Mathematics - 2017-11-29 (page 1 of 6)

1:47 AM

Hey everyone, I have a question about personal statements for Ph.D. applications. I'm getting a tension between two different kinds of letters -- one that embodies the 'personal' aspect (giving personal motivation for applying, why field X is interesting, etc.) as opposed to one that embodies the research aspect (describing past research projects, citing papers, etc.).

I'm torn between the two, and I get mixed signals depending on who I talk to. Should I just write two and guess at what school would prefer what letter, or just stick to one and hope for the best, or something else entirely?

2:11 AM

how do i expand

a function to make it continuous

say 1/x

make that function continuous at 0

say give it 2 branches

the constant ill give f(0)=c

@SantanaAfton If you are talking about statements of purpose, they should be direct and describe your mathematical interests and experience.

but satisfy the ε,δ limit

There are a couple schools that ask for "personal statements" as well as "statements of purpose" and if that is the case I don't think the former are read.

If a school only asks for a "personal statement" you should write a statement of purpose.

And that statement of purpose is the research-oriented letter, you're saying.

yeah

2:15 AM

i need to find a function that is not bounded from [0,1/2) can you help me?

In mathematics I think its typical to describe things you have learned preparing to focus on research as well as your actual research.

I didn't have any actual research when I applied.

Like if you did a reading with a professor in some advanced graduate topic or whatever.

How deep in the weeds should I get in my discussion of any projects I've done?

not very

I think the quality sops are probably less than a page.

There's some good examples online

I have no clue where mine is.

It should NOT be an essay.

actually googling samples now I'm having trouble finding a decent one.

can some1 describe me a function that blows up on a constant ? continuous?

like 1/x

but on a constant like 1/2

Yeah, I'm definitely keeping mine to a page maximum. Not sure why most of the examples online aren't stellar.

2:26 AM

Your statement should include:

Exactly what you want to study, and why you want to study it.

Why you are an awesome choice. This means specific examples of stuff related to math that you've done. If you don't have research, list courses or your GPA. List stuff you've read. Don't list some lady you talked to one time in high school.

Some very brief conclusion that explains how you will save the world once you have your PhD.

Furthermore, it should include only these things. Every fact about your background that you include should be in the interests of 1 or 2. If they are not, you should …

I found this advice

which seems correct.

I remember being able to find decent samples when I went through this 5 years ago.

Mm, thank you! That is good advice.

Probably don't list your gpa

but talking about higher level courses you are interested in is a good idea.

Okay, that's good to know. I've been struggling with how exactly to present my interests and why my research implies I'm a quality choice. Seems like talking about what classes I want to take and why is a good specific point to include toward that end.

Thanks for your help!

You may want to ping @Ted since he's probably the only one here who's actually been on committees.

2:42 AM

hi chat

3:09 AM

Hallo

3:46 AM

yoooooo

How's it going?

4:11 AM

Feels weird man! Dunno why.... just a weird-feeliong night

What about yourself?

I'm doing alright. Unfortunately I may not be able to do algorithms next quarter so I need to figure something out just in case

Ya contingencies... always gotta have those

In fact that why IM in this difftop course

the backup plan

I'm considering doing difftop as well but I may not get permission, and it might be too much this quarter

Ive really enjoyed it, though its been verry challenging for me given my background

But Im sure most people arent in the same spot

I mean so one of the prereqs for this class is grad algebraic topology

Undergrad difftop was all well and good but grad might be pushing it

4:34 AM

Oh I see. The structure is very different here. The prereqs were all undergrad, analysis, linear algebra

So the way things work here is that you have all the undergrad classes

Then first year grad students have to take 3 classes: Geotop (AT, DT, DG), Analysis (Real, Functional, Complex), and Algebra (Rep theory, Comm/AG, "topics in algebra")

Fawad

⁣

Difftop is the second quarter in the grad sequence, so students there typically already did AT

GFauxPas

4:54 AM

good eeveening gentlemen

5:05 AM

yo

@BalarkaSen got something to bounce off of you when you have a chance

I have a question about compactness in Rudin

Thm: Suppose $K \subset Y \subset X$. Then $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$.

One direction is fine. I have problems with the other direction:
Suppose $K$ is compact relative to $Y$. Let $\{G_{\alpha}\}$ be a collection of open subsets of $X$ which covers $K$, and put $V_{\alpha} = Y \bigcap G_{\alpha}$. Then $K \subset V_{\alpha_{1}} \bigcup ... \bigcup V_{\alpha_{n}}$ for some $n$. And since $V_{\alpha} \subset G_{\alpha}$, we have $K \subset G_{\alpha_{1}} \cup ... \cup G_{\alpha_{n}}$.

Suppose $K$ is compact relative to $Y$. Let $\{G_{\alpha}\}$ be a collection of open subsets of $X$ which covers $K$, and put $V_{\alpha} = Y \cap G_{\alpha}$. Then $K \subset V_{\alpha_{1}} \cup ... \cup V_{\alpha_{n}}$ for some $n$. And since $V_{\alpha} \subset G_{\alpha}$, we have $K \subset G_{\alpha_{1}} \cup ... \cup G_{\alpha_{n}}$.

We know for every set open relative to Y, call it E, there exists a set G such that G is open relative to X and $E = G \cap Y$. But, this just shows there exists such a G, not that if I intersect an arbitrary G, I'll get an open set.

5:24 AM

@orbit-stabilizer so if $(G_i)$ covers $K$ in $X$, then there is an induced cover $(E_i)$ of $K$ in $Y$, of which there is a finite subcover $(E_i')$ of $K$ in $Y$, which is then transferred back to a finite subcover $(G_i')$ of $K$ in $X$

Right, I understand that part

so which direction is difficult for you?

It's that we can only conclude $K \subset V_{\alpha_{1}} \cup ... \cup V_{\alpha_{n}}$ for some $n$ if $V_{\alpha}$ is an open cover of $K$. How do we know that?

which direction is that?

The direction you are doing. Assuming K is compact relative to Y and showing it is compact relative to X.

5:27 AM

$K$ is the same set...

Suppose $K$ is compact relative to $Y$. Let $\{G_{\alpha}\}$ be a collection of open subsets of $X$ which covers $K$, and put $V_{\alpha} = Y \cap G_{\alpha}$. Then $K \subset V_{\alpha_{1}} \cup ... \cup V_{\alpha_{n}}$ for some $n$. And since $V_{\alpha} \subset G_{\alpha}$, we have $K \subset G_{\alpha_{1}} \cup ... \cup G_{\alpha_{n}}$.

yes

That's the proof. I'm stuck on where we set V_alpha = Y intersect G

how do we know that V_alpa is open

that's how you define open

Subspace topology

In topology and related areas of mathematics, a subspace of a topological space X is a subset S of X which is equipped with a topology induced from that of X called the subspace topology (or the relative topology, or the induced topology, or the trace topology). == Definition == Given a topological space ( X , τ ) {\displaystyle (X,\tau )} and a subset S {\displaystyle S} of X {\displaystyle X} , the subspace topology on ...

Oh right. It's an if and only if. Gawd.

5:29 AM

lol

Reading comprehension - I must work on it. Thank you :)

are you learning topology from wikipedia?

Oh no. This is from Rudin. I simply misread a previous theorem.

theorem?

Suppose $Y \subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some open subset $G$ of $X$.

5:32 AM

why is it a theorem?

what is the definition then?

$E$ is open relative to $Y$ if $\forall p \in E $, $\exists r > 0 $ such that if $q \in Y $ and $d(p,q) < r$, then $q \in E$.

oh, you should do point-set topology without metric :P

Gotta go by the book :P

Point-set in general can be a bit boring

@Daminark you get fun things like the line with two origins

5:38 AM

Hey, 5/6 stars on the right were first starred by me!

I feel so involved.

yesterday, by Balarka Sen

That's the only one I didn't first star :(

Wait, I don't think I did 5/6. I did 4/6.

5:59 AM

How important is compactness? I really should be learning chapter 5 (differentiation), but I'm back to chapter 2 (basic topology) since I never learned compactness

Its used a lot

But its not like theres some really deep thing going on

Every open cover has a finite subcover. Thats what Ive used consistently

I see, thanks

@orbit-stabilizer check out Liouville's theorem (for complex analysis)

every bounded entire function must be constant

and "compactness" is a criterion for bounded

6:12 AM

ah I see. thanks

another use of compactness if one-point compactification

which is used in manifolds a lot

CP1 is the OPC of C

(Riemann sphere)

S^3 is the OPC of R^3

Interesting...

Want to see a hard compactness question that came up on my midterm?

sure

We define a metric space (L_∞, d) as follows. An element of the set L_∞ is a sequence x = {x_n} (n ∈ N) of real numbers x_n with the property that sup{|x_n| : n ∈ N} < ∞.
The metric is defined by d(x, y) = sup{|x_n − y_n| : n ∈ N} (you are not required to
prove that this obeys the defining properties of a metric). Prove or disprove: (L_∞, d) is compact.

interesting

start from the zero sequence, denote it as 0

let $\tau_n = \{x \mid d(0,x) < n\}$

$\tau_n$ is open for each $n$ and contains more element than the previous $n$

so $(\tau_n)$ is a countable cover that does not admit finite subcover

@orbit-stabilizer right?

6:20 AM

Huh, at first glance, that seems like it should work.

This is the given solution:

Let e_k be the element of L∞ which has 1 in the kth position and 0 everywhere
else. Then d(ej, ek) = 1 for all j != k. Let E = {e1, e2, . . .}. We claim that this
infinite set E ⊂ L_∞ has no limit point, which implies that L_∞ cannot be compact.

Suppose x ∈ L_∞ were a limit point of E. Then there must be a sequence of elements
of E that converges to x, and in particular this sequence must be a Cauchy sequence.
However no sequence of points from E can be a Cauchy sequence, because all points
in E are separated by distance 1.

note that "closed and bounded implies compact" is not true in general

(in general "bounded" means nothing)

Are you talking about the given solution or Louivilles theorem

given solution

but it is true for metric spaces in general so you might not need to care

wait, what is the logic of your solution?

ah, the other direction

"compact implies closed and bounded"

Just closed I believe - don't need boundedness

43

Q: Compact sets are closed?

I feel really ignorant in asking this question but I am really just don't understand how a compact set can be considered closed. By definition of a compact set it means that given an open cover we can find a finite subcover the covers the topological space. I think the word "open cover" i...

6:26 AM

"It is true, however, that compact sets in Hausdorff spaces are closed, though a bit of work is required to establish the result."

Thanks

I still do not follow the logic

of course you can find an infinite set in a compact set that has no limit point

how does that show that the set is not compact

Compact implies closed in metric spaces

hell, consider $\{\frac 1 n \mid n \in \Bbb N_{>0} \} \cup \{0\}$

this set is compact

wait, this doesn't work

No...

@orbit-stabilizer yes it is compact

hmm, my thinking was wrong then. looks like every infinite set does have a limit point

interesting

can you prove it?

6:30 AM

Every infinite compact set has a limit point?

no

every infinite subset of a compact set has limit point

Isnt it one definition of compactness that every sequence has a convergent subsequence?

Hmm.

Sequential compactness?

Oh wow. It's a theorem in Rudin. Just noticed

IF E is an infinte subset of a compact set K, then E has a limit point in K

oh ya maybe that ends up being a different notion

what's the proof? @orbit-stabilizer

6:34 AM

Suppose there was no limit point of E. Then each $q \in K$ would have a neighbourhood $V_q$ which contains at most one point of $E$.

So no finite subcover of $\{V_q\}$ can cover E, and the same is true for $K$ since $E\subset K$, which contradicts the compactness of $K$.

subcollection*

fair enough

Reading it, it seems so obvious, but coming up with the proof. agh.

well you just need to unfold enough definitions until it becomes trivial

Abhishekstudent

hi guyz

hi orbit

hello

Abhishekstudent

6:42 AM

how r u doing?

bad. got my analysis/algebra finals next week and im really behind

Abhishekstudent

oh

good luck

thanks

Annelise t.

how to multiply (1,-6) and a 2x2 matrix?

7:33 AM

Yo @Alessandro and @Eric!

Eric Silva

7:52 AM

Sup

Not too much, how about you?

8:09 AM

Hi @Dami

How's it going?

8:23 AM

Quite well, I've got a lot of uni work to do as usual but that's fine

Lmao, yeah

What about you?

I just finished writing up my review sheet in biology so I'm gonna do that test, turn in a complex pset, and then it'll be finals

And also gotta figure out what to do next quarter since algorithms might be full

What are the options available?

In the math department, there's ODEs and point-set topology, which I definitely have the prereqs for

Then there's grad differential topology which is gonna be... a push

I originally wanted to take commutative algebra but (for probably good reason) the department head said no

8:42 AM

Ah, that's a pity

It's weird that you can do point set now after AT

Lol, my AT wasn't a standard course, we jumped around a lot and I'm still gonna need to learn the standard material

That's even weirder then :P

True

But yeah it was like, simplicial sets/finite spaces for the first part, which was all good

Are there cool options outside the math department you can take?

8:58 AM

And then we just kinda hopped around, talking about (co)fibrations, cohomology, spectral sequences, some stuff about characteristic classes that I still don't understand, last class we did Poincare duality

And I guess I could look into compsci, or maybe something completely different

Linguistics, philosophy, econ, etc. I haven't thought about that much so far

BAYMAX

9:58 AM

.

Any help on Conjugate Gradient Method?

Maneesh Narayanan

10:12 AM

Can I use polar-coordinate to prove the continuity of functions?I used this for continuity at origin $f(x,y)=\frac{xy}{|x|}$$(x,y) \ne(0, 0)$ and$ f(x,y)=0, (x,y)=(0,0)$.

I got the function is continuous. Am I correct?

Mary Star

Iwant to calculate an integral over $\Sigma$, where $\Sigma$ is the boundary of the bounded space $D$ that is defined by $0\leq z\leq 1-x^2-y^2$ ans has such an orientation that the perpendicular vectors have direction to the outside of the space $D$.

Do we use here cylindrical coordinates?

10:54 AM

@Semiclassical Right, I see. That's a very interesting question.

I feel like the answer should be definitely a no.

@LeakyNun Not quite. One-point compactification of most manifolds is not a manifold.

You need the end space to be a sphere for that to happen, I believe

@Semiclassical Shoot!

Pfff all manifolds are compact

11:09 AM

Non-compact manifolds are useful and beautiful

jublikon

11:36 AM

hey all :) anyone here that could help me with a small singular value decomposition question ?

from here:

how does he make the matrix "unit length"?

so, how does he come from $\begin{bmatrix}- \sqrt{10}&2 \sqrt{10}\\ \sqrt{10}& \2 \sqrt{10}\end{bmatrix}$ to $\begin{bmatrix}\frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}$ ?

12:49 PM

@BalarkaSen not sure what this is supposed to mean

A sphere is defined by a specific degree 2 inequality in x,y,z. If I take the intersection of two spheres, I don’t in general get another sphere

@Semiclassical Oh that was a reply to Leaky

Ohhh

You can see why I thought it was in reply to mine, though

Yeah it was sandwiched between two of my pings to you

Sorry about that

Np

Here’s the line of thinking I had

abenthy

How do I show that a semisimple Lie group $G$ is perfect, i.e. $G = [G,G]$? Semisimple means that $G$ is connected and has no connected abelian normal subgroup other than $\{e\}$.

My idea was to use that $G/[G,G]$ is semisimple too (since $G$ is) and its abelian, so it must be trivial and it follows that $G=[G,G]$, but this seems strange to me because I don't know if the trivial group is to be considered semisimple.

1:01 PM

my set of interest is $K=\{(x,y.z)\in[0,1]^3:f(x,y,z)>0\}$ where $f(x,y,z)=4xyz-(x+y+z-1)^2$

I want to know whether K can be represented by an intersection of polynomial inequalities of degree at most 2

Hrm, I’m seeing a hole in the logic I wanted to apply. So I’ll withdraw it for now

(The above set is not compact unless I change the > to >=, and they pokes a hole in how I wanted to use the Positivstellensatz)

I still think one can do differential geometric arguments, subtler than the ones I made for cylinders, for this

Like if you could do this $f(x, y, z) = 0$ would have to be a locally quadratic surface

@Semiclassical I don't think you need semialgebraic geometry for this. If you wrote $K$ as $K_1 \cap \cdots \cap K_n$ where each $K_i$ is the region given by $f_i(x, y, z) > 0$ where $f_i$ is a quadratic polynomial, then $\partial K$, which is given by $f(x, y, z) = 0$, must be a subset of $\partial K_1 \cup \cdots \cup \partial K_n$, which is just the union of the algebraic varieties cut off by $f_i = 0$, $i = 1, \cdots, n$, right?

Because $\partial K = \partial (K_1 \cap \cdots \cap K_n) \subseteq \partial K_1 \cup \cdots \cup \partial K_n$

So that would mean at least one of $f_1, \cdots, f_n$ vanishes at any point on $\partial K : f = 0$. i.e., $f_1 \cdots f_n$ always vanishes on $f$.

That is to say, $Z(f) \subseteq Z(f_1 \cdots f_n)$. Now we can apply the algebraic geometric Nullstellensatz, plain and simple

1:30 PM

Neat

I guess that says, what, $(f_1 \cdots f_n)^r$ is a multiple of $f$ for some $r$. That would be impossible, I believe

A real multiple?

Well, no, polynomial multiple. I guess it takes some argument to see why that should not be possible.

$f_1 \cdots f_n$ has degree $2n$, right? So you'd want to have $2nr = 3k$ for some $k$, because $f$ has degree $3$ and $k$ is degree of the multiple.

Um

Hmm

Is $f$ irreducible?

1:33 PM

you could take some of the factors to be linear, though

Dunno

My guess is yes?

If so, you could say, if $f$ divides $(f_1 \cdots f_n)^r$, $f$ divides $f_1 \cdots f_n$, contradiction because $3$ does not divide $2n$

Heh. There's a small issue. We're doing real algebraic geometry.

Nullstellensatz is only for algebraically closed fields

No matter, let everything happen in $\Bbb C[x, y, z]$. Real polynomials are complex, after all

"$f$ is irreducible" is a much harder thing to verify in that case

Well, f is degree three in x,y,z. So if it were reducible it’d have to contain a linear factor?

Ah, yes, 3 = 2 + 1. Classic fact.

But f=0 doesn’t contain a plane, so

Careful! I was going for that argument about the zero set having a single connected component, but we're looking at the graph in $\Bbb R^3$

We have to look at the graph in $\Bbb C^3$

Which we can't, of course, so there's some argument

1:40 PM

Bleh

It's an annoying point, yeah

The argument I wanted to make was with Schmüdgen’s P-satz

“If p(x) is strictly positive on K = {x ∈ R^n | f_i (x) ≥ 0}, and K is compact, then p(x) ∈ cone{f1,...,fm}.

What's a cone

.,.good question.

:P

1:46 PM

Source: ocw.mit.edu/courses/electrical-engineering-and-computer-science/…

That one isn’t talking about sums of squares though. Hrm

See this one instead: math.ucsd.edu/~njw/Teaching/Math271C/Lecture_22.pdf

omg not njw

ah, wrong njw

boo

Lol

Yeah OK this just looks complicated to me

I don't see how or why this should solve your problem

To me looking at the boundary seems to be the best possible approach. You want to prove a certain cubic hypersurface is not cut out by a even degree hypersurface. Sounds pretty obvious.

And definitely Nullstellensatzable

I had hoped to make some argument like “the only want to get K from a set of polynomial inequalities is to include a cubic inequality “

Hey everyone!

1:52 PM

Really the only thing you need is irreducibility of $f$, because if it factors you get a quadratic component, in which case you can cut out that component by an even degree hypersurface.

Heya @BalarkaSen, you down for some covering space stuff in a few hours time?

If you're not busy that is

Right

@Perturbative Sounds good to me

What’s funny is that I managed to find a set of quadratic inequalities which do a very good job of bounding K

“Very good” meaning that the volume of the bounding set is only 1.5% larger than that K itself

(Huzzah for Mathematica)

Huh, interesting

Can I see a picture?

1:56 PM

Once I’m on my laptop, sure

Cool

To get that number I took the set of inequalities defining each region, used each to define an ImplicitRegion, and then used the Volume command to compute the volume of each region numerically

They work out to be about 0.625 and 0.619 respectively

So the set between them is small but finite

That's pretty interesting

Yeah

Ok ok ok ok what do I do now. I have ten million things I want to do

Maybe I'll read this exposition on exotic spheres

2:29 PM

@BalarkaSen Which exposition? If you have a link I'd love to see it

Here you go : homepages.math.uic.edu/~kauffman/Exotic.pdf

Thanks!

Ah damn, I donno characteristic classes yet, I'll take a look at it again in a few months

I only know the bare minimum, but that exposition talks about the prereqs to char classes

Do you know what book it's from?

hmm, no, not sure