Mathematics - 2020-05-29 (page 1 of 2)

Bubbler

00:21

@ArnoldFernández Without computer, you can e.g. find the global minimum of x^8-x^5 and x^2-x separately (which is -15*cuberoot(25)/256 and -1/4 respectively, you can find it by differentiation), and show that sum of the two plus 1 is higher than zero. Therefore the global minimum of the original function must be positive too.

Arnold Fernández

Whitot differentation?

Thorgott

01:27

I just read that every compact, connected 2-manifold can be covered with only 3 charts. Insane.

Bubbler

02:00

@ArnoldFernández You can convert the function to the sum of squares (x^4-x/2)^2 + (x/2-1)^2 + x^2/2, which can't be negative. To show it cannot be zero, show that x/2-1 and x cannot be zero at the same time (which is trivial).

Arnold Fernández

Thanks

user10478

I'm confused by en.wikipedia.org/wiki/… when it says, "Sometimes, it is more convenient to deduct backwards, proving the statement for $n - 1$, given its validity for $n$. However, proving the validity of the statement for no single number suffices to establish the base case; instead, one needs to prove the statement for an infinite subset of the natural numbers."

I was trying to find out if mathematical induction can work by (using the ladder analogy) proving an arbitrary rung $r$, proving that an arbitrary rung $k$ above $r$ entails $k + 1$, and proving that an arbitrary rung $j$ below $r$ entails $j - 1$. Basically, sawing the (possibly bidirectionally infinite) ladder into two ladders such that $r$ is both the bottom of one and the top of the other. But I don't think that's what it's saying?

Arnold Fernández

02:22

@Bubbier how you get that idea?

Mike Miller

02:50

@Thorgott Why insane? Take the polygonal representation of surfaces; use the interior as one chart, and a chart containing the vertices and half of the edges for the second, and a chart containing the remaining half of the vertices third, or something like this. You can draw the picture for the torus or genus 2 guy without much difficulty

Stan Shunpike

What area of mathematics are polytopes most commonly found in?

Is that something one would learn studying mathematics in undegrad or graduate school? or is that a niche topic?

Yuvraj

03:18

@robjohn it is a bit of complex!

and how did you derive that?

loch

03:44

@StanShunpike you'd see polytopes showing up in toric geometry

Balarka Sen

04:58

@Thorgott A modified version of Mike's atlas that I like is throwing away a wedge of circles from a genus g surface, which gives a chart, throwing away another distinct collection of wedge of circles, which gives the second chart - you now miss 2g points, which you join by a path and take a tubular nbhd

Three charts

The minimal number of R^n charts needed to cover is called the Lusternik-Schnirelmann category

(Or LSC - 1, there's some convention)

This number is still 3 for a non-orientable compact surface

Only 2 for S^2

Balarka Sen

05:15

It's still 3 for an orientable surface of infinite genus.

Balarka Sen

05:30

Take this 2D complex $X$, and consider folding it to a 2-simplex according to the folding scheme indicated. This gives a map $X \to \Delta^2$ which has 3 fold lines (the Z shape in the interior) and two cusp points (the two kinks of Z).

Taking a cone on this we obtain a simplicial map $\Delta^3 \cong CX \to \Delta^3$ from a subdivision of the 3-simplex to the 3-simplex which folds along a $C(\mathsf{Z})$, a fold surface consisting of two cuspidal lines that meet at a point $v$, the cone point of $CX$.

This is the local simplicial model of a swallowtail singularity

This provides a triangulation near the worst singularity of $f : \Bbb R^3 \to \Bbb R^3$, $f(x, y, z) = (x^4 + yx^2 + zx, y, z)$, say, and it's easy to extend this triangulation to a triangulation of the whole mapping $f$.

This is the Thom-Boardman singularity $\Sigma_{1, 1, 1}$ if I am not wrong

Arnold Fernández

05:59

How do you put a pciture?

robjohn

06:41

@Yuvraj I noticed a mistake. It might be more complicated, or possibly simpler. I am working through the correction.

Thorgott

09:45

I appreciate the explanation even though I don't really get it

Balarka Sen

Do it for the torus, eh? Let $m_1, m_2$ be two distinct meridians, $\ell_1, \ell_2$ be two distinct longitudes. $U_i := T^2 \setminus (m_i \cup \ell_i)$, $i = 1, 2$ are both homeomorphic to $\Bbb R^2$, and $U_1 \cup U_2 = T^2 \setminus \{p_1, p_2\}$ where $p_1 = m_1 \cap \ell_2$, $p_2 = m_2 \cap \ell_1$

Let $\gamma$ be a smooth path joining $p_1, p_2$, and let $V$ be a $\varepsilon$-neighborhood of this path, which is gonna be homeomorphic to $\Bbb R^2$. $\{U_1, U_2, V\}$ is an open cover of $T^2$

Alessandro Codenotti

10:06

@Balarka I did my homework, around a month ago you mentioned that every compact metric space of dimension $n$ embeds into $\Bbb R^{2n+1}$. In fact compact is overkill, separable metrizable of dimension $n$ is enough. This is due to Menger–Nöbeling, the idea is that they all embed into the Nöbeling space, the subspace of $\Bbb R^{2n+1}$ with at most $n$ rational coordinates

Balarka Sen

Oh shoot

Alessandro Codenotti

There's also some nonsense result that all strongly metrizable metric spaces of weight $\kappa$ and dimension $n$ embed into $\kappa^\omega\times N_n$ where $N_n$ is the Nöbeling space in $\Bbb R^{2n+1}$

Balarka Sen

what is weight

Alessandro Codenotti

Where strongly metrizable means something weird like metrizable with a basis consisting of a countable union of star finite coverings or metrizable+completely paracompact

Balarka Sen

Ugh

Alessandro Codenotti

10:15

@BalarkaSen smallest cardinality of a base

Balarka Sen

Ah OK

Alessandro Codenotti

Yeah that's just ugly

But I want to go through the proof for the separable metric case

Balarka Sen

Tell me the key ideas if you understand it

Alessandro Codenotti

Sure

Thorgott

oh, I see, that makes sense

Balarka Sen

10:21

Can you prove $T^2$ cannot be covered by two Euclidean charts

Thorgott

cutting along a meridian + longitude essentially undoes the gluing done when you construct the torus from a square

Balarka Sen

Yep

Alessandro Codenotti

Oh no, the proof for the nice case is to do the general case, embedding into $\omega^\omega\times N_n$ (since separable implies second countable for metric spaces) and then showing that the embedding built in the proof of the general case is actually constant in the first coordinate in the special case

I need to find a different book that does only the special case lol

Balarka Sen

Lmao

Balarka Sen

10:34

@Thorgott Follow-up (Challenging): Prove that there is a function $f : T^2 \to \Bbb R$ with exactly three critical points (Why is it a follow-up?)

There is a function with exactly 4, the height function on an upside down donut - classic Morse theory picture

You cannot improve to 2, by Reeb's theorem :)

Alessandro Codenotti

Engelking has a much saner proof. Actually it seems that embedding into $\Bbb R^{2n+1}$ rather than this Nöbeling space is much easier and a BCT trick

Balarka Sen

Gotcha

Alessandro Codenotti

It's lunchtime now but I think I can tell you more later

Balarka Sen

That'd be great!

Enjoy

linda Oiladali

Please I need help how to prove that $\lim_{x\to-\infty}x^n\exp(x)=0$

I need idea

maths student

10:46

Is this vector field conservative ?

I tried to draw closed loop around center to see work done to reach initial point to again same point .

0

Q: vector field property using figure

Is this vector field conservative ? I tried to draw closed loop around center to see work done to reach initial point to again same point .This is same as checking line integral along closed path is 0 or not.

Alessandro Codenotti

11:14

Still here? @Balarka

Thorgott

I thought I figured something out, but I just realized you wanted 3 critical points, not values..

urgh, I think messing around with projections isn't enough to do this

Balarka Sen

11:36

Yeah it's quite hard

@Alessandro Yup

Alessandro Codenotti

Ok so to prove that all separable metric spaces with $\dim X\leq n$ embed into $\Bbb R^{2n+1}$ we need three ingredients, one is easy, one is technical, one is needed to go from compact to separable

A function $f:X\to Y$ between metric spaces is called an $\varepsilon$-map if $\mathrm{diam}(f^{-1}(y))<\varepsilon$ for all $y\in Y$

So now let's assume that $X$ is compact so that $Y^X$ is a metric space with the metric induced by the sup norm. The easy ingredient is that for all $\varepsilon$ the set of $\varepsilon$-maps $X\to Y$ is open in $Y^X$

Balarka Sen

Ya OK

Alessandro Codenotti

The technical ingredient is that if $\dim X\leq n$ and $H$ is an $n$-dimensional linear subspace of $Y=\Bbb R^{2n+1}$, then the set of $\varepsilon$-maps that miss $H$ is dense in $Y^X$

(We don't really care about $H$, we only care about the set of $\varepsilon$-maps being dense)

Balarka Sen

Ah so it's the same trick as the manifolds thing

Alessandro Codenotti

But now we are done, let $X$ be compact with $\dim X\leq n$, for every $i$ consider the set $U_i$ of $1/i$-maps $X\to\Bbb R^{2n+1}$, by the previous two things every $U_i$ is open dense in $Y^X$, which is baire, so $\bigcap U_i$ is dense (in particular nonempty), but every element of the intersection is a continuous injection $X\to\Bbb R^{2n+1}$, hence an embedding

So not only there is an embedding $X\to\Bbb R^{2n+1}$, there's plenty of them, they form a dense $G_\delta$ set in the space of all continuous functions $X\to\Bbb R^{2n+1}$ (if $X$ is compact)

Balarka Sen

11:48

Hm, not quite

Cool proof

Alessandro Codenotti

To get the separable case there is a result saying that a separable metric space $X$ with $\dim X\leq n$ has a metrizable compactification $cX$ with $\dim cX\leq n$

@AlessandroCodenotti The hard part is showing this

The $n$-dimensional subspace thing is there to get embeddings into the Nöbeling space. If $N_n$ is the subspace of $\Bbb R^{2n+1}$ of points with at most $n$ rational coordinates then we can write $\Bbb R^{2n+1}\setminus N_n$ as a countable union of $n$-dimensional linear subspaces (those given by equations $x_{i_1}=q_1,\ldots,x_{i_{n+1}}=q_{n+1}$ for rational $q_j$) and use as $U_i$ the $1/i$-maps missing the $i$-th element in an enumeration of those linear subspaces

So $N_0$ is the space of irrational numbers, which is then universal for $0$-dim separable metric spaces. The Cantor space $2^{\Bbb N}$ is also known to be universal for $0$-dim separable spaces, and just like the Nöbeling spaces generalize the irrational there is also a family of higher dimensional universal spaces generalizing the Cantor space due to Menger

(That's where the Menger sponge comes from)

Mike Miller

Ew

Alessandro Codenotti

12:03

lol

Balarka Sen

This will take nontrivial effort for me to understand. I get the outline you explained, though. Thanks!

Weierstrass preparation theorem seems like a technical result

I have been staring at a proof without getting anything out of it for last 2 hours

Alessandro Codenotti

I got the impression that every result in several complex variables is a technical result

Mike Miller

Skip Weierstrass and just prove Malgrange

Balarka Sen

that's what I am interested in but it seems that they borrow ideas from Weierstrass

maybe thats an illusion

Mike Miller

Do you actually need Malgrange for something

Balarka Sen

12:14

Apparently its crucial for classification of singularities of map-germs

Mike Miller

Huh

I guess that makes sense

Balarka Sen

The algebra formulation of Malgrange preparation theorem is that if $f : (M, p) \to (N, q)$ is a smooth map-germ, $A$ is a finitely generated $C^\infty_p(M)$-module, then $A$, under basechange by $f$, is a finitely generated $C^\infty_q(N)$-module iff $A/\mathfrak{m}_q(N)A$ is a finite dimensional real vector space

Mike Miller

weird

Thorgott

13:43

@Balarka can that 3 critical point function be Morse?

I'm wondering whether it's possible to build a torus by two handle attachments

Balarka Sen

Nope :)

Good question though

You're correct to wonder about what having 3 critical points would mean in the sense of handle attachments

Thorgott

I'm wondering what it means generally, one ought to be a global minimum, the second a global maximum and the third is ???

Balarka Sen

You know the Morse lemma, right?

Alessandro Codenotti

@Balarka apparently the 2n+1 bound in the dim X=n comes out of the fact that the nerve of a cover with multiplicity n can be realized as a simplicial complex in $\Bbb R^{2n+1}$

Balarka Sen

@Alessandro That makes sense, actually. What is the homeomorphism, though?

Alessandro Codenotti

13:57

Which homeomorphism are you talking about?

Balarka Sen

You need to embed $X$ in $\Bbb R^{2n+1}$

Alessandro Codenotti

@AlessandroCodenotti it's here, any element of the intersection will do

Balarka Sen

So far I can see a Lipschitz map to the nerve by squishing stuff in the open sets to points and edges and faces etc, which embeds in $\Bbb R^{2n+1}$

Alessandro Codenotti

The thing is that functions in the intersection are $1/i$-maps for every $i$, so they must be injective, and continuous injection from compact to Hausdorff is an embedding

Balarka Sen

I didn't see the nerve in the outline you wrote

Thorgott

14:01

that's only for non-degenerate critical points though, right

Balarka Sen

@Thorgott Ah but you were trying to decide if the function can be Morse, right?

Thorgott

though I guess the third one could be non-degenerate and the global min/max degenerate

Alessandro Codenotti

@BalarkaSen No that comes up to prove the technical bit

Balarka Sen

I see

Alessandro Codenotti

To prove it you use that map, but you need to argue that you can place the vertices of the simplices close enough to get an $\varepsilon$-map

Mike Miller

14:02

@BalarkaSen Surely that's relevant for constructing these 1/i-maps

yeah

Alessandro Codenotti

And you also argue that you can place them in a way that misses an $n$-dimensional linear subspace apparently

Balarka Sen

I didn't pay too much attention because I didn't understand it, but makes sense

I only vaguely get it anyway

Mike Miller

What's the convex hull of the twisted cubic

Alessandro Codenotti

I haven't read all the details, it's a longish proof relying on a bunch of previous lemmas, so I'm still working my way through it

Balarka Sen

@MikeMiller Twisted cubic is what you get if you blowup $y^2 = x^3$ at the origin. So the image of everything in the inside of that curve by the inverse (well-defined away from origin), right

That might give a geometric way of understanding why segments with distinct pairs of endpoints are disjoint

The endpoints always have different direction data or something

Thorgott

14:14

Hmm, I guess it comes down to whether one can glue together two spheres to get a torus (and that shouldn't be possible)

Yuvraj

@robjohn any improvement?

Balarka Sen

@Thorgott This is in reference to if you can cover it by 2 R^2's?

robjohn

@Yuvraj it turns out that the weights are the reciprocals of what I had before. Just checking to make sure.

Yuvraj

@robjohn it will be very kind if you can help me in understanding it's derivation , once you have finished?

Thorgott

Eh, I confused things. It's in reference to handle attachments. I start with a 0-cell and since a graph can't be a 2-manifold, there's gotta be a 2-cell involved. The first attachment can be done in only one way in either case. So it comes down to a) can we glue a disk along the boundary to a circle to get a torus or b) can we glue a disk along its boundary to another disk to get a torus.

Balarka Sen

14:24

@Thorgott I'm confused about your setup. You have a function $f : T^2 \to \Bbb R$, $f$ has three critical points?

Each critical point contributes to a handle/cell - so you should be playing around with three handles/cells

Assuming $f$ is Morse and trying to get a contradiction from there, that is to say

Thorgott

isn't the first attachment just the 0-cell

no wait

this is nonsense

Balarka Sen

No that's correct. A Morse critical point of index k corresponds to a k-cell.

For k = 0 (i.e., the minimum) you add a 0-cell

(The standard height function T^2 -> R has four critical points, of index 0 (min), index 1 (first saddle), index 1 (second saddle), index 2 (maximum) - which thus says there's a CW structure on T^2 with a 0-cell, two 1-cells and a single 2-cell. That's correct!)

This naive analysis does not say how the cells are attached, but you indeed can get a T^2 from attaching two 1-cells to a 0-cell (only way to do this here - gives you $S^1 \vee S^1$) and attaching a 2-cell to this $S^1 \vee S^1$ - you don't know what this attaching map is from naive analysis but you know gluing by $aba^{-1}b^{-1}$ gives $T^2$

Agreed?

Thorgott

agreed except I'm not sure what you mean by "gluing by $aba^{-1}b^{-1}$". Is this in terms of fundamental groups?

Balarka Sen

Exactly, that describes how the 2-cell is glued to $S^1 \vee S^1$ completely. The 2-cell is a disk, whose boundary is a circle, so it must be glued to $S^1\vee S^1$ by some map $S^1 \to S^1 \vee S^1$. In fact, (fact:) homotopic maps give homotopy equivalence CW complexes upon attachment, so it suffices to specify the homotopy class of this map, an element of $\pi_1(S^1_a \vee S^1_b) = \langle a, b \rangle$, the free group on two generators. I claim this element is $aba^{-1}b^{-1}$

This can be seen from the square identification picture, because $S^1_a$ is the image of one edge of the square and $S^1_b$ is the image of the adjacent edge.

The 2-cell is the interior of the square, which is thus attached to $S^1_a \vee S^1_b$ by exactly the gluing scheme of the square, which is $aba^{-1}b^{-1}$

Thorgott

14:46

oh wow, that sounds like a great picture

Knight

Thorgott, a function $f$ doesn’t have any local maxima and minima, then can I say the function is monotonous ?

Balarka Sen

A monotonous function is boring

Knight

Really?

So, my above conclusion is true?

Balarka Sen

Thorgott

I'm doubtful of that tbh

Knight

14:55

About Balaraka’s joke or my statement?

Thorgott

I feel there should be a counterexample, but it would be really ugly

Knight

@BalarkaSen If I were a rude man I would have flagged it by saying “you have called me a joke” :-)

Balarka Sen

Lmao

That's not what the gif is saying

Hilarious

Knight

@Thorgott So, what special property the function $f$ have ? (Given that it doesn’t have any local maxima and minima)

Balarka Sen

It's the joke_over_you.gif going over you

Knight

14:57

@BalarkaSen LOL

Thorgott

oh wait, the Dirichlet function works as counterexample

Knight

Balaraka did you study Loney’s Elements of Coordiante Geometey ?

Thorgott

wait nvm

urgh, this is ugly

$f(x)=x$ if $x$ rational and $f(x)=1-x$ if $x$ is irrational

Balarka Sen

Just consider a piecewise function which goes up, stays constant in $[-1, 1]$, dips down

Knight

Well, my problem is that it is written that “solutions to Laplace’s Eqaution results in a function that have no local maxima and minima”

Thorgott

15:00

that has a lot of local minima/maxima

we're not specifying them to be strict

Balarka Sen

Local max/min means strict local max/min to me. Knight needs to clarify

Knight

What is strict maxima/minima ? Maixima/minima are the places where derivative is zero

Balarka Sen

Lol, i give up

Thorgott

uhhhhhhhhhhhhhhhhhhhhhhhhh

no

that is not what a maximum/minimum is in any sensible way of thinking about it

however, the statement that a differentiable function R->R with non-zero derivative everywhere is monotonous is true

cause the derivative has to have constant sign by Darboux

no local minima/maxima + continuous also implies monotonous on connected domains

Alessandro Codenotti

15:17

@Balarka if $M$ is a compact smooth manifold, what does $C^\infty(M,\Bbb R)$ look like as a subspace of $C(M,\Bbb R)$?

Balarka Sen

~~Like you~~ Dense

Alessandro Codenotti

Right, by Weierstrass

Is it $G_\delta$ though?

I'm wondering whether Whitney embedding also gives a result similar to the dimension $\leq n$ embedding, meaning is there some result of the form "the set of smooth embeddings $M\to\Bbb R$ is dense/G_\delta/somewhat big in $C^\infty(M,\Bbb R)$"

Balarka Sen

Hm

Alessandro Codenotti

Nah I don't think it's $G_\delta$

I think $C^1$ already isn't. $C^1(M,\Bbb R)$ should be already $\prod^0_3$ in the Borel hierarchy as a subspace of $C(M,\Bbb R)$

Yashas

15:33

Is there any compuationally efficient invertable mapping form {1, 1, 2, 1, 2, 3, 1, 2, 3, 4, .....} to {1, 2, 3, 4, 5, 6, ...}?

You can think of it as mapping a triangular matrix to it's the memory efficient form.

I need a fast way to compute the [i][j] index in the matrix from an element's linear memory location.

The matrix is stored as a[0][0],a[1][0], a[1][1], a[2][0], ... linearly. Assume it's a lower triangular matrix.

computational efficiency order: additions > multiplication >>> division = modulus

Mike Miller

You should really write in your first thing {(1,1), (2,1), (2,2), (3,1), (3,2), (3,3), (4,1), ...}, otherwise who knows how to distinguish between all the different things you call "1"? In this notation, your map is (m,n) -> (m-1)(m-2)/2 + n.

How large are your matrices?

To go from k to the corresponding (m,n), you first need to identify the largest triangular number $T_{m-1}$ less than $k$, at which poinst $n = k - T_{m-1}$

A pretty hackish approach (I know nothing about computational algebra) would be to store a list (m, (m-1)(m-2)/2) --- then you could compare k to the triangular numbers on this list, and you've already stored the information of what m corresponds to that triangular number

Balarka Sen

@AlessandroCodenotti Huh how do you prove this? I have no clue

Mike Miller

@AlessandroCodenotti If $N$ is a smooth manifold and $\dim N > 2\dim M$, then the space of smooth embeddings is dense in $C^\infty(M,N)$

Of course embeddings aren't dense in $C^\infty(M,\Bbb R)$ for $\dim M > 0$ --- there are no embeddings for dimension at least 2, and in dimension 1 you would be saying "maps with f' > 0 are dense in all maps", which is preposterous

Alessandro Codenotti

Yeah sorry, $\Bbb R$ was supposed to be $\Bbb R^{2n+1}$

Yashas

@MikeMiller They aren't matrices actually. I just rephrased my actual problem as a matrix storage problem thinking it would be easier to understand. I am writing a GPU program for performing non-max suppression. I have an array of boxes and I need to compute intersection area of every pair of boxes. Intersection area of two boxes is commutative. So I need to compute n(n - 1)/2 areas only. To do this I need to map linear thread ids to two box ids.

Alessandro Codenotti

15:46

@MikeMiller But this answers it, do you have a reference where I can read about this?

Mike Miller

I'm sorry to tell you these words are mostly meaningless to me. :) Does my comment help you in your situation at all? (I'm not even sure if it would be an efficient approach in the matrix case.)

@AlessandroCodenotti Hirsch, but I could also just explain without much difficulty

Alessandro Codenotti

If you want to and have the time do so I'll be very happy to listen

Mike Miller

Open dense, I should have said

Yashas

@MikeMiller I need a fast inverse computation. Need to go from (m - 1)(m - 2)/2 + n to (m, n). You can lay it out in any fashion: {(1, 2), (2, 2), (3, 1)...}. I need a one-to-one mapping from every pair of boxes to a linear index. I need a really fast way to compute (m, n) from the linear index.

Mike Miller

Got it, very clear. Thanks.

I assume "list off the triangular numbers ahead of time and check which is smaller than (m-1)(m-2)/2 + n" is an awful idea, then.

Balarka Sen

15:50

@Alessandro A modification of the argument here will prove it

I didn't realize that was your question, I got thrown off by $\Bbb R$

Alessandro Codenotti

@BalarkaSen Yeah that made no sense, sorry

Uh a long answer, let me read it

Mike Miller

No don't read it

@AlessandroCodenotti A smooth immersion is a map $f: M \to N$ so that $Df: M \to f^*T^*N$ has no zeroes; a smooth embedding is a smooth immersion so that $(f,f): M \times M \to N \times N$ misses the diagonal.

As soon as $\dim N \geq 2\dim M$, the first is equivalent to: "$Df$ is transverse to the zero section". As soon as $\dim N > 2\dim M$, the second is equivalent to: "$f \times f$ is transverse to the diagonal."

Yashas

Let me define it formally. S = {1, 2, 3, 4, ..., N} is an index set. I need a one-to-one function that maps every pair of indices in S to another index set Q = {1, 2, 3, ...., N(N - 1)/2}. That is, I need a function f : (m, n) -> k where m, n belongs to S and k belongs to Q. Ideally, m != n but it's ok if m = n (these happen relatively less often).

Alessandro Codenotti

@BalarkaSen let's do $X=C([0,1],\Bbb R)$. The trick is that a function $f\in X$ is $C^1$ iff for all rational $\varepsilon >0$ there are intervals $I_0,\ldots, I_{n-1}$ such that for all $j<n$ and for all $a,b,c,d\in I_j\cap\Bbb Q$, $|(f(a)-f(b))/(a-b)-((f(c)-f(d))/(c-d)|<\varepsilon$

Mike Miller

Transversality conditions are open and dense conditions by the usual arguments, so in these dimension regimes, the properties of being a smooth immersion or smooth embedding are open dense properties

Balarka Sen

15:53

@MikeMiller You need immersions are dense before this, but that can also be achieved

This is the summary of my argument in the answer

Ah you dealt with immersions being dense

Mike Miller

I am being a bit sloppy because $Df$ depends on $f$ and is not an arbitrary map

Balarka Sen

You need 1-jet prolongation of $f$, mapping to the jet space

Mike Miller

But the usual arguments still go through

Alessandro Codenotti

So if $A_{j,\varepsilon}$ is the set of $f\in X$ for which that inequality holds for all $a,b,c,d$ in $I_j$, it turns out that $A_{j,\varepsilon}$ is closed and $$C^1=\bigcap_{\varepsilon\in\Bbb Q^+}\bigcup_{n\in\Bbb N}\bigcup_{I_0,\ldots,I_{n-1}}\bigcap_{j<n}A_{j,\varepsilon}$$

Mike Miller

Gotcha

Wait no what's wrong with the outline I presented

Alessandro Codenotti

15:54

That was awful to type, it's an example in Kechris

Mike Miller

Oh

Alessandro Codenotti

This shows that $C^1([0,1])$ is $\prod^0_3$ in $C([0,1])$, I don't know how to show it's not already a simpler set

Mike Miller

honestly what's right with the outline I presented

Balarka Sen

I was referring to you being sloppy with $Df$. You need to move $Df$ around somewhere; I am saying you should move in in the jet space

It's essentially correct

Alessandro Codenotti

Hm I'm not very familiar with transversality, but I see the idea, thanks!

Thorgott

15:56

beautiful formula

Balarka Sen

@AlessandroCodenotti Ahhh I see this is great

That's a nice trick

I don't know how to show it's not somewhere below the Baire hierarchy either

Alessandro Codenotti

Those kind of arguments are often trying to replace the condition that defines your set with countably many uglier conditions

Balarka Sen

Yeah

I have used these tricks before in trying to investigate points where a function is not continuous, never farther up the Baire hierarchy though

This is nice

Thorgott

I've seen such things pop up in measure theory

Alessandro Codenotti

Right that's also a classic example, instead of saying "$f$ is continuous at $x$" you have the countably many conditions "$f$ has oscillation below $1/n$ for every $n$ at $x$"

To get continuity on a $G_\delta$

Mike Miller

16:01

Where in the B heirarchy do open dense sets lie

Alessandro Codenotti

they are the lowest level, $\Sigma^0_1$

$\Sigma^0_1$ is open, $\Pi^0_\xi$ is complement of sets in $\Sigma^0_\xi$, $\Sigma^0_\xi$ is sets $X$ which can be written as a countable union of sets $X_i\in\Pi^0_{\beta_i}$, where $\beta_i<\xi$ for all $i$.

Thorgott

you can also express convergence that way

from there it follows that the set of points where a sequence of measurable functions converges is measurable

Alessandro Codenotti

So for example $\Sigma^0_1$ is open sets, $\Pi^0_1$ is closed sets, $\Sigma^0_2$ is $F_\sigma$ sets, $\Pi^0_2$ is $G_\delta$ sets

Yashas

How do I write a cartesian product which doesn't have reflexive elements in set notation? I have an index set S. I need the set that consists of all pairs from S but not (a, a) elements.

Thorgott

it also appears in the proof of LLN

Alessandro Codenotti

16:05

Oh and $\xi$ goes up to $\omega_1$

In Polish spaces that is

@Thorgott LLN?

Balarka Sen

@Thorgott Yeah that's another B hierarchy but a different B - Borel instead of Baire

Thorgott

the set of tuples of the form $(a,a)$ is usually called the diagonal of that product and denoted $\Delta_S$, so what you want can be called $(S\times S)\setminus\Delta_S$ @Yashas

@Alessandro law of large numbers

Alessandro Codenotti

Oh, weird

@AlessandroCodenotti something similar should also work for $C^2$ right? Write more "nested incremental ratios" to approximate the second derivative

Mike Miller

@Yashas Well, what I was suggesting was a perfectly well-defined function, just not one given in terms of addition and multiplication etc. A perhaps faster approach than previous is that you know that $m$ is either $\lfloor \sqrt{2k} \rfloor$ plus one or plus two. Then you can check which it is by actually computing m(m-1)/2.

Thorgott

It's essentially just $\{\omega\colon\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^nX_i(\omega)=0\}=\bigcap_{m\ge1}\bigcup_{N\ge1}\bigcap_{n\ge N}\{\omega\colon\left\vert\frac{1}{n}\sum_{i=1}^nX_i(\omega)\right\vert\le\frac{1}{m}\}$

Balarka Sen

16:11

Yeah convergence in probability should be written explicitly like that

Thorgott

then you can complement and apply Borel-Cantelli to the inner two expressions

well, depending on how many finite moments you presuppose

the most general case requires more technicalities

Balarka Sen

SLLN is pretty terrible to prove

You go to like fourth moment, and then truncate shit

Thorgott

if you suppose finite fourth moment, it's not bad

Balarka Sen

I never understood the truncation trick I think

No I am not supposing anything

Thorgott

you should suppose finite first moment

cause that's an iff

Balarka Sen

16:14

Expectation is finite, otherwise the statement is nonsense :P But in general you truncate so that everything has finite fourth moment

See here, section 2

Thorgott

right

Balarka Sen

Never understood that

Horrible estimates

Thorgott

I've looked at the general case once, but I never fully grokked it

Yashas

@MikeMiller Turns out that is a difficult problem and people write papers on it :D

Balarka Sen

it was the definitive indication for me that i will never be able to do probability

Thorgott

16:18

probability is weird

Balarka Sen

I don't understand how people estimate man

How do you do that

Thorgott

reminds me, did you ever end up taking a look at the Janson-Suen inequality

Balarka Sen

Oh fuck yes

Thorgott

that was some real "how do people estimate"

Balarka Sen

it's like LLN for weakly dependent RVs, or am i misremembering?

Thorgott

16:21

it was like a version of Chernoff for dependent RVs

Balarka Sen

gotcha

Thorgott

with a bunch of super weird hypothesis that I'm forgetting

something about dependency graphs

Balarka Sen

Lmao

its coming back now

There was that weird shit wait

Lovasz local lemma

Thorgott

I did work through the proof fully, but, as you are witnessing now, I didn't retain much of it

Balarka Sen

WTF

Alessandro Codenotti

16:22

Thorgott why you gotta give him PTSD attacks like that

Balarka Sen

@Thorgott that happens to me, that's why i gave up on probability

Thorgott

it's in my nature to make you all suffer

I am now equipped to say something more about the categorical construction from yesterday

Alessandro Codenotti

Probability is a very efficient and effective way to reach that goal

So is category theory lol

I'm leaving. (I actually needed to leave anyway, the timing is just a fitting coincidence)

Balarka Sen

i remember pictures, i can't remember $\Bbb P(|S_n/n - \mu| \geq \varepsilon) \leq \Bbb E(|X|^2/n \Bbb{1}_{|X| \leq M})/(\epsilon/2)^2 + \Bbb E(M \Bbb 1_{|X| > M})$

Thorgott

haha see you

Balarka Sen

16:25

Cya @Alessandro

Thorgott

I wlll say something about colax-slice 2-categories some day

that bound looks familiar

I think you use this to prove Weierstrass approximation probabilistically or sth

Balarka Sen

Weierstrass approximation follows from LLN type things yeah

LLN for $f(X)$ where $X$ is binomial or something

Thorgott

yeah, that gives the Bernstein polynomial in expectation iirc

we had to work that out as homework once in my probability course, it wasn't too bad

Balarka Sen

yeah

something man

Research leading to these inequalities was motivated by a ground-breaking proof of B. Bollobás [a2], who, in order to estimate the chromatic number of a random graph, used martingales (cf. also Graph colouring; Graph, random; Martingale) to show that the probability of not containing a large clique is very small. Bollobás presented his proof at the opening day of the "Third Conference on Random Graphs (Poznań, July 1987)" . By the end of the meeting S. Janson found a proof of inequality (a1) based on Laplace transforms (cf. also Laplace transform), while T. Łuczak proved a related, less exp…

These nutcases

Thorgott

I think you need that trunctuation trick in proving CLT too

The CLT proof I learned was smart, but simultaneously horrible

Balarka Sen

16:29

CLT is actually fine

Thorgott

I should learn how to do it using Fourier transforms

Balarka Sen

It's just Fourier theory yeah

Mike Miller

@Yashas Are any of the algorithms comprehensible to someone without much knowledge of computational math? It's an interesting question

Thorgott

probabilistic graph theory is a really weird topic, but it's pretty cool in some ways

Yashas

17:00

@MikeMiller Not really. I looked at a bunch of papers. They suggest something similar to the sqrt method you suggested. Most of the paper deals with hacky solutions for fast sqrt computation and placing work items in blocks of threads. Barely any math. It's all computer science.

Mike Miller

Fair enough. Glad you were able to find something, though.

Yashas

This is one but it requires some rudimentary understanding of CUDA (or GPU programming in general) to make sense out of it.

Alessandro Codenotti

I'm confused @Balarka, surely $C^1([0,1])$ is completely metrizable, by the norm $\|f\|=\|f\|_\infty+\|f'\|_\infty$, right?

maths student

{(X,y)|X not equal to 0} is connected , open and simply connected or not ?

Thorgott

^draw a picture

robjohn

17:14

failed again :-(

Decapitated Soul

Hello

Arnold Fernández

Hi!

Decapitated Soul

How to say 4" x 9/2" x 8ft? Can anyone offer some insight?

Arnold Fernández

If x = [2;3>

Decapitated Soul

If I want to say that on phone, how would I say that?

Arnold Fernández

17:20

(x^3+4)/(2x-1)=?

Sorry, I don't speack English

Decapitated Soul

Ohh

No problem :)

Tobias Kildetoft

@DecapitatedSoul 4 inches by 4 and a half inches by 8 feet

Decapitated Soul

@TobiasKildetoft, Thank you so much. :)

Arnold Fernández

17:40

If x = [2;3>

(x^3+4)/(2x-1)=?

Knight

@WhyWhatWhereWhenHow Where are you brother? Long time no see :-)

Give us a come back

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$Mathematics$ Mathematics