You mean why that sequence is exact at the end copy of $C_n(X; \Bbb Z_2)$?

why is $C_n(X,Z_2) \rightarrow C_n(\bar{X},Z_2$ surjective ?

Every simplex lifts to cover by map lifting, as simplices are simply connected.

@Adeek That's the same as saying it's exact at the end.

I don't understand can you explain more ?

I guess this is by path lifting lemma ?

@Adeek Huh, wait, what? $C_n(X, \Bbb Z_2) \to C_n(\overline{X}, \Bbb Z_2)$ is not surjective. It's injective.

Did you make a typo?

I meant to say the other side why is surjective ?

I meant to say the other side

I can't tell my rights from my lefts sorry

$p_\#$, right?

yeh

Well, I just told you. Map lifting.

Every simplex in $X$ lifts to one in $\overline{X}$.

why is that true ?

Any map from a simply connected space lifts to cover. Look at Hatcher chapter 1.3.

It's called the map lifting lemma.

oh I see

I should review that section really quick

Have you studied 1.3. before?

no

Then I don't think it's possible to do it quickly.

I have done munkres

or atleast this part

You should probably just read the proof of map lifting lemma.

yeah

@BalarkaSen So... my man.

What did you do last night?

@PedroTamaroff Depends on what last night means.

It's 5:30 here, so would what I did today from 12 to 5:00 count as what I did "last night"?

Or do you mean what I did the night the day before?

Damn. You must sleep.

I guess. What're you upto?

@BalarkaSen the following is an exercise e in Hatcher: calculate a simply connected covering of a $2$-sphere plus a diameter.

Consider the canonical universal covering of $S^1\wedge S^2$ with a slight modification.

Label the endpoints of the diameter as $a,b$.

Now place the sphere on the real line alternately at integer points at $a$ and then at $b$.

Mhm.

what is the proposition ?

Now draw arrows in opposite directions in each alternating interval.

That tells you how to collapse the thing.

I say that works.

Sure.

Awesome.

@BalarkaSen I can't find that map lifting lemma what proposition is it ?

@Adeek Let me have a look and tell you.

Prop. 1.33.

oke good thank you

Can anyone answer math.stackexchange.com/q/1722834/231595 ?

@crocket Please avoid using the chat to ask for answers. You might use it to discuss the question, say, or to discuss answers to it. But please avoid using it exclusively for the promotion of the question.

Is this a correct natural deduction proof?

What about the picture below?

Which one is correct?

In other words, can I derive anything from absurdity by reductio ad absurdum(RAA)?

the definition of n-sheeted cover is that it is a cover and the inverse image of a point is union of n points right ?

@MikeMiller, you around?

@Adeek Yes, preimage of some point is a cardinality n set. Try to prove that preimage cardinality of a covering space is constant over the base.

If you don't know it already that is.

(i.e., notion of "n-sheeted" is well-defined)

Can I derive $\phi$ by reductio ad absurdum without having $\neg\phi$ as an assumption to discharge?

I see

@BalarkaSen

Hey @EricStucky.

hello

@PedroTamaroff Mm?

I don't understand a construction in Hatcher. I mean, he says. Take $I^3$, and he says to identify opposite faces "by means of a unit translation followed by a quarter rotation." I am not sure what he means by "unit translation".

WTF, Balarka is still here?

hi mr @Pedro

it means you look at the two faces, translate one face to the other in $\Bbb R^3$ ("unit translation") then rotate a quarter.

That's the equivalence relation on the cube you're defining.

Oh, I know that problem.

Then you quotient by it to get your space.

Ah, well. That's slightly redundant.

@TedShifrin It's a cool one.

One of our grad students used all my colored chalk drawing that one on my blackboard.

Many moons ago.

Hah.

I always have difficulties visualizing (even with self intersections) the geometry of such spaces.

@TedShifrin That's so poetic.

glares at Mr Pedro for his sarcasm

I have never tried to visualize the actual space, @Pedro.

@PedroTamaroff: We know you're an algebraist. You got algebra built into you, alge-bro.

@BalarkaSen Does he ask to find the fundamental group of that space?

What's a fundamental space?

When I got to know mr Pedro, he was more of an analyst. Who knew I was going to waste all those years? :D

I know about fundamental domains.

He talked a lot about algebra even then, IIRC. But yes, probably he talked more analysis. I don't remember, I was too busy being a prat.

I don't think you were around when I first was summoned to chat by Pedro.

@PedroTamaroff Yes, if I remember right.

It was either fundamental group or homology. I've forgotten.

He asks to compute homology later on in 2.2 :D

Yes, he asks to prove that the fundamental group is the quaternion group.

Mhm.

oh, right, yes, that's what we did.

hi @TedShifrin

hi Karim

@BalarkaSen I am not really sure still, how the space is obtained.

Suppose I do this with only two faces. Shouldn't I get a "twisted torus"?

Draw a picture of a cube, @Pedro, and using arrows or colors decide which edges are identified according to the rules.

I am not sure where you are not sure.

@PedroTamaroff Yes.

It's homeomorphic to a solid torus if you do 2 faces.

Yes.

@BalarkaSen what do you mean constant over the base ?

If one point has preimage cardinality = n, all points do. I am of course assuming here that the base is connected.

@BalarkaSen But should I twist only one face and glue or twist both and then glue?

I see

One, then glue.

That is, does he mean to twist all faces by a quarter turn?

OK.

Take a pair of opposites, twist one face, and glue.

Do this with all pairs.

Well, one should first observe if this depends on the faces one chooses. I suppose he asks to do this with a right hand rule to obtain the same space regardless of the choice of face.

Yes, rotate a quarter clockwise for each.

Then it won't depend on the choice of the face you rotate.

If one does it with the right hand, it will be counterclockwise.

just a quick question at @BalarkaSen if we go back to the right hand side

why is it injective ?

@EricStucky: I'm in Indianapolis but I'll be carpooling with some people in 4 hours.

So it will be a while before I'm in Bloomington.

I define $\tau$ to be the map as follows

$\tau(\sigma) = \bar{\sigma_1} + \bar{\sigma_2}$

where $\bar{\sigma_1},\bar{\sigma_2}$ is the lifts of $\sigma$

oh I guess it is injective by definition

since there doesn't exist any other lifts.

No, it just follows from the definition. Unique lifting is used to prove well-definedness.

yeah

I see it comes from unique lifting lemma

this proof is very interesting

it like has homotopy and homology in it

OK. I drew the thing.

@PedroTamaroff It was a blobby mess, wasn't it?

I mean, I drew on the cube what is identified with what else.

Ah.

Thank god.

I thought you were drawing the quotient :P

I am not that crazy.

Dumb linear algebra question. Orthogonally diagonalizable mayrices are symmetric. The same proof doesn't go through in the complex case, as I see. Are there restrictions on unitarily diagonalizable matrices?

juw5 q w3d

hmm

just a sec

The theorem states that there is a lift, but what guarantees it to be unique ?

@BalarkaSen ?

I mean prop 1.33

@MikeMiller They are symmetric because the inner product is symmetric. The complex inner product is conjugate-symmetric.

That's not an official name.

I just made it up.

@Adeek Weird. Hatcher should have proved that. Anyway, maybe you should work it out.

Yes, I got that far. But the answer is not "Unitarily diagonalizable matrices are Hermitian."

He does prove it.

yeah he proves it

but after wards

So...?

ah, ok. i thought he would.

yeah

@MikeMiller Ah. Bummer.

Wait, Mike.

When you say orthogonally diagonalizable, you mean diagonalizable over an orthogonal basis?

Then diag. over an orthogonal basis is equivalent to being normal.

That is, commuting with the transpose.

Hermitian is a bit stronger than that, sure.

And when I say transpose I mean adjoint.

I don't agree in the real case. Are you only claiming this in the complex case?

Yes.

You need things to have eigenvectors.

Else the proof won't go through.

Ok, great.

I am happy.

I mean orthonormal basis.

My point is: if a matrix is diagonalizable over an orthonormal basis then it must be normal.

Conversely, say $A$ is normal.

Take any eigenvector, and consider its orthogonal complement.

You don't need to write out the proof.

OK.

Just show the orthogonal complement of an eigenvector is invariant.

Anything interesting here today?

@BalarkaSen So, if things are correct, I am getting four edges and two vertices. Is that right?

One thing I would like to clarify though @BalarkaSen one last thing I do agree that for each sigma inside $C_n(X,Z_2)$ which is a map from $\Delta^n$ into $X$ has a lift, however why does in particular has two lifts ?

$\bar{\sigma_1}$ and $\bar{\sigma_2}$

@PedroTamaroff Yep, seems right.

@BalarkaSen Did you solve this one?

@Adeek $\tilde{X}$ is a twofold cover.

Seems interesting.

Yeah, I did. The van Kampen exercises are all nice.

yeah however that is for the points in X not maps ?

I am confused

i.e for $x \in X$ $p^{-1}(\{x\}) = \{x_1,x_2\}$

Think about it.

ok

@BalarkaSen If I take a small contractible chunk of the inside, nice things happen.

I am not sure what you mean by that.

I'm trying to find an appropriate decomposition.

Well you have a decomp. One 3-cell, three 2-cells, four 1-cells and two 0-cells. You now need to find out how they are glued.

Eh. We just agreed this has four edges and two vertices.

It has three 2-cells, yes.

That's what I just wrote down.

You found it.

oh I see

so you se

the elements of $C_n(X,Z_2)$ are of the for

@BalarkaSen I was meaning to use Van Kampen.

form

$\sigma : \Delta^n \rightarrow X$

@PedroTamaroff Do it!

But you have to figure out how things are attached first.

so we have $\sigma(t) = x$

if we take inverse image of x with the covering map p.

we will have $p^{-1}(x) = \{x_1,x_2\}$

@BalarkaSen Well, my point was that if I take a ball interior to the cube, then the remaining set deformation retracts onto the boundary, so now I have a complex of dimension 2 to look at.

this means that we will have two lifts $\bar{\sigma_1}$ and $\bar{\sigma_2}$

right ?

@Adeek You haven't given me a proof yet.

@BalarkaSen In fact I am getting the known fact that $\pi_1(X)=\pi_1(X^2)$.

Sure. Attaching 3-cells or higher doesn't change $\pi_1$.

For a second I thought you were raised in Brooklyn.

Haha.

@BalarkaSen Well, now I will -- in an act of pure originality -- remove an open ball interior to a face.

Sure, as you will.

any hints @BalarkaSen ?

we know a lift exist

so in our case we know a lift $\bar{F} : \delta^n \rightarrow (\bar{X},x_0)$ exists

What are you trying to do @Adeek=

Wait no.

I am trying to see why $\sigma$ in $C_N(X,Z_2)$ lifts precisely to $bar{\sigma_1},\bar{\sigma_2}$ in $C_n(\bar{X},Z_2)$

where $\bar{X}$ is the covering space

of X

@PedroTamaroff

@PedroTamaroff No, there are two.

Yes, there are two.

we know such lift exist since $\Delta^n$ is simply connected

I am out of context here.

but why is it priscely 2 lift ? @PedroTamaroff

precisely*

He's trying to prove that a map from a disk has two lifts in a 2-sheeted cover.

@Adeek Do you understand what it means for a cover to have two sheets?

yes that the cardinality of $p^{-1}(x)$ is two

x

@Adeek You just read the unique lifting property. Why don't you apply that?

I proved something exists but I don't remember why I was bothering.

Uh-oh.

small x

I have no idea what you're talking about.

If you look at some interior point of the image of $\Delta^n$, you fix the preimage, then you apply the unique lifting lemma to prove that lift's unique. Then you look at the other preimage, similar logic shows that lift's unique too. So, there are precisely 2.

I have to go sleep now. Later, everyone.

Good choice, Balarka.

I see

just a sec

why does it have two pre-image ?

@PedroTamaroff maybe can you help me with this point ?

mainly I need to use this to prove Ulam Borsuk theorem

I think I cannot.

I understand the rest of the proof, however this part I don't understand.

@Adeek You don't need to get into such technical details to prove the Borsuk Ulam theorem.

If you have to, perhaps you're missing some bigger picture.

Try letting things sink in a bit more.

I am not limiting myself to dimension 2 @PedroTamaroff

Whatever dimension.

I am looking at the proof at hatcher which uses if we have a map from $S^n$ to $S^n$ such that $f(-x) = -f(x)$ then it must be odd.

OK.

$H_n(S^n)$ is isomorphic to $\mathbb{Z}$ and the image is of the form $n\mathbb{Z}$(as the image is a subgroup of $\mathbb{Z}$) here the degree of f is n.

Page?

175

@Adeek "$S^n$ is isomorphic to $\Bbb Z$" is not what you mean.

I mean

the nth homology of $S^n$

I just want to know why does it have exactly two lifts.

because he uses it to define $\tau$

I don't understand blarka explanation

I understand the rest of the proof

Well, you have to show that if $p:\tilde X\to X$ is a two sheeted covering and $f:K\to X$ is a map from a simply connected space $K$, then $f$ admits exactly two lifts.

Of course you have to use the fact that the covering is two sheeted.

yes

Well, try to prove it. Play with the definitions.

ok

well we know

since x is two sheeted then for $x \in X$ we have $p^{-1}(\{x\}) = \{x_1,x_2\}$

I am not sure how to proceed I will think about it

I guess I do see it intuitively

because the pre-image of an open set in U is of the form $p^{-1}(U) = \{U_1\} \cup \{U_2\}$

where $U_1,U_2$ is open elements in $\tilde{X}$

@Adeek Consider first this problem: in how many ways can you lift a constant map $K\to X$ if $\tilde X \to X$ is two sheeted?

two ways

oh I see

OK. And a map from a contractible space is homotopic to a constant map, and two homotopic maps can either be both lifted or not.

This gives, modulo details, what you want.

oh I see

I understand thanks a lot @PedroTamaroff

@MikeMiller: Actually, I'm not going to be in until 10:30 tomorrow... my flight got delayed :/ :/

(⊥E) is ⊥⊢ψ

RAA(Reductio Ad Absurdum) says If {Γ,¬ψ}⊢⊥, then {Γ}⊢ψ. Why do people say (⊥E) is the same as RAA?

@EricStucky: Shucks.

@PedroTamaroff why can two homotopic map either "both" be lifted or not

That would be the aforementioned "details", I think :)

i see

good good

that is great

math.stackexchange.com/questions/1724199/… Anybody can answer? Thanks in advance if you try.

what do you mean by your edit?

Do you mean you want like, technical definitions?

Also, you first say you want "all" the properties. But later you say you want "one factor".

@Victor

@Adeek Well, that's something you have to prove. Prove it.

@EricStucky - I need an explanation to one of the technical definition with an example that has great math detail

what does that mean

"great math detail"

haha

I know why now @PedroTamaroff

you said 'know', he said 'prove' :P

yeah I will

@EricStucky - Some typos, sorry, I mean logical and formal.

Thanks to spend time to edit my question @EricStucky

hey @EricStucky $H_{n + 1}(RP^n) = 0$ right ?

...

stand up

stretch

drink a glass of water

and then ask yourself the question again :)

lol

yeah I guess because $RP^n$ is a CW complex so it agrees with simplicial homology

if I have a matrix $A$ of rank $m$, am I correct to say that $(A^T A)$ is invertible?

no, look at $\begin{pmatrix}1 &0\\0&0\end{pmatrix}$

hey @BalarkaSen

@DanielFischer Thanks for the answer. Is it true that the spectrum of a multiplication operator $L^2\to L^2, f\mapsto mf$ is the set $\{\lambda\in \Bbb C: m= \lambda \text{ on a set of positive measure} \}$? Why can't this be the spectrum of a compact operator (excluding the trivial case $m\equiv 0$)? (Probably a dumb question)

Can someone give me example of function that cross its slant asymptote infinitely many times?

any integration lovers in? I know Chris's sister is very keen on tricky integrals

@s.

* @s.harp sorry I was not clear enough

I meant if $A$ is $n \times m$

basically its a full rank matrix

will $A^T A$ be invertible?

Yes, because $m≤n$ must hold from rank $m$. Then the map $A$ is injective. On the other hand if $A^T A$ is not an isomorphism you have there exists an $x \neq0$ so that $A^T A x = 0$. But then $x^T A^T A x = 0= |Ax|$ and $A$ cannot be injective.

@iwriteonbananas No, that's not the spectrum. If it were, a multiplication operator with an injective $m$ (e.g. $m(t) = t$) would have empty spectrum. The spectrum is the essential range of $m$, $\bigl\{ \lambda : \bigl(\forall \varepsilon > 0\bigr) \bigl(\mu(m^{-1}(\{ z : \lvert z-\lambda\rvert < \varepsilon\})) > 0\bigr)\bigr\}$. That can be the spectrum of a compact operator, but in the question we had $m(t) = t$, so an uncountable essential range (the interval $[a,b]$).

If the measure has atoms, there are nonzero compact multiplication operators. But for an atomless measure, a multiplication operator is compact if and only if it's $0$.

@Anush Chris's sister hasn't shown up here in quite a while ..

@DanielFischer Thanks, that's good to know

@BalarkaSen Have you seen this question? math.stackexchange.com/questions/1723832/…

$S$ is the right upper quadrant, $Y$ the lower left quadrant of $Z$ the line connceting $(0,-1)$ and $(0,1)$

Najib told me it isn't continuous but I don't really see why (surely I'm being stupid...)

Morning.

@MikeMiller Morning (well, afternoon here)

Morning Mike

morning

@iwriteonbananas UGH

Hi, by the way.

Hello gentlemen et al.

@DanielFischer In the proof on page 20 of that supplement (see here), the author assumes that $f(x)$ is continuous on $[0, \infty)$ and that $H(x) = \int_{0}^{x} e^{-s_{0}u}f(u) \, du$ is bounded for all $x \in [0, \infty)$. (Well, at least I think he's assuming the latter. There is clearly a typo.)

If he had instead assumed that $\int_{0}^{\infty} f(x) e^{-s_{0}x} \, dx$ converges, and defined $H(x) := \int_{x}^{\infty} e^{-s_{0}u} f(u) \, dx$ , wouldn't the proof have shown that $\int_{0}^{\infty} e^{-sx}f(x) \, dx$ converges uniformly on $[s_{0} , \infty)$ in the case that $f(x)$ is continuous on $[0, \infty)$?

@RandomVariable It would have needed a small modification. He couldn't just have used the bound $M$, but would have needed to use $\sup \{ \lvert H(x)\rvert : x \geqslant r\}$ to get the bound arbitrarily small uniformly in $s$, since $e^{-(s-s_0)r}$ doesn't do that any longer when we let $s$ come arbitrarily close to $s_0$ (or even take $s = s_0$). Note that the continuity of $f$ is only used for the integration by parts, which is valid more generally.

@s.harp thanks bud

@iwriteonbananas: Najib misunderstood the space in the OP.

@DanielFischer Exactly. That modification is key.