Mathematics - 2016-01-11

Mike Miller

12:19 AM

@Balarka: Isn't it past your bedtime?

Oh, the last message was an hour and a half ago. Oops.

Michael

3:21 AM

what are the real-life applications of partial derivatives?

Clarinetist

@Michael Least-squares regression?

Michael

Wait really? I thought it was 3d only applications. I do my least-squares with matrices @Clarinetist

Quality

5:01 AM

Why is the degree of the zero polynomial conventionally undefined or called -infinity instead of just degree 0 since that is the degree of constant polynomials?

Stan Shunpike

6:05 AM

Hola

Mike Miller

6:54 AM

@Huy: Nice movie.

Huy

7:10 AM

I'm on the bus trying to stay awake

I need more than 6 hours of sleep

Mike Miller

why

Huy

Cuz I'm so tired

Have to teach statistics =_=

Mike Miller

no why arenyy awake oh ok

Huy

Also I need some serious breakfast

I'll order a pizza to my classroom

Mike Miller

7:30 AM

David bowie just died...

Stan Shunpike

8:00 AM

Holy crap wow

@MikeMiller :/

Mike Miller

I'm gonna try to sleep. This is messed up.

Balarka Sen

8:26 AM

@MikeMiller Yes, it was past my bedtime.

@Quality Given polynomial $p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots a_0$, $a_n \neq 0$, $p$ is said to have degree $n$. But then the degree $0$ polynomials are of the form $p(x) = a_0 x^0$ where $a_0 \neq 0$, and the $0$ polynomial cannot be of that form. So it's said to be undefined. Sometimes people denote it as $-\infty$ because $\lim_{k \to -\infty} x^k = 0$.

Note: $-\infty$ does not really mean as in extended reals here, although it's motivated from that. It's just a symbol for degree of the zero poly.

Evinda

10:40 AM

Hi!!! Could I ask you something?
We have $b=arc \cos \left( - \frac{1}{\alpha}\right)$ and $a=2 \pi -b$, $|\alpha|>1$. How do we deduce that $0< b < \pi$ ?

10:53 AM

@DanielFischer Do you maybe know?

Daniel Fischer

@Evinda It depends on the used branch of $\arccos$. Typically, if nothing else is stated, one uses the branch with values in $[0,\pi]$. That is necessarily used here, or you couldn't possibly have $0 < b < \pi$. Then you just have to see that $b$ is none of $0$ and $\pi$, i.e. that $-\frac{1}{\alpha}$ is none of $1$ and $-1$.

Evinda

With branch do you mean the range ? So we don't use the fact that $|-\frac{1}{\alpha}|<1$, right? @DanielFischer

@DanielFischer Also why can't it be in this case that a-b is a multiple of $2 \pi$ ?

Daniel Fischer

11:10 AM

@Evinda The branches of $\arccos$ - in the real setting - are determined by their respective ranges, so there picking a branch and picking a range are equivalent. But we do use that $\bigl\lvert -\frac{1}{\alpha}\bigr\rvert < 1$. If the modulus were larger than $1$, we couldn't apply $\arccos$ to it at all, and if the modulus were $= 1$, we'd have $b = 0$ or $b = \pi$.

@Evinda $a - b = (2\pi - b) - b = 2(\pi - b)$, and $0 < b < \pi$ implies $0 < \pi - b < \pi$.

Mary Star

Hello @robjohn !! Do you maybe have an idea about the following?

The hyperboloid of one sheet is
$$S=\{(x,y,z)\in \mathbb{R}^3 \mid x^2+y^2-z^2=1\}$$
For every $\theta$, the straight line
$$(x − z) \cos \theta = (1 − y) \sin \theta, \ \ (x + z) \sin \theta = (1 + y) \cos \theta $$
is contained in $S$, and every point of the hyperboloid lies on one of these lines.

How can we deduce that $S$ can be covered by a single surface patch, and hence is a surface?

Evinda

@DanielFischer I see... We have $(1+ e \cos a) \cos a=(1+ e \cos b) \cos b$ and $(1+ e \cos a) \sin a=(1+ e \cos b) \sin b$.

If $1+ e \cos a=0$ do we deduce that $1+ e \cos b=0$?

Or could it also hold that $\cos b=0$?

Also do you have an idea how we could show that $\gamma(t)=((1+e \cos t) \cos t, (1+e \cos t) \sin t )$ is a Jordan curve for $|e|>1$ ? @DanielFischer

Daniel Fischer

11:34 AM

@Evinda You have $X\cdot \cos b = 0$ and $X\cdot \sin b = 0$. Can you deduce $X = 0$ from that?

Evinda

No we deduce $X= \pm 1$, right? @DanielFischer

And can it be true that $1+ \alpha \cos a=-1- \alpha \cos b$ ? How can we check it? @DanielFischer

Daniel Fischer

@Evinda No. You can deduce $X = 0$ from that. Try to figure it out.

Evinda

@DanielFischer If $\cos b=0$ then $\sin b=1$ and so $X=1$. So for $b=\frac{k \pi}{2}$ we have X=0, right? @DanielFischer

Daniel Fischer

11:51 AM

@Evinda If $\cos b = 0$, then you don't necessarily have $\sin b = 1$, it could also be $\sin b = -1$. I have no idea why you wrote "and so $X = 1$", that doesn't follow at all.

Huy

hey @DanielFischer, just rereading one of your answers on MSE you sent me the other day about there being only one way to make a finite-dimensional vector space into a Hausdorff TVS. (here). you first argue that the identity map $(K^n, S_n) \to (K^n, \tau_V)$ is continuous, which is equivalent to saying that $S_n$ is finer than $\tau_V$.

@DanielFischer since the Hausdorff topology $\tau$ on $V$ was arbitrary, that means $S_n$ is finer than any topology making $K^n$ to a TVS obtained by a linear isomorphism from $(V, \tau)$. is it obvious every Hausdorff TVS must be obtained in such a way or why can you state that any TVS topology must be coarser than the standard one on $K^n$?

Daniel Fischer

@Huy By definition of a TVS, every linear map $K^n \to V$ is continuous. Try to show that using only the definitions of a TVS and the product topology with the universal properties of product topologies.

Huy

yes, that I know

hm

Huy

12:14 PM

@DanielFischer: what happens for infinite-dim VS? I'd guess there are several different topologies making the vector space to a TVS? given some continuous representation $G \to Aut(V)$, if I try to construct an inner product wrt. which the representation is unitary, what can I say about the induced topology now?

maybe for the second part assume $V$ to be Hilbert if that's more interesting

Balarka Sen

12:35 PM

"And the Raven, never flitting, still is Lie-ing, still is Lie-ing"

Huy

Balarka ......

aren't you like an algebra master

teach me Cartan subalgebra

Balarka Sen

I am going to make horrible jokes about Lie theory until you ask me that Heegaard decomposition question you never asked.

I am no master. I know a thing or two about algebra, but nothing about Cartan subalgebras.

Huy

then I will still be hearing Lie algebra jokes for years :P

Balarka Sen

Yep.

Huy

teach me some algebra @BalarkaSen

I forgot everything

Balarka Sen

12:40 PM

If you really want help (assuming I can give it to you), I think you need to be a bit more specific than that.

Huy

@BalarkaSen: solvability and its applications. I remember solvability of the Galois group being a big deal since you could relate it to solving a polynomial equation with radicals. in Lie theory, solvability seems to be a big deal too but it is to classify the Lie groups. so what kind of things can I "do" once I know a group is solvable?

Daniel Fischer

@Huy If you don't insist on TVSs being Hausdorff, the only space that has one unique TVS topology is $\{0\}$. If you write being Hausdorff into the definition, then indeed the non-uniqueness is only in the infinite-dimensional case. And then there are always several distinct TVS topologies. If you want an inner product with respect to which a representation is unitary, it is necessary that [the completion of] $V$ is "Hilbertable". So you would probably already start with a Hilbert space.

Note that the inner product you construct must induce the given topology on $V$, otherwise you are looking at a different representation of $G$, not the one you started with.

Huy

@DanielFischer: yes, I was thinking that too, I'd want to start with a Hilbert space already. just that I'm not misunderstanding you: as long as we require the TVS to be Hausdorff, there are different topologies to make an infdim-space into a TVS. However if I start with a Hilbert space, then any inner product making the representation unitary must induce the already given topology.

Balarka Sen

@Huy I pondered for a while, but actually I do not know of a good motivation for solvable groups. Personally I motivate them using the fact that they are formally dual to tower of field extension $K = K^0 \subset K^1 \subset K^2 \subset \cdots \subset K^n = L$ with each extension $K^{i+1}/K^i$ a Kummer extension (more generally, composition series is dual to a tower of extensions).

I haven't actually studied the "serious" group theory where composition series of groups and groups with special kind of composition series are studied. I have always thought solvability is just a certain class of groups you could look at to study the classification problem (Feit-Thompson does this). One intuition for why they would be easier to study might be that they are groups "built up" on abelian groups, and abelian groups have quite a concrete structure.

Huy

@BalarkaSen: yes, I think the building up thing is easier applicable to Lie theory

Balarka Sen

12:52 PM

Can you say in a few sentences how solvable groups come up in Lie theory?

Huy

@BalarkaSen: solvable Lie groups are easier than solvable groups, since you can relate them to upper-triangular matrices via Lie's theorem. I think the main idea for classification is that you look at the radical of some Lie group (the maximal solvable normal subgroup), and then the quotient G/rad is always semisimple, i.e. can be decomposed into simple Lie groups, which in turn are completely classifiable

@BalarkaSen: my problem is just that without Lie's theorem, I don't really know how to picture or think of solvable groups at all, and I don't think that's supposed to be so

Balarka Sen

Ah. Hm.

Oh, solvable Lie group means Lie groups with a solvable Lie algebra.

Huy

they are equivalent if $G$ is connected

Daniel Fischer

@Huy There are always different TVS topologies on infinite-dimensional spaces. If you require your TVSs to be Hausdorff, then finite-dimensional spaces carry a unique TVS topology. If you don't require them to be Hausdorff, then there are several distinct TVS topologies on every space of strictly positive dimension, finite or not.

Balarka Sen

But, see, the commutator eventually vanishes! If you have an upper triangular matrix $A$, $A^n$ vanishes for sufficiently large $n$!

Huy

12:57 PM

and we defined a solvable (Lie) group first

Balarka Sen

It's natural to expect there is a connection.

Daniel Fischer

Next thing: The topology on $V$ is part of the representation. A representation of $G$ on $(V,\tau)$ is a continuous homomorphism $\rho \colon G \to \operatorname{Aut} (V,\tau)$. If $(V,\tau)$ is Hilbertable, you may be able to make the representation unitary by changing the description of the topology.

But if you change the topology, you get either no representation at all [the $\rho(g)$ may not be all continuous with respect to the topology, i.e. $\rho(G) \not\subset \operatorname{Aut} (V,\hat{\tau})$, or the map $\rho \colon G \to \operatorname{Aut}(V,\hat{\tau})$ is not continuous], or a different representation (a representation on $(V,\hat{\tau})$ rather than on $(V,\tau)$).

Huy

@BalarkaSen: are you talking about nilpotency now?

Balarka Sen

@Huy Ah, alright.

Huy

ok, that's very clear now, thanks @DanielFischer

Balarka Sen

1:00 PM

@Huy I think so. What I said above is my immediate thought on hearing "solvable Lie group" and "upper triangular matrices".

But then I am not really sure if this is a sensible analogy. Maybe not.

@Huy: Sorry, don't think I have many helpful comments here. Maybe ask Mike.

I think Tobias could have given you good answer to all of this, but apparently he has left the chat.

Huy

1:19 PM

a women just got arrested in front of the window I'm sitting at

after she was arguing for about 30 minutes with the police

wondering what that was about

Balarka Sen

it was about getting arrested for arguing with the police for about 30 minutes, of course

Huy

no, they were just arguing

then they let her go

just a few minutes after they arrested her

not sure what she did

Balarka Sen

she argued for getting arrested. :P you're not getting the joke, i'm being fractal.

Huy

-.-

second police car arriving

I'm getting scared

ö.ö

Balarka Sen

i think they're supposed to arrest you.

Huy

1:24 PM

._______.'

Ted Shifrin

Oh oh @Huy

Balarka Sen

hi @TedShifrin

Ted Shifrin

greetings @DanielF, @Balarka

Huy

hi @Ted

Balarka Sen

@TedShifrin The isoperimetric inequality is cool.

Daniel Fischer

1:28 PM

@TedShifrin Morning. You're in the east now, or up terribly early?

Ted Shifrin

Back in GA this week, DanielF

Daniel Fischer

Visiting old friends?

Ted Shifrin

You're wandering again, @Balarka?

Yup, Daniel, and a cancer checkup.

Balarka Sen

@TedShifrin No, I'm just watching your lectures. You talked about it in the Fourier series lecture.

Ted Shifrin

Oh yeah, I did.

Balarka Sen

1:30 PM

The proof you gave in the lecture (which indeed was pretty smart) works for $C^1$ curves. Mike was asking how to prove this for $C^0$.

Ted Shifrin

Different proof in the geometry notes.

I don't care :)

Balarka Sen

:P

@TedShifrin Ah, let me have a look.

Ted Shifrin

Presumably it follows by approximating $C^0$ by smooth.

I'm not sure his question makes sense. Continuity won't guarantee finite length.

Balarka Sen

Right. Koch curve.

Ted Shifrin

Or $x \sin(1/x)$.

Balarka Sen

1:34 PM

ah, right.

Mary Star

Hello @TedShifrin !! Do you maybe have an idea about the following?

The hyperboloid of one sheet is
$$S=\{(x,y,z)\in \mathbb{R}^3 \mid x^2+y^2-z^2=1\}$$
For every $\theta$, the straight line
$$(x − z) \cos \theta = (1 − y) \sin \theta, \ \ (x + z) \sin \theta = (1 + y) \cos \theta $$
is contained in $S$, and every point of the hyperboloid lies on one of these lines.

How can we deduce that $S$ can be covered by a single surface patch, and hence is a surface?

Balarka Sen

By the way, here's a combinatorial version of the isoperimetric ineq (prof told me about this yesterday). $G$ be a group, $X_G$ be it's presentation complex or whatver it's called, i.e., $\pi_1(X_G) = G$. Define loops in $X_G$ to be loops in the 1-skeleton traversing along the circles (1-cells) in the wedge. Set length of each 1-cells = 1 and area of each 2-cell = 1.

Then given a loop of length $\ell$, what's the maximal area of the combinatorial disk it bounds? The function setting to each $n$ the maximal area a loop of length $n$ bounds is called the isoperimetric function. Gromov proved if this function is linear then $G$ is hyperbolic. Not sure if you care though.

Ted Shifrin

@MaryStar: Review parametrization of ruled surfaces. But I disagree with one patch, I think.

balarka, I know none of this geometric group stuff, sadly.

Balarka Sen

@TedShifrin I think he just means whether among all the $C^0$ curves of length $L$ the circle bounds the greatest area.

Ted Shifrin

Probably that follows from Steiner symmetrization, I bet.

Balarka Sen

1:38 PM

@TedShifrin Me neither :)

googles Steiner symmetrization

Szabolcs

What is the usual English name of the half-line described by the equation $x=y, x>0$ in the 2D plane? In my language we call it something that could be translated as "first angular bisector". Is there a standard English term?

Ted Shifrin

Nope

Huy

@Szabolcs: you'd probably call it something like the ray bisecting the first quadrant, but there's no specific common expression

Hungarian?

Szabolcs

thanks

yes

Huy

szia

Szabolcs

1:48 PM

chào anh

Huy

igen

Mary Star

@TedShifrin Could we maybe use the fact that every point of the surface lies on one of the lines and that every line is contained in the surface? Or isn't this related to that?

Huy

how do I remember which is the derived and which is the central series? ._.'

Ted Shifrin

@MaryStar: Of course the parametrization of a ruled surface uses the lines.

Mary Star

Using the fact that every point of the surface lies on one of the lines, can we not conclude that the whole surface is covered by these lines, so by a single surface patch? @TedShifrin

Ted Shifrin

2:00 PM

I do not see how to get a single patch.

Mary Star

Can we not take as the single patch the set of these lines? @TedShifrin

Ted Shifrin

Oh, I guess I do. Start with one hyperbola and use the lines through it. Not obvious this covers all points. You need to check

Huy

2:15 PM

@TedShifrin: do you know anything about the Cartan-Killing form?

Mary Star

So, every hyperbola should cover all the points of every line through it? @TedShifrin

user1323245

2:32 PM

Programming. Have cartesian system. Need to move along one axis one step at the time. Can easily be done by using a "direction coordinate (-1, 0)" and adding it to the current coordinate. Not sure that make sense though? Can instead add a vector (unit, basically the same thing, but maybe more semantically correct) to the current coordinate. Better? Make sense?

Huy

Incomplete sentences. Weird to read. Stop.

Balarka Sen

lol

user1323245

2:55 PM

Ah, sorry about that. New here, only saw short messages so thought to try to keep it short.

Now, if a coordinate in a cartesian plane represents a point, it doesn't seem to make sense to add one coordinate to another coordinate. Am I right about that?

Semiclassical

3:06 PM

morning chat

Semiclassical

3:21 PM

@user1323245: the way i'd put it is that, while you don't add points per se, you can identify a given point with the vector which points from the origin to it, and it's these vectors which can be added meaningfully.

but i'm not sure how important that distinction really is

@huy i feel like i'm hearing a telegram message

user1323245

@Semiclassical Right. I guess "adding coordinates" is ok depending on the context.

Semiclassical

right.

user1323245

I'll go down that path then since it'll make the code a lot easier to read and since it's ok. Thanks.

Huy

3:44 PM

@Semiclassical Telegram the original thing or Telegram the Whatsapp rival? :P

Semiclassical

given that i've never heard of the latter STOP i'll go with the former STOP

Evinda

6:54 PM

@DanielFischer How could we show that the minimum value of $x^2+y^2+z^2$ under the restriction $xyz=1$ is 3?

Daniel Fischer

7:04 PM

@Evinda Lagrange multipliers?

Evinda

@DanielFischer We set $g(\overline{x})=x^2+y^2+z^2, \phi(\overline{x})=1-xyz$.
We see that $g,f $ are differentiable.
$\nabla{g(\overline{x})}=2(x,y,z)$
$\nabla{\phi(\overline{x})}=(-yz,-xz,-xy)$

Thus if $\overline{x}$ minimizes $g$ over $\phi^{-1}(\{0\})$ it has to hold that $\nabla{g(\overline{x})}= \text{ multiple of } \nabla{\phi(\overline{x})} \Rightarrow 2(x,y,z)= \lambda (-yz,-xz,-xy), \lambda \in \mathbb{R}$.

Is it right so far? How can we use the above to determine that the minimum value is equal to 3?

Daniel Fischer

@Evinda Use that $xyz = 1$, then it's easier to see the critical point(s).

Evinda

7:26 PM

You mean that we find the partial derivatives of this equality? @DanielFischer

Daniel Fischer

Can you use $xyz = 1$ to simplify the right hand side?

Evinda

@DanielFischer So we can divide $(-yz,-xz,-xy)$ by $xyz$ and so we get that it is equal to $\left( -\frac{1}{x}, -\frac{1}{y}, -\frac{1}{z}\right)$ ?

Daniel Fischer

Good. Now take a look at what it means for both sides to be proportional.

Evinda

It has to hold that $(x,y,z)=C \left( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \right) \Rightarrow x^2=C, y^2=C, z^2=C$, right? @DanielFischer

Daniel Fischer

@Evinda Yes.

Evinda

7:41 PM

So $x^2+y^2+z^2=3C$. And how can we deduce now that the minimum value is 3? @DanielFischer

Daniel Fischer

@Evinda That's not the relevant thing. The relevant part is $x^2 = y^2 = z^2$.

Evinda

@DanielFischer How do we use this?

Balarka Sen

If $x^2 = y^2 = z^2 = C$, what is $xyz$?

Daniel Fischer

@Evinda If $x^2 = y^2$, what does that tell you about the relation between $x$ and $y$?

Evinda

$x= \pm y$ @DanielFischer

@DanielFischer So we have $ y= \pm x, = \pm x$. Then $xyz=1 \Rightarrow x( \pm x) ( \pm x)=1 \Rightarrow x^3=1 \Rightarrow x=1$. Right?

Daniel Fischer

7:53 PM

@Evinda Right. So up to signs, $x,y,z$ are equal at the critical points. Using $xyz = 1$, what are the possible values?

Evinda

@DanielFischer You mean of $x^2+y^2+z^2$ ? Isn't it always equal to 3?

Daniel Fischer

@Evinda Not quite, we could have any two of the coordinates $= -1$, and the thord $+1$, or all three $+1$. In either case, we have $x^2 + y^2 + z^2 = 1+1+1 =3$.

@Evinda So "Yup" to this. Now, you need to see that this is a minimum and not a maximum.

Evinda

From $x^2=y^2=z^2$ we get $y= \pm x, z= \pm x$.
$x^2+y^2+z^2=3x^2$
$xyz=1 \Rightarrow x( \pm x) (\pm x)=x^3=1 \Rightarrow x=1$, or am I wrong?
So $x^2+y^2+z^2$ under the restriction xyz=1 is always equal to 3, so 3 is both minimum and maximum. Or not? @DanielFischer

Balarka Sen

$y$ and $z$ need not have the same signs. $y = x, z = -x$ could might as well happen. In which case $xyz = x \cdot x \cdot (-x) = -x^3$.

Daniel Fischer

@Evinda $x$ could be $-1$, e.g. $(-1,1,-1)$ is a point on $xyz = 1$ at which the minimum is attained.

@Evinda What makes you believe that $x^2+y^2+z^2$ is always equal to $3$ under the restriction $xyz = 1$?

Look at $(2,1,\frac{1}{2})$.

Evinda

8:04 PM

@DanielFischer @BalarkaSen Ah I see...

@DanielFischer But then how can we deduce that $3$ is the minimum values of $3x^2$ ?

Daniel Fischer

@Evinda It's not the minimum value of $3x^2$. It's the minimum value of $x^2 + y^2 + z^2$ under the restriction $xyz = 1$.

Balarka Sen

$3$ is not the minimum value of $3x^2$. It's minimum is $0$, at $x = 0$.

Evinda

@BalarkaSen We have the restriction xyz=1

Balarka Sen

On $x^2 + y^2 + z^2$. $3x^2$ is not relevant here.

Evinda

We have deduced that $x^2=y^2=z^2$ @BalarkaSen

Daniel Fischer

8:08 PM

@Evinda For what situation have we got $x^2 = y^2 = z^2$?

Evinda

For a vector $\overline{x}$ that minimizes $x^2+y^2+z^2$ over $\phi^{-1}(\{0\})$ it has to hold that $x^2+y^2+^2$ @DanielFischer

But we want to deduce that $x^2$ must be equal to 1 in this case, right? @DanielFischer

Daniel Fischer

@Evinda More generally, for a critical point of $x^2+y^2+z^2$ on the surface.

Evinda

@DanielFischer A ok

Daniel Fischer

But a critical point could also be a local maximum, or a saddle point. We must rule that out.

Evinda

@DanielFischer How can we rule that out?

Daniel Fischer

8:28 PM

@Evinda The surface $\{(x,y,z) \in \mathbb{R}^3 : xyz = 1\}$ consists of four pieces, one with all coordinates positive, and three with one coordinate positive and the remaining two coordinates negative. We have found four critical points, one on each piece. Since all pieces look the same, we need only look at one piece, say the one with all coordinates positive.

Now sketch, or make a mental image of, that piece. You will see that the surface is unbounded, so the critical point cannot be a maximum. You can easily give a formal proof, for every $t > 0$ the point $(t,1,1/t)$ lies on the surface, and we have $x^2 + y^2 + z^2 = t^2 + 1 + t^{-2} > t^2$ there.

On the other hand, the part $\{(x,y,z) : xyz = 1,\, x^2+y^2+z^2 \leqslant 4\}$ of the surface is compact, so $x^2 + y^2 + z^2$ attains a minimum on that part of the surface. Since we have found a point with $x^2 + y^2 + z^2 < 4$, that is a global minimum. Since a global minimum is a critical point, and we only have one critical point (on each piece), the critical point must be the point where the minimum is attained.

Balarka Sen

^ That is exactly what I had in mind, but was not trying to comment since I didn't know what Daniel was thinking. An alternative way to do it is the second derivative test, but that's big machinery.

Well, I was thinking more in terms of $f(x, y) = x^2 + y^2 + 1/(x^2y^2)$, so didn't think of the geometry.

Evinda

Yes, I have also thought using the second derivative test. $\nabla^2 (x^2+y^2+z^2)=(2,2,2)>0$ so the critical point will be minimum.
Is this also right? @DanielFischer @BalarkaSen

Balarka Sen

I have no idea what that means. Second derivative test is about the Hessian, not Laplacian.

Evinda

Oh yes, right...

Balarka Sen

Have a look at what @DanielF said above. His technique is simple and useful.

Evinda

8:53 PM

@DanielFischer I got it now... Thank you!!!

Vrouvrou

hello, i have A and B two compact sets, i have to prove that $A\cup B$ i compact, then i let $A\cup B=\bigcup_{i\in I} U_i$, where $U_i$ are open , then As $A$ is compact $A\subset \bigcup_{k=1}^N U_{i_k}$ and $B\subset \bigcup_{l=1}^M U_{i_l}$

T hen i can say that $A\cup B\subset \bigcup_{j=1}^{\max{N,M}}U_{i_j}$ ?

Huy

@Vrouvrou: no

write it down cleaner. I think your idea isn't wrong in principle but you're trying to be too fast

Balarka Sen

You don't need to care about max N, M etc. That's too hard. Idea in your first message is ok and leads to a simple finite subcover for $A \cup B$.

Also, your indices are messed up. Just index it by some $I$ and leave it at that.

Vrouvrou

$A\cup B= \bigcup U_i$ and $A\subset A\cup B$ and $B\subset A\cup B$ as A and B are compact we cn deduce a sub cover of A and B , that is $A\subset \bigcup_{k=1}^{N} U_{i_k}$ and B\subset \bigcup_{l=1}^{M} U_{i_l}$ What is the sub cover of A\cup B

I don't know how to write it correctly

Huy

then learn it

Vrouvrou

9:03 PM

?

$A\cup B\subset (\bigcup_{l=1}^{M} U_{i_l})\cup (\bigcup_{k=1}^{N} U_{i_k})$

Balarka Sen

That's right.

Vrouvrou

so can i say that $A\cup B=\bigcup_{j=1}^S U_{i_j}$ where $S\leq max{N,M}$?

Balarka Sen

You don't need to say any max N, M crap. Just think about the right hand side of what you wrote in the message above.

Vrouvrou

it is a finit sub cover

Balarka Sen

Yes. Why?

Vrouvrou

9:09 PM

finite union

Balarka Sen

Sure. Why is it a subcover?

Vrouvrou

because it is from U_i

no?

Balarka Sen

Yes, that is right.

So you just gave me a finite refinement of $\{U_i\}_{i \in I}$ which also covers $A \cup B$. You're done.

Vrouvrou

thank you @BalarkaSen

we don't need to have eqality between A\cup B and the finit sub cover ?

Balarka Sen

There has to be an equality. Figure out why.

Vrouvrou

9:16 PM

there is some j such that $\bigcup_{j=1}^S U_{i_j} \subset (\bigcup_{l=1}^{M} U_{i_l})\cup (\bigcup_{k=1}^{N} U_{i_k})$

Balarka Sen

Again, whatever you are thinking, it's too hard. You have just proved $A \cup B \subseteq \bigcup_{j \in J} U_j$ ($J$ a finite subset of $I$). $U_j$ are subsets of $A \cup B$. Is it possible that a set is contained in union of it's subsets but is not equal to the union?

Vrouvrou

the union "A\cup B"?

Balarka Sen

No, I mean, is it possible a set is contained in union of it's subsets but is not equal to it?

Vrouvrou

yes it is posible

Balarka Sen

How so?

Vrouvrou

9:25 PM

$\bigcup_{k=1}^N U_{i_k} \cup U_{i_{M}}\subset (\bigcup_{l=1}^{M} U_{i_l})\cup (\bigcup_{k=1}^{N} U_{i_k})$

Balarka Sen

$U_{i\ell}$ for $\ell < M$ are not subsets of $\bigcup_{k = 1}^N U_{ik} \cup U_{iM}$, so that's not a valid example.

Vrouvrou

so we can't have equality ?

Balarka Sen

We can't, but that's not a valid example. Things inside your union has to be subsets. Here they are not.

@Vrouvrou This is set theory. Note that union of subsets of a set is also a subset. And if a set is contained in it's subset, they must be equal.

In this case, you have $A \cup B \subseteq \bigcup_{j \in J} U_j$ where $J$ is a finite subset of $I$. But $\bigcup_{j \in J} U_j \subseteq A \cup B$. So $A \cup B = \bigcup_{j \in J} U_j$

L33ter

9:50 PM

@BalarkaSen

can you give me some geometric intuition of something

I am having troubles visualizing it

Balarka Sen

Maybe.

L33ter

so for the second one

Huy

haven't you studied alg top already?

L33ter

here is what my intuition tells me what we do with it

I did it for my project @Huy not official

Huy

ah ok

L33ter

9:52 PM

and that was from munkres this book is allen hatcher

Balarka Sen

I don't know what's unclear there.

L33ter

I understand all of that stuff like the technicalities, but I want to get more geometric intuition @Huy.

Huy

that's odd, usually it's the other way around @L33ter

L33ter

I can't picture in my head how do we deform the second pic into $B_0$

Balarka Sen

@L33ter You also skipped a lot of exercises and theorems from Munkres.

Huy

9:53 PM

Karim isn't Rememberme though, right? :P

L33ter

yeah I did I only know point set topology until regularity

Balarka Sen

@L33ter Trace out the path in the first picture by your hand.

Now open up the place where it's self intersecting by moving it away from $x_0$.

@L33ter No, I mean you skipped stuff in part II.

L33ter

yeah I did

oh i see it now I see it in another way 2

is that is we pull the string away to the right

we will also get the same shape right ?

Balarka Sen

That's what I said above.

But yes.

L33ter

I see

@BalarkaSen I am very happy this semester it is very chill I can focus my time to understand algebraic topology

and analysis which I am taking this semester

Balarka Sen

10:04 PM

Good to hear.

Danu

Hullo

Balarka Sen

Hi @Danu.

Danu

You always want to hear what my topology lecture is like, right?

Balarka Sen

Uh, sure.

Danu

Today, we finished SVK I think.

Balarka Sen

10:07 PM

Cool.

Danu

After some discussion in the h bar, I found out that we constructed the pushout

(we called it the "generalized amalgamated product")

Balarka Sen

Yep. SVK says $\pi_1$ preserves pushouts.

Danu

@BalarkaSen preserves?

Balarka Sen

$\pi_1$ is a functor from the category of based topological spaces to the category of groups. SVK is equivalent to saying it sends pushout of based topological spaces to pushouts of groups.

Danu

@BalarkaSen Oh, I had no idea that the total space is also the pushout of the subsets; I guess it makes sense (but I should just unwind the definitions in my head for a sec)

Balarka Sen

10:10 PM

But don't worry about it too much. This perspective might be concise, but hasn't been more useful to me than the usual formulation.

Total space is pushout of the subsets in the open cover, yes.

Danu

Yeah it makes a lot of sense

@BalarkaSen Very nice; I like the categorical point of view, though of course I don't know anything about it.

Balarka Sen

Glad to hear. I have never been able to grok categorical things much.

Danu

What is there not to like about generalizing alllll the way?!

Balarka Sen

Can you elaborate on what you mean?

Danu

If you can make your statement meaningful in a broader way, then why not go for it?

Balarka Sen

10:18 PM

Generalizing by statement doesn't guarantee that it'd help me use my statement for what I want to do, or prove my statement, or do something specific.

Danu

No, but I see a lot of value in generalizing for the sake of it.

In my (historically-inspired) view, this usually turns out to be the way forwards.

How many fields have died off for "getting too general"?

Balarka Sen

I am not saying there's no value. I don't like to think about abstract generalization mostly because I do not know much math and like to stick to pictures :)

Danu

@BalarkaSen You really need to stop saying that; Ted called you out on it a while back and he was right to do so.

Balarka Sen

Oh, what did I say?

Danu

You often say you "don't know much math"

Even if you do think that, don't say it all the time.

Balarka Sen

10:23 PM

I brought that up to give an answer to your question on why I don't like abstract generalizations. :)

@Danu Unrelated, but here is a fact (corollary of SVK) that you might like. If $G$ is a group, then there is a CW complex $X$ such that $\pi_1(X) \cong G$. Not sure if you already know it.

Danu

@BalarkaSen I know you "mean no harm".

@BalarkaSen We didn't introduce CW complices (complexes?)

@BalarkaSen But it sounds very very nice!

Balarka Sen

Ok, replace "CW complex" by "topological space". And yes, complexes.

CW complexes are just a nice kind of topological space, the ones you could use to do homotopy theory.

Danu

I think the proper Latin would be complices, but oh well :P

Balarka Sen

"Complex" is a latin word?

Here's the idea. $G$ be an arbitrary group, and choose a presentation. Consider wedge of $|G|$-many circles, with each circle in the wedge labelled by a generator of $G$. Fundamental group of this wedge is the free group on $|G|$ generators. But $G$ also has "relators" $r_i = 1$ in the presentation. So now you have to "glue disks" to the circles "according to $r_i$" so as to make the loops corresponding to $r_i$ nullhomotopic.

Huy

@BalarkaSen: yes, comes from complexus

Balarka Sen

10:31 PM

The resulting quotient space is the desired topological space with $\pi_1 \cong G$.

@Huy Ah, didn't know that.

Huy

the plural would be complexus though

usus u-declination

Danu

@BalarkaSen I'm not sure how exactly the gluing disks implements relations, but once that is clear I guess it's not too bad ;)

Balarka Sen

Right, take an example. $\Bbb Z^2$, say. Consider wedge of two circles, labelled by $a, b$ respectively. Glue a square (disk) by gluing upper edge to $a$, right edge to $b$, lower edge to $a^{-1}$ (i.e., $a$ in the opposite way) and the last to $b^{-1}$. You have glued by $aba^{-1}b^{-1}$. The resulting thing is a torus, which has fundamental group indeed $\Bbb Z^2$.

The deal is, the boundary of the square is topologically a circle, and the gluing gives you a map from that circle to wedge of two spheres, i.e., a loop $S^1 \to W$, image of which corresponds precisely to $aba^{-1}b^{-1}$. But then the interior of the glued square is a disk bounding this loop, so you nullhomotope it through the interior.

Hence, $aba^{-1}b^{-1} = 1$ in the fundamental group. I.e., $ab = ba$. $\langle a, b | ab = ba \rangle \cong \Bbb Z^2$.

L33ter

hm if we have a loop that goes around circle A

and it intersect A how come can we unlink it from A ?

Danu

@BalarkaSen You mean integers mod 2, or integer lattice in 2D?

Balarka Sen

10:42 PM

@L33ter That's quite vague.

Danu

lattice right

yeah, nvm :D

Balarka Sen

@Danu Integer lattice in $\Bbb R^2$, yes. Pairs of integers, with addition pointwise, otherwise said.

Danu

How do I glue the parts of the boundary?

Along identity?

L33ter

suppose we have bunch of loop that intersects a circle A at a point A, then the book mentions we can unlink it @BalarkaSen if it intersects circle A at a point

Balarka Sen

@Danu Yes, along identity.

@L33ter I can't make any sense of that, and probably you're misreading. Unlink in this sense means nullhomotope in complement of $A$, so loops intersecting $A$ makes $0$ sense.

Nevermind, I saw the paragraph. Hatcher is being intuitive and nonrigorous. You can "unlink" (whatever that means) it from $A$ by pulling it away.

L33ter

10:48 PM

yeah I was picturing that in my head

Balarka Sen

Bad statement, to be honest. In fundamental group, you look at loops away from $A$ in the first place.

L33ter

what do you mean away from $A$?

Balarka Sen

In the complement of $A$. In $\Bbb R^3 - A$.

L33ter

I see

Balarka Sen

@Danu It must be hard to see how to glue the square to the wedge of circles in the way I said. But it's really the usual way to make a torus out of a square (ref). If you give your square a clockwise orientation, then the corresponding orientation of the wedge of 2 circles in the torus becomes $aba^{-1}b^{-1}$.

Danu

10:53 PM

@BalarkaSen I know how to make a torus out of a square; that's fine.

Balarka Sen

I know you do, but I was posting just in case you couldn't figure out how I was gluing.

If you can, ignore my message above.

Anyway, so, I just mentioned that every group appears as fundamental group of some space (of some 2-dimensional CW complex, even). It is true in particular that every finitely presented group appears as fundamental group of a 4-manifold.

This coupled with the undecidability of group isomorphism problem shows that classification of 4-manifolds upto homeomorphism is undecidable.

This is one side of the story for why dimension $4$ is harder than the dimensions below.

L33ter

hm

becko

11:44 PM

What's the best SE forum to ask information theory questions?

Hippalectryon

@becko Depends on what the question is. If it's heavily math oriented, then MSE (or MO but It's rarely necessary)

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$Mathematics$ Mathematics