The way I know how to show $S^ \infty$ is contractible is to show the identity is homotopic to the shift by 1 map (f(x_1,x_2,...)=(0,x_1,x_2,...) and show that map is homotopic to a constant map.

Ah, that's much better actually

A similar construction to that at least works for any $l^p$ (p \geq 1)

With the CW structure I mentioned I do it by producing a homotopy which contracts S^k in S^(k+1)

and then gluing it all carefully so that it's continuous at infinity

So I cannot deduce it from what I know

@PVAL-inactive Right.

@JeSuis my argument is completely elementary, and I figured it out within days of learning what a homotopy was.

and so?

so you should be able to fill in the gaps and deduce it.

(it's the same as in Hatcher book.)

Actually I think my argument and yours are secretly the same

yours just uses less heavy terminology

I was trying to prove it using the facts I have.

What's an example of a group for which the inner automorphism group is abelian ?

@Astyx S_3?

Thanks

You might want to check that

Oh yeah that's quite obvious, I was silly

@Astyx nevermind its wrong

inn(S_3)=S_3

Is it ?

Oh yeah my question changed in my mind upon reading your answer

Suddenly I was looking for a non abelian Inn

So my quesiton still holds

I do remember in algebra

learning that if Inn=G/Z(G) is cyclic, then it is trivial.

but I don't remember abelian examples.

Yup, but I read somewhere else that abelian is not excluded

I'd imagine if you plug in some basic examples you'd find one.

In fact, that any non cyclic finite abelian group is isomorphic to an Inn

@Astyx ça donne quoi en français ? ^^

Inn(Q_8)?

Un groupe fini abélien isomorphe au groupe des automorphisme intérieurs d'un autre groupe

What's $Q_8$ ?

that's gotta be order 4, right? so it's abelian

quaternions

Yeah G/(Z(G) is probably easier to see as order 4.

then actually thinking about automorphisms.

at least in my head.

yah

@JeSuis @Astyx o/

Hi @Hippa !

@Hippalectryon salut

Thanks @Balarka

no problem

T'as fini ton stage @Hippa ?

@Astyx encore 3 jours

Courage ! :p

merci :-D

On au eu un mail nous disant qu'il fallait préparer pleins de chèques pour le retour >.>

@BalarkaSen @PVAL Consider the action of $SO(3) \cong \Bbb{RP}^3$ on $S^2$. What's the subspace corresponding to $\Bbb{RP}^2$ and what's the map from it to $S^2$? (Is it degree 1 or 0?)

De chèques ?

It's the same as exponentials of traceless matrices with matrix operator norm pi

@Astyx Pour payer diverses charges

Mais c'est pas couvert ?

that's an interesting question

@Astyx La tangente, la KES, le self etc

@Astyx et les cautions

SO(3) is naturally diffeomorphic to the unit tangent bundle of S^2. Can I see RP^2 inside that?

Ah d'accord

@Balarka Good call. Then my map is just projection.

Right.

I think if you think about $SU(2)$ as $S^3$ the $\Bbb RP^2$ is the lift of an equator.

and it's pretty easy to identify explicitly SU(2) with S^3 and see what this equator is.

though I haven't thought about what happens to it under the covering map coming from quaternion conjugation

the $RP^2$ should be an image of an equatorial $S^2$ I mean.

If I think about RP^2 as S^2 with opposite points identified, isn't RP^2 in SO(3) like the rotation by pi angle around any axis?

Yeah that should be right.

I guess via the tangent bundle description that says the map to S^2 is degree 1

?

I think you just count preimages of a regular value

I am unsure about my picture but RP^2 in SO(3) is then just union of a point (corresponding to rotation by pi about the axis going through that point) on the circle in the fiber, indexed over all the fibers, right? Then by local trivialization that maps looks like identity on a chart to me

Which should be any point besides the original point of S^2

ah. that is true.

So there's certainly not one point of $RP^2$ in each fiber

or else $RP^2$ would be diffeomorphic to S^@

I think the critical values are probably x and its antipodal point at least.

(If we are thinking about the map coming from the action where we apply a rotation to x)

Yeah I agree

I'd think the critical set would be reasonably complicated but I'm not sure what it is.

Certainly [x] is in it, but is there really nothing else?

I don't understand how can there be anything else

The action is only "degenerate" near the points where the axis hits the sphere

So then

between any two non-distinct points

does there exist a unique axis

in which rotation by $\pi$ interchanges those two points

True. Just take a great circle passing through those two points

and take the central axis

because there is a unique great circle

No, sorry, take that back. How about two very close points on the sphere?

between any two points.

Yes there is a great circle passing between those points

and take the halfway point on the circle as your axis.

There is but how can you rotate by pi to take one to the other?

(doesn't matter which direction you go in)

oh, "interchanges". I'm happy now.

In the way I said.

The words mean the same thing

There is still a rotation to take on to the other.

Take a great circle between them and take a point on it equidistant to both.

Rotate along that axis.

Right, got it.

For some reason I was looking at the central axis to the great circle. I'm just dumb I guess

I don't buy it's deg 1. Most deg 1 maps RP^2 --> S^2 has critical set at least a circle.

I made a layman explanation of FAIL if anyone wishes to read

FAIL without @ symbols of course

FAIL is easily explained by an "F" on the math test

@BalarkaSen Is it that easy to explain? :P

@Balarka: I guess I'm having trouble seeing any (non-stoopid) maps $\Bbb P^2\to S^2$.

most F's should have some side notes

@TedShifrin Quotient P^1

Oh.

I was looking for even maps on $S^2$.

Compose it with a degree-2 map from the sphere to itself?

or, how i like to say that, pinch everything outside a chart

The usual degree-2 map I think of isn't even.

Yeah it leaves two points fixed

Oh. Whoops. Confused even degree map for even map

I should leave

Okay its critical set is a circle

The critical value is just the antipodal point

The only points which have non-unique great circles between them are antipodal points

AH

nice

and in that case there is an entire $S^1$'s worth of them

@TedShifrin You could take the even map that starts by contracting the equator to a point, giving $S^2\vee S^2$

So it really is the standard deg 1 map $RP^2 \to S^2$

killing the 1-skeleton.

and then maps them both to a sphere

Fun

Yeah, that's equivalent to what Balarka did, DogAteMy.

I knew this stuff a few years ago when I ran the summer topology qualifying exam study sessions ... I'm just brain dead :)

It's relatively easy to see (I figured this out on my walk) that if two points on S^2 are connected by a rotation then if you take an arc on a great circle from the axis to one of the points.

then the result after the rotation is an arc from the axis to the other point

so two such points are equidistant on a great circle passing through the axis.

so as long as there is a unique great circle between two points, there is a unique $\pi$ rotation between those two points.

@TedShifrin hi

hi @Liad

i got a small topo. question :)

yeah?

its about quotient space, we have $\Bbb R$ with the standard topo

and $[0,1]$ with the equivalent relation $x_1 ~ x_2 $ if $x_1,x_2 \in [0,1]$

So you're identifying everything in $[0,1]$ to a single point?

the qoutient space is $\{[0,1] , \Bbb R - [0,1] \}$

am i right so far?

Well, you can't write it that way.

hm, what do you mean?

So we have an equivalence relation defined on all of $\Bbb R$, identifying precisely all the points in $[0,1]$ and that's it?

yes

What you wrote is writing sets as if they were points.

So what do you think this looks like?

2 points doesn't it? one for points in $A$ and one for those that aren't

No, no.

All the points not in $[0,1]$ are still exactly what they originally were.

why? they are equivalent to one another

That's not what you told me.

I asked carefully.

$x_1$~$x_2$ if $x_1,x_2\in A$

NOT in $[0,1]$

$A=[0,1]$

what am i missing?

My computer is about to update ... so I'm leaving. You're not paying attention.

for when you're back. maybe the space is $\{[0,1] , \{x: x\notin [0,1] \} \} $ ? (this is what i meant earlier)

@Liad: So the points are $A$ (the one point you get identifying all of $[0,1]$ to a single point) and all the points of $\Bbb R-A$, yes. What topological space do you have, though?

23:37 reminds me of 1337 :D

Hey @Mike!

google.de/…

Hey, for the proof of the Brouwer's fixed point theorem via Sperner's lemma, why can't we just use the restricted version in which the triangulation is assumed to be the simple one with four triangles?

It's trivial to see that the lemma holds in that case.

this messed me up when it said 24 degree course and then bearing again?

so is the ship first going on a 24 degree bearing and then when it passes the lockheart it changes the bearing to 280?

Have a beatiful sleep folks

I'm going to bed now

@Alucard You too!

Cya pal