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00:00 - 17:0017:00 - 00:00

12:04 AM
Measure Lebesgue $[0,1]$ is equal to zero.
 
wat
 
question mark appears above head
 
@PawełKusz Because 0=1?
 
I'm a bit confused by a hint on a homework problem. Given an embedding $f: T^2 \to S^3$ and a closed 2-form on $S^3$, I want to show that $\int_{T^2} f^* \omega = 0$ and the hint is to homotope $f$ to a constant map and then use Stoke's theorem. But if we homotope $f$ to a constant map then the image of $f$ is just a single point in $S^3$ and I'm not sure how to apply Stoke's theorem to that setup
I don't see the boundary of anything anywhere
 
If a manifold has no boundary, the integral of anything over its boundary is $0$
 
12:17 AM
Thats what I though, but then why do I have the condition that the 2-form must be closed?
 
Good question... :thonk:
 
Somehow this has to become the integral of $d \omega$ over $S^3$ or some part of $S^3$ so the thing I homotope to should be the boundary of that corresponding thing.
Maybe I can think of the point as being the boundary of some $S^1 - \{p\}$ sitting in $S^3$
 
Maybe? This is quite strange honestly
 
12:44 AM
Oh I think I need that the form is closed because if $f \sim g$ then $f^*$ and $g^*$ are equal as maps.... on the appropriate DeRham Cohomology which only include closed forms
 
OH RIGHT
Yeah homotoping stuff requires closedness
 
1:17 AM
Octonions are like quaternion but instead with 3 imaginary parts they have 7 imaginary parts?
 
And they are not associative
 
@EnderLook the space of purely imaginary octonions is 7-dimensional
so, you specify the imaginary part of an octonion by using 7 real parameters
 
 
2 hours later…
2:58 AM
Hot take: MS Paint > MS Word > LaTeX
@Eric can confirm
 
i agree
 
Staying around for Thanksgiving?
 
yup
also may potentially doing a reading course with professor ngo next quarter
 
Wait really? That'd be sick, what on?
 
some kind of number theory
otherwise potentially might do something with amie or eskin
 
3:15 AM
I see. Chances are if I do a reading course with anyone, it'd be Nori
 
he's cool
 
Otherwise meaning, you're deciding? Or that Ngo might be too busy next quarter or something?
 
oh like a friend approached me to do a joint one w me
and these are the people who he has approached already
 
Ah, I see
 
How can I think about computing an integral like $\int_{M} g^* \omega $ where $g:M \to N$ is a constant map?
 
3:17 AM
Oh I guess there's also a chance I'll try to do a reading course with Marianna
Like a few people are doing descriptive set theory and I might hop in
Though thing is, it's between that, something with Nori, or civ
Probably not ready to do anything with Nori yet, Marianna would be an open option that's fun, civ is the responsible thing
That or I wouldn't do algorithms, I dunno
 
if i do a reading course im not registering for it
im taking it conjunction w 4 other things
 
Oh yeah I could just drop in on the descriptive set theory folk
 
ik some of my HA crew did an informal schlag-quest last year
 
Also turns out Nori's doing AG next quarter
Lol your group seemed to be quite close to Schlag
We've got a few people working with Soug, like he's teaching literacy in PDE but since enough (3-4) of us were thinking of a reading course with him, he was like yeah do literacy and I'll just focus on you guys
 
wait there's an AG class next quarter?
 
3:27 AM
Grad
Undergrad was last quarter
 
ahhhhh right grad ag is winter i forgot
 
Yup
 
we are also considering reaching out to benson to do some low dim top or MCG
 
Oh that could be fun
 
3:54 AM
Algebraic Geometry: A Geometric Approach
 
@user104729 Euclidean Geometry: A Geometric Approach
 
Approaching Geometry: A Geometric Approach
12
 
You know there are finitely many sub-fields of mathematics studied up to this point in histroy
So we could just build a bot that appends ': A geometric Approach' to every one of them and then never have to do that work again
 
@KevinDriscoll but where's the fun in that?
 
hi chat
 
4:05 AM
hullo
 
Geometric Approaches: A Geometric Approach
 
Yo
 
I guess that's not so different than what Leaky had, though
 
@Semiclassical hi
 
hollu
 
4:15 AM
Approaching Algebra: An Algebraic Approach
Approaching Number Theoy: A Number Theoretic Approach
 
Approaching Logic: A Tautological Approach
 
Approaching SemiClassical Physics: A SemiClassical Approach
 
i see what you did there
 
Approaching Baymax : Baymaxian approach
 
4:24 AM
Approaching -1/12: A Natural Number Approach
 
Oh no
 
Approaching Limits: A Limited Approach
 
Approaching Daminark: A Thonking Approach
4
 
Approaching the Limits of the Joke: An Exhaustive Approach
 
Approaching Insanity: Approached
 
4:28 AM
@LeakyNun You know what I'll take it
 
Approaching Approach: An approach0.xyz Approach
 
Approaching The Unapproachable: An Impossible Approach
 
Approaching the Disconnected: An Obstructed Approach
 
Approaching The Center of a Blackhole: An approach from which there is no return
 
4:38 AM
Approaching Light Speed: An Electromagnetic Approach
 
Approaching mathematics: A metamathematics approach
Approaching Ted: The Origin of Geometric Approaches
 
Approaching People: A Polite Approach
 
4:53 AM
Omae wa mou shindeiru: A NANI? approach
2
@Balarka
 
Approaching The Unknown: A Discovery Approach
 
Oh Ted's gonna have a time when he sees this
 
Imitation is a form of flattery
Approaching Flattery: An Imitation Approach
 
can someone please answer my question --> math.stackexchange.com/questions/2530908/…
And if it is not the correct chat forum. Please give me link of proper one
 
5:45 AM
It seems that recently the activity in functional analysis chatroom is bigger than before. And there is no shortage of users who ask questions there. It would be nice to get more users with good knowledge of functional analysis who could help with the question which remained unanswered.
Of course, there are many other chat rooms which could have more activity that they currently have: List of chatrooms
@Sadhu Maybe posting the question here in chat in the format which includes the title might increase the chances that somebody have a look.
If the title is included, users in chat can have at least a basic idea what the question is about without having to click the link. And if they see that it's a topic they are interested in, they might have a look.
In this case you might also try to make the title more descriptive. The phrase "following expression" does not say much about the question, maybe something like "modular computation with Fibonacci numbers and factorials" would be better.
Or you could simply include the expression $(\frac{\sum_{i=N}^M F_i\cdot i!}{K}) \mod p$ in the title.
 
Tips on formatting^ @Sadhu
Also, click on the "help" link in the bottom right corner :-)
 
hi chat
 
I have problem with one pdf on my ipad
it keeps crashing on only 1 book
the rest works fine
do you know if it can be fixed?
 
Delete it and reload?
 
I did that :/
Its heirstein book
17.2 mb
it is smaller than the others for some reason
that could be it but not sure
 
6:37 AM
You could ask here.
 
thanks :)
 
7:19 AM
Hey @Mathein!
 
7:35 AM
Hi
 
How's everything going?
 
Quite well. I have to prepare another talk I need to give, so I'm bit busy. And you?
 
Thanksgiving break, so I've got a few days to breathe
Hey @Alessandro!
 
7:53 AM
Mornin @Dami
On which topic @Mathei?
 
Galois cohomology
Infinite Galois theory, Profinite groups and Kummer theory
 
That sounds cool
 
It is, but it's also difficult
 
8:25 AM
I can imagine that
Also hey @Tasty!
 
Morning
 
How's everything going on your side?
 
Ah, nothing special. Hopefully a few hours of time for research today
 
Nice
 
And you? Bio finished yet?
 
8:34 AM
I got through that paper from earlier now
Probably not the best, like I'm pretty sure there were points where I was straight up rambly
But I'm done thinking about it now so yeah
 
you write papers in bio?
 
Yeah, we have just one paper to write
 
I see, they want you to stay "well rounded" :-)
Have you taken any Latin?
 
Nope, I'm not quite that well rounded :P
 
It comes in handy with nomenclature.
 
8:47 AM
Also you can read Euler and Gauss in the original Latin
 
let us not forget Newton :P
botany is full of Latin names
 
Meh... usually, when reading old maths texts, the language isn't the problem, the notation is
 
9:02 AM
Surely the notation can't be that bad... Can it?
Also hey @PVAL!
 
 
1 hour later…
10:12 AM
@Kevin @Daminark Yeah you need closedness because the argument is by Stokes
 
Hi @Balarka
 
Hey @Mathei
 
It's probably too algebraic for your taste, but group cohomology is really dope. You can do really awesome stuff, in particular when the group is finite
When the group is finite, you can patch together homology and cohomology to get one long exact sequence which is infinite in both directions
(You need slight changes in dimensions $1,0,-1$)
 
Oh? Please elaborate
I don't think I have seen this LES
 
It's called Tate cohomology
 
10:19 AM
Ah I see yes I have heard of this
 
The construction uses a particular description of the usual $H^0$ and $H_0$ which works only for finite groups
There's this awesome thing about Tate cohomology called dimension shifting $\hat{H}^n$ is the tate cohomology which is a regular cohomology group for $n\geq 1$ and a regular homology group (i.e. a $\operatorname{Tor}$ group). Then for any $G$-module $A$, you can define other $G$-modules $A^{\star}$ and $A_{\star}$ such that $\hat{H}^{n+1}(G,A)=\hat{H}^n(G,A^\star)$ and $\hat{H}^{n-1}(G,A) = \hat{H}^n(G,A_\star)$
 
@KevinDriscoll The pullback is zero. If $g$ is a constant map, $dg : TM \to TN$ is zero.
 
This means that if you want to prove something for all $G$-modules, you can just say without loss of generality $n=0$ (and for $n=0$ it's often obvious)
that's a pretty awesome proof technique
 
And you're pullbacking a multilinear form on each tangent space by the zero morphism, so...
my internet is so bad
@MatheinBoulomenos Hmm, I see
 
For example, it's pretty easy to prove that $|G|$ annihilates $\hat{H}^0$ from the construction of $\hat{H}^0$
Thus by dimension shifting $|G|$ annihilates all Tate-cohomology groups
If you combine this with the description of $H^2$ in terms of group extensions (and add a little more group theory), you get Schur-Zassenhaus
 
10:34 AM
Wow that's pretty cool
I wonder if there is a natural way to interpret Tate cohomology as co/homology of $BG$ though
 
I don't know about that
 
I have to go now but I'll return to this discussion in about two days
 
 
1 hour later…
11:55 AM
0
Q: Gram Matrices Induced by Vectors in Finite Dim. Inner Product Spaces

user193319I am hoping someone can corroborate my chain of reasoning. Let $P=[p_{ij}] \in M_n(\Bbb{C})$ be a gram matrix. The there exists a inner product space $(V,\langle, \rangle_V)$, which is possibly infinite dimensional, and vectors $v_1,...,v_n \in V$ such that $p_{ij} = \langle v_i, v_j \rangle_V$. ...

 
12:30 PM
0
Q: Free $C^*$-Algebras

user193319I am confused by a passage in the wikipedia article Universal $C^*$-algebra, particularly these two sentences This means that depending on the generators and relations, a universal C*-algebra may not exist. In particular, free C*-algebras do not exist. My confusion arises in connection with...

 
 
1 hour later…
1:53 PM
[Random]
The structure of the set $\omega$
{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,...}
 
there is a sandbox if you're interested
i only found out about it a few weeks ago
 
2:20 PM

Sandbox

Where you can play with chat features (except flagging) and ch...
There's an ACTUAL conversation in sandbox???
 
2:31 PM
Is wyzant good for Q&A?
 
2:43 PM
mdave: I will normally go to the sandbox. But today we have an emergency situation here
 
2:55 PM
(guessed \solved for now)
 
3:08 PM
so the group $\{x \in \Bbb Z[\sqrt d] \mid \| x \| = 1 \}$ is either $\Bbb Z_2$ or $\Bbb Z_2 \times \Bbb Z$?
 
Someone, please help me with this proof:
0
Q: Condition for 3 distinct normals from a point $(h,k)$ to the parabola $y^2=4ax$

Abcd What are the conditions in terms of $h,k$ to be able to draw three distinct normals to the parabola $y^2= 4ax$ ? Normal to the parabola from $(h,k)$ is given by: $am^3 +(2a-h)m+k=0$. This equation can yield three distinct slopes $m_1,m_2,m_3$ if $\Delta>0$ (source). The $\Delta$ of the eq...

I am done with half of it.
I am trying hard to prove the other half.
But unable to.
If someone wants details of how I proved the first part, I can send.
 
3:46 PM
Let $\pi_1(x)$ be the number of primes less than equal to $x$ which are $1 \pmod 4$, and similarly define $\pi_3(x)$. Littlewood proved that the inequality between these switches back and forth infinitely. Can someone provide a link for that one, or any modern source discussing this?
 
3:58 PM
A sphere with $r$ is doubling and its radius is being $2r$.
I wanna find what the increase of surface area $\pi r^2$ is.
Also, I'm getting $16 \pi r^2$ but it seems wrong.
According to my textbook, the answer is $12 \pi r^2$, Why?
 
@Lagranian They're not asking for the new surface area but for the increase:-)
 
Ohh
So what should we think?
I think that they're equal, right?
 
Alright so the increase is new surface area-old surface area right ? What's the new surface area, what's the old surface area ?
 
Okay, let's begin.
The surface area is $4\pi r^2$
and the new surface area is $16 \pi r^2$
 
Alright, so what's the increase in surface area ?
 
4:03 PM
That's $4$.
 
How do you get that ?
 
"increase" = how much more is the surface now than it was before
 
Yeah, that's $4$.
I think that's why I'm getting wrong answer.
Why isn't it $4$?
 
Can you explain in more detail how you got $4$ ?
 
From $4 \pi r^2$ and $16 \pi r^2$
$16$ is $4$ times of $4$.
 
4:06 PM
Ah I see the issue
The increase between two values isn't their ratio, it's their difference
The increase between $1$ and $5$ is $5-1=4$
 
However, I didn't get why textbooks says that $12 \pi r^2$.
 
@Lagranian Initial SA = $4\pi r^2$, Finally, SA = $16\pi r^2 \implies$ , increase in SA$= 12\pi r^2$
 
How did we get $12 \pi r^2$?
 
Where sa,= surface area
 
@Abcd Yes, it seems like $12$.
 
4:08 PM
@Lagranian it is 12.
 
(According to my textbook) that's why I don't get it.
Oh, hold on.
 
@Lagranian I have proved it to you.
 
I still didn't get why we get $12 \pi r^2$.
I got the equation perfectly.
 
@Lagranian 16-4 = 12.
2 mins ago, by Abcd
@Lagranian Initial SA = $4\pi r^2$, Finally, SA = $16\pi r^2 \implies$ , increase in SA$= 12\pi r^2$
 
@Abcd What a mistake I did
:/
 
4:10 PM
@Lagranian A silly error I guess.
 
@Abcd Yeah.
What about finding the increase of volume?
I think I'm gonna do same.
 
@Lagranian Don't immediately ask on SE. Try to figure out the mistake first and then ask, you'll improve your error eliminating skills.
 
@Abcd Yes, now I'm trying to work it out.
Hold on.
@Abcd I guess there are some errors.
:41320224
Look at your message.
 
SA = $4\pi (2r)^2$, (final SA)
Initial SA = $4\pi r^2$
 
Yeah, and from that, we get $12 \pi r^2$ =)
Now I'm trying to find the increase of volume.
Hold on
The volume of sphere is given by $\frac {4}{3} \pi r^3$.
 
4:15 PM
@Lagranian It's easy, you'll be able to do that yourself. And if the answer doesn't match, keep trying.
@Lagranian $\dfrac{4}{3}\pi r^3$
 
and now I'm gonna do same with SA
Initial Volume = $\dfrac {4}{3} \pi r^3$.
Finally Volume = $\dfrac {4}{3} \pi 2^3$
and im confused again.
 
why ?
 
I tried extracting both
but it seems wrong.
 
Og btw you're missing an $r^3$ in the final volume
 
Isn't radius being $2r$?
and it becomes $2^3$
 
4:22 PM
$(2r)^3=2^3r^3$
 
so is it $8r^3$?
 
yep, that's $(2r)^3$. So, what's the increase ?
 
I'm still getting wrong.
 
Break up your reasoning and write it here
 
$\dfrac {4}{3} \pi r^3 - \dfrac {4}{3} \pi 8r^3$
 
4:25 PM
It's the opposite, since increase is new-old
 
What do you mean?
 
The old volume is $\frac43\pi r^3$, the new (=after changing the radius) is $\frac43\pi 8r^3$, and the increase is $new-old$
 
Yes, I know.
 
you did $old-new$
 
Let me try it again then.
$\dfrac {4}{3} \pi 8r^3 - \dfrac {4}{3} \pi r^3$
From that, I'm getting $8\pi r^3$.
 
4:29 PM
How did you get $8\pi r^3$ from that expression?
 
$\dfrac {4}{3}$ and $\dfrac {4}{3}$ are gone.
 
That's not how differences work
 
Can you show where I'm going wrong?
 
$\frac43\cdot 3-\frac43\cdot2\neq1$
 
I'm safe till extracting.
Where did that $3$ and $2$ come from?
 
4:31 PM
It's a general example to show you that you're making a mistake
 
Then it becomes $4 - \dfrac {4}{3} . 2$
 
You're saying that we can "cancel" the $\frac43$ because they appear on both sides. That's not true.
 
I'm so confused right now.
then I got that we can't cancel. However, I still don't have any idea about extracting.
 
Alright so let's consider $\frac ab\cdot 3-\frac ab\cdot2$. Can you simplify that ?
 
I think I can't because don't know what to do.
 
4:35 PM
Alright, what happens if you factor $\frac ab$ on both sides ?
 
Hmm
lol my brain is firing
Any hint?
 
There no point in me doing the exercise for you, if your brain is tired rest a bit and come back to it later
 
What do you mean by factor?
or you mean $\dfrac {a}{b} . \dfrac {3}{1}$?
 
For instance if you have $ab+ac$, you fac factor by $a$ and obtain $a(b+c)$. That's what it means
 
Yes, because the both sides include $a$.
I got what you mean now.
If it been $ac$ + $bc$ then we could write it as $c (a+b)$
 
4:41 PM
Yep
 
Yup so you say that there's something we need to do with this
 
So, what of $\dfrac {4}{3} \pi 8r^3 - \dfrac {4}{3} \pi r^3$ ?
 
I see that the both sides include $\pi$ and $r^3$
 
What happens if you factor by $\frac43 \pi r^3$ ?
 
$\dfrac {4}{3}\pi r^3$ $(8 \pi r^3)$
 
4:46 PM
Wait why is there $\pi r^3$ twice ?
 
because you said that it's $(2r)r^3$.
 
I never said that ?
 
Hmm
Can we write
$\pi r^3$ $(\dfrac {4}{3} - \dfrac {4}{3} + 8)$
 
Alright so looks like you have so difficulties about factoring, it's a fairly essential concept if you want to solve your problem so I suggest you get more familiar with it before coming back to your problem
 
However, where I'm going wrong?
 
4:51 PM
You're going wrong on the factorization. What you write doesn't make sense.
 
Okay show me a hint
I see that $\pi r^3$ and $\dfrac {4}{3}$ are common.
 
I'm not sure that would help you though, it's aking to giving you the solution, you won't learn anything useful from it
 
I got everything about the solution.
The problem is to factorize.
You can give a hint about the commons.
 
As I told you, you can factorize by $\frac43\pi r^3$. That's the common part.
 
I'm still getting wrong.
 
4:55 PM
Show me your steps. I can't help you otherwise
 
As you said, I've tried factorizing by $\dfrac {4}{3} \pi r^3$
and my steps:
$\dfrac {4}{3} \pi r^3$ $(8 r^3)$
 
That's your solution. What I'm asking for is the steps, i.e. how you got to that solution
 
If $\dfrac {4}{3} \pi r^3$ is common
that's it.
 
I still don't have a clue how you get to your result.
 
Your hint doesn't work.
I don't get it.
 
4:59 PM
hii.. am learning derivatives concept now??? can anybody help ??? what does the exact meaning of derivate of x^2 = 2x
 
$\dfrac {4}{3} \pi 8r^3 - \dfrac {4}{3} \pi r^3\neq\frac43\pi r^3(8\pi r^3)$
@AchillesRamNakirekanti What exactly don't you understand in derivatives ? That's a vague question
 
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