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00:00 - 18:0018:00 - 00:00

12:00 AM
What's the connection?
no pun intended
@BalarkaSen do you lie?
Hm, $e^1 \otimes e^2 \otimes e^3$ is a F(TM)-valued 1-form, so I suppose this is the exterior covariant derivative theory for the frame bundle?
I need to understand what the right $d_A$ is... or what the connection is ;)
All the crummy differential forms suddenly makes more sense in the bundle formalism.
I'll sleep on it
12:29 AM
I wonder what $\operatorname{Spec}(\Bbb C[\sqrt0])$ looks like
12:46 AM
Any topologists in here tonight?
preferably low dimensional ones?
depends on what kind of topology you want
I'm trying to figure out some basic things, I think.
just ask
If I have a sphere $S$ that is properly embedded in $N=S^2\times [-1,1]$, then we (visually) know that one of two things is true
I refuse to visualize $S^2 \times [-1,1]$
12:48 AM
Either $S$ bounds a ball that is inside of $N$, or $S$ is homotopic to the boundary of $N$.
Thicken up a sphere
oh, right
yeah, luckily this isn't so whacky
and it is painfully obvious when you draw it on paper
err, the result is
I thought the boundary of $N$ has two components
I basically have 1 tool which is that any sphere in $\mathbb{R}^3$ bounds a ball (Alexander's Theorem)
You're right, I meant that $S$ should be homotopic to one of them
So we can embed $N$ in $\mathbb{R}^3$ which is nice, and therefore we have an embedding of $S$ and this embedding of $S$ in $\mathbb{R}^3$ will bound a ball by Alexander's Theorem. If that ball is in $N$, then we are done. So if it doesn't... then now we are up to the task of building some homotopy
Ooooo, you guys are talking about spherinders...
1:01 AM
hi @Secret
I wonder what happens if we do $S\times (1-,1)$ instead
That is still an uncapped spherinder, but now the boundary is open in the (forgot) topology
the boundary is the same
sigh. I'm finally tired of teaching these entitled brats. Time for a new job.
@ForeverMozart lets talk it out, am all ears
@ForeverMozart what kind of entitled brats?
1:13 AM
@ForeverMozart I tagged you just to continue the chain
Actually going back to cylinders, what is the difference in topology between S1 X [-1,1] vs S1 X (-1,1)?
freshman calculus students
so much whining and impatience
@Secret one is closed, the other open
@ForeverMozart well do you remember when you were a freshman ?
and how your teacher felt the same thing as you feel now
this is part of the job, teaching and learning are the beauty of being human, also the most annoying but that is a mistery
If am not making sense ignore me because am high
@LeakyNun Then the case for spherinders should be similar, since they can be built from a stack of cylinders along the 4th dimension?
1:18 AM
you don't need the fourth dimension
All of these things fit in $\mathbb{R}^3$
$S^2 \times [-1,1]$ perfectly fits in $\Bbb R^3$
woops sorry didn't mean to talk over you
@Prototank nah, you should have said "SNIPEDDDD"
as is the tradition of this room
I think there may be multiple applications of Alexander's theorem
since the boundary of $N$ consists of two more spheres
but I am very new to these arguments and I'm not taking a class or anything
1:20 AM
hmm... so $S \times [-1,1]$ is a spherical shell with some thickness?
user was banned for this post
1:21 AM
Hmm, so both the spherinder boundary and the spherical shell are defined by the same Cartesian product... I need to check...
In four-dimensional geometry, the spherinder, or spherical cylinder or spherical prism, is a geometric object, defined as the Cartesian product of a 3-ball (or solid 2-sphere), radius r1 and a line segment of radius r2: D = { ( x , y , z , w ) | x 2 + y 2 + z 2 ...
oh I didn't know what a spherinder was
it's a different thing then
it even has 4 dimensions
right, spherinder = $B^3 \times [-1,1]$
that looks right
its boundary (product rule) = $S^2 \times [-1,1] \cup B^3 \times \{-1,1\}$
with some gluing between the two objects
specifically, gluing $S^2 \times \{-1,1\}$ with $S^2 \times \{-1,1\}$
makes sense
it looks like $S^3$ to me
1:27 AM
the boundary?
Meanwhile, I have not found the set notation for spherical shell yet, but Prototank is talking about this shape:
In geometry, a spherical shell is a generalization of an annulus to three dimensions. It is the region between two concentric spheres of differing radii. == Volume == The volume of a spherical shell is the difference between the enclosed volume of the outer sphere and the enclosed volume of the inner sphere: V = 4 3 π R 3 − 4 3 π r ...
I think you are right
yes, we were talking about this object
one ball glued on the inside
the other glued on the outside
BAM $S^3$
1:29 AM
very amazing
So yeah, the theorem I'm trying to prove (or lemma, or whatever) is that if a spherical shell $N$ has a sphere inside of it, then that sphere either 1) bounds a ball inside of $N$ OR 2) the sphere can be wiggled around into being one of the boundary components of $N$.
let's just say $N = \{ \vec v \in \Bbb R^3 ~:~ 1 \le \|v\| \le 2 \}$
fair enough
1:32 AM
and what do you mean by a sphere?
an embedding (category $C^\infty$) $S^2\hookrightarrow N$
I imagine things can get quite complicated if it isn't a real sphere
we can control a lot of that with the categorical restriction
but yeah could still be pretty bizarre
I mean, there are a lot of endomorphisms of $S^2$ that aren't null-homotopic
you could imagine the embedding elongating the sphere into a tube with caps and then tying that up
every endomorphism of $S^2$ is homotopic to the identity or the antipodal map
1:35 AM
I think you meant automorphism
because afterall $\pi_2(S^2) = \Bbb Z$
that ain't right
sorry, fixed the subscript
Alexander's theorem does a lot of the leg work here, I think
whats a field extention ?
A field extension is when you have a pair of fields $K\subset F$ so that they share zero and one elements
for example, $\mathbb{R}\subset\mathbb{C}$
1:48 AM
We don't know if $\Bbb Q \subset \Bbb Q(\pi+e)$ is a field extension
Is... is that true?
nobody knows
my goodness
We don't even know if $\pi +e$ is irrational
2:05 AM
I don't even know if i am irrational
::thinking of a group made of irrationals such that some of these are orbit stabilisers...::
A: Irrationals: A Group?

Steven StadnickiTo speak to the spirit of your question a bit: the rational numbers are a so-called normal subgroup of the reals (since the reals form an abelian group, and all subgroups of an abelian group are normal), so we can talk about the quotient group of the reals by the rationals, $\mathbb{R}\ /\ \mathb...

2:25 AM
1 hour later…
3:40 AM
Hi all;
I'm solving AMC 10A 2016 P#10, and I had a question
I solved the question and got 3; I did by doing the following:
First, I represented each of the areas via 2l+2, 2l+10, and 2l+18 (from least to greatest area)
Then, we know the areas form an arithmetic series, and the average of an arithmetic series is it's last term,
So, I did (6l+30)/2=(2l+18)
This gave me an answer of 3; however, that is not one of the answer choices
If anyone could help that would be great
its 4
what did I do wrong? sorry for the naive question
let me read what u wrote.
well the total area should be 5(4+l)
3:52 AM
? Let me check
the area of the smallest rectangle should be 5l
less i am too brain fried to think correctly
No, it's 2l+2w, which is 2l+2
I'm pretty sure
whats l
and whats w
l for length, and w for width; Those are variables I use
well the total height of the rectangle is 5 and the total length is 4+l where l is the inside rectangle
so there product should yield the total area
then make a progression that makes sense no?
3:56 AM
That is true; I should consider this
I got it!
i dont think you have any idea what the total length is
but you do know the height
thanks for helping
4:20 AM
4:36 AM
does anyone want to check my rigorous-ish proof of an integral
(no fundamental theorem of calc)
is there ever a situation in maths where imaginary numbers is a valid x-intercept or do all mathematics consider that to mean there are no x-intercepts if the solution is imaginary
if thy dot mind
@WDUK not in the sense of "x-intercept"
but more generally as "roots"
i see
any $n$-th degree polynomial is guaranteed $n$ complex roots (real numbers are complex too!)
by the fundamental theorem of algebra
so if you look at something like $x^6 - 1 = 0$
4:40 AM
@MatheinBoulomenos here ?
is there a difference between roots and intercepts here?
you instantly thing $-1$ and $1$
but there are 4 more roots
in fact $x^n - 1 = 0$ have roots that form a circle in the complex plane
that's called the roots of unity
well intercept is usually taken to mean intersection with an axis
@Narcissusjewel maybe you can discuss this with me ?
since you can't plot complex numbers on the real plane it doesnt make sense to consider those roots as "intercepts"
ah yeah i briefly learnt roots of unity - kinda sucked at it, need to re-revise it at some point
ive seen imaginary numbers on a plane for number lines
guess thats bit different really
4:42 AM
that's because it was on the complex plane
if you graphed a real function on the complex plane it would just go back and forth on the real-axis
ok so if it doesn't specify its on a complex plane, i should assume intercepts are only valid on real numbers
as far as i know, yes, complex roots aren't considered $x$-intercepts
ok thanks for clarifying :)
no problem
5:04 AM
lol was looking at spivaks online and teds book appeared
2 hours later…
7:26 AM
Is it okay to write $\dfrac{dQ}{dT}= \dfrac{dU}{dT}+\dfrac{dW}{dT}$ in physics?
because $dQ= dU+ dW$
hello, i am trying to fix a question i've asked and i am wondering: how would you correctly say in english that a linear mapping(transformation) follow some rules?
abide maybe?
8:22 AM
what topological space is referred to as $T^n$ ?
@blat torus
i thought so, but it doesn't make sense in my script
could you show us?
@BeginningMath satisfy
@BeginningMath "satisfies," "such that," maybe some other phrases depending on what it is you're trying to say
my bad, it's indeed a torus. thanks :)
3 hours later…
11:52 AM
Can a function have exactly one right inverse and no left inverse?
12:04 PM
exists right inverse iff map onto
exists left inverse iff map one-to-one
if there is no left inverse, the map is not one to one, so there exist two things in the domain that map to the same element in the range, which implies there are at least two right inverses (i.e. sections)
Do we have to use Axiom of Choice in "onto implies right inverse"? @anon
@Silent yes
existence of right inverse
and we get "one to one implies left inverse" without using AoC, right?
@LeakyNun you should change your username to the "Leaky Sniper." :P
12:10 PM
thank you!
got'em again
completely unrelated, but as room owner when moving messages one gets literal crosshairs to select which messages to remove from the room
12:12 PM
The man with the golden gun?
@Silent it's equivalent to AC over ZF actually
12:35 PM
hello, i have $\ker g \subset \ker h$ why this means that there exists $\lambda$ such that $h=\lambda g$ ?
@AlessandroCodenotti @Silent: That's not correct. If S is non-empty and f injects S into T then the inverse relation to f is simply { (x,y) : x∈T ∧ ∃y ( f(y)=x ) }, which does not at all require AC. This inverse relation is already a left inverse of f.
@user21820 He is talking about onto, right invrse.
@Silent: Oh okay. @AlessandroCodenotti: Sorry I thought you were replying to Silent's last message.
@user21820 follow arrows :P
@Vrouvrou context?
X is a banach space and $h,g\in X'$ @LeakyNun
12:41 PM
@Silent: By the way, you accepted this answer but it is talking nonsense, because the author does not actually understand Godel's theorem. See also here.
@user21820 I am so sorry, I have unaccepted it. I fell in love with foundations and started reading somethings about it, but since i study on my own, there was a big stock of unanswered but (to me) very important questions, that haunted me every time i tried to move forward. I was mainly using Tourlakis' 2 volumes "Lectures in Logic and Set Theory", but soon it became unbearable.
@Balarka think I have the proof. Requires playing with Hall's marriage theorem
@Silent You are welcome to ask whatever you want about logic in the Logic room, where both I and LeakyNun will probably be there to discuss with you.
I am not free till June, but if I want to continue with Mathematical logic, which book is good to start?@user21820
@user21820 Wow!
@Silent No hurry. I'll give you some suggestions over in the other room now.
12:51 PM
Thank you! :)
or you could just talk here :-)
1:12 PM
Speaking of AC I asked a question that I discussed yesterday in chat about how much AC is needed for functional analysis, if someone knows an answer the question is here
1:47 PM
@Studentmath I don't know what that is.
Hi chat
I have to prove that f is continuous in $x_0 = 0$
$$f(x)=\frac{2x^3+x^2+x\cdot sin(x)}{(e^x-1)^2}$$

Well, if I check the limit for f it is like $\frac{0}{0} $(I have tried L'hopital but failed doing that).
Is it a good way to check if $\lim_{x \to x_0^+} = \lim_{x \to x_0^+} = L$ or is it best to use a different method here?
namely because it includes a statement of "Hall’s marriage theorem for hypergraphs"
I'll have to read it later
I don't know any serious combinatorics
1:52 PM
i have no idea about the marriage theorem myself either
if it's not easy to do with generating functions, I usually don't know it :P
anyone of here knows lens formula? (in physics)
1/v - 1/u = 1/f
anyone here?
i'm in trouble with that
Ask, don’t ask to ask.
if 1/v - 1/u = 1/f so 1/u - 1/v should be equal to 1/f?
but that doesn't happens to be
Interchanging the position of the object and the image has the effect upon the lens that it changes from being convex to concave (or the other way around)
Then either it doesn’t hold or you have the wrong values of u,v,f
2:00 PM
So, no, 1/u - 1/v would be 1/(-f)
Which is consistent with the formula
okay, let 1/v - 1/u = 1/f, so squaring both sides = (1/v -1/u)^2 = (1/f)^2, which is equal to (1/u - 1/v)^2 = (1/f)^2 cancelling 2 exponent both sides then its 1/u - 1/v = 1/f
One point re: the formula is that you never need to use it. You can always make the computations by appealing to ray-tracing diagrams and similarlity of right triangles
as (a - b)^2 = (b - a)^2
but that doesn't happens
(-2)^2 = (2)^2 hardly implies 2=-2...
I've only studied about this formula, other formula is not in my book @Semiclassical
yep! But, its true as my teacher said that square roots can't be negative
for iit preparation
square roots are always positive
2:04 PM
x^2 = y^2 does not imply x = y
I doubt that. Ray-tracing diagrams are what you use to obtain the formula
That is 8th grade algebra error
its given in my book, what's error?
x^2=y^2 implies |x|=|y|
Burn your book.
2:06 PM
(a - b)^2 = a^2 + b^2 + 2.a.b which is equal to (b - a)^2
@Semiclassical how?
(b - a)^2 also is equal to the previous one as a^2 + b^2 - 2.a.b
woops forgot - in 1st equation
If x^2=(-1)^2, what can you conclude about x?
minus symbol
@BalarkaSen Fahrenheit 451?
1 or -1 @Semiclassical
Right. x=1 or x=-1
2:09 PM
@Semiclassical so, you mean even exponents can't be cancelled?
Which is equivalent to |x|=1=|-1|
what is ||?
Absolute value
okay, omission of sign, you mean mods?
how you can use mods in practical gemoetery?
Can anyone give me an intuitive explanation of the Ultra Limit? I am failing to see how it relates to the standard topological/metric definition
2:13 PM
@Semiclassical is there any proof for Mods? I've not studied much about that.
anyone here knows how can I do IIT without coaching (specially mathematics)?
@Semiclassical I got that document to get sine and cosine values though calculation. That's a great idea, I've never thought of doing that, its great!, taking a triangle in 36, 36, 72 deg ratio and dividing them by bisecting... its a great idea!
2:52 PM
Sanity check: Let $(X,||\cdot||)$ be a Banach space, then any $A\subset X$ which is compact or weakly compact is bounded and has empty interior (but for different reasons in the weakly compact case), right?
3:16 PM
@Alessandro in the case of strong topology, then yeah, because if it had non-empty interior, it would contain a closed ball
That's a closed subset of a compact set, which would make it compact. But it isn't. Rip
In the case of the weak topology
I think that's just false
A Banach space is reflexive iff its closed unit ball is weakly compact (and weakly sequentially compact)
(That's a hard theorem in general but like, take the dual space, its closed unit ball is weak* compact by Banach-Alaoglu, so if that space is reflexive, it means weak* compact = weakly compact)
@Daminark but the closed unit ball has empty interior in the weak topology
weakly open sets are unbounded
... I'm an idiot
Okay maybe that's right, let's see
You're not, I realized this morning that weakly open sets are unbounded even though I should have noticed months ago so...
Yeah you're just right
Let's work with nbhds of $0$ because you can just traslate stuff otherwise. $0$ has a nbhd basis of the for $\bigcap\limits_{i=1}^k f_i^{-1}(-\varepsilon,\varepsilon)$ as $k,f,\varepsilon$ vary in $\Bbb N$, $X^\star$ and $\Bbb R$ respectively
3:23 PM
So in particular you're gonna contain the intersection of the kernels of those $f_i$
And each one has codimension 1, so in total if your space is infinite dimensional then yeah intersection of finitely many kernels is a subspace, so in particular it's unbounded
Okay that's dank
So not only open nbhds of $0$ are unbounded, but they actually contain closed, infinite dimensional subspaces
3:39 PM
(This works because weakly compact set are bounded wrt the norm on $X$, which is easy to prove)
4:04 PM
X is a banach space and $h,g\in X'$ i have $\ker g \subset \ker h$ why this means that there exists $\lambda$ such that $h=\lambda g$ ?
everyone here
is there a example to show that maths doesn't behaves just like it, everytime?
as it goes numerically but not practically
@Vrouvrou The kernels have codimension $1$ so if one is contained in the other they must actually be the same I think
4:59 PM
@AlessandroCodenotti why the codimension is 1?
my question is about the proof of this lemmr
5:47 PM
Hey guys. Is there a non-trivial function which has a zero with infinite multiplicity?
The keyword is "bump functions"
They wouldn't be analytic, of course, but can be $C^\infty$
So those are zero because outside of the bump they are defined zero?
Indeed it's a non trivial function but the idea seems kinda like cheating.
Yes, but they need not be. Eg $f(x) = \exp(-1/x^2)$ vanishes upto infinite order at $x = 0$.
But is this function defined for $x=0$ since $\frac{1}{x}$ is not defined for $x=0$?
Let me be precise. $f(x) = \exp(-1/x^2)$ for $x \neq 0$ and $f(0) = 0$
This is a smooth function and all of it's derivatives at $0$ vanishes.
5:51 PM
good morning
Ok fair, indeed a non-trivial example! Thanks
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