@TedShifrin Haha, cool!

Wu's question seems awful! Ow

Crazy that I remember that after 46 years.

yeah, Wu's question sounds ridiculous

he means any affine connection on the tangent bundle, yes?

It's a standard result. My silly answer was that all manifolds are paracompact.

Hahah

i dont know where i would start with the other three questions though, what is $M_{n}(\mathbb{R})$?

lol

nxn real matrices, @porridgemathematics

oh doh, of course

@TedShifrin I actually didn't know what he meant when he asked for a "simple" example of a manifold with $\pi_1 \neq 0$ and $H_1 = 0$. I started saying Poincare homology sphere and asked if I should give the construction but he didn't like it.

then what else

I mean, what's particularly simpler

Haha, I took a 2-complex with perfect fundamental group, embedded it in some huge dimensional Euclidean space and took an open nbhd

in the connected case, we would need a manifold that has fundamental group whose abelienization is trivial, but the group isn't itself trivial

Yeah, there are no examples in dimension 2, @porridgemathematics.

Connection on frame bundle makes its tangent bundle trivial. This leads to Riemannian metric, so the frame bundle is metrizable, blah blah blah.

Ah OK @Ted

That's kind of cool

Crazy.

are you an undergraduate @BalarkaSen ?

just wondering, these questions seem really difficult

Still so, yeah. I get done with it in a month.

@BalarkaSen this was my first idea too, not sure if it's particularly "simple", but yeah

wow

Yeah these were atypical questions just because they know I have studied some topology.

what does 4. have to do with topology

That was just for fun everyone said they have run out of questions lmao

horrible sense of "fun"

@Thorgott You can actually upgrade that idea to give a compact example, and I wonder if you know this.

Balarka, the level of this exam makes it clear they knew what they had decided before even starting. Nice job to you!

Haha, I think so. Thanks!

was this an exam you needed to do to get your undergraduate degree?

This was grad school interviews actually

oh okay, that makes more sense

We don't do those in this country.

Yeah, too many candidates.

Here it's still doable. I liked it, it was a cozy over-Zoom interview.

They usually do it in person, but not this year, clearly.

Way too many. And backgrounds from 0 to 100.

Even at Berkeley, for example, some people super advanced, others not very well-prepared.

Yeah.

@BalarkaSen no, I can't even argue why the normal example works, I only know the idea

You're stuck as to why the normal guy embeds in $\Bbb R^n$ at all, yeah?

that and why there is an open nbhd which deformation retracts to it

this is the statement that finite complexes are ENRs, right

Yes, correct.

yeah, horrible

Hatcher appendix is where dreams die

I have a group $G=\{f^rh^s: r=0,1;s=0,1,2\}$, where $f$ is map that reflects $(x,y)$ about y axis. $h$ rotates $(x,y)$ anti-clockwise by $2\pi/3$. A formula is to be found that expresses $(f^ih^j)*(f^sh^t)$ as $f^ah^b$

First you have to triangulate the CW complex. This can be done precisely because the complex you get by attaching 2-cells to a wedge of circles according to relators is a nice CW complex; the attaching maps are simplicial.

maybe first focus on writing h f as f^{0 or 1} h^something. this is a well studied class of groups but i won't spoil the surprise.

So if you barycentrically subdivide it enough, you can glue polygons to a graph (a barycentric subdivision of the 1-skeleton)

I have verified that $f^2=h^3=1$

And that $fh=h^{-1}f$

Next, any simplicial complex of dimension $n$ embeds in $\Bbb R^{2n+1}$. This is by using the twisted cubic, consider the curve $(t, t^2, t^3, \cdots, t^{2n+1})$. Pick any two different collection of $n$ points. Their convex hulls are always disjoint.

if you know what to do to push one f from the right of an h to the left of an h, you just iterate that. if that makes sense.

So finally you have a simplicial complex in $\Bbb R^N$. Find an ENR neighborhood.

@Thorgott The compact version is actually easier, trust me.

twisted cubics, ridiculous

By noting that: $f(x,y)=(-\frac{x+\sqrt 3 y}{2}, \frac{\sqrt{3}x-y}{2})$

@leslietownes Hi

I considered cases 1) 2|s 2) 2 doesn’t divide s

And then on i

I was wondering if there’s any shorter method

oh. it might be simpler to do it without coordinate formulas for f and h. or maybe jsut the single one you arguably need to move f past h (even then, arguable)

Here's what it is. Let $G = \langle g_1, \cdots, g_n | r_1, \cdots, r_n \rangle$ be any finitely presented group. Take $(S^1 \times S^3) \# \cdots \# (S^1 \times S^3)$, $n$ copies, indexed by $g_1, \cdots, g_n$. This is a manifold with fundamental group $F_n = F\langle g_1, \cdots, g_n \rangle$, because connected sum is along $S^3$'s, which do not contribute to $\pi_1$ @Thorgott

work in terms of powers of f and h as much as possible.

Now, pick a smooth embedded loop $\gamma$ representing $r_1$ in the fundamental group. Take a regular neighborhood of this, which is $\nu(\gamma) = S^1 \times D^3$. Throw it's interior out. This leaves a boundary of $S^1 \times S^2$; but plug a $D^2 \times S^2$ along it.

This is known as surgery, as you're aware. This gives a new manifold with fundamental group $\langle g_1, \cdots, g_n | r_1 \rangle$. Repeat with $r_2$, but you have to make sure that the embedded loop representing it does not hit the disk $D^2$ that $\gamma$ bounds in the first surgery operation. This is easy, as 1 + 2 = 3 < 4 quick maths.

So by transversality, you can keep repeating the argument, to get a 4-manifold with fundamental group $G$. This is exactly a closed handlebody version of the nbhd of 2-complex thing.

man, I'm not smart enough for this

why does the nbhd look like S^1xD^3

@leslietownes: yeah I’ll try pushing h to left by $fh=h^{-1}f$. I think that’s the only way.

Tubular neighborhood theorem. It has to be disk bundle of a rank 3 vector bundle over S^1, but there are only two such things, one is nonorientable.

But the ambient manifold is orientable, so it has to be the trivial rank 3 bundle.

S^1 = \gamma

For a while I thought Balarka Sen was answering my question when they said let $G$ be any finitely presented group. I had also written presentation of G above :)

is this a normal level of assumed knowledge for someone applying to graduate school?

Not for the grad school I am going to definitely, @porridgemathematics

They just asked this stuff for fun

They happen to know me from before as I had been visiting on and off

ok, so you just happen to be really knowledgable and they know this

@BalarkaSen this is what I'm missing, is the normal bundle of an orientable guy in an orientable guy always trivial?

@Thorgott Not trivial, but orientable.

Because if $V$ is orientable, $W$ is orientable, any complement of $W$ in $V$ gets a natural orientation

The only orientable rank 3 bundle over S^1 happens to be trivial

ah, that's the order in which we argue, ok

what would be an easy way to compute the homology of all subsimplices of dimension $\leq k \leq d$ (the $k$-skeleton) of some standard $d$-simplex $\Delta^d$?

I think it's sick that $\partial(S^n \times D^{m+1}) = S^n \times S^m = \partial(D^{n+1} \times S^m)$.

One of my favorite math facts I would say

is there something nice that this object is homotopy equivalent to?

specifically for the case $1 \leq k \leq d-2$

I believe it's harder to see than to compute.

this is a wild construction, more reason to learn surgery

Yeah I really love it.

We could read Sanders Kupers, "Lectures on Diffeomorphism Groups"

After I am done with my exams in 2 weeks

@koro it seems like from what they're asking you want to push h to the right. but your fh = h^{-1} f formula will get you there. left multiply by h [getting hfh = f] then right multiply by h^{-1} [getting hf = fh^{-1}, or as h^3 = 1, hf = fh^2]

i think if you iterate on that, you're there

it's crazy how different our semesters are

mine's only started a month ago

I'm currently precoccupied with not understanding Lie groups in any case

i understand the circle group

I probably don't even understand the circle group

@BalarkaSen could you give me a hint on how I could compute it?

@porridgemathematics what could you say about the homology groups off the bat?

hm, it seems like they would be zero for $n > k$, and for $1 < n < k$

by thinking about simplicial homology

Accurate.

And?

What can you say about $H_k$?

hmm, it is definitely non zero, and it would be generated by the $k$-simplices?

That's right. So can it have torsion?

@leslietownes: I solved that. Thanks a lot:)

i want to say it cant

@koro en.wikipedia.org/wiki/Dihedral_group see "other definitions"

will look familiar

Can anyone speak to my matrix?

hi matrix! how are you? it's nice to see you today.

i said that so nobody else has to.

is there a reason for the matrix formalism? it seems you're really defining some way in which R (via s) acts on tuples of positive reals

I think it only understands communication in terms of matrices

@leslietownes $s$ is a real parameter and $p,t$ are primes

they aren't primes after s comes around the corner

Just wanted to understand the periodic orbits

so i guess it would just be free abelian on the $k$-subsimplices?

@leslietownes my point is that the matrix acts on the lattice of all points for prime $p$ and $t$

I don't think it matters if $p$ and $t$ are prime, they could be integral

basically as the parameter is tuned, the matrix "drags" the lattice along for the ride

and the lattice clicks back into its original configuration every so often

@BalarkaSen oh and it would be zero for $n = 1$ as well, so would the answer be $H_k$ would be free abelian on the $k$-subsimplices, and $H_0$ would be $\mathbb{Z}$?

not free abelian on the $k$-subsimplices. Think $\Delta^3$, 2-skeleton

hm, I can see it would have to be $\mathbb{Z}$ in that case, but there are four two simplices..

so $\mathbb{Z}^{d - k}$ in general? I don't see why though

Here's a hint. Compute Euler characteristic.

ohhh

hello is anyone here good with geophysical fluids?

@leslietownes can you explain this more?

@Yep Does whiskey count as a geophysical fluid?

@XanderHenderson If theres enough of it sure haha

@Yep In that case, I am very good at geophysical fluids.

@XanderHenderson I think these days everyone is

@geocalc i'm not sure i understand the context. i note that your use of the term 'lattice' is slightly out of sync with another use of this term, although i guess i kind of understand it. i don't see much 'snapping into place' as s grows, i just see points being pushed down to the x axis as their x coordinates get bigger.

@BalarkaSen thanks, I think I figured it out!

@leslietownes see how it snaps back into place?

what's does that have to do with your context above? i understood the initial set (before being acted upon by s) to be {(e^p, e^t): p and t prime} per chat.stackexchange.com/transcript/message/57924837#57924837

e.g., first quadrant only

@leslietownes just trying to explain the periodic part using an example

i do get the general vibe of how a set of points can periodically be deformed back into itself, or a subset of itself

"periodically" being used in the informal sense of 'from time to time'

here I found a short reference and some context: golem.ph.utexas.edu/category/2014/04/…

in the section labeled Modular flow

I'm actually kind of stuck at the beginning, where they define $A$ as an integer matrix being an element of the special linear group. I'm not sure how to translate this info if the lattice in my question doesn't have co-volume equal to 1

the thing you have (with the primes and exponentials in it) isn't a lattice in the sense of that treatment

fine

the action they define does need the more restrictive definition of lattice

from my understanding what I described is a diffeomorphism of an integer "lattice" $\Bbb Z^2$

@leslietownes: I saw that Dihedral group of order $n$ link you shared. I knew that already.

What is the meaning of topology? Can I say that it is the study of open sets, closed sets, compact sets, perfect sets etc.?

geocalc: it is the image under a diffeomorphism of a subset of a lattice of Z^2. these details start to matter if you want to consider groups (particularly matrix groups) acting on them

In Layman terms, what is the meaning of topology?

@leslietownes thanks for letting me know the proper way of saying that

just to reiterate for myself: it's the image under a diffeomorphism of a subset of a lattice of Z^2

the sets of prime pairs, in Z^2, are not a lattice (e.g. because it is dramatically not closed under addition: if (p,q) is in there, (p,q) + (p,q) = (2p, 2q) will not be)

yeah that's why I changed it to integers instead of primes

I'm sure the one parameter matrix group is correct, and I have the: "image under a diffeomorphism of a subset of a lattice of Z^2"

so I think I have everything set up to attempt to calculate the periodic orbits

I think one other thing I need is to transport the group structure of SL(2,Z) under the diffeo so I can define an analogous matrix to $A$

and I've calculated that to be $[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}](x,y)=(x^ay^b,x^cy^d)$

for $a,b,c,d \in \Bbb Z$

oh wait should have said that that is the action of SL(2,Z) on

the image under a diffeomorphism of a subset of a lattice of Z^2

it doesn't really 'act on' that subset if it doesn't send that subset to itself. it just moves it from one place to another. i'm also not convinced that your definition plays nice with matrix multiplication.

what's the end goal here?

note, if you just want to compute the image of a R under a map, which you might regard as an "orbit" of f(1) if you regard the parameter t as time, you don't need groups or matrices to begin that project

i'm not sure what the purpose is, let alone whether you need the matrix stuff to do it

I mean after the diffeo, you're computing a flow on the "lattice" in the positive quad.( i.e all points taking the form $(e^n,e^k)$ for integer n,k)

I guess the flow is akin to the modular flow acting on the Z^2 but it's purely in quad. 1

again, just for purposes of reconciling this with what you may be reading, 'flow on X,' 'acting on X' and related terms often require that X be mapped to itself for all inputs, which is not happening here

that's just terminology but i figured i would point it out

yeah I understand that

I kind of see what you're saying

you're saying that the matrix doesn't always map "lattice" points $(e^n,e^k)$ onto "lattice" points

I think the interest is when the matrix does map lattice points to lattice points

@geocalc33 hypnotic

hey chat, evening

i have a vague question about semidirect products of groups

Not as it stands.

also, we usually do not want $H$ and $K$ to be 'completely unrelated', so when you have the conditions for direct product, it's cool