« first day (2396 days earlier)      last day (2626 days later) » 

10:00 PM
@Ted I also really don't have anyone here to work with at this point.
I feel like my area has really diverged from the other grad students here.
and the other topologists excluding my adviser.
 
@PVAL: I don't mean people there. I mean people you get to know at conferences ... for the future!
One line, @Zach?
 
for $c_1 = 0$ or the whole thing?
 
To do the whole thing, you change $1$ to $i$ in the one-line proof.
 
oh :P
 
@Ted Proof : This is left as an exercise for the reader.
 
10:02 PM
But I find that it always helps to be concrete to see what's going on.
 
Hi @Sophie
 
@Astyx: I've been nice and avoided smacks because you were sick. But, no more.
 
Hi
 
Hi @Sophie
 
Meh, I'm just trying to be funny .. (and failing miserably apprently)
 
10:03 PM
@Astyx: Sorta like @Balarka trying to be punny.
 
i got it!
:P
 
Too easy?
I should send you linear independence proofs to practice.
 
we have $c_0\vec{v}_0 + \dots = 0$. Dot by $\vec{v}_0$ to get $c_0 = 0$
 
Well, it started at $1$, but yes.
 
is there a closed form for $\displaystyle\sum_{j=0}^\infty\sum_{k=0}^\infty (j,k)x^jy^k$? Where $(j,k)$ is s the GCD of j,k
 
10:06 PM
because $\vec{v} \cdot \vec{v} = 1$
 
Works for any set of mutually orthogonal nonzero vectors.
What is $(j,k)$? gcd?
Ah.
You mean gcd of $j$ and $k$.
Don't use $x$ and $y$ when they're already used in the expression.
 
actually
 
Thanks, @Sophie :)
 
there we go
 
you'd get something like $||\vec{v}_0||^2 c_0 = 0$
 
10:08 PM
I've never seen such a thing.
Not something like. Precisely that.
 
sorry, i meant $\vec{v} \cdot \vec{v} = ||\vec{v}||^2$
 
Huh?
 
that wasn't that hard, i just had to not be stupid :D
 
@Zach: I worked very hard teaching the "right" linear independence proofs when I taught linear algebra (dozens of times).
 
@TedShifrin Well, I hope that I'll take a rigorous LA course in college...
too bad my calculus and multivariable calculus won't be rigorous because i will have to take them at my high school... :/
 
10:10 PM
You can still work on Spivak and my book :)
And in the latter you'll learn linear algebra right :P
 
I haven't read Spivak, or your book.
Should i order both of them? :P
 
Perhaps eventually.
There are some interesting questions later in that section relating to the projective geometry stuff. Don't skip 'em.
 
by the way, diagonalizing these quadratic forms is just like making them have the form $ay_0^2 + by_1^2 = c$, right?
 
Yup.
 
sweet
 
10:13 PM
This comes back to the stuff you were thinking about before about trying to decide which linear fractional transformations resulted in parabolas, ellipses, hyperbolas. Ages ago :)
 
hyperbola if the diagonalized form has a negative coefficient of $y_0^2$
 
Well, if you're in $\Bbb P^2$, remember that you have three variables.
Dunno whether you're thinking in $\Bbb R^2$ or in $\Bbb P^2$.
 
the former
 
Spivak is quite expensive
 
But the conics we were obtaining weren't centered at $0$, so you have to be a bit careful, but only a bit.
 
10:16 PM
I don't advocate illegally downloading books
 
Really, @PVAL? He didn't used to be.
 
I also don't advocate not illegally downloading books.
Maybe I mean a different book of his
one sec
 
I know my books are ridiculously overpriced ... I've fought with the publishers and given up.
 
oh nevermind spivak is quite affordable
 
Whew.
And there's enough in there for more than a one-year university course, typically.
 
10:17 PM
heh, i'm actually making money before legal age to work. kids are so desperate for candy in school that i can sell a 2 dollar bag in small amounts to make a total of 10 dollar+
 
apostol is insane
 
But virtually no universities teach it anymore :(
ROFL @Zach ...
Go sell some to the GOP :)
 
let me see what investment i can make
 
I was apparently without health insurance for the month of february until today
 
apparently 1 pound = 75 candies, so 5 pounds is 375
 
10:19 PM
though I think that I had it added on Feburary 1st retroactively
 
each is a quarter so thats 93 dollars... and the 5 lb bag is only $20
what a profit
 
Your high school principal may arrest you if he finds out, @Zach.
 
MIDDLE SCHOOL. :D
 
Oh, right, I keep forgetting. Even worser.
 
don't beat yourself up, i'll be in high school soon
 
10:20 PM
You can make money at any age can't you ?
 
@Astyx yeah but
 
Schools don't generally like this behavior.
 
this makes a lot of money. unfortunately, i'm kind of taking advantage of kids
 
Yes but you cannot run a business (even an informal one like a lemonade stand) at a public school.
 
What PVAL said.
 
10:21 PM
it doesnt have to be a business... and now i sound shady as fuck
 
Generally selling is against the rules
 
covers virgin ears
 
don't tell the popo, @Ted
or i'll call over Akiva and Balarka to beat you up
 
On the one hand it sounds like a terrible idea.
On the other hand at that age, I'm sure I did worse.
 
@Ted they'll come around the corner of an alley like those old motorcycle gangs, snapping their fingers with greased up hair
 
10:24 PM
John Travolta from many years ago can play the role in the movie.
 
"We'll make them an offer they can't refuse .."
 
Isn't it "make him"
 
Might be
 
because he's refering to the assassination of a certain singular character
 
10:35 PM
Hi.
Don't ask to ask ? I'm gonna ask to ask to ask.
 
no, you can't ask to ask
 
That is still asking to ask
 
No, I'm asking you permission to ask you to ask a question.
 
Guys, I'm reading the proof that a graph $G=(V,E)$ without isolated vertices is an Euler graph if and only if $G$ is not empty, connected, and all vertices have even degree.

At some point in the proof they assume $G$ is connected and each degree is even. Let $W=(v_0,v_1,\dots,v_n)$ be the longest possible walk where each edge occurs at most once. Then we have $v_n=v_0$. Otherwise we would have an odd number of edges in our walk. Because the degree of $v_n$ is even, there is a line $e\ni v_n$ that is not in the walk, say $e=\{v_n,u\}$.
 
hi
 
10:39 PM
I don't understand why there is such a line $e$
because we already have that $v_n=v_0$ has 'even edges' in the walk
So what goes wrong if this edge $e$ doesn't exist??
 
Is there a simple inverse to y=3*x^2-2*x^3? I've been trying to find a simple polynomial inverse but haven't had any luck so far. The closest I got was having maxima spit out an expression involving sin(atan2()/3) and cos(atan2(same)/3), but it behaves oddly (sometimes when I graph inv(poly(x)) it draws the proper line, sometimes it draws something completely different)
 
hey @PVAL-inactive I was wondering do you know about cylinder neighborhoods ?
 
@Zach maybe?:P I need this theorem for the Koningsberger problem!
 
@Adeek Is that the same as a tubular neighborhood?
 
I am not sure it is the following
no it is not the same as tubular neighborhoods but it seems to be very useful
to get HEP.
The way I understand this is that the cylinder neighborhood gives us a controlled way in which we can extend a homotopy onto the whole space.
Is there like different conceptual intuition in regards to this as well ?
 
10:46 PM
We say that a sequence $(a_n)$ satisfies the property $(L)$ if there exists a real number $q\gt 1$ such that $\frac{a_{n+1}}{a_n} \ge q$, for all $n\in \Bbb N$. $\\$ We say that $(a_n)$ satisfies the property $(S)$ if there exists a real number $r \gt 1$ such that the interval $(x,rx)$ contains at most one of $(a_n)$'s terms, for all $x\in \Bbb R^*_+$. $\\$ Prove that if $(a_n)$ satisfies $(L)$, then it satisfies $(S)$. $\\$ If $(a_n)$ satisfies $(S)$, does it satisfy $(L)$ ?
 
So its saying there is a neighborhood $N$ and another close subspace $B$ so that $(N,A\cup B)$ is homeomorphic to a mapping cyllinder f:B \to A (M_f, A\cup B)
 
@Sha What do you mean by walk ?
 
Maybe try and work out what the map f is in the example they drew.
 
That problem is indeed, literally, a problem.
 
@adeek see above.
 
10:49 PM
yeah
oh
 
I mean the main idea
 
yeah I should try that out.
 
Here's the intuition
if B deformation retracts onto A inside the space
you should expect such a neighborhood
 
@Astyx It's a sequence of points $(v_0,v_1,\dots,v_k)$ such that $v_{i-1}$ and $v_i$ are connected for each $i=1,\dots,k$.
 
yes right and such neighborhood is achieved through the homotopy ?
the homotopy which gives us the deformation
 
10:51 PM
So they don't have to be different, and we also don't necessarily need $v_0=v_k$.
 
oh okay yeah cool I see it
yeah that is cool.
 
though one needs homotopy to be one to one
away from the boundary
 
yes
 
as far as I can tell from what Hatcher wrote
 
10:52 PM
yeah that is right
 
though at the boundary things are becoming squished together
like the circle is getting pulled into that arc
 
yeah
 
For manifolds tubular neighborhoods are the typical construction of these things.
 
Link to the proof ? @ShaVuklia
 
@PVAL-inactive oh I thought you were talking about tubular neighborhood which is used in some proof in general topology. But, you mean things like this.
 
10:55 PM
I don't mean to intrude, I have a quick Q. Does this (see upcoming image) mean finding... limit as n approaches infinity, sum(k starts at 1 ends at n, |y(k)|)?
 
@Astyx I will have to translate it for you, because it's in Dutch
 
Huh
 
@PVAL-inactive I was windering is it true that if we have (X,A) having HEP then is $q : (X,A) \rightarrow (X/A,\{*\}$ is is a homotopy equivalence ?
 
I'll translate the part that deals with this even-ness, ok?
because the proof consists of different parts
 
10:56 PM
I know it is homotopy equivalence whenever A is contractible. But, to me intuitively that this should be always a homotopy equivalence
 
@Adeek That should be true for any nice space
I don't think you need X,A having HEP
oh wait
yeah that's probably true from the HEP
 
The reason we need HEP is in order to extend a homotopy from $I \times A$ into $I \times X$
 
@Adeek Do you know about CW-complexes or simplicial complexes?
 
The way I visualize this is for nice spaces we can think of A as the more fit version of X.
haha
@PVAL-inactive yeah
 
@Astyx Do you really think you need the other parts of the proof? Because they're not significantly related
This is the 'relevant' part:
 
10:59 PM
It's really important you have examples in mind you can construct of things satisfying these properties
 
"Let $W=(v_0,v_1,\dots,v_n)$ be the longest possible walk$^{(*)}$ where each edge occurs at most once. Then we have $v_n=v_0$. Otherwise we would have an odd number of edges in our walk. Because the degree of $v_n$ is even, there is a line $e\ni v_n$ that isn't part of our walk, say $e=\{v_n,u\}$. This contradicts our assumption that $W$ is the longest possible walk. We conclude that $W$ is a closed walk."
(I added the conclusion to it, which I initially left out)
In the first part they prove that $G$ is connected
 
Oh
 
like subcomplexes in either category.
 
Bad formulation
 
yeah @PVAL-inactive I agree. It helps makes ideas clear in the head.
 
10:59 PM
yea, maybe bad translation XD so you know what's wrong?
 
They mean that if $v_n \ne v_0$ there's a line not in the walk
And that's absurd
 
Ohhhh
 
So $v_n = v_0$
 
they're still arguing that $v_0=v_n$!
I thought they were continuing
instead of proving that
 
Yup, sentence breaks is confusing here
 
11:00 PM
Haha yea!
Alright, I'm also tired :P I think that explains it too
 
Anyway, time for me to go to bed
Bye all
 
I'm going as well! Bye everybody
 
brb thanks @PVAL-inactive
cya all
 
Hi guys, good night/morning. I have one question. Given velocity(Vx, VY) and coordinates(x0, y0) of point. How can I find coordinates of point after t seconds? Velocity is static, a = 0. Thanks.
 
I don't mean to intrude, I have a quick Q. Does this (see upcoming image) mean finding... limit as n approaches infinity, sum(k starts at 1 ends at n, |y(k)|)?
 
11:18 PM
@PedroTamaroff I find it hard to self-study more advanced topics. The courses go faster and the professor usually teaches you what's most important, while self-reading will lead one into traps of details. For example, if I want to pick up some algebraic K-theory, maybe I will choose Weibel's K-book, but maybe it will take me 10 years to read the whole stuff and to do all the exercises.
 
11:38 PM
@PedroTamaroff If I hadn't taken a course in homological algebra, I wouldn't have learned the machinery of derived categories. Now I'm in trouble: there is no course which covers $\infty$-categories, and I need to read Lurie's Higher Topos Theory, 800 pages, very intimidating.
 
11:58 PM
Why do you need to read his book?
 

« first day (2396 days earlier)      last day (2626 days later) »