$K = \{(x,y,z) \in R^3 | x^2 + y^2 - z^2 = 1\}$ , How do I show that it equals to those 3 vectors: $(1,0,0),(0,1,0), (1,1,1)$? @Studentmath

You want to show it equals -only- to these 3 vectors, or to show these three vectors are part of K?

Since the first isn't true - the latter is easy, you stick each of these vectors in and it works.

$1^2+0-0=1, 0+1^2-0=1, 1^2+1^2-1^2=1$

@Studentmath That's easy to show that way. I don't think in the exam they will accept that. Will they? Don't I need to show that somehow in another way? Although in the solution they didn't even bother explaining what you just wrote.

I'm not sure what the question is, but if you just want to show that these 3 vectors are part of K, it's sufficient and necessery to show they uphold $x^2+y^2-z^2=1$

So if I want to show that those 3 vectors $\in K$ it's enough to explain as you wrote? I don't need to expand somehow the equation $x^2+y^2-z^2=1$ in some way to get those 3 vectors algebrically

No need for that, but not sure what the question is, if you do not bring up a reason as to why you thought of these three vectors one might think you are cheating.

Heyas @Ilan and @Studentmath

@TedShifrin Heya :D

GRRR bad @Hippa

Did @GTR tell you my proof of the polynomial problem?

@TedShifrin Have a done a new bad thing ? xD

@TedShifrin no

not that I know of, but you probably have ...

Oh wait is it

mathb.in/17695

No, that was his proof

Hey Prof. @Ted

@Studentmath I would just say the way you explained :P

@TedShifrin I don't know yours then

Huh? @Ilan

@TedShifrin I thought about saying $1^2+0-0=1, 0+1^2-0=1, 1^2+1^2-1^2=1$

@TedShifrin Is that enough?

That's a whole surface, called a hyperboloid.

Yeah, enough to say (imo) "We can notice that these 3 vectors upholds the requirements"

Oh yeah, equals is really bad word-chosing, @Ilan.

@Studentmath Alright. :P

You should say includes.

@Studentmath Yup

@TedShifrin Let $K$ be the closed disc, with two smaller open discs deleted. [So that it looks like (o o)] Show that if $f: K \rightarrow K$ is a homeomorphism without a fixed point, then $f$ reverses orientation and cyclically permuted the boundary circles.

(I'm stuck near the end of it, but it's a fun problem.)

Dunno @Mike

@TedShifrin If I can show that at least one of the boundary discs must move to a different one, I'm done

@Hippa: Suppose $z$ is a complex number satisfying $\sum \dfrac 1{z-c_j} = \dfrac 1a$, where $a, c_j\in\Bbb R$. Show that $z$ must be real.

Also, I wasn't asking for a tip, I was telling you the problem... :)

That's one of topologists' favorite shapes, @Mike. It's called a pair of pants.

@TedShifrin I'm not supposed to use some theorem i don't know, right ?

Ah. I was thinking of it as the sphere with three discs deleted, @TedShifrin

No, you're just supposed to understand the geometry of complex arithmetic slightly, @Hippa.

Of course, they're all the same.

Easiest example to illustrate things can be cobordant without being homotopy equivalent, @Mike.

The view I'm using makes it easy to see that such a map exists

You removed awfully quickly.

@TedShifrin Is the sum finite ?

@TedShifrin OK, now I really do have an easier one.

Yes, yes, @Hippa. The $c_j$ are the roots of $P$, sorry.

Yes? @Mike

@TedShifrin Well, nowhere in the definition is it required that the cobordism be connected...

so any finite discrete set is cobordant to another one, as long as the difference in their cardinalities is even :)

Um, that isn't interesting, though.

Ah, but you never said easiest interest example.

braces for slap

You know me by now, @Mike.

Anyway, if I can show that at least one of the boundary circles moves to another one, I can show that they cyclically permute: assume they don't; then one of them is sent to itself (in a manner without fixed points) and the other two are swapped. Collapse the other two boundary circles to two points; this induces a map from the disc to itself without fixed points, which is nonsense.

So they cyclically permute. In that case, collapse all the boundary circles, each to a distinct point; so the map $f$ induces a self-map of the 2-sphere without fixed points, which thus must have degree -1 and be orientation-reversing.

(To see that an automorphism without a fixed point exists, imagine the three deleted open discs as placed on a sphere, each of whose center is placed equidistant from the others on some great circle, all of which have the same radius; rotate the sphere $2\pi/3$ in such a manner that the great circle is mapped to itself (I'm not sure what the word for this axis of rotation is), then reflect it across the plane that contains the great circle.)

Ah, nice, @Mike.

But I imagine that seeing that they must move at all is the hard part, @TedShifrin

So what do you get if you go back to your sphere-collapsing trick?

@Hippa: You got it?

@TedShifrin No :c

If $z$ is not real, then the imaginary part of $z-c_j$ is positive or negative and the same for all $c_j$.

And ?

@TedShifrin Yeah, I've thought of that; you get an endo with exactly three fixed points. I'm not sure how to use that. (I'm aware that there's a formulation of Lefschetz that counts fixed points, or something? But I don't know it.)

So the reciprocals have imaginary parts of all the same sign?

Yes, @Mike, certainly with a smoothness hypothesis ... that's back to intersection theory stuff you still have to learn.

Reciprocals ? you mean complex conjugate ? or .... ?

Reciprocal = $1/\text{blah}$.

Planar graphs will be the end of me..

@Studentmath: You keep saying everything is the end of you, and yet — here you still are :P

@TedShifrin Well, by Whitney any map is homotopic to a smooth map, but I'm not assuming my homeomorphism is smooth, and I certainly don't want to homotope since I lose all my fixed point info

Well, even topologically, Lefschetz number is an intersection number (that's what cup product in cohomology means) ...

@TedShifrin Well they all have opposite signs then for their img part

The issue comes with non-transverse intersection. Then we don't know what intersection number means topologically, unless you make things PL or something.

Right, @Hippa, but the sum definitely is NOT zero.

@Ted I think I am addicted to complaining :P

@TedShifrin I know that .

So we're done, @Hippa.

@TedShifrin uh ??

$z$ has to be real @Hippa. $z$ is a root of $P-aP'$.

I don't get it at all

Okay, the im part of $1/z$ has a sign opposite to $z$'s and the sum of $1/(z-c_j)$ isn't $0$... and ?

It's the same beginning as @GTR's solution. We write $P(x)=(x-c_1)\cdots (x-c_n)$, and note that $\dfrac{P'(x)}{P(x)} = \sum \dfrac1{x-c_j}$.

@TedShifrin I didn't look at @G.T.R 's one

the sum of the imaginary parts of $1/(z-c_j)$ can't be $0$, but yet that sum is $0$ because $\sum 1/(z-c_j) = 1/a\in\Bbb R$. !!!!

Why can't the sum of the imaginary parts of $1/(z−c_j)$ be $0$ ? (i thought you were talking about the global sum earlier)

Me too @Studentmath :P I'm complaining about your complaining.

I guess @Ilan disappeared.

If $z\notin\Bbb R$, it won't be $0$, @Hippa.

@TedShifrin why ?

That was the argument you made.

Where ?

All the terms have the same sign on the imaginary part, hence can't sum to $0$.

oooooooooh

Sure

Formidable :P

But

Don't "but" me

Why can we write $\dfrac{P'(x)}{P(x)} = \sum \dfrac1{x-c_j}$ ?

Logarithmic derivative! Or, just do it.

@TedShifrin OK, I looked up Lefschetz-Hopf FPT. Is it true that any fixed point has nonzero index?

Ahh, nevermind, that attempt won't work.

Why not?

Well to me $\dfrac{P'(x)}{P(x)} = k\dfrac{(x-d_1)\cdots (x-d_{n-1})}{(x-c_1)\cdots (x-c_n)}$

No, @Hippa. Take $\log P$ and differentiate ... or use the product rule.

You're correct, if the $d_j$ are the roots of $P'$, but you're missing a coefficient in front, anyhow.

I've never used the log derivative other than to approximate functions :/

When you take complex variables, you'll use it zillions of times, @Hippa. But it's very useful.

@TedShifrin Ah! It does work. Is it true that any fixed point has nonzero index?

Just differentiate $P$ by the product rule, @Hippa, and you'll see.

Ah! No, it doesn't work. Who knows?!

If you have a reasonable definition of index, sure. I'm wiggling the function if I have nontransverse intersection with the diagonal, and adding up the contributions of the wiggles. This all makes sense PL.

Why doesn't it work?

@TedShifrin I want to use a parity argument: $\sum \iota(f,x) \neq 0 \pmod 2$ because the $\iota$ have to be $\pm 1$ or something. But I don't have the background to assert that the $\iota$ have to be $\pm 1$.

$P'=\left((x-c_1)\cdots (x-c_n)\right)'=\sum\limits_{k=1}^n(x-c_1)\cdots(x-c_{k-1})(x-c_{k+1})\cdots(x-c‌​_n)$

(The Lefschetz number must be 0 or 2, because the degree of a homeomorphism must be -1 or 1.)

I'll have to think, @Mike.

Right, @Hippa. Now divide by $P$.

@TedShifrin If that's true, then no sphere has an automorphism with an odd number of fixed points. Not sure if I buy that.

No, it's certainly not. Consider the extension to $S^2$ of, say, $z \mapsto z+1$. The only fixed point is $\infty$.

$\frac{P'}P=\frac{\sum\limits_{k=1}^n(x-c_1)\cdots(x-c_{k-1})(x-c_{k+1})\cdots(x‌​-c‌​_n)}{(x-c_1)\cdots (x-c_n)}=\sum\limits_{k=1}^n1/(x-c_k)$

Ooooooooooooooh

Magic

I believe some dark force is at work here

:c

$\log$ wins @Hippa

@TedShifrin dar too

@TedShifrin What we're trying to prove is false. That method won't work.

OK.

Hmm... I lied! I think it will.

<--- gives up talking to @Mike.

My counterexample didn't work. I was going to suggest $z \mapsto z^2$ since the extension would have 3 fixed points. But it's also not even close to an automorphism.

Yeah, I think the index has to be $\pm 1$ when the map is a homeo away from the fixed point (locally).

Certainly true for smooth maps (exercise).

Even if it doesn't nail the problem, understanding stuff with extra hypotheses is not a bad thing to do.

OK, I'm outta here ... Time to drive home and do some work.

This is a really tasty problem.

@Ted what do you think about the proof I linked ?

oh, he's out, my bad

Sorry @G.T.R, I chased him off

Hehe, I can't not be amused by the fact that this was one of the Community Bulletin linked posts just now ;)

I'll admit I chuckled

@Ted think he went back to studying, as he should.

Oh he left.

Haha @G.T.R

Consider the mapping $re^{i\theta}\mapsto \frac{r}{1+r}e^{i\theta}$ which maps the complex plane onto the open unit disc, preserving the argument of each point. This is a homeomorphism. Now we can take the closure of the disk in $\Bbb R^2$ and obtain a compactification of the complex plane which effectively adds one point at infinity for each value of $\theta\in [0, 2\pi)$. Does this object have a name?

It is sort-of analogous to the two-point compactification of $\Bbb R$.

@TedShifrin cool proof

Maybe this is just a projective plane.

@MJD Certainly it's not just the projective plane, since it's topologically very different...

Unfortunately “projective plane” refers to a number of different things.

Any of them assign a single point at infinity to each line though, correct?

(Rather than two.)

It's not the manifold that results from identifying the antipodal points on the boundary of a closed disc, if that's what you mean.

I don't think I'll be much help, since I don't think I quite grasp what you're asking.

I would guess that my construction adds to each line two points at infinity, one in each direction.

And that parallel lines intersect at both those points.

Well, do you understand the homeomorphism I described?

yes

yeah, the closure of any two parallel lines is going to be those lines plus the two points at infinity

which I guess would best be interpreted as saying that the two lines intersect there

How do one prove:$$\int_0^\infty \frac{sin(u)}{\sqrt{u}}du = \sqrt{\frac\pi 2}$$

@cc contour integration

3-0

france

@G.T.R :D

This is a nice one:

Let $G$ be a $4$-regular planar graph. Prove or disprove: $G$ has a cycle of length $3$.

And the proof is amazingly simple!

If I have it right at least.

5-0

@G.T.R oh my

this is mass murder.

OK, question : Is $e^\pi \approx \pi^e$ just a coincidence?

maybe because $\pi \approx e$

Nah.

What do you mean by $\approx$ in maths anyway

$2^4 = 4^2$, @cc

@Hippalectryon approximately equal

@BalarkaSen Yes but how do you define that ?

0 5 loool

@BalarkaSen $100000000000 \approx 950000000000$ but $10 \text{ not }\approx 50$

@Hippalectryon The error is less than $1$

@BalarkaSen That's just like .... so arbitrary :c

$2 \approx 2.9$

@TedShifrin Welcome back

@Studentmath Error is equal to 1

Not allowed.

Merci beaucoup, @Hippa. I take it you understand the proof now. I should not have given it away to you. I won't ever do that again!

@Studentmath $\gamma \approx 1/\sqrt{3} \approx \pi/2-1$, you cant explain that

@TedShifrin Don't you ever dare doing it again :D

@Balarka fixed.

Hi again @Studentmath. Howdy @N3, @Balarka.

Heya Prof. @Ted

Hello Professor

@N3b what is this I don't even...

Pas de problème, @Hippa.

@Studentmath EM constant

STAHP

@TedShifrin Your proof is seems much easier than @G.T.R 's :c

Too much going around

damn it 5-1

Yeah, @Hippa. I don't like all the belabored algebra in his.

@Studentmath :p

belabored ? :D

Yes, one has to check torturous algebra details ...

Oh i thought that actually was a typing mistake xD

@Ted chat.stackexchange.com/transcript/message/16188961#16188961

BTW, if you're alleging a dark force, I already told you I was satan. No, not a typing mistake.

contour integral looks like a particular case of Stoke's theorem

@TedShifrin I joined the dark side long ago :D

@Hippa Go meme.

LOL @DanielF ... I don't think I'll be sending him any love notes.

@cc Other way around

Any line integral is Stokes's Theorem waiting to happen, @cc

@cc Not quite

@TedShifrin Aye sir

@Hippa: I will seriously stop talking to you if this continues.

6 goal avreage on that stadium, impressibe

@Hip I stay with the chemistry 0.05%

Fun is fun ... but ...

@TedShifrin You're the one inciting me :P (ok ok i stop)

regarding $\approx$

@Studentmath ooh

Oh darn, my proof has a flaw..

@DanielF ... any idea who he is, anyhow?

@TedShifrin You mean in real life? Nope.

I've corrected him more than one time on reasonably serious errors ... But he does love to give everything away to everyone.

2-5

@N3buchadnezzar So random -_-

@N3b I can't say I am unhappy.

damn... 5 minutes left anyway

@Studentmath but $\gamma$

Yes, @Ted, so did I. On the whole, his hit vs. miss record is better than Mhenni's though.

Stop it!

@N3buchadnezzar Know this one

Oh, way better than Mhenni's. And Mhenni is more destructive than constructive.

One year is exactly 10! seconds

@N3b I really like his comics

Good for you. Kool story bro

@TedShifrin Interesting enumeration problem : Draw the diagonals of a regular $n$-gon. How many interiors are triangles?

Elementary, but very hard in general.

How many interiors are triangles?

6-2

I think something like this is a famous counting problem that starts out with a plausible inductive formula, which then goes kerflooey at $n=6$ or so.

nah'

@G.T.R Such randomness

The enclosed areas formed by diagonal segments, @TedShifrin

5 - 2

@TedShifrin Precisely.

This is the one.

Yeah, I was shown this by a colleague pretty much shortly after I arrived at UGA ... so 30+ years ago.

@G.T.R Bon on est more pour les prochains matchs, on a épuisé notre quota annuel de buts :3

@TedShifrin Nice stuff, isn't it?

@Hippa: But I said no more "but"s! :P

@Balarka: Once again, this has never been my taste in mathematics.

@TedShifrin >8c

@TedShifrin I know.

Just sayin'

@G.T.R It went back to 5-2 ???

@Hippalectryon A negative goal

yeah. The match ended before he scored

@Hipp The match ended as he kicked.

By the teams in the mirror universe.

Eww

So didn't count.

@Hippalectryon they didn't count the very last ball going into the net

Ewwwwwwwwwwwwww

Negative goals. I love it.

@Studentmath Ah sad

Better so.

@TedShifrin Bows

7 goals, thats more than 6.

wow

More like wOow

@Studentmath Who kicked?

Benzema?

WO.ow

@BalarkaSen Switzerland kicked the bucket :p

@N3buchadnezzar that looks like a mad scientist with funny hair :3

@Hippalectryon ah

@Balarka the French.

WO.ow @N3buchadnezzar

Who da hell dared star my stupid message

WWWWO.owwwwww

Such hair

Hi @Ted

@Studentmath OK, which player?

No idea, didn't notice

trying to think of a way to explain why a simple 4 regular graph must have at least 6 vertices

I gave up talking to you, @Mike.

@TedShifrin Congrats for being the source of 70%+ of the stars here :3