
 
12:21 PM
@EdwardEvans fanks
gonna listen soon
 
fookin bangers on that album
also
someone tell me what Hodge theory is
 
what do u need it for
 
idk there's just a course on p-adic Hodge theory this semester (I probs won't take it, looks like I don't have the prereqs) but idek what Hodge theory is
lol
 
eek
 
Gonna start a band called Matrix specifically so that I can form a supergroup with Vektor and call it Tensor
 
12:24 PM
I can tell you what Hodge theory on Riemannian manifolds is, but I think that's not what you want
 
@Lelouch Fuchsian groups were studied in the late 1800s, certainly before uniformization was proved!! (It was conjectured around the same time, though)
 
Grothendieck didn't know $\pi_1 \Sigma$ is a Fuchsian group
 
it's somehow related to the algebro-geometric Hodge theory, but I don't know any of that
 
fair
I'll probs just stare at the wikipedia page until I start crying
 
therefore we conclude Lelouch is Grothendieck
 
12:25 PM
Somehow I don't think p-adic hodge theory uses the Laplacian
which makes one wonder: the fuck do these guys think hodge theory is
 
@BalarkaSen It's easy to forget such concrete and obvious things when you work at a high level of abstraction
 
reminds me of how if you say the fourier transform to an algebraic geometer they think you mean...
 
Lmao yeah wtf is Fourier-Mukai
 
oh god wtf
 
lol idk, which MSC does Hodge theory fall under?
 
12:27 PM
Algebraic geometry is cancelled due to hurting my brain and heart
 
DiffTop?
 
classical Hodge theory is DiffGeo
(I think)
 
I think I have to learn ultralimits
welp
 
Thanks @Ed I'll check it out
Also Gojira dropped a new song
 
npnp
orly
 
12:41 PM
damn
 
listening
 
@BalarkaSen noice, from which source?
 
dunno
just read the definition from wikipedia
 
A btw @Balarka @Thorgott I asked the topologically/point transitive question on MSE earliee
 
did u check if that weird example works
 
12:42 PM
Hopefully it'll get an answer math.stackexchange.com/questions/3781838/…
@BalarkaSen I thought about it, but I couldn't convince myself that the action is topologically transitive
 
let me see
it should work
 
nice
 
@Alessandro well it was certainly a Gojira song
 
Most definitely
 
say $f : X \to X$ is topologically transitive, look at the $\Bbb R$-action on $T_f$. Pick any two open subsets $U, V \subset T_f$ intersecting the meridian nontrivially say, so $U \cap X, V \cap X$ are open subsets of $X$, we know by transitivity there exists some $n$ such that $f^n(U \cap X) \cap (V \cap X) \neq \emptyset$.
so there is some $p \in U \cap X$ and $q \in V \cap X$ such that $f^n(p) = q$. That means the $\Bbb R$-orbit of $p$ in $T_f$ contains $q$ as well
so the $\Bbb R$-orbit of $U$ intersects $V$, no? what am I missing
why isnt this trivial
 
12:52 PM
Wait which direction is the meridian? The one with slices like $X$ or the other one?
 
the one which slices like $X$
the first return map of the $\Bbb R$-action wrt a meridian is $f$
this is for any two open subsets $U$, $V$ which intersects a common meridian, but you can always flow $U$ a little bit so that they do
 
Hm wait I don't think I understand the construction anymore
 
The construction is $T_f := X \times \Bbb R/(f(x), r) \sim (x, r+1)$ if you want me to write it like this
$\Bbb R$ acts on $T_f$ by $s \cdot (x, r) = (x, r + s)$
 
No ok but how do you flow backwards from points with many preimages through f
 
yeah, its still a semigroup flow if $f$ is not a homeo
thats annoying
but it's definitely topologically transitive
 
1:05 PM
Ok I agree with that now
Those semigroups are annoying...
Want to think about ultralimits instead?
 
Troll
Blocked
 
Lmao
Dude how do you make an action
 
Balarka said he wants to learn about them lol
 
Why am I making semigroup actions
 
What's really nice is when you don't have forward or backward flow either
 
1:09 PM
thats me
 
that's right... it's $e^{-tD}$ for $D$ dirac time
 
Can you make $\Bbb R_+$ act on $T_f$ by $r \cdot (x, t) = (x, rt)$? lol
Nah
This is too hard
 
1:37 PM
Is an infinite ``descend" of radicals possible in a nontrivial number field? To be more clear: If $\alpha$ is algebraic (and not equal to $0, 1$) over $\mathbb{Q}$, then does there exists infinitely many integers $n$ for which there's $\beta$ algebraic with $\beta^n =\alpha$, and $\mathbb{Q}(\beta) = \mathbb{Q}(\alpha)$ ?
nevermind, of course not.
 
@Alessandro Take the $\Bbb Z$-action on $\{0, 1\}^{\Bbb Z}$ by shift, which is topologically transitive, and consider the subspace of all functions $\Bbb Z \to \{0, 1\}$ which is $n\Bbb Z$-invariant for some $n$. This is dense in $\{0, 1\}^{\Bbb Z}$ so the restriction is also topologically transitive
But every orbit is periodic
 
Is there any direct way to see the above without using unit theorem ?
 
@BalarkaSen what do you mean? Sequences that have some $n$ values at the "beginning" and then repeat those?
 
there is no beginning, it's 2-sided. yes, $n$-periodic
look at all periodic sequences
that just works, fullstop, right?
 
Hmm let's see
 
1:48 PM
its the same as the doubling map on the circle, that one's semiconjugate to the shift operator on $\{0, 1\}^{\Bbb N}$
and the dyadic rationals are similarly periodic
but it's a fucking $\Bbb N$
 
What's the problem @BalarkaSen ?
 
It's obviously dense in the full space
 
@Lelouch find example of topologically transitive group action which has no dense orbit
 
I think that works. I need to leave for a moment but I'll think about it soon. Seems good to me though
 
Yeah I think this is ok
 
1:50 PM
@BalarkaSen topologically transitive means for any two open sets $U$, $V$, there's an elemeng $g$ such that $gU \cap V$ is nonempty, right ?
 
lmgtfy but yes
 
2:00 PM
yeah, why you can't take the integers with cofinite topology and the identity map
 
Hausdorff man
 
when did you say Hausdorr ?
*Hausdorff
 
I didn't
 
oh
 
Alessandro wants Hausdorff + separable examples
 
2:01 PM
oh shit
 
Hausdorff+separable+Baire forces it to be true
so he wants to understand if something weaker works
Who the fuck does topological dynamics on the cofinite topology dude
Are you nuts
 
Well that too with the trivial group acting on it:P
 
lmao
Lelouch = Grothendieck confirmed
 
lol
Ok, I guess this works, not too sure. Consider the interval $[0,1]$ with the dyadic rationals removed.
The basis for the topology is $(\frac{l}{2^m}, \frac{l+1}{2^m})$
And the functions which are acting are the permutations of length $2^{-m}$ open intervals $(0, \frac{1}{2^m}), (\frac{1}{2^m}, \frac{2}{2^m}), ...$
each element has finite orbit ig
 
What the hell is this do you mean $2x \mod 1$ or what
 
2:11 PM
@BalarkaSen what ?
 
you mean your function is $f(x) = 2x \pmod{1}$
you're writing it in a stupid way
 
no !
 
Then what is your function lol
 
for example, you take the four intervals $(0, .25), (.25, .5), (.5, .75), (.75,1)$ and any permutation of them is an valid fucntion
 
This is godawful I dont understand your example
@Alessandro posted
0
A: Is a topologically transitive action on a second countable space point transitive?

Balarka SenTake the natural $\Bbb Z$-action on $\{0, 1\}^{\Bbb Z}$ by shift, and consider the subspace $X$ of periodic $2$-sided sequences of some unspecified period length. Clearly the original action is topologically transitive, and $X \subset \{0, 1\}^{\Bbb Z}$ is dense. Thus, the $\Bbb Z$-action on $X$ ...

 
2:13 PM
@BalarkaSen uhh.... Let me be clear: Fix any permutation $\pi$ of $\{0, \cdots, 2^m-1 \}$, and for $x \in (\frac{l}{2^m}, \frac{l+1}{2^m})$, send it to $x - \frac{l + \pi(l)}{2^m}$
that's one function. Now, do it for all permutations and all $m$
these functions forms a group
 
Ok, it's some awful group, like direct limit over $S_m$'s or some garbage. What is your claim?
 
The action is topologically transitive, but hte orbit of any element is finite, and thus not dense
 
Maybe that works
Seems like a pain
 
both are clear to see, unless I'm mistaken
 
A shift operator is easier to me :P
 
2:35 PM
@Lelouch After some thought I agree what you say should work in principle. Your group is something horrifying like the rooted automorphism group of the binary $2$-tree, but it should be topologically transitive + finite orbit.
The point is you can do it with just 1 self-homeomorphism of a space, which is what a $\Bbb Z$-action is.
Anyway good example, if it works.
 
2:57 PM
small question: lim n/n = 1 as n approaches \infty, and lim (n/n)= lim (1/n + 1/n + .... + 1/n) = 0 + 0 + 0 + .... + 0 as n approaches \infty. Is this possible?
 
@BalarkaSen accepted, I'm convinced
Now I have another awful dynamical system to think about lol
 
3:16 PM
@BalarkaSen ok, nice; I see. thanks.
@user777 you can't swap limits and infinite sum without getting weird results.
essentially what you're doing is writing $1 = \sum_{i \geq 1} a_{j_i}$, where $a_{j_i} = \frac{1}{j}$ if $i \leq j$, otherwise $0$, and taking the limit as $j$ goes to infnity. Swapping inifinite sum and limit can be done in some cases, but not always.
 
@AlessandroCodenotti Uh oh
 
4:08 PM
@BalarkaSen lol
I'm trying to understand this homeomorphism of a Cantor like space
 
fun fact: René Descartes didn't call pullback squares Cartesian diagrams
 
4:28 PM
@LeakyNun If you don't mind, I want to ask an irrelevant question : Are you a 2nd year undergrad?
 
no
I probably forgot to update something
 
So, can you please share your current level of academic education?
 
I just finished year 3
 
Nice and congratulations
 
thanks
 
4:39 PM
:)
 
GUYS
I have a question
is anyone here?
 
TechnoKnight? :(
 
lol same name
Anyway, my question is
There is a plane, let's call it Q
and there's another plance called P
and there is a line formed by their intersecting, let's call it D
My question is, if we put the equation of Q = equation of P. Will we find the equation for the line D?
 
@TechnoKnight What is the equation of a place?
 
I mean plane
There is two planes in space, P and Q
 
4:47 PM
and now, how can you set two equations equal to each other?
 
what do you mean?
the equation of Q is something like
x + y + z + c = 0
And the equation of P is something like
x +y + z + c = 0
too
I mean, a linear equation
then
we put
x + y + z + c = x + y + z + c
and we find the equation of the line D
Like x' + y' + z' + c = 0
Is that correct or I'm wrong in doing something?
 
Did you try a few examples?
 
what do you mean?
 
Try putting in some numbers and see what the result is
 
how that's gonna help me?
 
4:53 PM
The same way doing examples always helps think about math problems
 
dude
how the heck that's gonna help me?
Do you know what my problem is?
I'm telling you if I set Q = P
will I find the equation of D?
 
And I am telling you to try an example and see what you get
 
what the
are you kidding?
is there anyone else here?
 
chill
 
I don't have much time.
 
4:58 PM
In the time it took you to complain you could have written down an example
 
$x^p-x$ has p distinct zeros in $\mathbb{Z}_p$ for p prime. I know this follows easily from Fermat's, but I want to show it using the fact that the ideal generated by a polynomial $p(x)$ in a field is maximal iff $p(x)$ is irreducible. This should be possible, yes?
 
What exactly do you have in mind?
 
@Lozansky Hmm, I don't see any direct way using that
 
I mean, $(x^p-x)$ clearly isn't maximal, because it is contained in $(x)$, but that only gives us reducibility and says nothing about the roots
 
5:15 PM
@Thorgott Good point
 
I can think of an abstract argument that doesn't rely on Fermat to prove this, but not using Fermat honestly just seems needlessly obtuse
 
We have to prove that
$$
f(x) = \begin{cases}
0 & if~x~is~irrational \\
1/q & if~x~is~of~the~form ~p/q (p , q \in \mathbb Z, q\neq 0)
\end{cases}
$$
$0 \lt x \lt 1$
have a limit zero at any $0 \lt a \lt 1$.
I'm not able to understand the green boxed thing.
(p/q is in the lowest form)
 
Which part do you not understand?
 
I don't know why he took $n$? what is it's role?
I understand for all irrational $x$, $f(x)$ will be zero and hence the inequality $|f(x)| \lt \epsilon$ will be satisfied. It's only the rational number that will cause the problems
And he is representing the rational numbers in the form of $p/q$ as
$$
1/2, 1/3, 2/3, 1/4, 3/4 ....
$$
But what's the role of $n$ here?
 
5:33 PM
Yes, and for a rational number $p/q$ (in lowest terms), $f(p/q)=1/q$, hence $|f(x)-0|=1/q$ and thus we want to know for which natural numbers $q$, $1/q<\epsilon$
 
Yes, clear that far.
 
No, $1/q\ge\epsilon$.
 
okay, maybe thats what the book (Mr. Spivak) is doing.
 
I knew that page looked 1000% familiar.
 
What he does next? Why to introduce $n$?
 
5:36 PM
Because fractions are built out of natural numbers.
 
@TedShifrin Do you know Michael in real life?
 
Yes.
That's how my name (and contributions) show up in his book
 
I cannot find acknowledgements in the book
 
Ok, so the point now is that $1/q<\epsilon$ for all large enough $q$, hence there are only finitely many $q$ with $1/q\ge\epsilon$ and consequently only finitely many $x$ in the interval $[0,1]$ for which $|f(x)-0|\ge\epsilon$. So, in any small enough neighborhood of $a$, there won't be any of these $x$ except possibly $a$ itself.
 
Thats a very clear explanation, thanks.
@TedShifrin Now, I realize how cold it is to say "Calculus by Spivak" when people like you (whom I know) have enough contributions in it.
 
5:41 PM
Not at all. It's perfectly correct to say it's his book.
If you read the preface to the third or fourth editions, I'm in there. I contributed hundreds of problems and (to the fourth edition) some substantial changes to the text.
 
Did you get to know about Mr. Gilbert Strang?
 
Yes, I know Strang quite well. Interestingly, when I taught at MIT for two years, I lectured the 350-person multivariable class two times, and one time he was a TA for me :)
I also credited him in my linear algebra book for influencing the way I taught the material and wrote the book.
 
Wow!
@robjohn Sir did you have any conversation or something like that with Steve Jobs?
 
OK, I'm done name-dropping for a while.
I think Spivak's book is exceptionally clear and well-written. Which edition do you have? The chapter on limits and $\delta$-$\epsilon$ proofs changed noticeably in the fourth because of my complaints to him.
 
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