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user228700
05:00
Now, I split the mod again to get:
user228700
$f(x) =$
Either $(5-x) ; 3 \le x \le 4$
Or $(1-x) ; 0 \le x \le 1$
Or $(x-) ; 1<x<3$
Why $x \leq 4$? What you said works for all $x \geq 3$
Xam
Xam
@Adeek yeah sure, but do I have to write my email here? xD
The tophat hat is so swanky looking.
you can delete it after you post it.
05:01
@BalarkaSen Remind me what the arithmetic genus is?
user228700
@BalarkaSen I'm sorry, I've edited it to indicate that the original function has a domain (not equal to $R$)
user228700
The problem is that my textbook's answer varies ever so slightly, in that it is this: (the first "piece" is the same)
user228700
$(1-x) ; 0 \le x <1$
or $(x-1) ; 1 \le x <3$
user228700
Notice that the $\le$ and $<$ signs are switched.
user228700
Can anybody be so kind as to point out what mistake I have made, please?
05:04
@MikeMiller I don't really know a workable definition; the one I understand is as the leading term in the Hilbert polynomial of a proj. variety. But it's probably the Euler characteristic in the sheaf cohomology wrt the structure sheaf or something.
That's part of why I want to keep an eye on the question.
user228700
Is anybody interested to help me out..? If no, I'll leave now...
It should indeed be $x - 1$ for $1 \leq x < 3$, because $|x - 1| = x - 1$ even when $x = 1$.
Also, have patience. Give the people more than 2 seconds to read your question...
I'd star that last remark, but uh...
Y'know.
2 and 3 are both integers so just graph all the intervals [0,1], [1,2], [2,3], [3,4] instead of thinking about the intervals too deeply
05:10
@Semiclassical lol
user228700
@BalarkaSen I'm sorry. It's just usually everybody's super busy around here and these last few times, I've had to leave after waiting around for like, 10 minutes. I guess I should learn how/where to draw the line.
If people are interested, they'll reply to your message and thereby ping you.
user228700
@BalarkaSen But isn't the case for 1 ambiguous? It works for both expressions 'cause u end up getting 0.
brb
@Kaumudi Ah I see. Yeah, both your and your textbook's answers are correct.
I didn't stop to notice what you wrote for the previous piece
I just write \leq for both terms to avoid ambiguity
05:15
@Kaumudi.H just post your problem and activate desktop notification?
heck you see your pings even when offline, soo..
user228700
@BalarkaSen ::facepalm:: Okay, thanks :-)
Request to reopen. This question, has been closed saying that $f^{-1}$ does not exist x. Now that I have edited and defined the domain and co-domain for $f$, we can find $f^{-1}$ and I would be interested to see how can this problem be approached in different ways.
user228700
@Null Then again, some people don't ping back, but I guess that's only in rooms where there's no activity and this place is usually buzzing with messages.
Who's ever offline anymore?
Oh, you mean not on the site.
Can someone explain how this works?
Can i get help understanding this question?
Stahl is absolutely correct. — Noah Schweber 4 mins ago
05:24
@TripleA Stahl sounds right to me. The new proof Stahl gave is a proof by contradiction, sure.
What's your issue with it?
@JessyCat people who turn off their devices from time to time
@Null I don't trust people like that.
haha
I dont know
stahl doesnt seem right to me
stahl's esplanation* ;)
explanation*
That's not at all a valid refutation. What doesn't seem right?
@MikeMiller oops
05:43
@BalarkaSen It's the Euler characteristic of the sheaf cohomology for the structure sheaf.
Yeah, so I thought.
@PVAL-inactive Do you know an elementary construction of a diffeomorphism that induces [1,1;0,1] on the second homology of $S^2 \times S^2$?
@MikeMiller That sends a homology class with square 0 to one with square 1.
@Kaumudi.H i came up with $f(x)=-x+1$ when $x\leq 1$, $f(x)=x-1$ when $1<x\leq 3$, $f(x)=-x+5$ for $3<x\leq 4$. which seems like yours, but you got a typo in (x-"1");1<x<3 i think (the missing 1)
Yeah you're right.
What's the automorphism group of the intersection form?
05:47
of H? hmm
It's big enough to send (1,0) to any (p,q) .
1,0 has to map to either (k,0) or (0,k) and similarly with (0,1)
can you send (1, 0) to (1, 1)? the former has cup square 0, latter 2.
oh no i am thinking of the diagonal
ignore that
Fucking hell
I'm a fucking idiot
If you assume orientability your left with ID
and i
05:49
Yes I agree.
Worthless answer I spent a ton of time writing is wrong.
so Z/2Z
Actually i^2=-1
so
why does (1, 0) have to always map to (k, 0)?
it can also map to (0,k)
those are the only elements with square 0
ah I'm being silly as well
you get i,-i,-1,1
those all preserve orientation.
I kept forgetting it was symmetric and not antisymmetric.
Me too!
crap
05:53
You don't claim to be a low dimensional topologist
yeah, so I have excuse to be silly
so do you get Q_8 if you don't assume orientability?
the thing I said about (1, 1) was right then
it does have self-intersection 2
05:55
lol, oh well
Of course MCG(S^2\times S^2) is still completely untractible at this point.
Topologically it should be Q_8. I don't know the right paper of Wall to quote though.
You know, some people ask questions on SE, don't provide any kind of attempt at a solution and everybody jumps in to help. Me? I provide some details, but say I'm having a lot of trouble, and I get downvoted...
Makes me mad.
Marcia, Marcia, Marcia!
@MikeMiller So if Aut(H)=Q_8 whats Aut(H^n)?
What's H here?
06:07
The standard hyperbolic form on $\Bbb Z^2$ [0,1;1,0]
ah, ok.
I bet Aut(H) is actually D_4
and aut(H^n) is D_4n
but I'm too lazy to compute this myself
0
Q: Direct product of two nilpotent groups is nilpotent and direct product of two solvable groups is solvable

Jessy CatLet $G = G_{1} \times G_{2}$. I need to prove the following two things: If $G_{i}$ is nilpotent of degree $n_{i}$, $i = 1, 2$, then $G$ is nilpotent of degree $n = \max \{ n_{1}, n_{2} \}$. If $G_{i}$ is solvable (some people call them soluble) of degree $n_{i}$, $i = 1,2$, then $G$ is solvabl...

Actually its bigger than what I said
It has S_2n as a subgroup (and A_2n in the orientable case).
thats still not right, I give up.
@PVAL Wall has calculated all of these things.
06:26
@MikeMiller I feel like Serre or Milnor or someone else probably computed these things way before Wall used them for more cobordism applications.
Nah he has two papers explicitly computing automorphisms of quadratic forms.
06:54
Anyway the total space of that bundle is $S^2 \times S^3$ but now I have no even remotely explicit argument.
@JessyCat you have one downvoted question?
one as in exactly 1
I found multiple
last one was late november though
07:33
Somebody was asking about elliptic functions a few hours ago. They certainly are fun to plot in the complex plane.
I know that's supposed to look like a parallelogramic tiling, but that has threefold symmetries about the poles. Strange.
@BalarkaSen Yes, I designed this "random" elliptic function to have threefold symmetry.
@Balarka You should really see if you can find an explicit proof of the thing in that question.
@MikeMiller The $S^3 \times S^3$ thing, you mean?
(You'll see the period parallelograms if you stare at it hard enough.)
07:37
@J.M. Ya I see it now
What does the Z/3 symmetry mean, mathematically?
Yes
Might require understanding quaternions but you'll survive
@BalarkaSen That would mean it's invariant under a multiplication of its argument by $$\exp(\pm 2\pi i/3), up to a change in sign.
@J.M. Ah, gotcha
Are there restrictions on the fundamental domain? Are there non-trivial functions on $\Bbb C^2 / \Bbb Z^2$ with "threefold symmetry"?
13
A: elliptic functions on the 17 wallpaper groups

Jim BelkThis is an old question, but it's still interesting. Here is a more precise version of it: Definition. Let $f\colon\mathbb{C}\to\widehat{\mathbb{C}}$ be a meromorphic function. A Euclidean isometry $g\colon \mathbb{C}\to\mathbb{C}$ is a symmetry of $G$ if $f(g(z))=f(z)$ for all $z\in\mathbb{C...

07:42
There are not even a nontrivial function on $\Bbb C^2/\Bbb Z^2$, @PVAL-inactive.
Jim explained better than I could've in that answer.
AFAIK you need to remove two points from the fundamental square to have a doubly periodic function. This one seems to have multiple so that's an interesting q.
08:00
That's obvious. Bleh.
08:11
didn't see the question
I wasn't recognizing the fiber of the bundle in Rosen's description as RP^3.
What's your question again? You seem to have recognized it as S^2 x S^3 by your classification of RP^3-bundles.
08:33
hlo i need serious help from all mathmatics
is math confusing ? I did not think so .. but i think that due to prepositional logic conversion
Hi @Alessandro
i don't know when to use ^ and when to ->
What is the first order predicate calculus statement equivalent to the following?

"Every teacher is liked by some student"

∀(x)[teacher(x)→∃(y)[student(y)→likes(y,x)]]

∀(x)[teacher(x)→∃(y)[student(y)∧likes(y,x)]]
∃(y)∀(x)[teacher(x)→[student(y)∧likes(y,x)]]
∀(x)[teacher(x)∧∃(y)[student(y)→likes(y,x)]]
@Alessandro whats up
Eh... Not much, still studying numerical analysis
08:37
Keep doing the good work
ny one will reply?
@sittian ^ means and. -> means implies.
A implies B is equivalent to not(A and not B), which is equivalent to (not A) or B.
for all teacher, there exists a student who likes the teacher
what is answer?
and how it is ?
08:51
for all x, x is teacher implies that exists y where y is student and y likes x
ny better way to understand thse things
?
09:30
no idea
This is the smallest number of people I think I've ever seen in here.
[Division by zero] Constructing division by zero algebra (all types, including the not interesting ones) is actually very easy. You just take any group, and then pick any element so that it become a one sided additvie identity, and then fill in the addition structure based on the dominance and idempotence properties of zero terms.

Doing this will guareentee you get something that is distributive and associative. The important question ,however can be summarised in the following conjecture:

Group based division by zero problem: Are all associative division by zero algebra where the multipl
09:54
Hello
Let $f:A\to C$ and $g:B\to C$. Sorry if this is stupid, but $ \left\{ (a,b)\in A\times B\mid fa=gb \right\} \overset{?}{\cong}\coprod_{c\in C}(f^{-1} \left\{ c \right\} \times g^{-1} \left\{ c \right\})$
10:05
@Arrow, yes.
Thanks. Could you by any chance explain how this comment fits this?
0
Q: Trouble understanding Latin squares and group theory

RyanThis is more of a theoretical question, but I'm having trouble understanding why it is that Latin squares are generalizations of a group? I kind of arrived at this question trying to figure out why independent vectors were so important in linear algebra. To phrase the question differently, wh...

ok, so that means it is no for all finite groups
@Arrow I think the problem is that the fiber-product of topological spaces also gets a topology. Your statement is true for sets, but with topological spaces it may not be true (as the comment says).
Ahh, the counterexample shows it doesn't hold even for covering maps.. I see now. Thanks!
In mathematics, the Latin square property is an elementary property of all groups and the defining property of quasigroups. It states that if (G, *) is a group or quasigroup and a and b are elements of G, then there exists a unique element x in G such that a*x=b, and a unique element y of G such that y*a=b. The Latin square property receives its name from the fact that for a finite group (G, *), it is possible (in principle) to draw a Cayley table, which gives the element a*b in the row corresponding to a and the column corresponding to b; the Latin Square property says that this table will be...
10:13
@BalarkaSen Every Riemann surface admits meromorphic functions. I don't understand your comment.
Ok, its now settled (although I have no idea how to proof the uncountable case)
the answer is no in general for groups
@PVAL-inactive Oh, meromorphic, sure. I was thinking of holomorphic functions; were you not asking how many poles such a Z/2 symmetric meromorphic function must have?
*Z/3
@BalarkaSen I was asking what kind of tori admit Z/3 symmetric meromorphic functions.
Say a class of functions $\Sigma$ has the left cancellation property if $g,g\circ f\in \Sigma\implies f\in \Sigma$.

Does one of the following classes have left cancellation? Closed maps, continuous closed maps.
11:01
do you want $g\circ f\in \Sigma \forall g\in \Sigma\implies f\in\Sigma$? Because if not take $g$ to be the constant map.
if yes, then $g=\mathrm{id}\in\Sigma$ and $f\in\Sigma$ is one of the conditions
 
1 hour later…
12:23
just a downvoted post, unrelated
morning
12:38
Let $A\in\mathbb{R}^{4\times 2}$ and $b\in\mathbb{R}^4$ with $b\not=0$.
Is it possible that:
$Ax=b$ has exactly one solution AND $Ax=0$ has infinite solutions.
I think not, but I am bogged down on this.
Can you give me a hint?
infinite solutions for $Ax=0$ means, WLOG of generality: every line is (a -a) to begin with.
assuming we use (b,b) as our $x$
mmh, more precise. let $x=(a,b)$
then every line of the matrix is such that we get 0s
13:02
Is root 2 * root 3 = root 6??
@Abcd yes
Thankss !!
13:26
What is a compact metric space supposed to mean? In my understanding, compactness is defined for a subset of a metric space, not a metric space itself.
@Il-seobBae 한국인이에요?
The space itself is a subset of itself is it not
In mathematics, and more specifically in general topology, compactness is a property that generalizes the notion of a subset of Euclidean space being closed (that is, containing all its limit points) and bounded (that is, having all its points lie within some fixed distance of each other). Examples include a closed interval, a rectangle, or a finite set of points. This notion is defined for more general topological spaces than Euclidean space in various ways. One such generalization is that a space is sequentially compact if any infinite sequence of points sampled from the space must frequently...
Actually you usually define "compact" for topological spaces first and then generalize it to subsets of topological spaces by saying that they are compact if they are compact spaces with the relative topology
13:41
I wonder what the state of the Atiyah's 6-sphere paper is.
@DHMO 네.
@Il-seobBae 다른 한국인 알아요.
@DHMO I don't quite understand what you're saying. Do you want me to introduce other Koreans, or simply state the fact that you know another Korean?
@Il-seobBae the latter.
In fact he is called @kayak
Oh god.
@DHMO
13:49
@kayak hi
@BalarkaSen I found a recent discussion on that
Want me to look it up?
@balarka follow the links here
@DHMO If that's the case, that's nice. Thank you for letting me know.
and here, plus those in the comments
I know nothing about that by the way, I just saw that there was some discussion when this question got on the front page a few days ago
13:51
Ah yes, that seems to be what I have read as well
This as well
@Alessandro I am aware of the MO post
and the one in the reddit as well, @Krijn
@AlessandroCodenotti Thank you for your answer. I'll keep digging.
I'm afraid I can't help you then
Such a bizarre move by Atiyah
The reddit post seems to be mostly nonexperts sounding way too judgmental
14:13
hey @BalarkaSen can I ask a quick question about category theory ?
Not really interested, @Adeek, sorry
I just don't like to think about category theory
alright.
14:33
@arctictern would you like to discuss the following question ?
we're on a verge of a political emergency in this part of the world. lol.
Let A, resp B be sets, endowed with equivalence relations $\sim_A$, resp $\sim_B$. Define a relation $\sim$ on $A\times B$ by setting $(a_1,b_1) \sim (a_2,b_2)$ iff $a_1 \sim_A a_2$ and $b_1 \sim_B b_2$. Prove that $(A \times B)/ \sim \cong (A/\sim_A) \times (B/\sim_B)$ using universal property of quotients and products.
Hello, is anybody interested in discussing this question: math.stackexchange.com/questions/2080142/… ?
@arctictern I would like to check my solution with you. If it is possible.
@BalarkaSen why ?
it's rather complicated
we're having lots of political and religious riots
14:38
oh
Balarka, where ?
religion is a joke @BalarkaSen.
where i live
@Adeek religion is not a problem; the politics behind it is
The problem though that some people follow religious text word for word. They don't use their mind to analyze if what they are doing is logical.
hmm something is fishy in my argument.
14:58
@BalarkaSen not good.
@TedShifrin hmm
so was my impression, but i didn't really hear anyone talk about it recently
Hi @Alessandro — I'm leaving the doctor, heading to a coffee house. Back soon.
Enjoy the coffee
15:00
Balarka, I checked with authorities.
@TedShifrin Does that include your friend RB?
Ah. I have heard that he's an expert in this area.
Hey @TedShifrin
what question are you guys talking about ?
15:01
Do you think it's a good idea to talk about writing for proofwiki on college applications?
Would they care?
I want to somehow get across that I practice writing proofs regularly
15:13
Hello, @TedShifrin
@GFauxPas hi
@AkivaWeinberger shalom/hola
@GFauxPas: Admissions people don't know what a proof is:)
hi @TedShifrin @AkivaWeinberger @DHMO
It is really cool we can define fibered products and fibered co products.
Hi :) DogAteMy
Hi Karim
Hi @Akiva
I mean, DogAteMy
15:16
LOL
Stealing yet again!
I haven't done the $|x|^\alpha|y|^\beta/(|x|^\gamma+|y|^\delta)$ problem yet…
I never did this ...
I'm actually proud of it. :)
what is that problem ?
@Adeek For what values of the exponents does it have a limit at $(0,0)$.
15:19
cool
@TedShifrin plagiarize! / let no one's work evade your eyes / remember why the good lord made your eyes
And then when is it diffable at 0 with $f(0)=0$.
You found Tom Lehrer! Yay!
I actually found him quite some time ago.
Much better than anything Ted :)
@BalarkaSen So don't shade your eyes!
15:21
@Ted Sometimes. Not always. :)
I have grown a distaste for satire through the years, but I still like him
Puns, but no satire?!
@Ted what about the people reading my recommendation letters?
This is for grad
Ohhhh, I misinterpreted. Sorry @GFauxPas. For grad it's math profs. Put something in your statement of purpose. I also recommend writing things as specific as you can — hard problems you've liked working on, etc.
Cam you proofread my statement of purpose please?
It's one page
@TedShifrin puns are innocent. there's just too much misery to be laughed off as satire.
15:28
I'm on my phone, @GFauxPas, but I'll try.
BTW, I learnt Pontryagin's point of view of homotopy groups of spheres the previous day
worked out $\pi_3 S^2$
morning chat
Interesting point, Balarka.
Worked out how?
Hi @Semiclassic.
by classifying framed cobordism classes of 1-manifolds in S^2
So, by the substitution $x\mapsto|x|^{2/\delta}$ and $y\mapsto|y|^{2/\gamma}$, as long as the exponents were positive, I can assume $\gamma=\delta=2$? And then I can replace it with $r$ in polar coordinates
15:29
Trying to figure out if a calculation I wrote out this morning is 1) valid, and 2) useful.
it essentially boils down to doing so with a circle, which are classified by homotopy classes of maps $S^1 \to SO(2)$
Oh, excellent. I did that when I taught diff top at MIT.
so $\pi_3 S^2 \cong \pi_1 SO(2) \cong \Bbb Z$
@TedShifrin Ahh. Cool.
Precisely.
Funny thing is, all I'm actually doing here is a first-order linear system of ODEs.
15:31
Smart move, DogAteMy. Just remember to unwind at the end.
And then it's $r^{\alpha+\beta-1}|\cos(\theta)|^\alpha|\sin(\theta)|^\beta$ I think
namely, $i\dfrac{d\Psi}{dt}=H(t)\Psi(t)$ with $H(t)=A+Bt$
It seems the map $\pi_{n-k} SO(k) \to \pi_n S^k$ given by sending an framed cobordism class of an embedded $(n-k)$-sphere in $S^n$ to the corresponding homotopy class of maps $S^n \to S^k$ is called the $J$-homomorphism.
I even have $B$ being a diagonal matrix and $A$ having no diagonal parts.
So I need $\alpha+\beta\ge1$? Assuming $\gamma$ and $\delta$ are positive @TedShifrin
15:33
But the fact that it's matrices makes it still really hard :/
♫Nothing really matrix♫
Oh, without positive I think it's hopeless. So what's the final verdict, DogAteMy?
(the technical issue being, $H(t_1)H(t_2)\neq H(t_2)H(t_1)$ unless $t_1=t_2$.)
@TedShifrin One or more could be $0$
For $n = k + 1$ this is an isomorphism by the same reasoning. Google says it's an isomorphism for $n = k + 2$ too but I don't know how to prove that. It should fail hard for $n = k + 4$.
15:34
♫To meee♫
Lots of 4-manifolds not cobordant to the 4-sphere, let alone framed cobordant inside something.
You need the matrix exponential, Semiclassic. You need the Jordan form of A.
I actually need to go now. See you later
Matrix exponential won't cut it, @Ted. That only works when $H(t)$ commutes with itself at all times $t$.
15:36
Yeah, right.
There's still an exponential representation of the solution---namely, the Magnus series---but it's not simple.
I should also say that I have a specific $A,B$ in mind, but I don't want to ramble too much.
Too hard for Ted.
can someone give me a hint for what might the fibered product for sets might be ?
I think I got it hm
Fibered over what?
Here's the interesting thing about it. Suppose we write the solution as $\Psi(t)=U(t)\Psi(0)$, with $U(t)$ serving as the time evolution operator.
15:40
The fibered category is defined as follows:
I think it is
Cx(AxB)
No way.
In particular, we can get the behavior at large times from $\Psi(\pm T)=U(\pm T)\Psi(0)$. If we get rid of $\Psi(0)$, we have $\Psi(T)=U(T)U(-T)^{\dagger}\Psi(-T)$.
it should be the subset of A x B of pairs (a, b) such that f(a) = g(b), f and g are maps A --> C and B --> C
oh
The claim is that $U(T)U(-T)^\dagger$ has a fairly simple form in the $T\to\infty$ limit.
15:43
Pseudoinverse?
I was thinking of AxB at first but never thought about the subset issue yeah I agree with you @BalarkaSen
Eh, I just meant $\dagger$ as Hermitean conjugate.
thanks @BalarkaSen
I don't actually know how $U(-T)$ relates to $U(T)$ at this point, though. I presume it's determined by symmetry.
How do you get rid of $\Psi(0)$ without an inverse? I don't have paper here.
15:46
@BalarkaSen I guess for sets they don't exist always unless they agree at least in one pair.
Hm. I guess I'm assuming that $U(t)$ is unitary.
Ohhhhhhhh ....
Which sounds right to me on physical grounds, but is admittedly sloppy.
Well, now I don't feel like a doofus.
lool
15:51
Hello everyone! Can I ask a question?
If that's true, one furthermore has $(U(t)U(-t)^\dagger)^{\dagger}U(t)U(-t)^\dagger = U(-t)U(t)^\dagger U(t)U(-t)^\dagger=1$.
So $U(t)U(-t)^\dagger$ is itself unitary.
@DavidDavid: Just ask.
uhm, how do I upload a pic?
Hmm, I just realized that on my phone the upload button is absent.
If you have upload privs, then there'll be a button for that next to the text box.
I don't recall what's required for that, though.
15:55
there isn't but i ll post a link
what is lambda_chi_A
in the definition of Y
$\lambda$ is the particular value you're using as a reference point for the inequality. It's some arbitrary real number.
XY-pic is really awesome for drawing diagrams.
It's $\lambda$ times the indicator function.
$\chi_A$ is the indicator function for the event $A$.
ooooooh it s a product
15:57
It's sorta annoying how $\chi$ by itself looks like a subscript.
that subscript
confused me
thank you
Justifiably so :)
$A_\chi$ versus $A\chi$.
hmm, $A_{\big{\chi}}$
Nuts. Probably can't use \big in a subscript.
for uhm, random variables, we can define them as we want in general
is that right?

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