I think we are misunderstanding each other

My American Dream is to be Uncle Sam, haha

@Huy Good morning! Bye...

my definition of the solenoid is as the inverse limit of the system above

call it X.

It's 5 am here

the $p$-solenoid is the space $\displaystyle{\lim_{\longleftarrow}(S^1\stackrel{\times p}{\leftarrow} S^1\stackrel{\times p}{\leftarrow} S^1\leftarrow \cdots)}$ yes?

then the short exact seq would be $1 \to \check{\pi_1}(X) \to \check{\pi_1}(X/\mathbf Z_p) \to \mathbf{Z}_p \to 1$

@DanielRust mine is homeomorphic to that

Your arrows are backwards, @DanielRust

they're essentially equivalent

sorry yeah

@Hakim My idea would be to start from the right side, and a good starting point might be this identity

@Hakim ^^^ (so, yes, I have a starting point for it)

There's a natural map given by mapping each $S^1$ to itself via the $\times p$ map.

@DanielFischer Do we have to find the sum of the elements from the position low till the position p and then compare it with the sum from the elements from the position low+i till the position p+i, for $i \leq \frac{m-3}{2}$? Or is there a better way?

@DanielRust yeah but that is homeomorphic to mod p map by stretching each circle by p, what i did above

@BalarkaSen ah right, I see.

you have yet to explain 1 --> Z_p --> Z_p --> Z/pZ --> 1 though

what?

@BalarkaSen I thought there was a natural $p$-fold covering map of the $p$-solenoid onto it self, but I think I'm mistaken.

i believe the p-solenoid is cech simply connected

i haven't proved it however, but it looks pretty obvious. pick up a nice cover with no three open sets intersecting at the base S^1. lift up to the solenoid (fiber product).

you get a covering with nerve being just a 1-simplicial complex.

I still don't know what the cech fundamental group is, so I can't confirm :P

I can tell you the Cech cohomology is $\mathbf{Z}[1/p]$.

(first cohomology)

@DanielRust take a connected space with a cover, construct nerve, compute the fundamental group of the nerve. the pro-group is the inverse limit of all these fundamental groups with limit running through open covers

i am working with the first one (just the group, with a fixed open cover).

i don't want to meddle in the business of profinite fundamental groups, it'll only mess stuff up

then yes, I think the solenoid is cech simply connected

as you have the limit $\mathbf{Z} \leftarrow \mathbf{Z} \leftarrow \cdots$ with all maps $\times p$.

@DanielRust It's a fanatic group from Czechoslovakia.

LEL

What is LEL?

KEK

laughed out enormously loud

ok, so if 1 --> pi_1(X) --> pi_1(X/Z_p) --> Z_p --> 1 holds, then this would imply cech pi_1 of X/Z_p is Z_p

I have exactly zero intuition for how $\mathbf{Z}_p$ acts on $X$ in this case.

would it be induced by some permutation of the Cantor fibers?

well translate every S^1 by an integer.

no, the cantor set is left invariant under this action

the set may be invariant, but an action will be induced on the set right?

not sure what you are referring to.

the action is the obvious action of Z_p on \prod R/p^iZ by termwise addition.

restrict to the solenoid, done

so if you further restricted that action to a fiber, what would it look like?

as a homeomorphism of the cantor set

oh. an elt of Homeo(Cantor set)

right

@evinda Sorry, internet connection troubles. You look at the elements next larger and next smaller than what has been selected so far. You compare which one is closer to the median. If the smaller, you move the small-pointer one forward, if the larger, the large-pointer one towards the end of the array. Stop when you have $p$ elements.

For instance, I can imagine the action cyclicly permutes the obvious clopen subsets depending on the level of the profinite group the element lives in.

wait.

@robjohn are you around? I wanna let you a message privately ...

why would it even permute the fibers.

@BalarkaSen Well if it fixed every fiber pointwise then it would just be the trivial action. So if it fixes the fibers set-wise, then the individual points inside the fibers must be permuted.

@teadawg1337 HAHA

@DanielRust take an elt (x_0, x_1, x_2, ...) of the solenoid. take an elt (y_0, y_1, y_2, ...) of Z_p embedded inside the solenoid. the action is (x_0 + y_0, x_1 + y_1, x_2 + y_2, ...)

@Hippalectryon ?????

@teadawg1337 Your comment there has a wonderful typo

Lol

................................................................................‌​..................

LEL @Hippa

I didn't even notice.

@BalarkaSen ok I can believe that. It's difficult for me to visualise because I don't normally think of solenoids in this way as quotients of the reals.

@Hippa There, I corrected it

ah i see

@teadawg1337 Aww too bad

i am thinking of ... --> R/p^2Z --> R/pZ --> R/Z @DanielRust :)

@Hippa I'm not deleting the original one, though xD

tbh I normally think of these kinds of spaces as suspensions of cantor-set homeomorphisms :P

in this point of view, Z_p nicely embeds inside

Holy cow, the downvotes are flying on that question now.....

If $h\colon C\to C$ is a homeomorphism of a cantor set, the suspension of $h$ is $(C\times [0,1])/{\sim}$ where $(x,1)\sim (h(x),0)$.

@Balarka I saw that......

@DanielRust ah!

and so I guess in this way the solenoid is just the homeomorphism given by addition with $1$?

yes

$h(x) = x + 1$

Ok, that makes me more comfortable :P

actually this identification-type interpretation gives some hope of thinking about monodromies in this context.

sure, it gives you a natural fibering over the circle.

but path-lifting isn't really unique in fiber products

so perhaps Cech lifting...?

I am a big fan of Cech methods :P

well given a base point in the solenoid, you can lift paths in the circle uniquely.... but that's only because the solenoid is foliated by disconnected copies of the reals.

@Chris'ssis I am here now.

no that doesn't seem right. you can lift uniquely iff you give the cantor set fiber discrete topology @DanielRust

in which case the map is just a covering map

@BalarkaSen the monodromy action is very interesting though. The orbit of a point inside the fiber under addition will follow a countable, dense path in the cantor set.

yes, indeed. i think the monodromy group is Z_p

Why would you need to give the cantor set the discrete topology?

it's a totally disconnected space

so you are claiming that X \to S^1 is a covering map?

@robjohn OK, I wanna explain a bit that one line proof :-)

Not a covering map, but it has unique lifts

hrm

@BalarkaSen Unique path-lifting does not imply covering map. The other way around does. It's not a biconditional.

i don't see why it has unique path liftings

@evinda No, first you determine the median, that's either A[(m-1)/2] if m is odd or 0.5*(A[m/2-1]+A[m/2]) if m is even. Then you set left = (m-3)/ and right = m/2+1, and count = 1 if m is odd, count = 2 if m is even. Then while(count < p) { ++count; if (left < 0){++right; continue;} if (right >= m){--left; continue;} if ((median-A[left]) < (A[right]-median)) {--left;} else ++right;}

pick a point on the solenoid, a path which goes once around the circle just lifts to a path which follows the leaf to which the basepoint belongs 'once around' until it hits the fiber again.

erm sure

sorry, leaf might be a dynamicists term :P

the path connected component of the base point.

know what a leaf is. heard of foliated spaces from prof.

Ah, cool

but only "heard" ;)

for the solenoid it's the same as a path connected component

wait i think i see it

@DanielRust I think lots of (most? I dunno) topologists/geometers run into distributions and foliations at some point.

consider the the natural map f : X \to S^1

pick a point x_0 of S^1

make a loop $\sigma$

@MikeMiller You're probably right, it was certainly Bill Thurston's subject and he took part in most parts of geometry.

now pick a $y_0$ in $X$ such that $f(y_0) = x_0$

pick up a small nbhd $U$ around $x_0$ and lift uniquely to $X$ such that it contains $y_0$

Anyway, I'm glad we have two people in here who like messed up spaces now. Relief for me!

lift $\sigma |_{U}$ onto $X$

this is unique.

the inverse image of $U$ will be a cantor set crossed with an interval

(if $U$ is small enough)

@DanielRust i mean pick an open set which is in the preimage of U

pick it so that y_0 is in it

i.e., pick an $U'$ around $y_0$ such that $f(U') = U$

sure

now lift $\sigma |_U$ onto $U'$.

@DanielFischer Is it left=(m-3)/2 ?

this is unique

it should still look like $C\times (0,1)$

Welp, my first answer after misreading a question.

Lel, I should've gotten more sleep. Oh well

@evinda Yes. Compiler would have caught the typo.

@robjohn sent it.

and continue lifting pieces of $\sigma$ by lifting it to small small nbhds. as these liftings are unqiue in the small nbhds, it is unqiue in general

does that work?

My stupidity is high today

@BalarkaSen Yep that would be the general method of a proof

ok, cool

so monodromy is well defined

question is, what the monodromy group is

@Chris'ssis Do you have a proof of the summation identity that doesn't use complex representation of $\cos(nx)$?

@BalarkaSen Intuitively, $y_0$ belongs to a path connected component which is 'morally' the same as the real line (they're not homeomorphic of course) and the lift from the circle corresponds.

@robjohn Sure, an elementary one. I have that somewhere. Actually, let me show you another proof I showed you in the past.

this looks a slightly better approach than the messed up cech fundamental group stuff i was thinking of

@DanielRust makes sense

@MikeMiller Heh.

I will stop bothering you then

@BalarkaSen You might like to look up the Poincare first return map, as it's the same as the monodromy action on cantor suspensions

@robjohn the same style as here

If you restrict yourself to the fiber that is.

@robjohn take it

googling

but it's incredible that the monodromy group is well defined @DanielRust! it means that a sense of \pi_1(X) is hiding somewhere.

You assign a flow to your space according to the natural action of $\mathbf{R}$ and you pick a fiber to be your transversal.

seeing that the classical monodromy group is the rep of pi_1(X) onto Aut(p^{-1}(x_0))

@DanielRust "flow" :?

dunno what a flow is.

A flow is basically a continuous $\mathbf{R}$ action

ok

@Chris'ssis It would be nicer presented backwards, just so someone doesn't have to go back through the proof to make sure all the steps are reversible.

@Chris'ssis but it is a nice proof

@BalarkaSen In some sense the monodromy action isn't enough because it only has countable orbits (by definition), whereas the fiber is uncountable.

@robjohn Yeah.

hrm

@BalarkaSen In particular, almost every point is missed in the orbit.

that makes sense

You could pick a basepoint in every leaf, but then you're making uncountably many choices, all of which are important to the action.

Actually maybe not

then loops would look like S^1 cross cantor set no?

the solenoid is simply connected

well no

cech simply connected

$\pi_1=0$

non-cech

oh yeah sure

it's not path connected

whoops so i don't see how you are defining path lifting again

haha, loops only lift to paths, not necessarily loops

but it's not path connected

that's why you pick a base point

oh right yeah